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http://math.stackexchange.com/questions/138465/algebrically-independent-elements/138940	# Algebrically independent elements [duplicate] Possible Duplicate: Why does K->K(X) preserve the degree of field extensions? Suppose $t_1,t_2,\ldots,t_n$ are algebrically independent over $K$ containing $F$. How to show that $[K(t_1,\ldots,t_n):F(t_1,\ldots,t_n)]=[K:F]$? - [K:F] is a finite extension – Jishnu Ray Apr 29 '12 at 16:03 4 – Zev Chonoles♦ Apr 29 '12 at 16:13 ## marked as duplicate by Arturo Magidin, Gerry Myerson, Qiaochu YuanJun 2 '12 at 1:26 This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question. ## 1 Answer Using the answer in link provided by Zev, your question can be answered by simple induction over $n$. For $n=1$ we proceed along one of the answers shown over there. Assume we have shown the theorem for some $n$. Then we have $[K(t_1,\ldots,t_n):F(t_1,\ldots,t_n)]=[K:F]$, and by the theorem for $n=1$ we have also $[K(t_1,\ldots,t_n,t_{n+1}):F(t_1,\ldots,t_n,t_{n+1})]=[K(t_1,\ldots,t_n)(t_{n+1}):F(t_1,\ldots,t_n)(t_{n+1})]=[K(t_1,\ldots,t_n):F(t_1,\ldots,t_n)]=[K:F]$ which completes the proof by induction. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 10, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9568194150924683, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/116977-complex-numbers-theory-question.html	# Thread: 1. ## Complex numbers theory question If $z = a+bi, w = c+di$ and $arg(z+w) = 90^{\circ}$ then which of the following is true? a) z + w = 0 b) \|z + w\| = 0 c) a = -c d) b = -d I hate theory questions, I can't get them for some reason. Are there any properties that would help me with this question? Here's how I interpret it, If arg(z+w) is 90 degrees, then z is perpendicular to w. I'm not sure what else to do for this kind of question. I know the answer, but I don't know why Can someone point me in the right direction? 2. If you can't get "theory" questions, it's because you don't know the basic definitions! Go back and review the definitions. Don't just get a general idea of what they mean, learn them by heart. You use the precise words of definitions in proving theorems and solving problems. Saying that a complex number has "arg 90" means that the line from the point in the complex plane to 0 make a 90 degree angle with the positive real axis and so it lies on the positive imaginary axis. That is, z+ w is pure imaginary. What does that tell you about its real part?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9546936750411987, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/49865/quantum-superposition-of-states-experimental-verification	# Quantum superposition of states: experimental verification How can somebody demonstrate the quantum superposition of states directly by other means than the double slit experiment? And why can't macroscopic objects like a pen be in superpostion of states? Will it ever be possible to have an object like a pen to be in superposition of more than one state? - Others have given good answers, but it all depends on what you mean by "directly." You simply can't "see" a wavefunction. There are many experiments showing quantum superpositions, but they all rely to some degree on interference since that is how you measure the relative phase of different components of the wavefunction. – Michael Brown Jan 11 at 9:54 1 – Emilio Pisanty Jan 12 at 22:44 So, the reason why macroscopic objects like pen can't be in superposition of different because of decoherence. But will it be ever possible to "remove" decoherence and demonstrate the quantum of superposition of large objects? – user774025 Jan 13 at 13:06 ## 4 Answers This reference describes an experiment in which an ion is prepared in a superposition state for which the relative phase between the component states is controllable, and ulimately measurable. The obstruction to having objects like pens in superposition states is the inability to isolate its parts from the environment. Interactions between the environment and the system (the atoms in the pen) act through a process called decoherence, which has the effect of suppressing these relative phase terms incredibly quickly. - There are experiments with cold atom gases in optical traps that deal with matter's essential quantum nature. Cold atom gases' is actually quite a hot field (no pun intended). In an attempt to answer your question, I will try to explain how I understand quantum superposition experiments work in the most general possible way I can. A superposition state $\left\|\Psi\right\rangle$, under this most general description possible, is a linear superposition (that is precisely the reason why they are called that way) of the different quantum states that the system may have: $$\left\|\Psi\right\rangle=\sum_i c_i\left\|\tilde{\Psi}_i\right\rangle$$ Where the sum is performed over all possible $\left\|\tilde{\Psi}_i\right\rangle$ states in the chosen basis, each one of them weighed by a complex factor $c_i$. The common idea in every experiment dealing with quantum superposition is simple: to have a system whose initial state, which is given by the $c_i$'s, can be prepared and controlled (that is called a coherent state), and in which a certain observable quantity can be measured (associated with an operator branded $\hat{A}$). At the end of the day, the result of every observable measurement can be understood in terms of how the operator $\hat{A}$ acts over the different states in the basis, how each one of these states in the particular basis of choice evolve in time and which are the weights $c_i$. What happens in the double slit experiment, trying to explain it in these terms without delving into further details, is that the different time evolutions of the (initially coherent) electrons moving through the experimental setup introduce phase factors that keep adding up to the $c_i$'s. When the position of the electrons is recorded on a screen (which constitutes the measurement), the mean value of the position observable has these different phase factors interfering with each other, giving rise to an interference pattern in the screen as a result. The key idea in these experiments is always the following: To start with a coherent state (or at least as coherent as it can be managed in practice), to let interactions and time evolution make their respective jobs, and later on to see what happens when the system is looked at. What happens with a macroscopic system? The abridged answer is: 'Way too many things'. For starters, the number of components would be of the order of $N_A=6'022\cdot10^{23}$ when not greater, so the number of states in our basis for a complete, many-body description would be absolutely humongous. On top of that, atoms and electrons are subject to many interactions on a microscopic level, so any coherence that may be momentaneously achieved will be destroyed in a quick, uncontrolled way before any measurement may be performed... ... Unless what we are dealing with is a system of a very particular kind, either with not too many particles (cold atom gases) or some specific interaction which favours the appearance of a coherent state even on a $N_A$-level macroscopic scale (as in conventional superconductivity and superfluidity). - In my opinion, the concept behind the famous Schroedinger's cat is exactly this: to Schroedinger, it appeared paradoxical that a macroscopic object -- even a living being -- could be in a superposition of states. However, cases in which the radiation inside a cavity was prepared in a superposition of two macroscopically distinguishable states have already been observed experimentally. For example, this was one of the result for which the Nobel Prize this year was awarded, see here. In the case cited, decoherence was also observed. Even better: it was observed a slow, gradual process of decoherence. - Others have covered the usual microscopic systems that are currently amenable to controlled quantum manipulations, but there are in fact "macroscopic" objects (to some definitions of the word) that can be placed in quantum superpositions. These fall broadly within a field known usually as cavity optomechanics, which has a reasonable wikipedia page. The essential idea is to couple the oscillations of light to those of one of the mirrors of a cavity; this allows 'quantumness' in the state of the light to translate into the state of the mirror. In this and usually all macroscopic examples, it is only one degree of freedom - the centre-of-mass position in this case - that gets put in a superposition; all other degrees of freedom are left in classical, often thermal, states. This is nevertheless quite enough to get quantum behaviour. For example, you can obtain double-slit interference patterns using buckyballs (C$_{60}$) using their centre-of-mass positions, while maintaining the rotational and vibrational degrees of freedom (which are merely a better way of accounting for all motional degrees of freedom apart from the average) in (fairly cold) thermal, classical states. Another quantum superposition of a macroscopic object - (just) visible to the naked eye! - is the placing of a microwave oscillator in a superposition state using other microwave sources en lieu of laser beams on an atom. This is fairly well explained by Aaron O'Connell in his TED talk. For more formal references try the UCSB Cleland group page; good popular science articles are listed in O'Connell's wikipedia page. Finally, I would try macroscopic quantum phenomena" in wikipedia. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.940044105052948, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?t=223264&highlight=newtons+bucket	Physics Forums Thread Closed Page 1 of 10 1 2 3 4 > Last » Recognitions: Gold Member ## Newton's Bucket Hi. First post here. I have no formal math or physics training, but read popular books on physics and am pretty well read as far as that goes. Now for the question. I'm fascinated by the Newton's Bucket problem and fortunately for me it's cleared my head of the 2 brothers paradox (one on earth, one in ship, ship ages) with regard to which one is considered moving and which is stationary. For a description of Newton's Bucket, here's a good one: http://www-groups.dcs.st-and.ac.uk/~...on_bucket.html I've never liked the traditional idea that the brother that is considered moving (and therefore aging) is the one that is accelerating away because once acceleration stops and the ship continues at near light speed, the aging process continues yet the ship is only moving relative to the Earth and not accelerating away from it. Newton's Bucket solves that problem by inferring that the ship is moving near light speed relative to either the stars or some universal fabric that is static or almost static relative to the stars. Newton's bucket implies that if the universe were empty (I suppose this would include dark matter and energy) except for the bucket and a single observer, the bucket would seemingly have to behave strangely. For example, if the observer were spinning around the bucket (and the bucket around the observer) but both in the same direction as far as the two axis of rotation are concerned, the bucket could not be said to be spinning and therefore would not exhibit inertial forces or the resultant concave water. If the observer and bucket were spinning opposite to each other, then what? Would the water then become concave relative to the velocity of the observer? Or is a greater mass (or something else altogether) required such as massive galaxies? And if either or both are causing the water to become concave, then what exactly is causing it. I realize the simple answer is inertia, but this paradox implies that inertia would cease to exist in an empty universe and with the observer and bucket moving in the same direction or possibly in different directions as well. Inertia would have to cease to exist in an empty universe that contained only a bucket of water and a single observer moving in the same direction around it as there would be absolutely no frame of reference with regard to acceleration. With no inertia, one could not feel any effects of acceleration so if the bucket exploded, or the observer sneezed, which would move relative to the other, and which one would age when applied to the two brother paradox. Glad to have found this forum. PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug Blog Entries: 6 The answer you seek is the article you linked to. ".....in simple terms, in a universe with no matter there is no gravity. Hence general relativity reduces to special relativity and now all observers agree when the rock system is spinning (i.e. accelerating). " In other words relativity says rotation is detectable even with one object in an empty universe. Of course this is hard to prove with an experiment, as we do not have a spare empty universe to try it out in :P Tha article also tries to lend some support to Mach's views (that all inertia is relative to the fixed stars): "In 1985 further progress by H Pfister and K Braun showed that sufficient centrifugal forces would be induced at the centre of the hollow massive sphere to cause water to form a concave surface in a bucket which is not rotating with respect to the distant stars. Here at last was a form of the symmetry that Mach was seeking. " A counter argument is this: Rotate a bucket clockwise (when looking from above) so that the water contained within it has a concave surface. Define the bucket as stationary and atribute the concave surface of the water to the gravitational influence of the all the universes stars orbiting anti-clockwise around said bucket. Now place another rotating bucket alongside the first bucket while the water within it is still spinning. If the first bucket is exactly at the axis of the spinning universe, then the second bucket is not and yet the lowest point of the water in the second bucket is exactly at the centre of its spinning surface. Mach's principle seems to fall apart as soon as we introduce a second bucket. Recognitions: Gold Member Science Advisor Welcome to these Forums Buckethead, glad too that you found us! Mach's Principle might not rely on just gravitational influences, as it would in GR. In the Brans Dicke theory an extra scalar field coupled to matter endows fundamental particles with inertial mass. Thus introducing the second bucket proves that Mach's Principle is incompatible with GR but it may not be incompatible with an alternative gravitational theory. Garth ## Newton's Bucket "Newton's Bucket" only works in the presence of gravity as kev pointed out. That said - the pressure in the water increases linearly from 0 to $\rho gh$ no matter where you check from top to bottom. When the bucket/water is spinning uniformly, a new force is added to keep the water from travelling along a linear path. This new force creates another linear pressure gradient that starts from the center of the bucket and increases as you move away from the axis of rotation. The product of the two orthogonal linear pressure gradients leads to a parabolic pressure profile at any fixed height. The water surface assumes a parabolic shape to support both linear pressure gradients simultaneously. Regards, Bill Recognitions: Gold Member Quote by kev The answer you seek is the article you linked to. ".....in simple terms, in a universe with no matter there is no gravity. Hence general relativity reduces to special relativity and now all observers agree when the rock system is spinning (i.e. accelerating). " In other words relativity says rotation is detectable even with one object in an empty universe. Of course this is hard to prove with an experiment, as we do not have a spare empty universe to try it out in :P But the article just a little before that quote also states that Einstein said that Mach's view was in complete agreement with GR so this conclusion in the article confused me. I'm also confused about how observers could agree that the bucket is spinning. Because the water would go concave? Again, why would it go concave in an empty universe? Tha article also tries to lend some support to Mach's views (that all inertia is relative to the fixed stars): "In 1985 further progress by H Pfister and K Braun showed that sufficient centrifugal forces would be induced at the centre of the hollow massive sphere to cause water to form a concave surface in a bucket which is not rotating with respect to the distant stars. Here at last was a form of the symmetry that Mach was seeking. " Yes, I was trying not to bring this up too soon, but logically, I'm in agreement with this. A counter argument is this: Rotate a bucket clockwise (when looking from above) so that the water contained within it has a concave surface. Define the bucket as stationary and atribute the concave surface of the water to the gravitational influence of the all the universes stars orbiting anti-clockwise around said bucket. Now place another rotating bucket alongside the first bucket while the water within it is still spinning. If the first bucket is exactly at the axis of the spinning universe, then the second bucket is not and yet the lowest point of the water in the second bucket is exactly at the centre of its spinning surface. Mach's principle seems to fall apart as soon as we introduce a second bucket. If the first bucket were indeed at the very axis of the spinning universe and by definition not spinning, then the concaveness of the water would be due to a force (gravity or otherwise) from the stars pulling equally at all sides of the water causing it to rise up the sides of the bucket. (One could no longer state inertia being the cause as the bucket is "not spinning") A second bucket placed off center would also feel this same "pull" and it's water would also rise, but one side would rise higher then the other, having a stronger "pull" on that side. Because of the scales the offset would be infintesimally small, perhaps a plank length. In a universe with non-rotating stars (the real universe) one cannot say that two spinning buckets side by side have their dips in the absolute center of the bucket or that the two buckets have their dips in the same location.[/QUOTE] Recognitions: Gold Member Quote by Garth Mach's Principle might not rely on just gravitational influences, as it would in GR. In the Brans Dicke theory an extra scalar field coupled to matter endows fundamental particles with inertial mass. Thus introducing the second bucket proves that Mach's Principle is incompatible with GR but it may not be incompatible with an alternative gravitational theory. Garth My gut tells me that gravity can't play much of a part in Mach's Principle as the stars are simply too far away. Doesn't gravity eventually diminish to a single planck value at which point gravity can be said to not exist at all? Of course there is still the sun and a spinning bucket in our universe may be under it's sole influence. Is there anyway to determine this or is there any theory indicating this? Also, I am a bit confused by the terms tensor vs scaler. Wikipedia didn't help me much here, can you explain this in simple (non-math) terms? Recognitions: Gold Member Quote by Antenna Guy "Newton's Bucket" only works in the presence of gravity as kev pointed out. It does seem that gravity is the most suspect reason for Mach's principle, but how about a situation of an empty universe with one bucket of water and one observer. If the observer were to grab the bucket and spin it then one of three things would happen. 1) The water would go noticably concave (and simultaneously the observer would also feel a centrifugal force on it's own body) due to both spinning relative to a (non local) absolute space, 2) the water would stay flat even though it was spinning relative to the observer because there is no absolute space. or 3) there would be an infintesimally small inertial force on both the bucket of water (causing it to go ever so slightly concave) and the observer due to both spinning relative to each other and because the delta between the masses of the two objects define an absolute space that is moving more slowly relative to the more massive object then it is to the less massive object. Blog Entries: 3 Recognitions: Gold Member Buckethead: It does seem that gravity is the most suspect reason for Mach's principle, but how about a situation of an empty universe with one bucket of water and one observer. What would keep the water in the bucket ? The water would form into a sphere and freeze. It's been pointed out to you that the parabolic surface is due to a combination of lateral and vertical forces, so talking about the surface of the water in your scenario isn't realistic. I would expect any spinning object to experience stresses because of the spin, and this would happen in any sort of universe, regardless of gravity. Recognitions: Gold Member Quote by Mentz114 What would keep the water in the bucket ? The water would form into a sphere and freeze. It's been pointed out to you that the parabolic surface is due to a combination of lateral and vertical forces, so talking about the surface of the water in your scenario isn't realistic. I was using the bucket in the spirit of a thought experiment for it's ease of visualization. It is a totally impractical object to use in a real experiment, but the point of my original post is that you can use any practical object here with the same effect. For example two spheres tied together with a string and spun around the axis of the center of the string, or an elastic sphere which would bulge at the center and so on. The actual object is not important here, only the fact that there is centrifugal forces acting on that object. I would expect any spinning object to experience stresses because of the spin, and this would happen in any sort of universe, regardless of gravity. Not so if Mach's Principle were true. In an empty universe there would be no stresses on a spinning object because there would be way to know what that object was spinning in reference to, or in other words, whether it was spinning at all. This does have the deeper implication that in an empty universe what we know of as inertia would cease to exist altogether. For example, if you were in a spaceship in an empty universe and flipped the switch to start the rocket engine, it would fire (maybe), but there would be no sensation of forward thrust, the accelerometer onboard would not show any change, you would not feel any G force, and in essense Newton's 3 laws of motion would break down. Please realize though that I am also trying to figure out here what is an "empty universe". Is it simply a universe void of matter? Of dark matter and dark energy? Of virtual particles? Also, I'm not completely convinced that it's matter that is the real reference point for a spinning object and it's associated stresses (acceleration). It could also be that even an empty universe has some kind of inherent frame of reference that defines that it is static and not moving regardless of whether or not it contains matter, dark matter, and/or dark energy. If this is the case, then I would think a spinning object would still show rotational forces acting on it even in a massless universe. But if this is the case, then it would turn the physics world upside down I would think. Blog Entries: 3 Recognitions: Gold Member Buckethead: Not so if Mach's Principle were true. In an empty universe there would be no stresses on a spinning object because there would be way to know what that object was spinning in reference to, or in other words, whether it was spinning at all. This does have the deeper implication that in an empty universe what we know of as inertia would cease to exist altogether. For example, if you were in a spaceship in an empty universe and flipped the switch to start the rocket engine, it would fire (maybe), but there would be no sensation of forward thrust, the accelerometer onboard would not show any change, you would not feel any G force, and in essense Newton's 3 laws of motion would break down. OK, from your earlier remarks I can see we are on the same playing field now. I will try and refute the bit I've quoted above. Firstly, rotation can only be defined for an extended object. A point cannot rotate. So the parts of the extended object have proper spatial relationships with each other and provide a frame in which to define rotation independently of any external reference. I can choose the centre of the rotation as the origin of a frame, and then define a tangential velocity of a piece away from the centre. The same argument might well do for the acceleration case, but you should bear in mind that your one single object in the universe can only accelerate by ejecting some matter, in which case we have more than one object and the argument short circuits. Re-reading this, I'm not 100% convinced by my logic, it would be interesting to hear other views. Quote by Buckethead The actual object is not important here, only the fact that there is centrifugal forces acting on that object. "Centrifugal force" is an artificial construct used to balance the centripetal force (i.e. that exerted by the bucket wall) acting on an object that would otherwise travel in a straight line. The closest approximation to a "centrifugal force" would be the tendency of like charges to repel one another - which isn't the sort of thing that keeps a mass rotating at constant radius. Regards, Bill Blog Entries: 6 Hi, I am aware that Einstein himself concluded that Mach's principle is incompatible with GR as demonstrated by this quote: "This certainly was a clever idea on Einstein's part, but by June 1918 it had become clear that the De Sitter world does not contain any hidden masses and is thus a genuine counterexample to Mach's principle. Another one of Einstein's attempts to relativize all motion had failed. Einstein thereupon lost his enthusiasm for Mach's principle. He accepted that motion with respect to the metric field cannot always be translated into motion with respect to other matter." from this article http://science.jrank.org/pages/11027...elativity.html However, after further reflection Mach's principle is not dismissed by the simple counter example I gave. In that example the second bucket would appear to be rotating along with the distant stars from the point of view of an observer stationary with respect to the water in the first bucket. The second bucket would not therefore be submitted to the "spiralling spacetime" that the water in the first bucket is subjected to, because the second bucket is comoving with the spiralling spacetime/ gravitational field. A clearer (and fairer) example would be to place the first bucket at the centre of a large rotating turntable. An observer on the turntable could place a second bucket near the rim of the turntable and observe that the water in the second bucket is at rest with with respect to the water of the first bucket and that the water in the second bucket is piled up asymmetrically on the side furthest from the centre of the turntable. If the water in the second bucket is spinning then the centre of the concave depression would indeed be offset from the centre of the bucket. In this fairer second example, Mach's principle does not fail. Can anyone think of a simple example (that is easy to visualise), where Mach's principle fails? Quote by Mentz114 Re-reading this, I'm not 100% convinced by my logic, it would be interesting to hear other views. Consider that a fixed volume following a curved path will have different velocities at different points on/within that fixed volume. If the differential velocities become too great, the object flies apart. The spherical blob of water you mentioned only remains so because of surface tension. If that blob of water were to rotate about some axis, there would have to be more surface area in a plane perpendicular to the axis of rotation to keep the forces in equilibrium - leading to an ellipsoidal shape. Oddly, a spherical blob of water travelling at a significant fraction of the speed of light would also look like an ellipsoid to a stationary observer - but for a different reason. Regards, Bill Blog Entries: 3 Recognitions: Gold Member In this fairer second example, Mach's principle does not fail. Can anyone think of a simple example (that is easy to visualise), where Mach's principle fails? It fails on Occams razor, surely. There's nothing to explain. All rotating phenomena are accounted for by present dynamics without need for a cosmic frame. Or am I missing something deep here ? Blog Entries: 6 Quote by Mentz114 It fails on Occams razor, surely. There's nothing to explain. All rotating phenomena are accounted for by present dynamics without need for a cosmic frame. Or am I missing something deep here ? General Relativity can explain any motion including accelerated motion in a straight line in terms of no motion and and complicated gravitational spacetime. For example, if you turn on your rocket motor and accelerate from a standstill to 0.8c, it can be explained in terms of a gravitational field that springs up the instant you turned your rocket motor on and draws the universe towards a black hole behind you while your rocket motor resists the gravitational "pull". When you drive to work, accelerating and breaking at junctions and experiencing "centrifugal force" as you go round corners, the whole journey can be explained in terms of gravitational fields and complicated accelerations of everything in the universe while you have remained stationary throughout the entire journey. Now this point of view is necessary or we have to accept a notion of absolute motion which is incompatible with Relativity. Occam's razor and even considerations of conservation of energy are not strong enough arguments to support a notion of absolute motion or acceleration. Recognitions: Gold Member Quote by Mentz114 Firstly, rotation can only be defined for an extended object. A point cannot rotate. So the parts of the extended object have proper spatial relationships with each other and provide a frame in which to define rotation independently of any external reference. I can choose the centre of the rotation as the origin of a frame, and then define a tangential velocity of a piece away from the centre. Let's take two bricks tied together by a rope and define that the bricks are not spinning (one face of each brick always faces the other). If there is tension on the rope, then one can say the bricks are revolving about each other. But in an empty universe, this would mean the system would be revolving relative to absolute space. If there is no absolute space, then there could be no tension on the rope since the objects are not rotating relative to anything (not even to each other if their faces are stationary) The same argument might well do for the acceleration case, but you should bear in mind that your one single object in the universe can only accelerate by ejecting some matter, in which case we have more than one object and the argument short circuits. The rocket is a matter/anti-matter engine and all exhaust is converted into energy. Blog Entries: 3 Recognitions: Gold Member Kev: Now this point of view is necessary or we have to accept a notion of absolute motion which is incompatible with Relativity. Well, I don't see at all how that follows from your argument. I can accept absolute rotation, because of the extended object argument, and I think acceleration can always be detected so it's got nothing to do with absolute motion. Thread Closed Page 1 of 10 1 2 3 4 > Last » Thread Tools \| \| \| \| \|--------------------------------------\|-------------------------------\|---------\| \| Similar Threads for: Newton's Bucket \| \| \| \| Thread \| Forum \| Replies \| \| \| Special & General Relativity \| 7 \| \| \| Introductory Physics Homework \| 2 \| \| \| General Discussion \| 18 \| \| \| General Physics \| 5 \| \| \| Introductory Physics Homework \| 10 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9541715383529663, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/99570/a-question-from-titchmarshs-riemann-zeta-function-textbook	# A question from Titchmarsh's Riemann Zeta Function textbook. I have one query, concerning the newest edition of this monograph. 1. At page 7, section 1.2, at the bottom of the page, it's written that: " It is easily seen that $\zeta(s)=2$ for $s=\alpha$, where $\alpha$ is a real number greater than 1..." I am quite sure that for $\alpha=2$ which is real and greater than 1, $\zeta(s)=\frac{\pi^2}{6} \neq 2$, what do you think did Titchamarsh meant here?! Thanks in advance. - Irrelevant comment: I looked for Titchmarsh on Google Books, and they appear to have it filed under "Architecture > History > General"! – David Loeffler Jan 16 '12 at 16:03 ## 1 Answer I will take a stab at this. I think he is saying that for some real number $s > 1$, $\zeta(s) = 2.$ We know that on the interval $(1,\infty)$ the $\zeta$ function is continuous and monotonically decreasing. Since $\zeta(t) \to \infty$ as $t \downarrow 1$ and $\zeta(2) = \pi^2/6 < 2$, there must be some $s \in(1,2)$ with $\zeta(s) = 2.$ His writing somewhat obscures the existential nature of the statement he is making. - 1 In short, T is saying $\zeta(\alpha)=2$ for some $\alpha$, not claiming $\zeta(\alpha)=2$ for every $\alpha$. – Gerry Myerson Jan 16 '12 at 20:44 1 I wonder why T is cheap with words. Thanks. – MathematicalPhysicist Jan 17 '12 at 9:59	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 18, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9559115767478943, "perplexity_flag": "middle"}
http://mathhelpforum.com/pre-calculus/102340-average-rate-change.html	# Thread: 1. ## Average rate of change I am unsure as to which forum to put this in. I took a shot and put it here, apologies if wrong. Okay, to the question I am having trouble with. "Find the average rate of change of $f(x) = \sqrt{x + 11}$ with respect to x from $x = 5$ to $x = 5 +h$. [4 marks]" My attempt to answer: $f(5) = \sqrt{5 + 11}$ $f(5) = \sqrt{16}$ $f(5) = 4$ $f(5 + h) = \sqrt{5 + h + 11}$ $f(5 + h) = \sqrt{16 + h}$ $f(5 + h) = 4 + \sqrt{h}$ I am unsure if I am correct in square rooting the 16, so having 4 + the square root of h. Then I use the $\frac{y2 - y1}{x2 - x1}$ formula to get the slope of the secant rate of change: $\frac{4 + \sqrt{h} - 4}{5 - 5 + h}$ $= \frac{\sqrt{h}}{h}$ That's my final solution. My problem with this solution is when I sub a number in for h, I get a different answer from running the numbers through, and simply going square root of the number divided by the number. Example: Let's pretend h is 20. So I have $f(5) = 4$. I can sub in 20 for h: $f(5 + 20) = \sqrt{5 + 20 + 11}$ $f(25) = \sqrt{36}$ $f(25) = 6$ Run that through the slope formula: $\frac{6 - 4}{20 - 5}$ $= \frac{2}{5}$ $= 0.4$ Then, using my average rate of change thing: $\frac{\sqrt{h}}{h}$ I get: $\frac{\sqrt{20}}{20}$ $= 0.223606797$ which is a completely different answer. So I must be doing something wrong here. Okay, I tried using 25 for h in the $\frac{\sqrt{h}}{h}$ and it gives the same ansewr as running the math through. So it is not technically h, but it is h + 5, so is this actually right??? I see that as: $\frac{\sqrt{h+5}}{h+5}$ which I think is different, please correct me, I am so confused. EDIT:There, I have successfully turned this into something readable using latex 2. Originally Posted by Kakariki I am unsure as to which forum to put this in. I took a shot and put it here, apologies if wrong. Okay, to the question I am having trouble with. "Find the average rate of change of $f(x) = \sqrt{x + 11}$ with respect to x from $x = 5$ to $x = 5 +h$. [4 marks]" My attempt to answer: $f(5) = \sqrt{5 + 11}$ $f(5) = \sqrt{16}$ $f(5) = 4$ $f(5 + h) = \sqrt{5 + h + 11}$ $f(5 + h) = \sqrt{16 + h}$ $f(5 + h) = 4 + \sqrt{h}$ I am unsure if I am correct in square rooting the 16, so having 4 + the square root of h. Then I use the $\frac{y2 - y1}{x2 - x1}$ formula to get the slope of the secant rate of change: $\frac{4 + \sqrt{h} - 4}{5 - 5 + h}$ $= \frac{\sqrt{h}}{h}$ That's my final solution. My problem with this solution is when I sub a number in for h, I get a different answer from running the numbers through, and simply going square root of the number divided by the number. Example: Let's pretend h is 20. So I have $f(5) = 4$. I can sub in 20 for h: $f(5 + 20) = \sqrt{5 + 20 + 11}$ $f(25) = \sqrt{36}$ $f(25) = 6$ Run that through the slope formula: $\frac{6 - 4}{20 - 5}$ $= \frac{2}{5}$ $= 0.4$ Then, using my average rate of change thing: $\frac{\sqrt[h]][h]]$ I get: $\frac{\sqrt{20}}{20}$ $= 0.223606797$ which is a completely different answer. So I must be doing something wrong here. Okay, I tried using 25 for h in the $\frac{\sqrt{h}}{h}$ and it gives the same ansewr as running the math through. So it is not technically h, but it is h + 5, so is this actually right??? the wrong thing as what you said you can't separate 16 from the square root $f(5+x)=\sqrt{16+h}$ you can't simplify it more than that $s=\frac{\sqrt{16+h} - 4 }{h}$\ EDIT:There, I have successfully turned this into something readable using latex CONGRATULATIONS LATEX is simple and useful	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 49, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9501042366027832, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/9402/pedagogical-question-about-linear-algebra	## Pedagogical question about linear algebra ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Last semester I taught a linear algebra class that is intended to introduce young students (at a sophmore-junior level) to "abstract mathematics". It seems that a major conceptual hurdle for many of the students is understanding the definition of a vector space. More specifically, a vector space is some set of things to which we can perform the operations of addition and scalar multiplication. Despite an enormous amount of effort on my part, many of the students insisted that it makes sense to do things like "take the real/imaginary part" of a vector or look at the components of a vector. What strategies have you found useful for getting students to understand this type of definition? I made this community wiki -- please edit it if the question seems badly phrased. - Did you introduce the definition of a group or a field? – Qiaochu Yuan Dec 20 2009 at 4:58 2 Does MO have a policy yet on questions about teaching questions? We certainly will need one. PS this question isn't yet community wiki PPS sophomore-junior=2nd/3rd yr of university? 3rd year seems a little late to me to be learning what a vector space is. – Peter McNamara Dec 20 2009 at 4:59 I've community wikified the question, since it looks like the OP wanted to but was having trouble with it. @Peter McNamara: What kind of policy did you have in mind? Why don't you start a new thread about it at meta.mathoverflow.net and post a link to it here. – Anton Geraschenko♦ Dec 20 2009 at 5:45 4 Peter- Probably quite a few of the people taking this course are not math majors. I don't think there's a problem if economists wait for their 3rd year to learn about vector spaces. Also remember that American students are often not very focused in their first two years. Just taking the equivalent of 18.01,18.02, and 18.06 in the first two years would not be unusual, even for a math major (that was basically what I did). – Ben Webster♦ Jan 21 2010 at 0:17 ## 13 Answers I have addressed this issue at a slightly higher level, when teaching second or third quarter of abstract algebra for juniors and seniors. I had in mind a similar but not identical purpose, which was getting the students to truly understand the difference between a real and complex vector space. The best solution that I thought of was to teach something beyond that strictly requires an understanding of the difference. Given a real vector space $V$, I define its complexification $V_\mathbb{C}$, and given a complex vector space $V$, I define its realification $V_\mathbb{R}$. Of course the conventional quick way to set up the former is with a tensor product, but without that scary idea, you can prosaically define $V_\mathbb{C}$ as $V \oplus iV$, two copies of $V$ with a defined multiplication by $i$. Meanwhile $V_\mathbb{R}$ is the same set as $V$, but with restricted scalars. These definitions can really get the students to think. They can consider that complexification inherits bases and does not change matrices, but does change the set of vectors. On the other hand, realification does not change the set of vectors, but doubles the dimension and requires extended bases. These issues are developed in very similar terms in Arnold, Ordinary differential equations, although in different notation. (I got the idea from Milnor and Stasheff.) Arnold has the very nice exercise of computing the realification of a complex matrix, and you can likewise ask what happens to the trace and the determinant. Another pair of ideas that is helpful for overthrowing the idea of a set basis is, (a) the vector space of formal linear combinations of a set, and (b) a quotient of such a vector space by relations. A particular example is the color vector space and the reduced color vector space: $$\mathbb{R}[\{\text{red}, \text{green}, \text{blue}\}] \qquad \mathbb{R}[\{\text{red}, \text{green}, \text{blue}\}]/(\text{red}+\text{green}+\text{blue}).$$ The students cannot choose a basis of the second vector space without breaking a symmetry that they would like to keep. Expressing the same linear transformation of the reduced color vector space in different bases is enlightening. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. I think linear algebra is not a good topic to start with if you want to introduce students to abstract mathematics: because all n-dimensional vector spaces (over R, say) are isomorphic to R^n, it is not easy to say what has been gained by the abstraction. Of course, something definitely has been gained, but that something is hard to explain. With finite groups, by contrast, the role of abstraction is much more obvious: all you need to do is present two rather different groups (such as S_4 and the group of rotations of the cube) and show that they are isomorphic. - IMO it's maybe not the best 1st course but it does make a good "redundancy" course for students to build-up confidence in their comprehension of abstract mathematics. – Ryan Budney Dec 20 2009 at 8:17 4 I agree. I think seeing what a group is, and what a homomorphism is before seeing what a vector space is, is a good way to go. I think that seeing group homomorphisms before working with vector spaces might help students understand linear maps, and especially help them to understand how to work with them without choosing a basis. – Grétar Amazeen Dec 20 2009 at 15:26 1 Gowers' remark is spot on. The gain from the abstraction - greater generality, no coordinates - becomes apparent when one studies, say, boundary value problems or the hairy ball theorem. I'd prefer to attempt a description of those things to beginning linear algebra students than take the disingenuous line of some textbooks, which claim that one gains from the abstract approach when all their proofs work equally well for subspaces of $\mathbb{R}^n$. – Tim Perutz Dec 22 2009 at 12:44 3 As I comment in my answer, I think that linear algebra is one of the topic where one has the biggest gain by abstraction. The very fact that for a given dimension you only have an example up to isomorphism means that with the minimun geometric effort (devising a proof for R^n) you get the maximum reward (everything you have done applies to polynomials, operators, matrices, solutions to linear differential equations, field extensions and so on). Many of these examples can be understood at the level of first year undergraduate. – Andrea Ferretti May 27 2010 at 23:47 I would suggest the approach Tom Apostol takes in his linear algebra book. In chapter 1, after introducing abstract vector spaces, he goes on to Gram-Schmidt, and then immediately to best approximations. At the end of the first chapter, he solves questions like: "find the polynomial of degree three $p(x)$ which approximates $\sin(x)$ best over $[2,3]$ in the sense of minimizing the error $\int_2^3 (sin(x)-p(x))^2 dx$. When I first read this, I was amazed. Prior to this, I only knew high school mathematics plus basic calculus - no abstract math at all. The problem of approximating one function by another seemed completely unsolvable given the mathematics I knew at the time. And yet here it had a simple solution. Even more amazingly, the solution was right in front of me all along. If you had asked me how to approximate the vector $(1,2,3)$ by a vector whose last coordinate was $0$ - I would have immediately said $(1,2,0)$. I knew a little bit about geometry problems, and the problem of finding the closest point in a plane seemed "easy" and "natural" to me. And yet this this is all the solution of this problem required - all I needed was just to think about "vectors" or "points" a little more abstractly. I was completely sold on the benefits of the abstract approach. - My understanding is that this is how calculators actually compute the sine, too. – Qiaochu Yuan Jan 20 2010 at 21:54 2 Qiaochu, I don't think that's actually true; that's just a lie we tell to calculus students to convince them that Taylor series are worth learning! Apparently the CORDIC algorithm (en.wikipedia.org/wiki/CORDIC) , which is based on calculating the sine or cosine by applying rotation matrices to the vector (1, 0), is what's actually used. – Michael Lugo Jan 20 2010 at 22:16 Interesting. (Taylor series and L^2 approximations aren't the same, but I see your point.) – Qiaochu Yuan Jan 20 2010 at 23:12 I think the best way to appreciate abstraction is to actually see examples of vector spaces which are not $\mathbb{R}^n$ in any obvious way. For instance all polynomials of degree $k$ such that $p(0) = 0$ or all symmetric $3 \times 3$ matrices. For a more subtle example, subset of a finite set $X$ are a vector space over $\mathbb{Z}/(2)$ taking as $+$ the symmetric difference (until one realizes that this is just $(\mathbb{Z}/(2))^n$ in disguise). Students should be offered many exercises with these vector spaces, so that they become familiar. When finite dimensionality is not necessary, one can make even better. For instance it is very nice to see the derivative as an example of a linear operator, and if one wants to have a finite dimensional example one can take the vector space of all solutions to some constant coefficients linear differential equation. Even a particular one, like $4f''' + 2f'' -f' -2 = 0$ will do. The derivative of a solution is a solution, hence we have a very natural linear endomorphism of this space. And by the way some linear algebra (for instance Jordan decomposition or even less) can be used to actually solve the equation. Moreover I think that the fact that all vector spaces are isomorphic to $\mathbb{R}^n$ shows the power of abstraction at its best. Geometrically we only need to have the intuition for one very simple case; but then the proofs we give will apply to a plethora of other unexpected cases. - I completely and utterly agress,Andrea.I personally think mathematicians use too many proofs and not enough examples in teaching-and if they don't give proofs,they turn the whole course into a Moore method problems course. I think a much more effective teaching philosophy is to give many,many detailed examples and state all but the most difficult proofs as either exercises or sketchy proofs. Theorums are generalizations of many examples and this method will emphasize this aspect of mathematics.It will also encourage them to build on this stock of examples,a very important skill for research. – Andrew L May 28 2010 at 0:20 Yes, spending some time looking at the spaces of polynomials, matrices, and later their subspaces, describing bases, the dimension, doing Gram-Schmidt orthogonalization, etc really helps! I, too, like to do examples with symmetric/skew-symmetric matrices, as well as even/odd functions (or polynomials of bounded degree), and one can illustrate things like the kernel and image, and the rank-nullity theorem using the symmetrization map $A\mapsto (A+A^t)/2.$ – Victor Protsak May 28 2010 at 5:20 4 In addition to ODEs, which are fairly conventional (I even taught some courses with both LA and DE in the title), I'd like to advocate the use of $\textit{difference}$ equations: e.g. consider all sequences $(a_n), n\geq 0$ s.t. $a_{n+2}=a_{n+1}+a_n.$ They form a 2-dimensional vector space $A$ with shift operator $T: A\to A,\ (Ta)_n=a_{n+1}.$ Show how the Binet formula for Fibonacci sequence can be obtained "naturally" from the diagonalization of $T$. The book I used last fall (David Lay) spends some time on linear discrete dynamical systems, another good application of the same idea. – Victor Protsak May 28 2010 at 5:29 To understand a definition, show the students (a) lots of examples, (b) lots of non-examples and why they don't work, (c) misconceptions related to the definition (e.g., coordinates and real/imag. parts have no intrinsic meaning in an abstract vector space) and (d) applications which use the definition in a productive way. In addition to showing a class how concepts they thought make sense in concrete settings (e.g., the first coordinate of a vector, or even that a vector has all positive coordinates) do not make sense in the abstract setting, show them that other things they have heard about do make sense abstractly, e.g., the determinant, characteristic polynomial, and eigenvalues look the same using two different bases, and those are the really important concepts. Otherwise they get the idea that all of their previous education in linear algebra doesn't work anymore. You can't expect the students to catch on to the definition of an abstract vector space right away, but only over time, based on what you do with it. In the original question there was no indication of what was actually done with abstract vector spaces. A definition on its own will inspire few people, particularly a typical class of math majors with varying abilities. One nice application of linear algebra over Q is rationalizing a nonquadratic denominator (e.g., 1/(1 + 2^(1/3) + 34^(1/3)) and one nice application of linear algebra over Z/2 is the quadratic sieve factorization algorithm. These are not basis-free applications, but they serve to illustrate how the ideas of linear algebra can be used in settings that are not directly about "concrete vectors". - This is something I could get behind, provided the coordinate free approach is stressed. (Not at the level of, say, Axler, who takes it way too far and somehow manages to lose the determinant entirely until the very end of the book.) – Harry Gindi Jan 19 2010 at 18:04 I learned abstract vector spaces in a coordinate free fashion straight away in my first linear algebra course-but after we spent a month doing concrete calculations with matrices and thier determinants. I totally agree with KConrad in the importance of COUNTER-examples. They are not utilized enough in lower level courses and thier importance cannot be overestimated. – Andrew L May 28 2010 at 1:14 You could try giving the following example: the set of all positive real numbers, considered as a vector space over the field R, with vector addition given by multiplication and scalar multiplication given by taking exponents. As a first step, you could verify that this satisfies a few of the vector-space axioms, and then let students check the rest of them (say, as homework). Then, you could ask questions like, "what is the dimension of this vector space?" or, "give an example of a (nontrivial) linear transformation from this space into R^3." - 1 I never saw the point of unnatural examples of vector spaces like this one, when in fact there are many natural ones: functions, polynomials, subspaces of other spaces, $\mathbb{C}$ as a real vector space, etc. This is just an example of "definition for definition's sake" mentioned elsewhere on this thread. – Victor Protsak May 28 2010 at 4:52 5 I disagree strongly with Victor. I think one of the problems I had when learning linear algebra was boredom: people claimed great generality of results but they were always applied to objects which always had extremely obvious representations as lists of numbers. Polynomials, especially if you're not going to multiply them together, are nothing but the list of their coefficients. Matrices are big lists of numbers. Everything that could "now" be treated just like R^n already looked exactly like R^n. SNORE. The multiplicative example is nice because it doesn't look "linear" at first glance. – Pietro KC May 28 2010 at 7:49 Like Pietro says, by the time I could prove that all $n$-dimensional vector spaces "are" $\mathbb{R}^n$, this seemed apparent from the examples. To be honest, I was nevertheless impressed that our abstract definition led to such a concrete characterisation, and one matching the intuition, but this example would have seemed to give the theorem more content, and led me earlier to the interesting idea that $\mathbb{R}$ "is" $\mathbb{R}^+$. – Max Nov 22 2010 at 22:31 1 I'm not at all convinced that this is an unnatural example- is this example not where ln comes from? You can motivate exp and ln this way- Cartier does so in mat.univie.ac.at/~slc/wpapers/s44cartier1.pdf%20 (Thanks Tom Copeland!). You can argue, as Cartier does, that ln IS the vector space isomorphism from $\mathbb{R}^+$ to $\mathbb{R}$. This is the best mathematical motivation of ln that I know- "rate of growth" isn't mathematical, "integral of 1/x" is not compelling... but "converts to + and exponentiation to scalar multiplication" is gold. – Daniel Moskovich Apr 20 2012 at 5:46 There is a substantial set of people who understand the notion of an interface in computer science, but don't understand abstraction in mathematics. For those people, it's worth pointing out that these are in fact the same thing (eg. if this is a linear algebra course for CS students). Many programming languages (eg. the commonly taught Java) support the notion of an interface that is separate from an implementation. An example at wikipedia is that of a Predator interface shared by many different types of predator. CS students generally get the idea (or at least they should) that if they write for the Predator interface then their code can be reused with any predator, but that if they use implementation specific details of a particular predator then their code cannot be reused. The situation is identical in mathematics. If you write mathematics for the "vector space interface" then your theorems can be reused for any vector space. But if you use specific knowledge of an underlying implementation (eg. about the specific set $\mathbb{R}^n$) then you lose the ability to reuse those theorems. In fact, even if the class isn't being taught to CS students it's worth a brief mention as any large enough class of mathematics students is bound to contain a few who are computer savvy. - Unfortunately, this still doesn't help too much if the student knows that every fin.-dim. vector space is isomorphic to some $\mathbb{R}^{n}$. To give a CS analogy, this is like writing a program that first determines what the specific implementation it is being applied to, and then uses implementation specific details to proceed. If it contains a comprehensive list of all possible implementations, then it can work quite well. Though it's a bit more subtle in the linear algebra case - we don't have access to the implementation, so we have to construct a virtual implementation isomorphic to it. – darij grinberg Jan 23 2010 at 16:21 But still this doesn't explain to the student why it is that important to work at the interface level only. And I have no idea how to explain this, other than by using modules over rings (where not every module is free anymore, but lots of linear algebra don't hold either), by categories (the category of fin.-dim. vector spaces is not equivalent to that of the $\mathbb{R}^n$'s, or is it?), or by experience (superficial extra structure obfuscates stuff and makes thinking more difficult, and experience shows that vector space bases are often superficial, though a student will hardly believe it). – darij grinberg Jan 23 2010 at 16:25 The category of f.d. vector spaces is in fact equivalent to the category whose objects are the non-negative integers and whose morphisms are matrices; the latter is the skeleton of the former. Categorically an important point to drive home is that in linear algebra there are related but different categories floating around, such as the category of f.d. vector spaces equipped with an inner product and so forth. (It is really really important to drive home that inner products constitute extra structure rather than a convenient calculation.) – Qiaochu Yuan Jan 23 2010 at 17:20 Okay, how do you construct the equivalence between Vect^(fin) and the category of nonnegative integers and matrices? There is a canonical functor from the latter into the former, but how do you get the other direction? By a sort of axiom of choice for classes? – darij grinberg Jan 28 2010 at 19:31 @darij: this is the well-known classification of equivalence of categories. an equivalence is just a fully-faithful essential-surjective functor. this uses axiom of choice for classes. – Martin Brandenburg Jan 29 2010 at 13:01 I can share what I did having a similar concern in mind, but it was for point-set topology, not linear algebra. I am not sure how much of this can be translated to linear algebra, since student's minds are already full of preconceptions about what a vector space, but not about what a topological space is. After many years of tutoring point-set topology, I observed that students systematically thought of all topological spaces as $\mathbb{R}^n$, and that they always wanted to use balls, even if the topology was non-metrizable. Hence, when I got to teach my own point-set topology course, I tried something a bit radical: I did not talk about metric spaces at all until later in the course. I started with motivation. On the second day, I defined the notions of topology, homeomorphism (but not continuous function), and convergence of a sequence. Then I did only small finite examples first. I gave the students the following exercise: 1) How many topologies can you define in {0,1,2}; 2) How many of them produce homeomorphic topological spaces?; and 3) In how many of them does the sequence $0,1,0,1,0,1, \ldots$ converge to $2$? Then I made sure to give students enough time (and guidance) to solve this exercise before moving to anything else. I wanted to force the students to accept the abstract notion of topology and to not be scared by it (and to realize that everything we do in point-set topology is logical). Also, in this example, there is no way a student is going to attempt to use balls (particularly when I have not talked about balls). I think it worked well. - 4 To be honest, I would be somewhat confused by finite-set topologies - after all, they are not the raison-d-etre of topology, and they are not a particularly interesting object either (I mean, not even their number is known; to me this is evidence that they are not really in the spirit of mathematics). I would probably start with a parallel treatment of "real" topologies (the kind one has on $\mathbb{R}^n$ and on manifolds), of Zariski-style topologies (very important) and of $\mathfrak{m}$-adic topologies (such as on power series rings). – darij grinberg Jan 23 2010 at 16:30 1 Munkres has five main counterexamples in his topology course: R with the lower-limit topology, [0, 1] x [0, 1] in the dictionary order, the first uncountable ordinal, R^J in the product topology, R^J in the box topology. It's especially helpful to have examples of unfamiliar topologies defined on familiar spaces. – Qiaochu Yuan Jan 23 2010 at 17:14 3 Except that in this case, it leads us nowhere, or are there interesting nontrivial facts on finite-set topologies? Besides, the subtle but important difference between arbitrary (as in: arbitrary unions) and arbitrary finite (as in: arbitrary finite intersections) is lost in this finite example. Anyway, topology is not the subject by which people should be introduced to mathematics... – darij grinberg Jan 28 2010 at 19:28 1 @darij Every finite CW complex is weakly homotopy equivalent to a finite topological space. For example there is a space with 4 elements which has all the same homotopy groups as the circle. – Steven Gubkin May 27 2010 at 14:48 7 I think I'd rather risk that students use balls when inappropriate than having them puzzled at the meaning of this strange axioms of topology when they do not have even the most natural examples. There is time to grasp and appreciate the complexity. It sounds to me like introducing schemes without having seen quasiprojective varieties: maybe you will think twice before making a geometric argument which is not backed up by commutative algebra, but... well, unless you dump algebraic geometry altogether. – Andrea Ferretti May 27 2010 at 23:39 show 3 more comments Every time that I've taught someone what a vector space is, I first spent some time on what a field is, including examples of finite fields. It's rather hard for someone to claim you can take the real and imaginary part of a vector if you say "Ahh, but the field I'm thinking of is $\mathbb{F}_2$, so what does that mean?" It's always seemed to help with what a vector space is to first see what a field is, and get some experience manipulating axioms that are more familiar. - 1 It might also be helpful to use counterexamples, e.g., $\mathbb{Z}$ is not a field under the usual operations...you don't have to say the words "ring", "module", etc., but I've personally always found that it helps to show where things can break. In mathematics you've always got the Scylla of showing that the objects of interest exist at all on the one hand, and the Charybdis of showing that said objects also aren't trivial or exhaustive. – Steve Huntsman Dec 20 2009 at 5:10 In Israel the two mandatory courses of first year undergrad math are real-analysis and abstract linear algebra (I think it's the same in Europe). You define fields before you define vector spaces, and you give as examples $\mathbb{F}_p, \mathbb{Q}, \mathbb{R}, \mathbb{C}$. Once you teach what a linear transformation is, you have several examples involving $\mathbb{F}_2$ coming from computer science; e.g. Hamming code. I'm not claiming that teaching first years abstract linear algebra is good (when I was an undergrad, half the students flunked first year math), just that if you do it you must have some non $\mathbb{R} / \mathbb{C}$ examples. - 1 That's the way it should be. – Harry Gindi Dec 20 2009 at 17:26 12 I'm not sure that's the way it should be (I'm not sure it shouldn't either). I taught linear algebra for science majors in Princeton; after one semester they've seen (with only partial proofs, and only over R/C) SVD, FFT and finite elements. Who knows more linear algebra: the student who knows what fields are and saw all the proof but not a single real life example, or the students who does not know what a field is, saw only few proofs, but knows important examples ? – David Lehavi Dec 20 2009 at 17:58 1 Real life examples are misleading. – Harry Gindi Dec 22 2009 at 17:57 14 @Harry: Bad real life examples are misleading. Good real life examples are e.g. the two body problem and pendulum (which started differential equations), heat conduction on a rod (which started Fourier transform), length of the lemniscate (which - via elliptic integrals - was the beginning of both algebraic geometry and a lot of complex analysis), three body problem (which started ergodic theory). They are anything but misleading. – David Lehavi Dec 22 2009 at 20:38 5 Time and again one meets people who have come up with the "proper way" to teach undergrads by introducing topoi before integers, and as many times one finds undergrads completely unresponsive to these methods. One suspects that, if such people got hold of a time machine, they would be utterly unable to teach their past selves anything. – Pietro KC May 28 2010 at 8:08 show 1 more comment This is partly redundant with previous answers: one can present the students with a vector space that does not have a natural basis. What I would like to stress is that the simplest is probably the best, at least for a first example, and that the simplest is to take a vectorial plane in $\mathbb{R}^3$. What would be the two coordinates of a vector in $V=\{(x,y,z) \| x+y+z=0 \}$? I confess that students could be trying to think of these vectors in $\mathbb{R}^3$ rather than in $V$, so that this example maybe better to explain the need of a definition for a basis. - This example actually makes a nice introduction to resolutions of vector spaces, since we get to keep the symmetry at little cost. – John Wiltshire-Gordon Jul 2 2011 at 12:42 Maybe I overlooked it, but I didn't see, in the previous answers, anything about a really geometric view of vectors. When introducing vector spaces, I like to use 2-dimensional vectors (arrows drawn on the blackboard, with the understanding that only length and direction matter, not the location on the board), with geometric definitions of addition and scalar multiplication. It is, of course, easy to explain that these geometric vectors are "really the same" as 2-component algebraic vectors (i.e., elements of $\mathbb R^2$), and also that the sameness depends on the choice of a coordinate system. This approach provides me with a lot of analogies for more complicated things that come up later in the course. - The fixed ideas you describe probably originated from earlier calculus courses where students were exposed to "vectors" without any reference to vector spaces. You could try some decontamination by first introducing groups, rings, fields and modules, before proceeding with vector spaces. Of course I do not suggest to turn the course into abstract algebra by going deep into group theory or ring theory; just giving a few definitions, plenty of examples and some immediate results, should be sufficient. I see the following advantages: • the students meet something new right in the beginning, so they are less likely to fall into the "Oh, I already know this"-mode. • later definitions e.g. of a vector space can be build on previous ones and grouped into meaningful parts. • results like e.g. the homomorphism theorems can be given several times in slightly different situations. The main problem with this approach is the danger of running out of time. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 60, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9515283703804016, "perplexity_flag": "middle"}
http://www.physicsforums.com/showpost.php?p=3264559&postcount=5	View Single Post Hi Tom, You're after a unified description of scalar, fermion and gauge fields… very ambitious. But don't forget the gravitational spin connection and frame. Let $$A$$ be a 1-form gauge field valued in a Lie algebra, say spin(10) if you like GUTs, and $$\omega$$ be the gravitational spin connection 1-form valued in spin(1,3), and $$e$$ be the gravitational frame 1-form valued in the 4 vector representation space of spin(1,3), and let $$\phi$$ be a scalar Higgs field valued in, say, the 10 vector representation space of spin(10). Then, avoiding Coleman-Mandula's assumptions by allowing e to be arbitrary, possibly zero, we can construct a unified connection valued in spin(11,3): $$H = {\scriptsize \frac{1}{2}} \omega + \frac{1}{4} e \phi + A$$ and compute its curvature 2-form as $$F = d H + \frac{1}{2} [H,H] = \frac{1}{2}(R - \frac{1}{8}ee\phi\phi) + \frac{1}{4} (T \phi - e D \phi) + F_A$$ in which $$R$$ is the Riemann curvature 2-form, $$T$$ is torsion, $$D \phi$$ is the gauge covariant 1-form derivative of the Higgs, and $$F_A$$ is the gauge 2-form curvature -- all the pieces we need for building a nice action as a perturbed $$BF$$ theory. To include a generation of fermions, let $$\Psi$$ be an anti-commuting (Grassmann) field valued in the positive real 64 spin representation space of spin(11,3), and consider the "superconnection": $$A_S = H + \Psi$$ The "supercurvature" of this, $$F_S = d A_S + A_S A_S = F + D \Psi + \Psi \Psi$$ includes the covariant Dirac derivative of the fermions in curved spacetime, including a nice interaction with the Higgs, $$D \Psi = (d + \frac{1}{2} \omega + \frac{1}{4} e \phi + A) \Psi$$ We can then build actions, including Dirac, as a perturbed $$B_S F_S$$ theory. Once you see how all this works, the kicker is that this entire algebraic structure, including spin(11,3) + 64, fits inside the E8 Lie algebra.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 16, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8983510136604309, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/tagged/plane-curves+bezier-curve	# Tagged Questions 1answer 29 views ### Understanding the Spiro Spline My name's Wray. This is my first time here. Firstly, I like curves. I've been keeping a pet project for a long time that would implement a delightful new curve-interpolation algorithm named the Spiro ... 1answer 104 views ### An almost straight curve with infinite curvature? I played around with computing the curvature of some curves, and found this weird example that is driving me nuts. Consider the following (Bézier) curve (on a plane, the first point is $[-1,0]$): ... 1answer 95 views ### Can an involute gear profile be modeled with a Bézier curve? In the context of a game, I want to draw gears. The most common curves available on the platforms I'm using are third degree Bézier curves. Is there an exact representation of the involute gear ... 1answer 383 views ### Bézier approximation of archimedes spiral? As part of an iOS app I’m making, I want to draw a decent approximation of an Archimedes spiral. The drawing library I’m using (CGPath in Quartz 2D, which is C-based) supports arcs as well as cubic ... 1answer 146 views ### Continuous curve interpolating a list of points I need a function (a curve -- preferably a simple one) that, given $n$ points of a 2D space ($R^2$) passes (interpolates) through all points in a smooth/continuous way. Found out that what I need is ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9323148727416992, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2011/02/25/classes-of-manifolds/?like=1&source=post_flair&_wpnonce=c8d80d4293	# The Unapologetic Mathematician ## Classes of Manifolds As we discussed the dimension of a manifold yesterday, we passed by an interesting construction that we want to look at in more detail. Let $(U_1,\phi_1)$ and $(U_2,\phi_2)$ be two coordinate patches on an $n$-manifold $M$. We can restrict each coordinate map to the intersection of the two patches — if the intersection is empty this is trivial, but not wrong — and get a function from one open region of $\mathbb{R}^n$ to another. Specifically, if the coordinate patches have images $V_1=\phi_1(U_1\cap U_2)$ and $V_2=\phi_2(U_1\cap U_2)$, then we get a function $\phi_1\circ\phi_2^{-1}:V_2\to V_1$. We call this the “transition function” from one coordinate patch to another, and often write it $\phi_{2,1}$ So what do we know about these transition functions? Well, $\phi_{2,1}$ is a homeomorphism from $V_2$ to $V_1$, meaning it’s continuous with a continuous inverse. But remember that $V_i\subseteq\mathbb{R}^n$, and we know a lot about real $n$-dimensional space. We know all kinds of extra structure beyond just its topology, and in particular we know what it means for a function from $\mathbb{R}^n$ to itself to be smooth. Now, as I’ve said before, “smooth” ends up behaving like a term of art. Really what we know is what it means for a function from $\mathbb{R}^n$ to itself to be differentiable, and for that differential to be continuous, and for there to be a second differential, and so on. We even introduced classes of functions to describe this whole tower, where $C^0$ consists of continuous functions, $C^1$ consists of continuously differentiable functions, $C^2$ consists of continuously twice-differentiable functions. This pattern continuous on through $C^k$ for continuously $k$-times-differentiable functions, $C^\infty$ for functions with arbitrarily many continuous derivatives, and $C^\omega$ for “analytic” functions which are the limits of their own Taylor series. One case we didn’t explicitly mention is $PL$: the class of “piecewise linear” functions. These are continuous functions that can be defined by breaking up their domains into a finite number of chunks and giving an affine function — a linear transformation plus a constant offset vector — on each chunk so that the borders line up continuously. These functions aren’t generally differentiable, since they turn sharp corners at the boundaries of chunks, but they’re pretty nicely behaved anyway. The class $PL$ sits between $C^0$ and $C^1$. Anyway, for each class of functions we can define a corresponding class of manifolds. If all the transition functions between all the coordinate patches are piecewise-linear, we say that we have a “PL-manifold”. If they’re all $C^1$, we say that we have a $C^1$-manifold or a “differential manifold”. This goes up through $C^k$-manifolds, $C^\infty$-manifolds, and analytic manifolds. In practice, “smooth” ends up meaning $C^\infty$, so that we can always take as many derivatives as we want, but really we rarely need more than a few. Still, we’ll say “smooth” and not worry about it. And from here on all our manifolds will be smooth unless explicitly stated otherwise. ### Like this: Posted by John Armstrong \| Differential Topology, Topology ## 10 Comments » 1. In “These functions aren’t generally continuous”, you meant “These functions aren’t generally differentiable”,right? Comment by \| February 25, 2011 \| Reply • Ah, right. Thanks for catching me. Comment by \| February 25, 2011 \| Reply 2. [...] Differentiable Structures It’s high time we introduced the “standard” smooth structures on real vector spaces, which are (of course) our models for all other smooth [...] Pingback by \| March 2, 2011 \| Reply 3. [...] that we know what it means for coordinate patches to be compatible — for various definitions of “compatible” — we can define atlases on a [...] Pingback by \| March 2, 2011 \| Reply 4. [...] usual, we’re going to want our objects of study — smooth (or differentiable) manifolds — to be objects in a category. And a category means we need [...] Pingback by \| March 2, 2011 \| Reply 5. [...] sheaves of use in differential topology. Given a smooth manifold — for whatever we choose smooth to mean — we can define sheaves of real algebras of real-valued functions for every [...] Pingback by \| March 23, 2011 \| Reply 6. [...] all this means is that we take the transition function , take the th component, and take the th partial derivative of that function. And this is precisely [...] Pingback by \| April 1, 2011 \| Reply 7. [...] are the components of the Jacobian matrix of the transition function . What does this mean? Well, consider the linear [...] Pingback by \| April 13, 2011 \| Reply 8. [...] we can cancel off the two second partial derivatives because we’re assuming that is “smooth”, which in this case entails “has mixed second partial derivatives which commute” in any [...] Pingback by \| June 2, 2011 \| Reply 9. [...] for smooth structures on and , we define them exactly as usual; real-valued functions on a patch of containing some [...] Pingback by \| September 13, 2011 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 32, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9318819642066956, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2011/03/07/open-submanifolds/?like=1&source=post_flair&_wpnonce=c1c9c39881	# The Unapologetic Mathematician ## Open Submanifolds Eek! None of these drafts went up on time! In principle, we know what a submanifold should be: a subobject in the category of smooth manifolds. That is, a submanifold $S$ of a manifold $M$ should be another manifold, along with an “inclusion” map which is smooth and left-cancellable. On the underlying topological space, we understand subspaces; first and foremost, a submanifold needs to be a subspace. And one easy way to come up with a submanifold is just to take an open subspace. I say that any open subspace $S\subseteq M$ is automatically a submanifold. Indeed, if $(U,\phi_U)$ is a coordinate patch on $M$, then $(U\cap S,\phi_U\vert_{U\cap S})$ is a coordinate patch on $S$. The intersection $U\cap S$ is an open subset, and the restriction of $\phi_U$ to this intersection is still a local homeomorphism. Since the collection of all coordinate patches in our atlas cover all of $M$, they surely cover $S$ as well. As a quick example, an open interval in the real line is automatically an open manifold of $\mathbb{R}$, and so it’s a manifold. Any open set $U$ in any $n$-dimensional real vector space is also automatically an $n$-manifold. More generally, it turns out that what we want to consider as a “submanifold” is actually somewhat more complicated, and we will have to come back to this point later. ### Like this: Posted by John Armstrong \| Differential Topology, Topology ## 2 Comments » 1. [...] we said before, the notion of a “submanifold” gets a little more complicated than a naïve, purely [...] Pingback by \| April 18, 2011 \| Reply 2. [...] we can conclude that is an open submanifold of , which comes equipped with the standard differentiable structure on . Matrix multiplication is [...] Pingback by \| June 9, 2011 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 15, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.918121337890625, "perplexity_flag": "head"}
http://mathhelpforum.com/algebra/31672-equation.html	# Thread: 1. ## The equation The equation $[x^{10}+(13x-1)^{10}=0$ has 10 complex roots $r_1, \overline{r_1}, r_2, \overline{r_2}, r_3, \overline{r_3}, r_4, \overline{r_4}, r_5, \overline{r_5},,$ where the bar denotes complex conjugation. Find the value of $\frac 1{r_1\overline{r_1}}+\frac 1{r_2\overline{r_2}}+\frac 1{r_3\overline{r_3}}+\frac 1{r_4\overline{r_4}}+\frac 1{r_5\overline{r_5}}$. 2. The solution is 850. Use the sum and the product of the roots of a polynomial rules.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.7431881427764893, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/116399/if-ell-a-mathfrakpn-infty-then-is-it-true-that-operatornamehom-a	# If $\ell_{A_\mathfrak{p}}(N)<\infty$, then is it true that $\operatorname{Hom}_A(N,E(A/\mathfrak{q}))=0$? Let $A$ be a Noetherian ring, $\mathfrak{p},\mathfrak{q}\subset A$ distinct prime ideals of the same height, $N$ an $A_\mathfrak{p}$-module of finite length. Then is it true that $$\operatorname{Hom}_A(N,E(A/\mathfrak{q}))=0,$$ where $E(A/\mathfrak{q})$ is the injective hull of $A/\mathfrak{q}$? The main trouble here is that $N$ may not be finitely generated over $A$. If it were, I would use formulas like \begin{eqnarray} \operatorname{Ass}\operatorname{Hom}_A(N,E(A/\mathfrak{q}))&=&\operatorname{Supp}_AN\cap\operatorname{Ass}E(A/\mathfrak{q})\ &=&V(\mathfrak{p})\cap\lbrace\mathfrak{q}\rbrace\ &=&\emptyset. \end{eqnarray} - Presumably $E$ stands for injective envelope? – Steve Mar 4 '12 at 19:04 @Steve Yes; Sorry, I edited that in. – ashpool Mar 4 '12 at 19:31 ## 1 Answer Suppose $\phi\in\operatorname{Hom}_A(N,E(A/\mathfrak{q}))$, and $\phi(n)=e\neq0$. We will derive a contradiction. Since $N$ is of finite length over $A_{\mathfrak{p}}$, $\mathfrak{p}^kN=0$ for some $k$. Since $\mathfrak{p}\not\subset\mathfrak{q}$, $\exists a\in\mathfrak{p}^k\backslash\mathfrak{q}$, so $a$ acts as a unit on $E(A/\mathfrak{q})$. Then $$0=\phi(an)=ae\neq0,$$ a contradiction. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.7926022410392761, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/tagged/periodic-functions+continued-fractions	# Tagged Questions 2answers 275 views ### Is the continued fraction of the square root of a base $\phi$ (golden ratio) number periodic when the continued fraction is expressed in base $\phi$? I have been looking at concise ways to represent irrational numbers using only integers. I was thinking about base $\phi$ (golden ratio base) and how it can represent the quadratic extension of the ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9313554763793945, "perplexity_flag": "middle"}
http://mathforum.org/mathimages/index.php?title=The_Golden_Ratio&diff=33504&oldid=33500	# The Golden Ratio ### From Math Images (Difference between revisions) \| \| \| \| \| \|-----------\|----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------\|-----------\|----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------\| \| \| \| \| \| \| Line 5: \| \| Line 5: \| \| \| \| The term '''golden ratio''' refers to any ratio which has the value phi. The image to the right is a warped representation of dividing and subdividing a rectangle into the golden ratio. The result is [[Field:Fractals\|fractal-like]]. This page explores real world applications for the golden ratio, common misconceptions about the golden ratio, and multiple derivations of the golden number. \| \| The term '''golden ratio''' refers to any ratio which has the value phi. The image to the right is a warped representation of dividing and subdividing a rectangle into the golden ratio. The result is [[Field:Fractals\|fractal-like]]. This page explores real world applications for the golden ratio, common misconceptions about the golden ratio, and multiple derivations of the golden number. \| \| \| \|ImageDescElem=[[Image:Monalisa01.jpg\|Does the Mona Lisa exhibit the golden ratio?\|thumb\|400px\|right]]The golden number, approximately 1.618, is called golden because many geometric figures involving this ratio are often said to possess special beauty. Be that true or not, the ratio has many beautiful and surprising mathematical properties. The Greeks were aware of the golden ratio, but did not consider it particularly significant with respect to aesthetics. It was not called the "divine" proportion until the 15th century, and was not called "golden" ratio until the 18th century. <ref>[http://en.wikipedia.org/wiki/Golden_ratio "Golden ratio"], Retrieved on 20 June 2012.</ref> \| \| \|ImageDescElem=[[Image:Monalisa01.jpg\|Does the Mona Lisa exhibit the golden ratio?\|thumb\|400px\|right]]The golden number, approximately 1.618, is called golden because many geometric figures involving this ratio are often said to possess special beauty. Be that true or not, the ratio has many beautiful and surprising mathematical properties. The Greeks were aware of the golden ratio, but did not consider it particularly significant with respect to aesthetics. It was not called the "divine" proportion until the 15th century, and was not called "golden" ratio until the 18th century. <ref>[http://en.wikipedia.org/wiki/Golden_ratio "Golden ratio"], Retrieved on 20 June 2012.</ref> \| \| - \| <br /> \| + \| <br /> [[Image:Finalpyramid1.jpg\|Markowsky has determined the above dimensions to be incorrect.\|thumb\|400px\|left]] \| \| \| Since then, it has been claimed that the golden ratio is the most aesthetically pleasing ratio, and claimed that this ratio has appeared in architecture and art throughout history. Among the most common such claims are that the Parthenon and Leonardo Da Vinci's Mona Lisa use the golden ratio. Even more esoteric claims propose that the golden ratio can be found in the human facial structure, the behavior of the stock market, and the Great Pyramids. <br /> \| \| Since then, it has been claimed that the golden ratio is the most aesthetically pleasing ratio, and claimed that this ratio has appeared in architecture and art throughout history. Among the most common such claims are that the Parthenon and Leonardo Da Vinci's Mona Lisa use the golden ratio. Even more esoteric claims propose that the golden ratio can be found in the human facial structure, the behavior of the stock market, and the Great Pyramids. <br /> \| \| \| However, such claims have been criticized in scholarly journals as wishful thinking or sloppy mathematical analysis. Additionally, there is no solid evidence that supports the claim that the golden rectangle is the most aesthetically pleasing rectangle.<ref>[http://www.math.nus.edu.sg/aslaksen/teaching/maa/markowsky.pdf "Misconceptions about the Golden Ratio"], Retrieved on 24 June 2012.</ref> \| \| However, such claims have been criticized in scholarly journals as wishful thinking or sloppy mathematical analysis. Additionally, there is no solid evidence that supports the claim that the golden rectangle is the most aesthetically pleasing rectangle.<ref>[http://www.math.nus.edu.sg/aslaksen/teaching/maa/markowsky.pdf "Misconceptions about the Golden Ratio"], Retrieved on 24 June 2012.</ref> \| \| Line 12: \| \| Line 12: \| \| \| \| ===Misconceptions about the Golden Ratio=== \| \| ===Misconceptions about the Golden Ratio=== \| \| \| Many rumors and misconceptions surround the golden ratio. There have been many claims that the golden ratio appears in art and architecture. In reality, many of these claims involve warped images and large margins of error. One claim is that the Great Pyramids exhibit the golden ratio in their construction. This belief is illustrated below. \| \| Many rumors and misconceptions surround the golden ratio. There have been many claims that the golden ratio appears in art and architecture. In reality, many of these claims involve warped images and large margins of error. One claim is that the Great Pyramids exhibit the golden ratio in their construction. This belief is illustrated below. \| \| - \| [[Image:Finalpyramid1.jpg\|Markowsky has determined the above dimensions to be incorrect.\|thumb\|400px\|left]] \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| In his paper, ''Misconceptions about the Golden Ratio,'' George Markowsky disputes this claim, arguing that the dimensions assumed in the picture are not anywhere close to being correct. Another belief is that a series of [[The Golden Ratio#Jump2\|golden rectangles]] appears in the ''Mona Lisa''. \| \| In his paper, ''Misconceptions about the Golden Ratio,'' George Markowsky disputes this claim, arguing that the dimensions assumed in the picture are not anywhere close to being correct. Another belief is that a series of [[The Golden Ratio#Jump2\|golden rectangles]] appears in the ''Mona Lisa''. \| \| \| However, the placing of the golden rectangles seems arbitrary. Markowsky also disputes the belief that the human body exhibits the golden ratio. To read more, [http://www.math.nus.edu.sg/aslaksen/teaching/maa/markowsky.pdf click here!] \| \| However, the placing of the golden rectangles seems arbitrary. Markowsky also disputes the belief that the human body exhibits the golden ratio. To read more, [http://www.math.nus.edu.sg/aslaksen/teaching/maa/markowsky.pdf click here!] \| \| - \| <br /> \| + \| : \| \| - \| <br /> \| + \| : \| \| - \| <br /> \| + \| \| \| - \| <br /> \| + \| \| \| - \| <br /> \| + \| \| \| - \| <br /> \| + \| \| \| - \| <br /> \| + \| \| \| - \| <br /> \| + \| \| \| - \| <br /> \| + \| \| \| - \| <br /> \| + \| \| \| - \| <br /> \| + \| \| \| \| ====''What do you think?''==== \| \| ====''What do you think?''==== \| \| \| George Markowsky argues that, like the ''Mona Lisa,'' the Parthenon does not exhibit a series of golden rectangles (discussed below). Do you think the Parthenon was designed with the golden ratio in mind or is the image below simply a stretch of the imagination? \| \| George Markowsky argues that, like the ''Mona Lisa,'' the Parthenon does not exhibit a series of golden rectangles (discussed below). Do you think the Parthenon was designed with the golden ratio in mind or is the image below simply a stretch of the imagination? \| \| Line 417: \| \| Line 407: \| \| \| \| \|Field=Algebra \| \| \|Field=Algebra \| \| \| \|InProgress=Yes \| \| \|InProgress=Yes \| \| \| \| + \| } \| \| \| \| + \| \|Field=Algebra \| \| \| \| + \| \|InProgress=No \| \| \| } \| \| } \| \| \| \|Field=Algebra \| \| \|Field=Algebra \| \| \| \|InProgress=No \| \| \|InProgress=No \| \| \| }} \| \| }} \| ## Revision as of 09:59, 9 July 2012 The Golden Ratio Fields: Algebra and Geometry Image Created By: azavez1 Website: The Math Forum The Golden Ratio The golden number, often denoted by lowercase Greek letter "phi", is ${\varphi}=\frac{1 + \sqrt{5}}{2} = 1.61803399...$. The term golden ratio refers to any ratio which has the value phi. The image to the right is a warped representation of dividing and subdividing a rectangle into the golden ratio. The result is fractal-like. This page explores real world applications for the golden ratio, common misconceptions about the golden ratio, and multiple derivations of the golden number. # Basic Description Does the Mona Lisa exhibit the golden ratio? The golden number, approximately 1.618, is called golden because many geometric figures involving this ratio are often said to possess special beauty. Be that true or not, the ratio has many beautiful and surprising mathematical properties. The Greeks were aware of the golden ratio, but did not consider it particularly significant with respect to aesthetics. It was not called the "divine" proportion until the 15th century, and was not called "golden" ratio until the 18th century. [1] Markowsky has determined the above dimensions to be incorrect. Since then, it has been claimed that the golden ratio is the most aesthetically pleasing ratio, and claimed that this ratio has appeared in architecture and art throughout history. Among the most common such claims are that the Parthenon and Leonardo Da Vinci's Mona Lisa use the golden ratio. Even more esoteric claims propose that the golden ratio can be found in the human facial structure, the behavior of the stock market, and the Great Pyramids. However, such claims have been criticized in scholarly journals as wishful thinking or sloppy mathematical analysis. Additionally, there is no solid evidence that supports the claim that the golden rectangle is the most aesthetically pleasing rectangle.[2] ### Misconceptions about the Golden Ratio Many rumors and misconceptions surround the golden ratio. There have been many claims that the golden ratio appears in art and architecture. In reality, many of these claims involve warped images and large margins of error. One claim is that the Great Pyramids exhibit the golden ratio in their construction. This belief is illustrated below. In his paper, Misconceptions about the Golden Ratio, George Markowsky disputes this claim, arguing that the dimensions assumed in the picture are not anywhere close to being correct. Another belief is that a series of golden rectangles appears in the Mona Lisa. However, the placing of the golden rectangles seems arbitrary. Markowsky also disputes the belief that the human body exhibits the golden ratio. To read more, click here! #### What do you think? George Markowsky argues that, like the Mona Lisa, the Parthenon does not exhibit a series of golden rectangles (discussed below). Do you think the Parthenon was designed with the golden ratio in mind or is the image below simply a stretch of the imagination? [3] ## A Geometric Representation ### The Golden Ratio in a Line Segment The golden number can be defined using a line segment divided into two sections of lengths a and b. If a and b are appropriately chosen, the ratio of a to b is the same as the ratio of a + b to a and both ratios are equal to $\varphi$. The line segment above (left) exhibits the golden proportion. The line segments above (right) are also examples of the golden ratio. In each case, $\frac{{\color{Red}\mathrm{red}}+\color{Blue}\mathrm{blue}}{{\color{Blue}\mathrm{blue}} }= \frac{{\color{Blue}\mathrm{blue}} }{{\color{Red}\mathrm{red}} }= \varphi .$ ### The Golden Rectangle A golden rectangle is any rectangle where the ratio between the sides is equal to phi. When the sides lengths are proportioned in the golden ratio, the rectangle is said to possess the golden proportions. A golden rectangle has sides of length $\varphi \times r$ and $1 \times r$ where $r$ can be any constant. Remarkably, when a square with side length equal to the shorter side of the rectangle is cut off from one side of the golden rectangle, the remaining rectangle also exhibits the golden proportions. This continuing pattern is visible in the golden rectangle below. ### Triangles The golden number, $\varphi$, is used to construct the golden triangle, an isoceles triangle that has legs of length $\varphi \times r$ and base length of $1 \times r$ where $r$ can be any constant. It is above and to the left. Similarly, the golden gnomon has base $\varphi \times r$ and legs of length $1 \times r$. It is shown above and to the right. These triangles can be used to form regular pentagons (pictured above) and pentagrams. The pentgram below, generated by the golden triangle and the golden gnomon, has many side lengths proportioned in the golden ratio. $\frac{{\color{SkyBlue}\mathrm{blue}} }{{\color{Red}\mathrm{red}} } = \frac{{\color{Red}\mathrm{red}} }{{\color{Green}\mathrm{green}} } = \frac{{\color{Green}\mathrm{green}} }{{\color{Magenta}\mathrm{pink}} } = \varphi .$ These triangles can be used to form fractals and are one of the only ways to tile a plane using pentagonal symmetry. Pentagonal symmetry is best explained through example. Below, we have two fractal examples of pentagonal symmetry. Images that exhibit pentagonal symmetry have five symmetry axes. This means that we can draw five lines from the image's center, and all resulting divisions are identical. # A More Mathematical Explanation Note: understanding of this explanation requires: *Algebra, Geometry # An Algebraic Derivation of Phi [Click to expand] How can we derive the value of $\varphi$ from its characteristics as a ratio? We may algebraically solve for the ratio ($\varphi$) by observing that ratio satisfies the following property by definition: $\frac{b}{a} = \frac{a+b}{b} = \varphi$ [Click to hide] Let $r$ denote the ratio : $r=\frac{a}{b}=\frac{a+b}{a}$. So $r=\frac{a+b}{a}=1+\frac{b}{a}$ which can be rewritten as $1+\cfrac{1}{a/b}=1+\frac{1}{r}$ thus, $r=1+\frac{1}{r}$ Multiplying both sides by $r$, we get ${r}^2=r+1$ which can be written as: $r^2 - r - 1 = 0$. Applying the quadratic formula An equation, $\frac{-b \pm \sqrt {b^2-4ac}}{2a}$, which produces the solutions for equations of form $ax^2+bx+c=0$ , we get $r = \frac{1 \pm \sqrt{5}} {2}$. The ratio must be positive because we can not have negative line segments or side lengths. Because the ratio has to be a positive value, $r=\frac{1 + \sqrt{5}}{2} = 1.61803399... =\varphi$. The golden ratio can also be written as what is called a continued fraction,a fraction of infinite length whose denominator is a quantity plus a fraction, which latter fraction has a similar denominator, and so on. This is done by using recursion. [Click to expand] [Click to hide] We have already solved for $\varphi$ using the following equation: ${\varphi}^2-{\varphi}-1=0$. We can add one to both sides of the equation to get ${\varphi}^2-{\varphi}=1$. Factoring this gives $\varphi(\varphi-1)=1$. Dividing by $\varphi$ gives us $\varphi -1= \cfrac{1}{\varphi }$. Adding 1 to both sides gives $\varphi =1+ \cfrac{1}{\varphi }$. Substitute in the entire right side of the equation for $\varphi$ in the bottom of the fraction. $\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{\varphi } }$ Substituting in again, $\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{\varphi}}}$ $\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{\cdots}}}$ Continuing this substitution forever, the last infinite form is a continued fraction If we evaluate truncations of the continued fraction by evaluating only part of the continued fraction, replacing $\varphi$ with 1, we produce the ratios between consecutive terms in the Fibonacci sequence. $\varphi \approx 1 + \cfrac{1}{1} = 2$ $\varphi \approx 1 + \cfrac{1}{1+\cfrac{1}{1}} = 3/2$ $\varphi \approx 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1} } } = 5/3$ $\varphi \approx 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1+\cfrac{1}{1}}}} = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{2}}} =1 + \cfrac{1}{1 + \cfrac{2}{3}} = 8/5$ Thus we discover that the golden ratio is approximated in the Fibonacci sequence. $1,1,2,3,5,8,13,21,34,55,89,144...\,$ \| \| \| \| \|-----------------------------------------------------------------------------\|-----------------------------------------------------------------------------\|-----------------------------------------------------------------------------\| \| $1/1$ \| $=$ \| $1$ \| \| $2/1$ \| $=$ \| $2$ \| \| $3/2$ \| $=$ \| $1.5$ \| \| $8/5$ \| $=$ \| $1.6$ \| \| $13/8$ \| $=$ \| $1.625$ \| \| $21/13$ \| $=$ \| $1.61538462...$ \| \| $34/21$ \| $=$ \| $1.61904762...$ \| \| $55/34$ \| $=$ \| $1.61764706...$ \| \| $89/55$ \| $=$ \| $1.61818182...$ \| $\varphi = 1.61803399...\,$ As you go farther along in the Fibonacci sequence, the ratio between the consecutive terms approaches the golden ratio. Many real-world applications of the golden ratio are related to the Fibonacci sequence. For more real-world applications of the golden ratio click here! In fact, we can prove that the ratio between terms in the Fibonacci sequence approaches the golden ratio by using mathematical Induction. [Click to show proof] [Click to hide proof] Since we have already shown that $\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{\cdots}}}$, we only need to show that each of the terms in the continued fraction is the ratio of Fibonacci numbers as shown above. First, let $x_1=1$, $x_2=1+\frac{1}{1}=1+\frac{1}{x_1}$, $x_3= 1+\frac{1}{1+\frac{1}{1}}=1+\frac{1}{x_2}$ and so on so that $x_n=1+\frac{1}{x_{n-1}}$. These are just the same truncated terms as listed above. Let's also denote the terms of the Fibonacci sequence as $s_n=s_{n-1}+s_{n-2}$ where $s_1=1$,$s_2=1$,$s_3=2$,$s_4=3$ etc. We want to show that $x_n=\frac{s_{n+1}}{s_n}$ for all n. First, we establish our base case. We see that $x_1=1=\frac{1}{1}=\frac{s_2}{s_1}$, and so the relationship holds for the base case. Now we assume that $x_k=\frac{s_{k+1}}{s_{k}}$ for some $1 \leq k < n$ (This step is the inductive hypothesis). We will show that this implies that $x_{k+1}=\frac{s_{(k+1)+1}}{s_{k+1}}=\frac{s_{k+2}}{s_{k+1}}$. By our assumptions about x1,x2...xn, we have $x_{k+1}=1+\frac{1}{x_k}$. By our inductive hypothesis, this is equivalent to $x_{k+1}=1+\frac{1}{\frac{s_{k+1}}{s_{k}}}$. Now we only need to complete some simple algebra to see $x_{k+1}=1+\frac{s_k}{s_{k+1}}$ $x_{k+1}=\frac{s_{k+1}+s_k}{s_{k+1}}$ Noting the definition of $s_n=s_{n-1}+s_{n-2}$, we see that we have $x_{k+1}=\frac{s_{k+2}}{s_{k+1}}$ So by the principle of mathematical induction, we have shown that the terms in our continued fraction are represented by ratios of consecutive Fibonacci numbers. The exact continued fraction is $x_{\infty} = \lim_{n\rightarrow \infty}\frac{s_{n+1}}{s_n} =\varphi$. ## Proof of the Golden Ratio's Irrationality [Click to expand] [Click to hide] Remarkably, the Golden Ratio is irrational, despite the fact that we just proved that is approximated by a ratio of Fibonacci numbers. We will use the method of contradiction to prove that the golden ratio is irrational. Suppose $\varphi$ is rational. Then it can be written as fraction in lowest terms $\varphi = b/a$, where a and b are integers. Our goal is to find a different fraction that is equal to $\varphi$ and is in lower terms. This will be our contradiction that will show that $\varphi$ is irrational. First note that the definition of $\varphi = \frac{b}{a}=\frac{a+b}{b}$ implies that $b > a$ since clearly $b+a>b$ and the two fractions must be equal. Now, since we know $\frac{b}{a}=\frac{a+b}{b}$ we see that $b^2=a(a+b)$ by cross multiplication. Foiling this expression gives us $b^2=a^2+ab$. Rearranging this gives us $b^2-ab=a^2$, which is the same as :$b(b-a)=a^2$. Dividing both sides of the equation by a(b-a) gives us $\frac{b}{a}=\frac{a}{b-a}$. Since $\varphi=\frac{b}{a}$, this means $\varphi=\frac{a}{b-a}$. Since we have assumed that a and b are integers, we know that b-a must also be an integer. Furthermore, since $a<b$, we know that $\frac{a}{b-a}$ must be in lower terms than $\frac{b}{a}$. Since we have found a fraction of integers that is equal to $\varphi$, but is in lower terms than $\frac{b}{a}$, we have a contradiction: $\frac{b}{a}$ cannot be a fraction of integers in lowest terms. Therefore $\varphi$ cannot be expressed as a fraction of integers and is irrational. # For More Information • Markowsky. “Misconceptions about the Golden Ratio.” College Mathematics Journal. Vol 23, No 1 (1992). pp 2-19. # Teaching Materials There are currently no teaching materials for this page. Add teaching materials. # References 1. ↑ "Golden ratio", Retrieved on 20 June 2012. 2. ↑ "Misconceptions about the Golden Ratio", Retrieved on 24 June 2012. 3. ↑ "Parthenon", Retrieved on 16 May 2012. # Future Directions for this Page -animation? http://www.metaphorical.net/note/on/golden_ratio http://www.mathopenref.com/rectanglegolden.html Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 117, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9245808720588684, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/38492/will-cosmological-gravitational-waves-be-weaker-or-stronger-than-astrophysical-o	# Will cosmological gravitational waves be weaker or stronger than astrophysical ones? Will gravitational waves of cosmological origin be weaker or stronger (higher amplitude $h \simeq\Delta L/L$) than those created from astrophysical sources? I'm having a real hard time finding the amplitude of cosmological gravitational waves (from inflation and from cosmic defects (strings etc.)) in terms of $h$, so that I can make a comparison. Any help on this would be great. - ## 2 Answers You probably should look here first for some relatively up-to-date predictions of signal-to-noise from cosmological sources. Astrophysicists are very confident that advanced LIGO will see signals from merging compact objects like neutron star binaries and black hole binaries, on the order of a few tens per year (see first link). It is far less certain whether gravitational waves from the very early universe will be detected by LIGO, eLISA, or even the Big Bang Observer if it ever gets built. The answer you get will be different depending on who you ask and how optimistic they are. Regarding signals from the end of inflation, the frequency at which the signal peaks depends on, among other parameters, an unknown inflationary energy scale. I'm afraid I can't provide you with the expected signal in terms of $\Delta L / L$, but you can see some predictions for this sort of signal plotted against the LIGO, LISA, and BBO noise curves in this paper, and how they vary with the energy scale. The chances of detection don't seem very high, and require the inflationary energy scale to be quite low, $< 10^9$ GeV, to be in a favorable frequency range. (Take these results with a grain of salt, though - they are based on highly simplified models and should not be viewed as definitive). - Hello and thanks for your answer. In third paper you cited (GW Production at the End of Inf.), it says that $\Omega_{gw}h \sim 10^{-11}$. This is the kind of information I was looking for. However, I do not understand this relation. If the Universe contains a larger contribution of GWs, then shouldn't the amplitude $h$ be larger, not smaller? Additionally, you say (rightly) say that the energy scale for the model in this paper is $\sim 10^{9}$ GeV, but how low is this (likely, unlikely, extremely unlikely)? – user12345 Sep 30 '12 at 10:39 2 The $h$ in the papers I've cited refers to the non-dimensional Hubble parameter, such that the dimensional Hubble parameter can be written 100 h km/s/Mpc, and practically $h$ can be thought of as a constant of around 0.7 . It is not $\Delta L / L$. – kleingordon Sep 30 '12 at 21:13 Regarding the energy scale, much of the original work on inflation postulated that it occurred at around the GUT scale, approximately $10^{16}$ GeV. There have been a variety of suggestions for how a lower energy scale might be possible, many of them cited in the papers in the answer. I'm afraid I'm not qualified to make statements on their likelihood. – kleingordon Sep 30 '12 at 21:18 there is chance to detect the waves Observable Spectra of Induced Gravitational Waves from inflation http://arxiv.org/pdf/1203.4663v2.pdf ...A recently discovered observational by-product of an enhanced power spectrum on small scales, induced gravitational waves, have been shown to be within the range of proposed space based gravitational wave detectors; such as NASA's LISA and BBO detectors, and the Japanese DECIGO detector... -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9379011988639832, "perplexity_flag": "middle"}
http://embuchestissues.wordpress.com/2009/04/06/cardboard-dodecahedron/	# Embûches tissues ## Cardboard dodecahedron 6 April 2009 at 12:01 am Leave a comment I made a cardboard dodecahedron for the needs of a talk. If you draw 5-coloured stars on all facets, by choosing smartly the colours, you can get five coloured cubes whose vertices are vertices of the dodecahedron. This trick can be used to show that the symmetry group of the dodecahedron is the alternate symmetric group $\mathcal A_5$: it replaces a star by a star with a different arrangement of colours. Since there are 12 facets and 5 ways of rotating each of them, 60 colourings can be seen by rotating the dodecahedron by direct isometries. However, it is NOT true that you can see the 120 possible colourings by allowing also reflections (the full isometry group of the dodecahedron is the symmetric group on five colours). An easy reason for this is that the colouring is invariant under symmetry through the central point (which is a determinant -1 transformation). You can also argue that reflections act as double transpositions of the colours of a star. People also talk about five tetrahedra in a isocahedron, which can also be obtained in the dodecahedron by choosing a tetrahedron in each cube in a consistent way. The tetrahedra have faithful action of the isometry group: there are two sets of five tetrahedron, which are exchanged under signature -1 transformations, and even permutations of the tetrahedra correspond to direct isometries. About these ads ### Like this: Like Loading... Entry filed under: english, geometry, group theory. Tags: handcrafted mathematics, icosahedral group. Trackback this post \| Subscribe to the comments via RSS Feed ## Recent Comments Rodney Price on Mathkaba : my tool for managin… Carmine Napolitano on Merry Christmas from a dying… remyoudompheng on Mathkaba : my tool for managin… Alberto on Mathkaba : my tool for managin… Gaetan Bisson on Mathkaba : my tool for managin… %d bloggers like this:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 1, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.914605438709259, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/208851/need-a-tip-hint-evaluating-a-limit?answertab=votes	# Need a tip/hint evaluating a limit I have the following limit: $$\lim_{x\rightarrow\infty}\left(1+\frac{a}{x^{1/2+\epsilon}}\left(1-\exp\left(-\frac{b}{x^{1/2+\epsilon}}\right)\right)\ln\left(\frac{a}{x^{1/2+\epsilon}}\right)\right)^x$$ where $0<b<a$. I care for the case where $\epsilon>-1/2$. I suspect that for $\epsilon>0$ this limit evaluates to 1, and for $-1/2<\epsilon\leq0$ it evaluates to 0. However, I am having hard time evaluating this. I have tried taking the log of the expression (moving the $x$ in the exponent down), substituting $y=1/x$ and then Taylor-expanding the log, but didn't get anywhere. Does anyone have any tips/hints that might help me evaluate this? - ## 1 Answer Since $c=1/2+\varepsilon\gt0$, one knows that $1-\exp(-bx^{-c})\sim bx^{-c}$ and $\log(ax^{-c})\sim-c\log(x)$ when $x\to+\infty$. Thus, the function to be evaluated is $$f(x)=(1-kg(x))^x,\quad k=abc\gt0,\quad g(x)\sim x^{-2c}\log x.$$ Since $g(x)\to0$ and $\log(1+u)\sim u$ when $u\to0$, $$f(x)=\exp\left[x\log(1-kg(x))\right]=\exp\left[-kxg(x)\cdot(1+o(1))\right].$$ Note that $xg(x)\sim x^{-2\varepsilon}\log x$ and recall that $k\gt0$. This yields: • If $-1/2\lt\varepsilon\leqslant0$, then $xg(x)\to+\infty$ hence $f(x)\to0$ when $x\to+\infty$. • If $\varepsilon\gt0$, then $xg(x)\to0$ hence $f(x)\to1$ when $x\to+\infty$. - So it turns out I made a mistake in the original definition of the problem: I flipped the plus sign after the first 1 to a minus sign when I was typing this up (I've edited the question since I saw your answer and my mistake). However, you solved the problem that I originally had in mind (otherwise the minus in $-c\log(x)$ would've flipped the minus in your re-definition of $f(x)=(1-kg(x))^x$ to a plus). Thank you, this confirms my intuition (which was backed up by numerical evaluations)! – M.B.M. Oct 8 '12 at 5:52	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 25, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9428955912590027, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/59313/evaluating-lim-limits-x-to-01x1-x/59322	# Evaluating $\lim\limits_{x\to 0}(1+x)^{1/x}$? How would I work out a limit of the form: $$\lim_{x\to 0}\;(1+x)^{1/x}$$ I know these types of limits have a solution based on $e$ but how do I find this solution? - 3 The limit is $1$. $1^0$ is not an indeterminate form. If you asked for $\lim_{x\to 0} (1+x)^{1/x}$, that would be a different story, – Grumpy Parsnip Aug 23 '11 at 20:43 $(1+0)^0 \to 1^0 \to 1$, no indeterminancy here. The indeterminancy appears in limits of the form $1^\infty$, eg $\lim\limits_{x\to0}(1 + x)^{1/x}$ – leonbloy Aug 23 '11 at 20:43 I asked the wrong question, I'm sorry. You were guessing for the right question.. could you provide me with a solution now? – Mats Aug 23 '11 at 20:48 ## 5 Answers Hint: The functions $y = \log x$ and $y = e^x$ are continuous, and continuous functions respect limits: $$\lim_{n \to \infty} f(g(n)) = f\left( \lim_{n \to \infty} g(n) \right),$$ for all continuous functions $f$, whenever $\displaystyle\lim_{n \to \infty} g(n)$ exists. Let $$L=\lim\limits_{x\to 0}(1+x)^{1/x}$$ be the limit which you to wish to find. Instead of finding $L$ directly, try on your own to find $\ln(L)$. - This is actually the best answer! You take a nice teaching approach! – Mats Aug 23 '11 at 21:24 @Mats: Thank you. Considering you said you had an exam tomorrow, I figured I'd explain the more general strategy :) – JavaMan Aug 23 '11 at 23:51 $\lim\limits_{x \rightarrow 0}\exp (x\ln (1+x))=\exp(\lim\limits_{x \rightarrow 0}(x\ln(1+x)))=\exp(0)=1$. $\lim\limits_{ x \rightarrow 0}\exp ( \frac{\ln (1+x)}{x})=\exp(\lim\limits_{x \rightarrow 0}(\frac{\ln(1+x)}{x}))=\exp(1)=e$. Use L'Hospital.. to see $\lim\limits_{x \rightarrow 0}\frac{\ln(1+x)}{x}=1$ - With exp(x) do you mean $e^x$? – Mats Aug 23 '11 at 21:01 @Mats:$\quad$yes! – Quixotic Aug 23 '11 at 21:06 Try writing this as $$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$$ The binomial theorem may be of some help then. Another way of looking at this is taking logs and using L'Hospital $$\log\left(\lim_{x\to0}(1+x)^{1/x}\right)=\lim_{x\to0}\frac{\log(1+x)}{x}$$ - This is not the right solution .. It was something with putting the whole limit as a power of $e$ .. but I forgot the correct solution method. – Mats Aug 23 '11 at 20:54 @Mats: I added another approach. However, both yield the same answer, but in different forms. – robjohn♦ Aug 23 '11 at 20:56 3 @Mats: The first suggested solution is fine, it just requires some work. There is no universal "right solution"..... – Eric♦ Aug 23 '11 at 20:59 @Eric: The Binomial Theorem can produce a series for the limit that converges pretty fast. The limit is usually the definition for $e$, and if that is the case, all that is left is to compute the value. – robjohn♦ Aug 23 '11 at 20:59 1 @Mats:This is a correct solution,put $\frac{1}{x} = n$ then as $x\to 0 \Rightarrow n\to\infty$ after this expand using multinomial/binomial theorem with rational index,arranging the terms and substituting the limit you will get this $1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots = e$ – Quixotic Aug 23 '11 at 21:02 show 3 more comments If we use the substitution $x=\frac{1}{y}$, since $\lim_{x\rightarrow 0}x=\lim_{y\rightarrow \infty }\frac{1}{y}$, we get $$\lim_{x\rightarrow 0}\left( 1+x\right) ^{1/x}=\lim_{y\rightarrow \infty }\left( 1+\frac{1}{y}\right) ^{y}=e,$$ which uses the result $$\lim_{n\rightarrow \infty}\left( 1+\frac{1}{n}\right) ^{n}=e.$$ - @AidenStrydom See last formula of your answer. – Américo Tavares Sep 10 '12 at 7:21 Dear sir - my apologies i miss read your answer. Please forgive me – Aiden Strydom Sep 10 '12 at 8:09 Sorry for uploading the image - i am new and have yet to figure out how to mark up the math The first line assumes you know that `if f(x) = ln(x) then f'(1) = (ln(x+h) - ln(x)) / h` ps `f'(1) = 1` -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 26, "mathjax_display_tex": 7, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9222148060798645, "perplexity_flag": "middle"}
http://simple.wikipedia.org/wiki/Pi_(mathematical_constant)	# Pi (math) (Redirected from Pi (mathematical constant)) $\pi = \frac {C}{d}$ (pi is equal to the circumference divided by the diameter). Pi is an endless string of numbers. Pi or $\pi$ is a mathematical constant, one of the first ever discovered. It is the ratio of the distance around a circle to the circle's diameter. This produces a number, and that number is always the same. However, the number is rather strange. The number starts 3.14159265... and continues without end. Such numbers are called irrational. The diameter is the longest line which can be fitted inside a circle. It passes through the center of the circle. The distance around a circle is known as the circumference. Even though the diameter and circumference are different for different circles, the number pi remains constant: its value never changes. This is because the relationship between the circumference and diameter is always the same.[1] ## Approximation A diagram showing how $\pi$ can be found by using a circle with a diameter of one. The circumference of this circle is $/pi$. Pi is often written formally as $\pi$ or the Greek letter π as a shortcut. Pi is also an irrational number, meaning it cannot be written as a fraction ($a \over b$), where 'a' and 'b' are integers (whole numbers). This basically means that the digits of pi that are to the right of the decimal go forever without repeating in a pattern, and that it is impossible to write the exact value of pi as a number. Pi can only be approximated, or measured to a value that is close enough for practical purposes.[2] A value close to pi is 3.141592653589793238462... A common fraction approximation of pi is $22 \over 7$, which yields approximately 3.14285714. This approximation is 0.04% away from the true value of pi. While this approximation is accepted for most of its use in real life, the fraction $355 \over 113$ is more accurate (giving about 3.14159292), and can be used when a value closer to pi is needed.[3] Computers can be used to get better approximations of pi. ## History Mathematicians have known about pi for thousands of years because they have been working with circles for the same amount of time. Civilizations as old as the Babylonians have been able to approximate pi to many digits, such as the fraction 25/8 and 256/81. Most historians believe that ancient Egyptians had no concept of π and that the correspondence is a coincidence.[4] The first written reference to it dates to 1900 BC.[5] Around 1650 BC the Egyptian Ahmes gave a value in the Rhind Papyrus. The Babylonians were able to find that the value of pi was slightly greater than 3 by simply making a big circle and then sticking a piece of rope onto the circumference and the diameter, taking note of their distances, and then dividing the circumference by the diameter.[6] Knowledge of the number pi passed back into Europe and into the hands of the Hebrews, who made the number important in a section of the Bible called the Old Testament. After this, the most common way of trying to find pi was to draw a shape of many sides inside any circle and use the area of the shape to find measure pi. The Greek philosopher Archimedes, for example, used a polygon shape that had 96 sides in order to find the value of pi, but the Chinese in 500 A.D. were able to use a polygon with 16,384 sides to find the value of pi. The Greeks, like Anaxagoras of Clazomenae, were also busy with finding out other properties of the circle, such as how to make squares of circles and squaring the number pi. Since then, many people have been trying to find out more and more exact values of pi.[7] A history of pi Philosopher Date Approximation Ptolemy around 150 A.D. 3.1416 Zu Chongzhi 430-501 AD 3.1415929203 al-Khwarizmi around 800 A.D. 3.1416 al-Kashi around 1430 A.D. 3.14159265358979 Viète 1540–1603 3.141592654 Roomen 1561–1615 3.14159265358979323 Van Ceulen around 1600 A.D. 3.14159265358979323846264338327950288 In the 16th century, better and better ways of finding pi became available such as the complicated formula that the French lawyer François Viète developed. The first use of the Greek symbol "π" was in an essay written in 1706 by William Jones. A mathematician named Lambert also showed in 1761 that the number pi was irrational; that is, it cannot be written as a fraction by normal standards. Another mathematician named Lindeman was also able to show in 1882 that pi was part of the group of numbers known as transcendentals, which are numbers that cannot be the solution to a polynomial equation.[8] Pi can also be used for figuring out many other things beside circles.[4] The properties of pi have allowed it to be used in many other areas of math besides geometry, which studies shapes. Some of these areas are complex analysis, trigonometry, and series.[6] ## Pi in real life Today, there are different ways to calculate many digits of $\pi$. This is of limited use though. Pi can sometimes be used to work out the area or the circumference of any circle. To find the circumference of a circle, use the formula C (circumference) = π times diameter. To find the area of a circle, use the formula π (radius²). This formula is sometimes written as $A = \pi r^2$, where r is the variable for the radius of any circle and A is the variable for the area of that circle. To calculate the circumference of a circle with an error of 1 mm: • 4 digits are needed for a radius of 30 meters • 10 digits for a radius equal to that of the earth • 15 digits for a radius equal to the distance from the earth to the sun. People generally celebrate March 14 as Pi Day because March 14 is also written as 3/14, which represents the first three numbers 3.14 in the approximation of pi.[2] Pi is frequently used in math jokes to make the pun about how the word 'pi' sounds like the word 'pie'. ## References 1. Lennart Berggren, Jonathan M. Borwein, and Peter B. Borwein (eds.), Pi: A Source Book. 2nd ed, Springer, 1999. ISBN 978-0-387-98946-4 2. ↑ Goodwin College of Professional Studies. (1994-2010). "About Pi" (html). Retrieved 2010-06-05. 3. "Pi to 4 Million Decimals" (php). Retrieved 2010-06-05. 4. ↑ Arndt, Jörg & Haenel, Christoph 2006. Pi unleashed. Springer-Verlag, 168. ISBN 978-3-540-66572-4 5. Beckmann, Petr 1971. A History of Pi. St. Martins Press, London. 6. ↑ Wilson, David (Spring 2000). "The History of Pi" (html). History of Mathematics. Rutgers University, New Brunswick. Retrieved 2010-06-06. 7. "Pi History" (html). August 2001. Retrieved 2010-06-05. 8. "PI" (htm). 2000-2005. Retrieved 2010-06-06.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 10, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9320783615112305, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/78379/how-to-prove-sum-dn-tau3d-left-sum-dn-taud-right2	# How to prove $\sum_{d\|n} {\tau}^3(d)=\left(\sum_{d\|n}{\tau}(d)\right)^2$ $\sum_{d\|n} {\tau}^3(d)=\left(\sum_{d\|n}{\tau}(d)\right)^2$, where $\tau(d)$ designates the number of positive divisors of d. Now I only know that both sides are multiplicative functions, could you tell me what I need to do next? - 2 If you have proved that both sides are multiplicative functions, it suffices to prove that the statement holds for powers of primes. In this case, it is straight forward to compute the value of each side. It may be useful to use the fact that $\sum_{i=1}^n i^3=(\sum_{i=1}^n i)^2$. – Aaron Nov 3 '11 at 0:06 ## 3 Answers If you've shown that both the LHS and the RHS are multiplicative functions, then you must now show it's true for arbitrary prime powers $n=p^r$. In doing so, use $\tau(n)=\sigma_1(p^r)=r+1$ and this. It then follows for all composite numbers by prime factorizing both sides through the multiplication. - Recall that if $f$ is a multiplicative function, and $$g(n)=\sum_{d\|n}f\left(\frac{d}{n}\right),$$ then $g$ is a multiplicative function. From this you can deduce that both sides are multiplicative functions. Thus to verify the identity, all you need to do is to verify it for $n=p^k$, where $p$ is prime. For $n=p^k$, the left-hand side is $1^3+2^3+\cdots +k^3$. The right-hand side is $(1+2+\cdots+k)^2$. The fact that the two are equal is a probably familiar fact. If it isn't, it can be proved by induction. - To get this result you need to show (a) it is true for prime powers, which you can show with $$\sum_1^n i^3 = \frac{n^2(n+1)^2}{4} = \left(\sum_1^n i\right)^2$$ (b) both the left and right sides are multiplicative, which you say you know -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.949040412902832, "perplexity_flag": "head"}
http://mathoverflow.net/questions/71270/highest-weight-representation-and-electromagnetic-fields/71350	## highest weight representation and electromagnetic fields ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) An electromagnetic field is given by 6 components (Ex,Ey,Ez,Bx,By,Bz). Now this is a 6-dimensional irreducible representation of so(1,3) which is a highest weight representation. So there should be a single function S(t,x,y,z) (a "superpotential") corresponding to the highest weight and such that all the other components are derived from it through lowering operators. A similar question can be asked for the electromagnetic 4-potential, (\rho,Ax,Ay,Az); there should be a single function and lowering operators to derive all components. Can someone provide a reference where this is is discussed - 1 There is something not quite right in this question. The value of the electromagnetic field at a point defines a vector in a six-dimensional real irrep of the Lorentz group, but the electromagnetic field itself defines a section through a homogeneous bundle over Minkowski spacetime associated to that irrep, and the space of sections is an infinite-dimensional representation which is not highest weight and hence I don't see how to arrive at this "superpotential". – José Figueroa-O'Farrill Jul 26 2011 at 0:03 ## 2 Answers The 6-dimensional representation corresponding to the electromagnetic field is the realification of the symmetric square of the defining representation of $\mathrm{SL}(2,\mathbb{C})$, whose Lie algebra is isomorphic to that of $\mathrm{SO}(1,3)$. The highest weight vector is the square of the highest weight vector of the defining representation. Any book which treats electromagnetism in the "spinorial" language should discuss this: perhaps Penrose and Rindler "Spinors and spacetime"? - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. I think I see the discrepancy. The 6 components of the field then do NOT transform as a 6 dimensional rep, even for a fixed spacetime point. The action is : F(x) -> M(g) * F(g' x) (F=field, M : 6x6 matrix) which is an infinite dimensional rep. I don't see any finite reps here at all; so the common statement "the electromagnetic field transforms as a 6 dimensional rep..." really isn't true. That being said, is there any way to adapt the well structured theory of highest weight representation to this infinite representation.... -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.915317714214325, "perplexity_flag": "head"}
http://crypto.stackexchange.com/questions/5761/construct-an-election-protocol-for-the-following-election-problem	# construct an election protocol for the following election problem..? [closed] I am something new to election protocols and I am trying to construct an election protocol for the following problem: Let $A= \lbrace a_{1} , . . . , a_{n} \rbrace$ and $B = \lbrace b_{1} , . . . , b_{m} \rbrace$ be two groups of voters, and $K_{1}$and $K_{2}$ be two candidates. One of the candidates is to be elected. A candidate wins the election if he gets at least 50% of the votes (absolute majority) and if he gets at least 10% of each group (veto of group A or B). You are not allowed to use or assume the existence of a trusted center. - 1 There are some holes present in your question. What must be kept private (individual votes, the exact percentages earned, the fact that someone even voted)? What assumptions can be made about the participating parties (are they all present at the same time, only some threshold will behave maliciously, they don't collude, etc)? – mikeazo♦ Dec 18 '12 at 16:00 only individual votes should be kept private – Sam Dec 18 '12 at 16:34 Can you say anything about what practical context this problem arose in? – D.W. Dec 19 '12 at 21:34 – D.W. Dec 20 '12 at 21:55 ## closed as off topic by D.W., mikeazo♦Dec 21 '12 at 12:25 Questions on Cryptography Stack Exchange are expected to relate to cryptography within the scope defined in the FAQ. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about closed questions here. ## 1 Answer A general rule of thumb when developing cryptographic solutions to a problem is to start with something that gets you as much as possible of what you want and make as few modifications as possible to make up the difference. In this case, I'd start with something like Practical Multi-Candidate Election System. It would give you tallys for $K_1$ and $K_2$. From there you would know if the candidate got 50%. From there, adding the necessary functionality wouldn't be too hard, but will require you to more formally define what is allowed, what is public knowledge (e.g., is a voter's group membership public). If, for example, the voter's group membership is public, then when they cast their ballot, the ballots can be separated by groups and a tally given for each group in addition to the overall tally. Then you'd know if the candidate gets 10% of each group. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9544220566749573, "perplexity_flag": "middle"}
http://mathhelpforum.com/statistics/156369-question-normal-distribution-2-a-print.html	# Question on normal distribution #2 Printable View • September 15th 2010, 09:51 PM mngeow Question on normal distribution #2 A manufacturer of electric light bulbs finds that his bulbs have an average life span of 1200 hours and a S.D of 200 hours.Assume that the distribution of lifetimes is normal. What should be the guaranteed life of the bulbs if the manufacturer is prepared to replace 5% of the bulbs sold? How should I go about doing this question? • September 15th 2010, 10:00 PM Unknown008 So, let X be the number of hours of life span of a bulb. X ~ N(1200, 200^2) P(X < x) = P(Z < z) = 0.05 Find the z value that corresponds to this probability, then convert it into hours using: $z = \dfrac{x - \mu}{\sigma}$ • September 15th 2010, 10:25 PM mngeow I tried doing that but the answer was wrong.The solution said 0.5-0.05=0.45 and find the correspoding Z value,they also said that this Z value would be negative.I don't really understand it. The answer is 870hours btw. • September 15th 2010, 11:17 PM MathoMan Quote: Originally Posted by mngeow A manufacturer of electric light bulbs finds that his bulbs have an average life span of 1200 hours and a S.D of 200 hours.Assume that the distribution of lifetimes is normal. What should be the guaranteed life of the bulbs if the manufacturer is prepared to replace 5% of the bulbs sold? How should I go about doing this question? X is normally distributed random variable $X\sim(1200,200^2)$ measuring the life span of light bulbs. You have to determine the value A (number of hours such) that the life span of 95% of light bulbs will last longer than A hours. That means that only 5% of light bulbs will have life span shorter than A and thus such A should be the guaranteed life of light bulbs. $P(X>A)=0.95$ $1-P(X<A)=0.95$ $P(X<A)=0.05$ $F_X(A)=0.05$ $\Phi(\frac{A-1200}{200})=0.05$ Now use the table values for standard normal distribution and you'll see that $\Phi(-1.65)\approx 0.05$ so you form the equation $\frac{A-1200}{200}=-1.65$ $A=870.$ • September 16th 2010, 01:25 AM mngeow Quote: Originally Posted by MathoMan X is normally distributed random variable $X\sim(1200,200^2)$ measuring the life span of light bulbs. You have to determine the value A (number of hours such) that the life span of 95% of light bulbs will last longer than A hours. That means that only 5% of light bulbs will have life span shorter than A and thus such A should be the guaranteed life of light bulbs. $P(X>A)=0.95$ $1-P(X<A)=0.95$ $P(X<A)=0.05$ $F_X(A)=0.05$ $\Phi(\frac{A-1200}{200})=0.05$ Now use the table values for standard normal distribution and you'll see that $\Phi(-1.65)\approx 0.05$ so you form the equation $\frac{A-1200}{200}=-1.65$ $A=870.$ I only understood until the P(X<A)=0.05 part,I got lost after that.What does Fx(A)=0.05 mean? and whats the funny symbol in the line after that? • September 16th 2010, 01:31 AM Unknown008 Quote: Originally Posted by mngeow I tried doing that but the answer was wrong.The solution said 0.5-0.05=0.45 and find the correspoding Z value,they also said that this Z value would be negative.I don't really understand it. The answer is 870hours btw. That means that you are asked to use a z table different from mine. So, from where I left: $P(Z < z) = 0.05$ Is your graph like this: http://p1cture.me/images/51394697395543890592.png Or like this: http://p1cture.me/images/88238005700832427228.png ? • September 16th 2010, 01:34 AM mngeow Its the one in the 2nd link,where the shaded area is on the right hand side. • September 16th 2010, 01:47 AM Unknown008 Ok. You need to find the z value for this area: http://p1cture.me/images/17640253637132137970.png But you don't have this in your table. The equivalent of this area is this area: http://p1cture.me/images/50823189974773945850.png However, this isn't in your table either. But you have this area: http://p1cture.me/images/30470114981255486093.png The probability of the last area is 0.5 - 0.05 = 0.45, okay? What is the value of z for this area? in your table, you should read z = 1.65. But when you look at the original distribution (first graph), you know that z must be negative as it's value is on the left of the mean, so, the z value is -1.65. From there, you can then work back the value of X, the maximum lifetime of the bulb for it to fall in the 5% worst quality. Is it clear now? • September 16th 2010, 01:52 AM mngeow Yes.I understand now.Thanks :D • September 16th 2010, 05:07 AM MathoMan Quote: Originally Posted by mngeow I only understood until the P(X<A)=0.05 part,I got lost after that.What does Fx(A)=0.05 mean? and whats the funny symbol in the line after that? $F_X(A)$ is the value of the distribution function of the random variable X at point A. $\Phi$ a uppercase Greek letter phi, commonly used to denote the distribution function of the standard normal random variable. If X is normal random variable with parameters called expectation and variance, $X\sim {\cal N}(\mu, \sigma ^2)$ then distribution function of that random variable is usually denoted with $F_X$ and it is defined as $F_X(a)=P(X<a)$ Standard normal random variable is normal random variable with parameters 0 and 1, and to distinct it from other normal variables its distribution function, although defined in the same manner, is usually denoted with $\Phi$: $X\sim {\cal N}(0, 1) , F_X(a)=\Phi(a)=P(X<a)$ All times are GMT -8. The time now is 05:24 AM.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 27, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9159683585166931, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/79583/when-does-every-point-in-the-volume-of-a-polytope-lie-along-a-chord-between-its-e	## When does every point in the volume of a polytope lie along a chord between its edges? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Consider the 3-simplex, or tetrahedron, in 3-space. Regardless of the positions of the vertices, every point in the volume defined by the triangular faces of the polytope's skeleton graph can lie along a chord between two non-adjacent edges. Or, equivalently, every interior point can lie along a straight line segment which intersects two non-adjacent edges. When is this property true of other convex (or non-convex) polyhedra? How does this property extend to the general $N$-simplex? - First reaction, written without much thought: what's the polar dual question? Is that easier? – Alexander Woo Oct 31 2011 at 4:26 @Alexander Woo, I have to think a bit more to understand what the equivalent formulation would be on the dual of a polyhedron, but I have no reason to believe this would simplify the problem. – UltraBlue06 Oct 31 2011 at 4:43 For my comfort, I restrict discussion to subsets of R^3. Since each pair of edges forms a (possibly degenerate) tetrahedron after taking its convex hull, the question may boil down to an (overlapping) decomposition of such a polyhedron into tetrahedra which have at least two edges in common with the polyhedron. I imagine Bill Thurston or Joseph O'Rourke will have something further to say. My sense is that all regular polyhedra do, and some (Csakar?) polyhedra will not, and it will be a combinatorial result. Gerhard "Ask Me About System Design" Paseman, 2011.10.30 – Gerhard Paseman Oct 31 2011 at 4:58 @Gerhard Paseman, I very much agree with your analysis, though I was hoping for an easier method of tackling the problem. – UltraBlue06 Oct 31 2011 at 5:05 I am also perfectly fine restricting treatment of this question to R^3. – UltraBlue06 Oct 31 2011 at 5:06 show 1 more comment ## 2 Answers The question asks whether every point $v$ in the interior of a 3-polytope $P \$ is on an interval between two edge-points. This is easy. Project the edges of $P$ onto a unit sphere centered at $v$. Call the resulting graph $G$ blue. Take the opposite $-G$ and call this graph red. Clearly red and blue graphs intersect, since otherwise one must lie in the face of another, which is impossible since $v$ is interior. Thus the line through the intersection point and $v$ is as desired. As for higher dimensions, this is clearly not possible already for dim-reasons. We are talking about 2-parametric family of intervals, which cannot possibly cover the interior of a $d$-polytope, for $d\ge 4$. - Nice answer. =) – UltraBlue06 Oct 31 2011 at 5:32 So the opposite graph $−G$ is constructed by mapping the vertices of $G$ to antipodal positions on the unit sphere (centered at $v$) and then connecting the vertices via the opposite or alternative edge path around the sphere. As such, $−G$ should be combinatorially isomorphic to $G$. Furthermore, when $G$ and $−G$ intersect, which is guaranteed if $v$ is internal to $P$, we know there exists a chord through that point of intersection which also intersects $v$ at the center of the sphere. – UltraBlue06 Nov 1 2011 at 20:54 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. I think that the proposed solution is slightly incomplete since the original question asks for nonadjacent edges of a 3-polytope. It can be easily fixed by studying the intersection of the graphs G and -G. Regarding possible n-dimensional versions of the problem, the following result can be proved. Let P be an n-dimensional polytope and k and m positive integers such that k + m = n + 1. For any point x in P there are two faces, F and G, of P such that dim F \le k - 1, dim G \le m - 1, and x is in conv(F U G). -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9417884349822998, "perplexity_flag": "head"}
http://mathhelpforum.com/algebra/39594-arithmetic-geometric-sequences.html	# Thread: 1. ## Arithmetic and Geometric Sequences Here is the question: In an old story, a king asked a peasant to come and work for him. They both agreed that the job to be done could be completed in a month (30 days). The peasant asked the king if he could be paid in the following way: "On the first day that I work, I would like to be paid one penny. Each day that I work, I would like you to double my income." The king thought this was a great deal and happily agreed. Analyze the deal that the king and the peasant made to determine whether or not the peasant made a good deal with the king. Now I know that the peasant is getting the better deal. And I believe I have to use the formula tn=ar^n-1. 2. You're correct. The formula is $2^{n}-1$ For instance, how much on the 5th day?. 16+8+4+2+1=31 $2^{5}-1=31$ So, on the 30th day, he would have $2^{30}-1$. A rather large number, even in pennies. That is \$10,737,418.23 The king may have him beheaded after seeing that.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9776217937469482, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/90129/orthogonality-in-non-inner-product-spaces/90136	## Orthogonality in non-inner product spaces ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I have come across a notion of orthogonality of two vectors in a normed space not necessarily inner product space. Two vectors $x$ and $y$ in a normed space are said to be orthogonal (represented $x\perp y$) if $\|\|x\|\|\leq \|\|x+\alpha y\|\|,$ for every $\alpha,$ a scalar. 1) What is the rational behind the definition above? I guess, it has got something to do with minimum overlap between $x$ and $y$. 2) Is this unique generalization of the concept of orthogonality from inner product spaces? Thank you. - Here's another paper (seems not to be referenced in the ones mentioned in Valerio's answer): projecteuclid.org/… – Ralph Mar 4 2012 at 22:31 ## 4 Answers Concerning question 1: The rational is that in an inner product space $$x\perp y \Leftrightarrow \forall \alpha \in K: \|\|x\|\|\leq \|\|x+\alpha y\|\| \qquad(K = \mathbb{R} \text{ or } K = \mathbb{C})$$ Now, if no inner product is available (but a norm), the idea is, to just take the right hand side as definition of orthogonality (call it $\perp_1$). Concerning question 2: No, there are other -non-equivalent - generalizations as well. As an example, note that in an inner product space over the reals $$\langle x,y \rangle = \frac{1}{4}( \|\|x+y\|\|^2 - \|\|x-y\|\|^2).$$ Hence $x\perp y \Leftrightarrow \|\|x+y\|\| = \|\|x-y\|\|$. So the definition $$x\perp_{\scriptstyle 2}\; y : \Leftrightarrow \|\|x+y\|\| = \|\|x-y\|\|$$ generalizes the orthogonality from an inner product space to any normed space (over the reals). Now let's show that $\perp_1, \perp_2$ aren't equivalent. Let $E = \mathbb{R}^2$ with norm $\|\|(a,b)\|\| = \max(\|a\|, \|b\|)$. Then $\qquad (0,1) \perp_2 (2,1)$ but not $(0,1) \perp_1 (2,1)\quad$ (take $t=-1/4$) $\qquad (1,1) \perp_1 (2,0)$ but not $(1,1) \perp_2 (2,0).$ - Follow up questions: (i) Are these two notions of orthogonality symmetric? $\perp_2$ certainly is, but what about $\perp_1$? (ii) Are these notions invariant under scaling? For example, if $x \perp_1 y$, then $x \perp_1 \beta y$ for any $\beta \in K$. Most importantly, (iii): In which contexts are these two notions of orthogonality useful? – Martin Mar 4 2012 at 1:43 Another follow up question I have is: Since $\perp_{1}$ and $\perp_{2}$ are equivalent in inner-product spaces, in what way normed spaces are deficient so that these orthogonality concepts don't agree? – Uday Mar 4 2012 at 2:07 Another question: Are there any more non-equivalent orthogonality definitions? – Uday Mar 4 2012 at 2:16 1 @Martin: (i) $\perp_1$ is not symmetric: From above $(1,1) \perp_1 (2,0)$, but not $(2,0) \perp_1 (1,1)$ (take $t = -1$). (ii) $\perp_1$ is invariant under scalar multiplication, as is aparant from the definition. $\perp_2$ ist not: Again from above, $(0,1) \perp_2 (2,1)$ but not $3 \cdot (0,1) \perp_2 (2,1)$. – Ralph Mar 4 2012 at 10:00 1 @Uday: Yes, there are more notions of orthogonality in normed spaces - see the survey article quoted by Valerio. In principle you can take any norm-expression that is equivalent to orthogonality in an inner product space and use it as definition of orthogonality in a normed space. – Ralph Mar 4 2012 at 10:04 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. The definition you gave is called Birkhoff-James orthogonality and the intuition is the following: suppose you have $x,y\in\mathbb R^2$ and construct a triangle with sides $x$ and $y$. Now let $x$ be fixed and consider the same triangle with $-\alpha y$ instead of $y$. Observe that $\|\|x+\alpha y\|\|$ is the length of the third side of this triangle. If you try to write down a picture, you figure out in a moment that the condition $\|\|x\|\|\leq\|\|x+\alpha y\|\|$ can be true for all $\alpha$ iff $x$ and $y$ are orthogonal (looking at the picture, if they are not orthogonal and the inequality is true for some $\alpha$, then it is false for $-\alpha$). Birkhoff-James' orthogonality is a tentative to capture orthogonality through this geometric property. Birkhoff-James' orthogonality is not the unique notion of orthogonality for normed space. Some references: In the following paper http://arxiv.org/pdf/0907.1813.pdf you can find some recent very easy application of BJ's orthogonality, as well, if you go through the bibliography, some references about other notions of orthogonality are given. In particular I suggest the paper of Diminnie Diminnie, C.R. A new orthogonality relation for normed linear spaces, Math. Nachr. 114 (1983), 192-203 and the survey by Alonso and Benitez http://dmle.cindoc.csic.es/pdf/EXTRACTAMATHEMATICAE_1989_04_03_03.pdf P.s. Bikhoff-James orthogonality is not symmetric in general. Some interesting remarks about symmetric orthogonalities can be found in the paper(s) by Partington in the bibliography of the arxiv paper cited above. - Concerning your follow-up question (iii) there is the following very nice result: For Birkhoff-James orthogonality it is easy to find examples where $y\perp x$ but $\left\\|x\right\\|/\left\\|x+\alpha y\right\\| > 1$ for some real $\alpha$, and so natural to investigate the largest such value $\left\\|x\right\\|/\left\\|x+\alpha y\right\\|$ over $X$. In "R. L. Thele, Some results on the radial projection in Banach spaces. Proc. Amer. Math. Soc., 42(2):484--486", it is it is shown that this quantity is exactly the Lipshitz constant for the radial projection onto the unit ball in this norm. - Well, it depends what do you need it for. You may also have a look at semi-inner-product spaces, which are natural generalizations of inner product spaces. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 50, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9034988284111023, "perplexity_flag": "head"}
http://physics.aps.org/synopsis-for/print/10.1103/PhysRevC.79.054304	# Synopsis: Getting closer to the Bohr model #### Bohr model as an algebraic collective model D. J. Rowe, T. A. Welsh, and M. A. Caprio Published May 5, 2009 In the algebraic collective model, the five variables define the quadrupole moments of a nucleus. A group theoretical approach is used to separate the variables into a “radial” coordinate ($β$) and four angular variables. The radial wave functions can be chosen corresponding to a specific mean deformation in $β$ while calculations involving the angular coordinates are made simple by group theoretical techniques. This leads to huge computational savings over calculations in a spherical basis ($β=0$), used in some previous models, for which a very much larger set of basis functions is required. The Bohr Hamiltonian can now be solved for virtually any assumed potential. In a paper appearing in Physical Review C, David Rowe and Trevor Welsh of the University of Toronto in Canada and Mark Caprio of the University of Notre Dame in the US demonstrate the practical utility of the algebraic collective model when applied to various well-known solvable limits of the Bohr model. In several cases, they find a substantial amount of centrifugal stretching (elongation of the nucleus with increasing angular momentum), which is neglected in adiabatic approximations to the Bohr model. They argue that, as in the case of the interacting boson model, the ease of carrying out collective model calculations for a wide range of Hamiltonians can be used to quickly characterize a large body of nuclear phenomena and test for limitations of the Bohr model. – John Millener ISSN 1943-2879. Use of the American Physical Society websites and journals implies that the user has read and agrees to our Terms and Conditions and any applicable Subscription Agreement.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8935654163360596, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/115921/laplace-transform-how-to-derive/115938	# Laplace transform. How to derive I have this integral related to a Laplace transform and I was wondering if anyone knows of a clever way to derive it. I know we usually look these up in a table, but this form is not in a table I have. The transform is: $$\mathcal{L}\left\{ \frac{\sin 1/t}{\sqrt{t}} \right\}(s)=\int\limits_{0}^{\infty}\frac{\sin(1/t)}{\sqrt{t}}e^{-st}\,\mathrm dt, \;\ s>0.$$ I ran it through Maple and it gave me $$\sqrt{\frac{\pi}{s}}\sin(\sqrt{2s})e^{-\sqrt{2s}}.$$ A rather simplistic looking solution, but how to derive? I did notice that the $$\frac{1}{t^{\frac{1}{2}}}=\sqrt{\frac{\pi}{s}},$$ since the Laplace of $$t^{k}=\frac{\Gamma(k+1)}{s^{k+1}}, \;\ k>-1.$$ This would mean $$t^{\frac{-1}{2}}=\frac{\Gamma(\tfrac12)}{\sqrt{s}}=\sqrt{\frac{\pi}{s}}.$$ Perhaps it is too difficult to do by hand. I just thought it was interesting how to derive this if possible. Thanks very much for your time, interest, and expertise. - – Mathlover Mar 3 '12 at 10:32 Expanding in power series doesn't help since negative powers of $t$ appears in the series expansion. – user17090 Mar 3 '12 at 11:20 Terminology nitpick: When you want to derive X, you start with some set of givens and conclude with X as your destination; what you want to do is evaluate the integral, not derive it. – anon Mar 3 '12 at 11:43 Sorry about the terminology. I will use 'evaluate' from now on. By derive, I meant do it by hand. – Cody Mar 3 '12 at 16:29 ## 2 Answers Let's set $t:=x^2$ then : $$I=2\int_0^{\infty} \sin(\frac 1{x^2}) e^{-sx^2}dx=i\int_0^{\infty} \left(e^{-\frac i{x^2}}-e^{\frac i{x^2}}\right) e^{-sx^2}dx$$ At this point you may use following formula : $$J(a,s)=\int_0^{\infty} e^{-\frac a{x^2}-s x^2} dx=\sqrt{\frac{\pi}{4s}}e^{-2\sqrt{as}}$$ (right and left terms both verify $J(a,s)=-\frac{\partial J(s,a)}{\partial s}$, $\frac{\partial^2 J}{\partial a \partial s}=J$ and $J(0,s)=\sqrt{\frac{\pi}{4s}}$, this formula is usually provided with $a\gt 0$, $s\gt 0$ but may be extended by analytic continuation) For a more serious justification of this formula of P. Laplace you may search it in Gradshteyn and Ryzhik's famous "Table of integrals, series and products" getting (3.325) and use Moll and other's heroic effort to justify them all! In this case see the chapter 5 of this paper of Albano, Amdeberhan, Beyerstedt and Moll. so that we get : $$I=i\sqrt{\frac{\pi}{4s}}\left(e^{-2\sqrt{is}}-e^{-2\sqrt{-is}}\right)$$ Use $\sqrt{i}=\frac{1+i}2$ and $\sqrt{-i}=\frac{1-i}2$ to get : $$I=i\sqrt{\frac{\pi}{4s}}\left(e^{-(1+i)\sqrt{2s}}-e^{-(1-i)\sqrt{2s}}\right)$$ $$I=\sqrt{\frac{\pi}{s}}\sin(\sqrt{2s})e^{-\sqrt{2s}}$$ Simple looking things are sometimes not so simple and some of the previous steps are not so easy to justify! (if you remove the $\sqrt{t}$ at the denominator or put a $t$ or something else there you won't get a simple answer!) - Well yes, if we have such a formula then the rest is easy. I've spent the last 2 hours trying to evaluate such an integral, I just got it! – Ragib Zaman Mar 3 '12 at 12:04 @RagibZaman: it is a classical integral from qft (path integrals). I am searching a reference... – Raymond Manzoni Mar 3 '12 at 12:07 Thanks much, Raymond, and thanks for the links. I have the error function paper, so I will have to look it over. – Cody Mar 3 '12 at 16:36 The Key Lemma: For all $a\in \mathbb{C}$ with $\Re(a)>0$ we have $$\int^{\infty}_0 \frac{1}{\sqrt{t} } e^{- a (t+ 1/t)^2 } dt = e^{-2a} \sqrt{ \frac{\pi}{a} } .$$ Proof: Let the integral be denoted by $I$ and $t=u^2$ so that $$I = 2 \int^{\infty}_0 e^{-a (u^2+1/u^2) } du= 2 e^{-2a} \int^{\infty}_0 e^{-a (u-1/u)} du.$$ By letting $u=1/t$ we see that $$\int^{\infty}_0 \frac{1}{u^2} e^{-a (u-1/u)^2} du = \int^{\infty}_0 e^{-a (t-1/t)^2 } dt$$ so then $$I = e^{-2a} \int^{\infty}_0 \left(1 + \frac{1}{u^2} \right) e^{-a (u-1/u)^2 } du.$$ Now letting $x= u-1/u$ gives $$I = e^{-2a} \int^{\infty}_{-\infty} e^{-ax^2} = e^{-2a} \sqrt{ \frac{\pi}{a} } .$$ Idea for the rest of the proof: (I will come back soon and complete it.) Prove that $$\int^{\infty}_0 \frac{1}{\sqrt{t} } e^{-st + i/t } dt = \sqrt{\frac{\pi}{s}} e^{-(1-i)\sqrt{2s} } dt$$ from which the result follows by taking imaginary parts of both sides. To do this, let $t= \frac{1-i}{\sqrt{2s} } x$ (This "substitution" can be made valid by integrating along a suitable sector in the complex plane). Then the resulting integral becomes of the form of our lemma (up to a constant multiple) and then we are basically done. - Thanks much Ragib. I would not have thought of that. I figured it was rather complicated even though the solution was ratehr simplistic-looking. – Cody Mar 3 '12 at 16:34	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 18, "mathjax_display_tex": 16, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9474182724952698, "perplexity_flag": "middle"}
http://mathforum.org/mathimages/index.php?title=Harter-Heighway_Dragon&oldid=2797	# Harter-Heighway Dragon ### From Math Images Revision as of 14:58, 29 May 2009 by Ryang1 (Talk \| contribs) Jump to: navigation, search Harter-Heighway Dragon Curve (3D- twist) The image above is an example of a Harter-Heighway Curve (also called Dragon Curve). This fractal was first described in 1967 by American Martin Gardner and is often referred to as the Jurassic Park Curve, because it garnered popularity after being drawn and alluded to in the novel Jurassic Park by Michael Crichton (1990). Harter-Heighway Dragon Curve (3D- twist) Fields: Dynamic Systems and Fractals Created By: SolKoll # Basic Description First 5 iterations of the Harter-Heighway Curve The curve itself is fairly simple, with a line as the base segment. Each iteration replaces each line with two line segments at an angle of 90 degrees (other angles can be used to make various looking fractals), with each line being rotated alternatively to the left or to the right of the line it is replacing. 15th iteration The Harter-Heighway Dragon is created by iteration of the curve described above. The curve can be repeated infinitely, so that the perimeter of the dragon is in fact infinite. However, if you look to the image at the right, a 15th iteration of the Harter-Heighway Dragon is already enough to create an impressive fractal. The perimeter of the Harter-Heighway curve increases by$\sqrt{2}$ with each repetition of the curve. An interesting property of this curve is that the curve never crosses itself. Also, the curve exhibits self-similarity because as you look closer and closer at the curve, the curve continues to look like the larger curve. # A More Mathematical Explanation Note: understanding of this explanation requires: *Algebra [Click to view A More Mathematical Explanation] ### Perimeter First iteration in detail The per [...] [Click to hide A More Mathematical Explanation] ### Perimeter First iteration in detail The perimeter of the Harter-Heighway curve increases by a factor of $\sqrt{2}$ for each iteration. For example, if the first iteration is split up into two isosceles triangles, the ratio between the base segment and first iteration is: $\frac{2s\sqrt{2}}{2s} = \sqrt{2}$ ### Number of Sides The number of sides of the Harter-Heighway curve for any degree of iteration (k) is given by $N_k = 2^k\,$. ### Fractal Dimension 2nd iteration of the Harter-Heighway Dragon The Fractal Dimension of the Harter-Heighway Curve can also be calculated. As seen from the image of the second iteration of the curve, there are two new curves that arise during the iteration and N = 2. Also, the ratio of the lengths of each new curve to the old curve is: $\frac{2s\sqrt{2}}{2s} = \sqrt{2}$, so e = $\sqrt{2}$. Thus, the fractal dimension is $\frac{logN}{loge} = \frac{log2}{log\sqrt{2}} = 2$, and it is a space-filling curve. ### Angle The Harter-Heighway curve iterates with a 90 degree angle. However, if the angle is changed, new curves can be created: Curve with angle 85 Curve with angle 100 Curve with angle 110 # Teaching Materials There are currently no teaching materials for this page. Add teaching materials. # About the Creator of this Image SolKoll is interested in fractals, and created this image using an iterated function system (IFS). If you are able, please consider adding to or editing this page! Have questions about the image or the explanations on this page? Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page. Retrieved from "http://mathforum.org/mathimages/index.php/Harter-Heighway_Dragon" Categories: \| \| \| \| \| \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 7, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9091458320617676, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/60235?sort=newest	## Overview of the interplay of Harmonic Analysis and Number Theory ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I'm kind of disappointed that the question here was never sharpened. The Laplacian $\Delta$ on the upper half-plane is $-y^{2}(\partial^{2}/\partial x^{2}+\partial^{2}/\partial y^{2}))$. Suppose $D$ is the fundamental domain of, say, a congruence subgroup $\Gamma$ of $Sl_{2}(\mathbb{Z})$. Eigenfunctions of the discrete spectrum of $\Delta$ are real analytic solutions to $\Delta (\Psi)=\lambda \Psi$ that are $\Gamma$-equivariant functions in $L^{2}(D, dz)$, where $dz$ is the Poincare measure on the upper half-plane. These eigenfunctions evidently carry quite a bit of number theoretic information. Frankly, this point of view on number theory sounds incredibly interesting... Question: Would someone please suggest a readable introductory account that tells this story? (I imagine that answers will include the words Harish-Chandra, Langlands, etc...) Also, if experts are inclined to write a short overview as an answer, that would also be much appreciated. - ## 4 Answers I think it is also fair to say that things like hyperbolic $n$-spaces (and other symmetric spaces), and arithmetic quotients of them, are primordial, one-of-a-kind (well, not quite) objects. The opposite of "generic" mathematical objects. Partly because of the abruptly greater technical complexity in discussing the harmonic analysis on them, they are much less familiar than Euclidean spaces or their familiar quotients, circles and products thereof. Apart from Langlands' program and direct, intentional discussion of $L$-functions, I find it provocative that the basic harmonic analysis of $SL_2(\mathbb Z)\backslash H$ is not merely far subtler than that of $\mathbb R^2$ or $\mathbb R^2/\mathbb Z^2$, but that those subtleties are directly related to profound unsolved problems. As a well-known example, while we easily understand the sup norm of exponentials in the harmonic analysis on the real line, sharp estimates on pointwise behavior of eigenfunctions for the Laplacian on the upper half-plane give Lindelof: e.g., the value of the Eisenstein series $E_s$ at $z=i$ is the zeta of the Gaussian integers (divided by $\zeta(2s)$). Continuing, unlike the fact that the product of two exponentials is an exponential (that is, the tensor product of two one-dimensional irreducibles is still irreducible), decomposition of tensor products of waveforms, or of the repns they generate, is very tricky, and, again, is connected to serious outstanding problems. (Iwaniec' book mentions such examples.) That is, apart from "big conjectures" about automorphic forms and L-functions by themselves, even-more-primitive number-theoretic things just arise unbidden when we try to do innocent, ordinary things that would be trivial in Euclidean space. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. That's how I see it. Please correct me if I am wrong: 1. Induce the trivial representation from a lattice $\Gamma$, e.g. $\mathrm{SL}_2( \mathbb{Z})$, to the group $\mathrm{SL}_2(\mathbb{R})$. 2. Then consider the $\mathrm{SO}_2(\mathbb{R})$-invariant subspace.(weights are actually associated to one dimensional representations of SO_2) This representation is isomorphic to the representation $L^2(D,dz)$, you construct above. How does the Laplace Beltrami operator enter the picture... It is the descent of the Casimir operator to this space, which is the generator of all invariant differential operators (= center of the universal envelopping algebra of \mathfrak{sl}_2(\mathbb{R})). So intuitevely the Casimir operator captures the $G$ structure on the Hilbert space. Why did we became interested in such constructions? Maass discovered that the Mellin transforms of Dedekind zeta function associated to a quadratic real fields are Maass forms. Now, since all interesting $L$ functions (ass. to Galois repr., elliptic curves,...) are conjectured to be associated to some representation to some group, it seems worthwhile: 1. To study if and how they are associated ... (e.g., Taniyama-Shimura conjecture) 2. To study their properties on either side and conclude about the other... Mappings between groups and comparing their L-functions give also nice information about them (functoriality), e.g. they generalized Ramanujan conjecture would be implied by certain "functoriality" conjectures betweens general linear groups . Perhaps I should conclude that with the adeles $\mathbb{A}$ the better picture is $$\mathcal{L}^{2} ( \mathrm{SL}_2(\mathbb{Q}) \backslash \mathrm{SL}_2(\mathbb{A}))^{\mathrm{K}(m)\mathrm{SO}(2)} \cong \mathcal{L}^2(\Gamma(m) \backslash \mathbb{H}),$$ where $\mathrm{K}(m)$ is the product over $\mathrm{K}_p(m)$, the group of elements in $\mathrm{SL}_2( \mathbb{Z}_p)$ which are the identy modulo $m$. This picture contains the Hecke operators more naturally... - Thanks for this answer. – Jon Bannon Apr 1 2011 at 11:18 Ten Lectures on the interface between analytic number theory and harmonic analysis, by Hugh Montgomery is also worthwhile. - This is good, thanks DJC! – Jon Bannon Apr 28 2011 at 21:22 I highly recommend Iwaniec's 1983 ICM lecture, which you can read here, and Peter Sarnak's article "Spectra of hyperbolic surfaces," which is here. - Thank you, David! – Jon Bannon Mar 31 2011 at 23:28 These are exactly the sort of thing I was looking for. Thanks again! – Jon Bannon Mar 31 2011 at 23:29 Actually, Sarnak has been focussing on this interplay for a long time. His PhD thesis was about a generalization of Selberg's Trace Formula. – Denis Serre Apr 1 2011 at 5:26 3 Wow, I did not know that all ICM talks are online (mathunion.org/ICM)! Thanks, David. – GH Apr 1 2011 at 8:52	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 30, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9296897053718567, "perplexity_flag": "middle"}
http://physics.stackexchange.com/tags/quantum-information/info	Tag info About quantum-information Quantum information is the study of the informational content of quantum states. The most common object of study is the "qubit", the information in a two-state quantum system such as spin-1/2 or photon polarization. Quantum systems have infinitely more information than the corresponding classical system. An example of a two-state classical system is a coin which can be either heads-up or heads-down. This corresponds to a single "bit" of information; we could code it as "0" or "1". The quantum two-state system or "qubit" is usually described by a normalized complex vector with two elements such as the "ket": $\left(\begin{array}{c} \alpha \\ \beta \end{array}\right)$ where $\alpha$ and $\beta$ are two complex numbers subject to $\|\alpha\|^2+\|\beta\|^2 =1$. An alternative is the corresponding "bra", which has the same information content: $\left(\begin{array}{cc} \alpha^* & \beta^* \end{array}\right)$ A less common representation of a qubit is the (pure) density matrix form: $\left(\begin{array}{cc} \|\alpha\|^2 & \alpha\beta^* \\ \alpha^\beta & \|\beta\|^2 \end{array} \right)$, in which the extraneous complex phase information has been removed. From this we see that there are only two real informational degrees of freedom in a qubit, for example, $\|\alpha\|^2$ and the complex phase of $\alpha\beta^$. Another way of representing these two degrees of freedom is with the Bloch sphere. Probably the most popular textbook for quantum information is Nielsen and Chuang's Quantum Computation and Quantum Information.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 8, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9195746183395386, "perplexity_flag": "head"}
http://www.reference.com/browse/exothermicity	Definitions # Spontaneous process A spontaneous process is the time-evolution of a system in which it releases free energy (most often as heat) and moves to a lower, more thermodynamically stable, energy state. The sign convention of changes in free energy follows the general convention for thermodynamic measurements, in which a release of free energy from the system corresponds to a negative change in free energy, but a positive change for the surroundings. A process that is capable of proceeding in a given direction, as written or described, without needing to be driven by an outside source of energy. The term is used to refer to macro processes in which entropy increases; such as a smell diffusing in a room, ice melting in lukewarm water, salt dissolving in water, and iron rusting. The laws of thermodynamics govern the direction of a spontaneous process, ensuring that if a sufficiently large number of individual interactions (like atoms colliding) are involved then the direction will always be in the direction of increased entropy (since entropy increase is a statistical phenomenon). ## Overview For a reaction at constant temperature and pressure, the change ΔG in the Gibbs free energy is: $Delta G = Delta H - T Delta S ,$ The sign of ΔG depends on the signs of the changes in enthalpy (ΔH) and entropy (ΔS), as well as on the absolute temperature (T, in degrees Kelvin). Notice that changes in the sign of ΔG cannot occur solely as a result of changes in temperature alone, because the absolute temperature can never be less than zero. When ΔG is negative, a process or chemical reaction proceeds spontaneously in the forward direction. When ΔG is positive, the process proceeds spontaneously in reverse. When ΔG is zero, the process is already in equilibrium, with no net change taking place over time. We can further distinguish four cases within the above rule just by examining the signs of the two terms on the right side of the equation. When ΔS is positive and ΔH is negative, a process is spontaneous When ΔS is positive and ΔH is positive, a process is spontaneous at high temperatures, where exothermicity plays a small role in the balance. When ΔS is negative and ΔH is negative, a process is spontaneous at low temperatures, where exothermicity is important. When ΔS is negative and ΔH is positive, a process is not spontaneous at any temperature, but the reverse process is spontaneous. The second law of thermodynamics states that for any spontaneous process the overall change ΔS in the entropy of the system must be greater than or equal to zero, yet a spontaneous chemical reaction can result in a negative change in entropy. This does not contradict the second law, however, since such a reaction must have a sufficiently large negative change in enthalpy (heat energy) that the increase in temperature of the reaction surroundings (considered to be part of the system in thermodynamic terms) results in a sufficiently large increase in entropy that overall the change in entropy is positive. That is, the ΔS of the surroundings increases enough because of the exothermicity of the reaction that it overcompensates for the negative ΔS of the system, and since the overall ΔS = ΔSsurroundings + ΔSsystem, the overall change in entropy is still positive. Another way to view the fact that some spontaneous chemical reactions can lead to products with lower entropy is to realize that the second law states that entropy of a closed system must increase (or remain constant). Since a positive enthalpy means that energy is being released to the surroundings, then the 'closed' system includes the chemical reaction plus its surroundings. This means that the heat release of the chemical reaction sufficiently increases the entropy of the surroundings such that the overall entropy of the closed system increases in accordance with the second law of thermodynamics. Just because a chemist calls a reaction “spontaneous” does not mean the reaction happens with great speed. For example, the decay of diamonds into graphite is a spontaneous process, but this decay is extremely slow and takes millions of years. The rate of a reaction is independent of its spontaneity, and instead depends on the chemical kinetics of the reaction. ## See also • Endergonic reaction reactions which are not spontaneous at standard temperature, pressure, and concentrations. • Diffusion spontaneous phenomena that minimize Gibbs free energy.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9408702254295349, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/9506/tunneling-rate-constant?answertab=active	# Tunneling Rate Constant I am trying to "decode"/derive an expression for the macroscopic rate constant for the tunneling of protons through a potential energy barrier that I read in a journal article: $$k_{\rm tun}(T)=(2\pi\hbar)^{-1}\int_0^{V_{\rm max}} Q(V,T) P_{\rm tun}(V)\ dV.$$ So basically: the authors say work out the probability of tunneling at each point up the potential energy barrier (from 0 to $V_{\rm max}$), multiply this by the Boltzmann factor ($Q(V,T)$), integrating over all energies (i.e. we are taking a Boltzmann average )and then convert from energy in J to rate by multiplying by $(2\pi\hbar)^{-1}$. The authors use $(2\pi\hbar)^{-1}$ to convert from an energy in J to a rate in per second. This implies that the relationship between energy and frequency being used is: $$E=\hbar\omega$$ I thought this just applied to electromagnetic radiation and free particles not in a potential. Is it OK to use this relation for protons tunneling in a potential? Or should you use some info about the shape of the potential? Many thanks N26 - Welcome! I fixed up the formatting of your post using the MathJax formatting standard that we use here. I think that, when you're editing a post, there's a link on the right that'll give you some formatting tips to get started. Also, if you open up this post for editing, you'll see the syntax I used. – Ted Bunn May 5 '11 at 20:48 you can also do the same (open up a post for editing to see the used syntax) even if its not your own post – lurscher May 5 '11 at 21:05 The language MathJax eats is basically LaTeX, there are many tutorials on the web. – dmckee♦ May 5 '11 at 22:40 ## 1 Answer $E=\hbar\omega$ is a totally universal formula that holds for all particles and everywhere in quantum mechanics. Schrödinger's equation guarantees that. The same question was being answered yesterday: Gravitational wave energy -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9223705530166626, "perplexity_flag": "middle"}
http://sbseminar.wordpress.com/2009/10/07/how-to-almost-prove-the-4-color-theorem/	## How to almost prove the 4-color theorem October 7, 2009 Posted by Noah Snyder in planar algebras, quantum topology. trackback Vaughan Jones often quips at the beginning of talks on Planar Algebras (see these lectures, for example) that the worst thing you can say about Planar Algebras is that they have not yet yielded a proof of the 4-color theorem. In this post I’ll sketch how a common “evaluation algorithm” (used by Greg Kuperberg and by Emily Peters, for example) almost proves the 4-color theorem. I believe this (failed) argument is due to Penrose, though I’m taking it from an article of Chmutov, Duzhin, and Kaishev and some notes of John Baez’s. There are some more elaborate attacks (by Kauffman, Saleur, Bar Natan, and probably others) that I won’t discuss at all. This is the second of what hopefully will be a short series of posts on “evaluation algorithms” (the first was on the Jellyfish algorithm). The outline of the post is as follows. First I’ll explain a standard reduction of the 4-color theorem to a question about 3-coloring edges of trivalent graphs. Second I’ll explain why 3-colorings of edges is a question about finding a positive evaluation algorithm for a certain planar algebra. Third, I’ll discuss “Euler characteristic” evaluation algorithms. Fourth I’ll explain how this technique almost answers the 4-color theorem. Suppose you have a planar graph and you want to 4-color the faces (note, I said faces, not vertices). First notice that you can reduce to the case of 3-valent planar graphs by “adding a new country at every higher valency vertex” (see Week 22 in Mathematical Physics for a delightful ASCII drawing). This observation is from the early days of the 4-color problem and due to Cayley and Kempe. The next reduction also comes from the 19th century, and is due to Tait. There is a bijection between “4-colorings of faces” and “4-coloring one distinguished face and then 3-coloring the edges.” The way to understand this bijection is to label the edges with “rules for changing face colorings.” Namely there are 3 ways to assign a pairing of the four colors. If you’ve chosen such a rule for every edge then you can propogate your single face coloring to the whole graph. It’s a simple exercise to check that this gives a bijection between valid colorings. The invariant of planar graphs “number of 3-colorings of edges” extends to a planar algebra/TQFT/tensor category by the recipe similar to that Ben outlined here. Any planar graph with boundary can be thought of as a functional on the space Span{labelings of the boundary}. Extending by linearity any linear combination of planar graphs with boundary can also be thought of as such a functional. Mod out by the equivalence setting two diagrams equal to each other if they give the same functional. The key fact is that these relations play well with gluing together diagrams and hence gives a planar algebra (or tensor category together with a chosen generating object). What explicitly do these relations look like? You can remove a bigon for a multiplicative cost of 2, a triangle for a multiplicative cost of 1, and a circle with no trivalent vertices for a multiplicative cost of 3. Furthermore there is a version of the I = H relation (I strongly encourage you to check yourself that this relation between number of 3-colorings of edges holds for any coloring of the boundary): So what do we need to do to prove the 4-color theorem? We need to give a manifestly positive algorithm for evaluating closed diagrams in this planar algebra! That’s it. One common technique for evaluating closed planar diagrams is to concentrate on the faces. Notice above we know that we can remove all bigons and triangles. What about bigger faces? Well the beautiful thing is that by Euler characteristic arguments all you need to do is prove that you can remove squares and pentagons because any planar diagram has a pentagonal or smaller face. In fact this technique and small variations on it have been remarkably successful at answer planar algebraic questions (once for quantum groups and once for an exotic subfactor). So let’s try that. Here’s the square (finding this formula using the above relations is kinda fun): Now there’s just one case left, the pentagon. Just find a positive formula there and you’d be done! ## Comments» 1. Greg Kuperberg - October 7, 2009 This is an important argument because it proves the “x-color theorem” for any real $x \geq 5$. This is of course using the chromatic polynomial to interpret a non-integer number of colors. It is still a conjecture that the chromatic polynomial is strictly positive for all real $x \geq 4$. Moreover, it’s an intriguing features of this question that it exactly corresponds to real values of the quantum parameter q. 2. Scott Morrison - October 7, 2009 This conjecture says “every planar network of 2-strand Jones-Wenzl idempotents evaluates to something positive” (at d=2 for the four colour theorem, but we might hope for all d>=2). That suggests a generalisation in a different direction. It’s plausible that every planar network of arbitrary Jones-Wenzl idempotents is positive. You could think of this as saying “Every history of (quantum) angular momentum interactions which isn’t forbidden is allowed.” Here forbidden means that your trivalent graph labeled by naturals satisfies the usual triangle inequalities and parity restrictions (but in the language of JW idempotents this is automatic) and allowed means positive “amplitude” or evaluation. 3. Noah Snyder - October 7, 2009 Greg, where I can I find a succinct explanation of the general relationship between U_q(so_3) and the chromatic polynomial? (Namely that not only does q=1 correspond to the chromatic polynomial specialized at 4, but in general that the quantum group corresponds to the chromatic polynomial specialized at q+2+q^-1?) 4. Scott Morrison - October 7, 2009 I know you’ve found this already Noah, but for everyone else: http://arxiv.org/abs/0711.0016 is excellent. 5. Noah Snyder - October 7, 2009 Is 5 the smallest value of x (other than 4) such that the x-color theorem has been proved? 6. David Speyer - October 7, 2009 It can’t be quite that easy. Take the dodecahedron and quotient by the antipodal map. So you obtain a tiling of $\mathbb{RP}^2$ by six pentagons. The edge graph is the Petersen graph, which is not 3-colorable. If you could deal with the pentagon, wouldn’t this show that any surface tiled with pentagons had a 3-colorable edge graph? In general, this is a heuristic for ruling out proofs of the 4 color theorem. Any proof must look at a sufficiently large neighborhood of the pentagon to see that we are not dealing with this tiling of $\mathbb{RP}^2$. 7. Noah Snyder - October 7, 2009 It’s not that easy, there’s a reason I called it a failed proof! 8. Greg Kuperberg - October 7, 2009 Noah: The succinct explanation is the Yamada polynomial. The Yamada polynomial is the link invariant of the adjoint representation of U_q(sl(2)). It is also an invariant of tangled trivalent graphs, because there is an invariant tensor in $V_2^{\otimes 3}$. If the graph is planar, then its Yamada polynomial is proportional to the chromatic polynomial of the face graph. One way to argue this is to check that the skein relations for n-gons are all the same when n ≤ 5. Of course to get started you have to match x with q. The easiest way to do this is to check the 0-gon, i.e., a closed loop. A closed loop contributes a factor of x-1 to the chromatic polynomial, and it contributes a factor of [3] to the Yamada polynomial. So $x - 1 = [3]$, or $x = [2]^2 = q^2 + 2 + q^{-2}$. I believe that x can be pushed down to $5-\epsilon$ in the skein proof of the x-color theorem. The way to understand the status quo is that the actual proof of the 4-color theorem is a lot like skein theory, except that it is skein theory with existential quantifiers rather than skein theory with enumerative equality. The extra idea (due to Heesch) is that the skein relations need some extra oomph which you can obtain by diffusing the local Euler characteristic. The diffusion of the local Euler characteristic (or the rule for diffusion) is called “discharging”. Probably some simple discharging rule implies the x-color theorem when x is close enough to 5. I suspect that there hasn’t been a whole lot done with this conjecture. 9. Noah Snyder - October 7, 2009 Thanks. I’d sorted that out from the reference Scott mentioned above. The point where I was confused was that when I learned about the 4-color version of this I thought the reduction from 4-coloring faces to 3-coloring edges was an important intermediate step. But it seems the general connection between the Yamada polynomial and the chromatic polynomial doesn’t actually go through an intermediate step of edge coloring. Hence my confusion. 10. Greg Kuperberg - October 7, 2009 That’s right, it doesn’t. The Yamada polynomial model implies an alternate model of x-colorings in terms of a certain sort of edge 3-coloring with one oriented color and one unoriented color for the edges. This is just using the weight basis for the irrep V_2. What’s remarkable, and not fully explained, is the fact that the edges always have 3 states even though the faces have x states. A closely related fact is that there is a functor from the fusion category of U_q(so(3)) to the fusion category of S_n, when n = [2]^2 = [3] + 1. The functor takes the adjoint representation to an n-1-dimensional irrep of S_n. Of course, S_n only has finitely many irreps while U_q(so(3)) has infinitely many, so irreps of the latter can’t all go to irreps of the former. Instead, most of them go to various reducible representations. Somehow there is an S_n sitting inside U_q(so(3)), or at least morally sitting inside. As I think I mentioned when we met, there is an analogue of this for U_q(sp(4)). For one particular value of q, there is a functor to the fusion category of the Higman-Sims group. This was discovered by the late Francois Jaeger, and I have a paper on that. Or if you like, you can think of it as a relation between the U_q(sp(4)) polynomial of a certain type of planar graph, and the number of “Higman-Sims colorings” of its faces. 11. Noah Snyder - October 7, 2009 Another way of talking about the relationship between U_q(so_3) and S_t is that U_q(so_3) is the free version of S_t (free in the sense of free probability). Similarly, the free version of O_t is U_q(sl_2), and the free version of GL_t is “oriented Temperley-Lieb.” See work of Banica and collaborators. Yup, you mentioned the connection between U_q(sp_4) and Higman-Sims when you were here in June. I’ve actually been thinking about some stuff related to this connection quite a bit over the summer, it’s not ready for the blog, but I’ve been meaning to email you about it at some point. 12. David Speyer - October 8, 2009 Noah: Fair enough. From your description, it wasn’t clear whether the problem is that the pentagon is not positive, or simply that no one can show it is positive. I think my argument shows that the problem is the former. 13. Noah Snyder - October 8, 2009 Yup, good point. It’s also not that hard to see that you can’t write the pentagon as a positive linear combination of forests directly. Notice that if there were a positive formula using the rotation action you could show that there was a positive formula which was also rotationally invariant. The space of rotationally invariant 5-leaf forests is just linear combinations of two things (1-tree and 2-tree forests). So you just need to show that those two vectors are linearly independent, and then find a non-positive formula. Showing linear independence could be done by computing some inner products. It seems plausible that somewhere in that inner product you end up with the graph that you mention, so perhaps this isn’t an independent argument. 14. Michael Welford - October 8, 2009 Finding the formula for the square does look like a fun challenge. Now I’ll have to leave this post up on my browser until I work it out. Of course, the whole ‘almost proof’ is just a slighty exotic variant of the original ‘almost proof’. Going from 4 colors for faces to 3 colors for edges can be taken another step. I assign what I call a polaritiy to a vertex based the order of edge colors when going around the vertex clockwise starting with red. Red-white-blue has opposite polarity to red-blue-white. There is a face 4 coloring just when for each face the number of vertices on that face of each polarity are equal mod 3. There. Now you have a fun challenge too. 15. dick lipton - October 10, 2009 Have you looked at our idea of approximate Tait colorings: at http://rjlipton.wordpress.com/2009/04/24/the-four-color-theorem/ 16. Noah Snyder - October 11, 2009 I hadn’t seen that before. Thanks for sharing it. Awesome blog you’ve got going there! 17. Gil Kalai - October 12, 2009 Two questions: 1) Is Dror Bar-Nathan’s approach to the 4CT related? (It can be found here http://www.math.toronto.edu/~drorbn/LOP.html ). 2) Does the approach here gives a proof (to the easy theorems) that every bipartite cubic graph (or even every paipartite planar cubic graph) is 3-edge colorable? This is a nice toy problems for new approaches to the 4CT. 18. Noah Snyder - October 12, 2009 My understanding of Dror’s approach is that he gives a new description of planarity that is suggestive in the context of this argument. In particular, the planar algebra I gave above is nothing but integer spin representation theory of the lie algebra sl(2). What Dror does is give a lie theoretic description of planarity in terms of sl(n) for large n. So if you believe “everything about lie algebras really comes from sl(2)” then you should believe the 4-color theorem. Since Dror’s approach uses actual lie algebras rather than quantum groups, at face value it only applies to 4-coloring not to x-coloring for x>=4. It would be interesting to see if his approach generalizes to the quantum group setting. 19. Carnival of Mathematics #59 « The Number Warrior - November 6, 2009 [...] The 4-color-theorem states that any map can be colored with at most four colors so no two adjacent regions share a color. It was proven in 1976 by Kenneth Appel and Wolfgang Haken using a computer to check 1,936 maps. Noah Snyder at the Secret Blogging Seminar almost but doesn’t quite prove the 4-color-theorem in a single blog post. [...] %d bloggers like this:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 8, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9242958426475525, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/173400/solving-for-x-the-following-linear-equation-is-the-solution-a-unique-rectangu/173413	# solving for $X$ the following linear equation, is the solution a unique rectangular matrix? I have an invertible matrix $A$ of size $n \times n$ and a matrix $U$ of size $n \times m$, for $m < n$. Matrix $U$ is orthonormal, meaning, the rows are orthonormal vectors. I also have an $m \times m$ diagonal matrix $\Sigma$ with positive values on the diagonal. I want to find the solution for the equation $$X A U = \Sigma.$$ Since $U$ is orthonormal, then $U^T U = I$, and therefore, for $X = \Sigma U^T A^{-1}$, we have: $$X A U = \Sigma U^T A^{-1} A U = \Sigma.$$ I would like to know if this is the only solution, and if so, how to show it. - Is $U$ in the first line the same as $B$? Is $\Sigma$ the $m \times m$ diagonal matrix and you are solving for $X$? – Ross Millikan Jul 20 '12 at 21:40 @Ross, terribly sorry. I fixed the question. Yes to both of your questions. – kloop Jul 20 '12 at 21:45 1 @copper, I am confused by your answer. I stared a bit at the equation and guessed the solution. My problem is that even though $U'U = I$, we don't have $UU' = I$ because $m < n$. So we can't multiply both sides by $U'$. – kloop Jul 20 '12 at 21:48 @kloop: OOps, sorry, I didn't check your calculation. I will delete my comment momentarily. – copper.hat Jul 20 '12 at 21:51 The title is rather extraordinary in that it's so general that it could be the title for almost any question, but it doesn't actually fit this particular question, since you're not in fact asking how to find a solution; you've already found one and want to know whether it's unique. Please try to choose titles that summarize the question. – joriki Jul 20 '12 at 21:55 show 6 more comments ## 2 Answers Since $\Sigma$ and $A$ are invertible, from $\Sigma^{-1}XAU=I$, we can characterize the solution set as the set of all matrices $X$ of the form $\Sigma VA^{-1}$, where $V$ is any matrix with $VU=I$. For $m\lt n$ there are matrices other than $U'$ with this property. - thanks. that seems correct. but why can we "characterize the solution set as the set of all matrices $X$ of the form ..." ? – kloop Jul 20 '12 at 22:22 Write $V=\Sigma^{-1}XA$. Then your equation is $VU=I$. Since $\Sigma$ and $A$ are invertible, there's a one-to-one correspondence between $V$ and $X$; being invertible, $\Sigma$ and $A$ merely "translate" between the two. So for any $V$ with $VU=I$, there's exactly one solution $X=\Sigma VA^{-1}$, and conversely, for every solution $X$, your equation requires $VU=\Sigma^{-1}XAU=I$. – joriki Jul 20 '12 at 22:35 Take $\Sigma = \begin{bmatrix} 1 \end{bmatrix}$, $A=\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$, $U = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$. Then the equation above becomes $$\begin{bmatrix} x_1 & x_2 \end{bmatrix} A U = \Sigma$$ and all solutions satisfy $x_1+x_2 = 1$. So the solution is not, in general, unique. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 48, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9539310336112976, "perplexity_flag": "head"}
http://mathhelpforum.com/algebra/177191-solving-rational-inequalties-application-problem-help.html	Thread: 1. Solving rational inequalties application problem help Here's the problem: An economist for a sporting goods company estimates the revenue and cost functions for the production of a new snowboard. These functions are $R(x)=-x^2+10x$ and $C(x)=4x+5$, respectively, where x is the number of snowboards produced, in thousands. The average profit is defined by the function AP(x) = P(x) / x , where P(x) is the profit function. Determine the production levels that make AP(x) > 0 What i did is : $-x^2+1x + 4x+5 > 0$ Is this the right inequality to solve the problem? 2. Won't profit be $\displaystyle P(x) = R(x)-C(x) = -x^2+10x-(4x+5) = -x^2+6x-5$ ? 3. Originally Posted by pickslides Won't profit be $\displaystyle P(x) = R(x)-C(x) = -x^2+10x-(4x+5) = -x^2+6x-5$ ? Yeh, so i factored out the negative and put the equation into the quadratic formula to find the x's which are 6.74 and -0.74. So -0.74 < x < 6.74 but the answer according to the book is 1 < x < 5	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9122304916381836, "perplexity_flag": "middle"}
http://mathhelpforum.com/differential-geometry/28006-loci-complex-plane.html	Thread: 1. Loci In complex plane Show that in an Argand diagram the equation $\arg (z-2) - \arg (z-2i) = \frac{3 \pi}{4}$ represents an arc of a circle and that $\frac{\|z-4\|}{\|z-1\|}$ is constant on this circle. Find values of z corresponding to the points in which the circle is cut by the curve given by $\|z-1\| + \|z-4\| = 5$ I am okay with getting the that the first arg equation describes the arc of a circle, but i am a bit stuck on how to interpret the next part. anyone willing to give me a hand? thanks, Bobak 2. Originally Posted by bobak I am okay with getting the that the first arg equation describes the arc of a circle, but i am a bit stuck on how to interpret the next part. anyone willing to give me a hand? thanks, Bobak The locus of $\|z-1\| + \|z-4\| = 5$ is an ellipse with centre at (5/2, 0), major axis of length 5 and vertices at (0, 0) and (5, 0). There are several ways to see this - the simplest is if you understand the locus definition of an ellipse ..... Alternatively, you can sub z = x + iy, but there are a couple of twists and turns that will catch the unwary. This gives a geometric picture of the lay of the land - there's only one intersection point. 3. Originally Posted by mr fantastic The locus of $\|z-1\| + \|z-4\| = 5$ is a line segment joining z = 1 and z = 4. There are several ways to see this - believe it or not, the simplest is if you understand the locus definition of an ellipse ..... Alternatively, you can sub z = x + iy, but there are a couple of twists and turns that will catch the unwary. Does this give you a nudge? My apologies - this is wrong. It's an ellipse, not a line segment. I misread. It's an ellipse, with centre at (5/2, 0) and major axis of length 5. Vertices at (0, 0) and (5, 0). I'm a bit short on time now - I'll drop by later unless someone beats me to the punch. 4. Originally Posted by bobak I am okay with getting the that the first arg equation describes the arc of a circle, but i am a bit stuck on how to interpret the next part. anyone willing to give me a hand? thanks, Bobak OK, from what I posted earlier you can hopefully see that there's only one intersection point of the arc with the ellipse. How to find it: You know from the earlier part that $\|z - 1\| = \lambda \|z - 4\|$, where you'll have found the value of $\lambda$. Sub $\|z - 1\| = \lambda \|z - 4\|$ into $\|z - 1\| + \|z - 4\| = 5$: $\lambda \|z - 4\| + \|z - 4\| = 5 \Rightarrow \|z - 4\| (\lambda + 1) = 5 \Rightarrow \|z - 4\| = \frac{5}{\lambda + 1}$. This is a circle of known radius $\left( \frac{5}{\lambda + 1} \right)$ and centre at z = 4. Now sub $\|z - 4\| = \frac{1}{\lambda} \|z - 1\|$ into $\|z - 1\| + \|z - 4\| = 5$: $\|z - 1\| + \frac{1}{\lambda} \|z - 1\| = 5 \Rightarrow \|z - 1\| \left( 1 + \frac{1}{\lambda} \right) = 5 \Rightarrow \|z - 1\| = \frac{5}{1 + \frac{1}{\lambda}} = \frac{5 \lambda}{\lambda + 1}$. This is a circle of known radius $\left( \frac{5 \lambda}{\lambda + 1} \right)\,$ and centre at z = 1. Now find the intersection of these two circles (but only keep one of the solutions, the one that is the intersection of the arc with the ellipse ......): $(x - 4)^2 + y^2 = r_1^2\,$, where $r_1 = \frac{5}{\lambda + 1}$ .... (1) $(x - 1)^2 + y^2 = r_2^2\,$, where $r_2 = \frac{5 \lambda}{\lambda + 1}$ .... (2) Equation (1) - Equation (2): $\, (x - 4)^2 - (x - 1)^2 = r_1^2 - r_2^2$ $\Rightarrow (x - 4 + x - 1)(x - 4 -[x - 1]) = (r_1 + r_2)(r_1 - r_2)$ $\Rightarrow 3(2x - 5) = \frac{25(\lambda - 1)}{\lambda + 1}$ $\Rightarrow x = ......$. Remember that you know the value of $\lambda$ from the first part. Substitute x into either equation (1) or (2) and solve for y: y = ...... Therefore the intersection point is z = x + iy = ........ Let me know if you have further trouble with this. 5. Thank you a lot for your response Mr Fantastic. I think it is only appropriate that I post all my working. For the first part. $Part (1)$ $\arg (z-2) - \arg (z-2i) = \frac{3 \pi}{4}<br />$ rewriting arg(Z) in arctan from where $z = x +iy$ $\arctan \left ( \frac{y}{x-2} \right ) - \arctan \left ( \frac{y-2}{x} \right ) = \frac{3 \pi}{4}$ .... $\tan \left (\arctan \left ( \frac{y}{x-2} \right ) - \arctan \left ( \frac{y-2}{x} \right ) \right )= \tan \left ( \frac{3 \pi}{4} \right )$ then using compound angle formula. $\frac{ \frac{y}{x-2} - \frac{y-2}{x} }{1 + \frac{y}{x-2} \cdot \frac{y-2}{x}} = -1$ $\frac{y-2}{x} - \frac{y}{x-2} = 1 + \frac{y}{x-2} \cdot \frac{y-2}{x}$ .... skipping some step giving the equation of he circular arc as $x^2 + y^2 = 2^2$ Now I am not too sure about how I should restrict the domain of the circle to give the equation of the arc. $Part (2)$ $\frac{\|z-4\|}{\|z-1\|} \Rightarrow \frac{\|x-4 + iy\|}{\|x-1 + iy\|}$ $\Rightarrow \frac{(x-4)^2 + y^2}{(x-1)^2 + y^2}$ $\Rightarrow \frac{x^2 + y^2 + 16 - 8x}{x^2 + y^2 +1 -2x}$ using $x^2 + y^2 = 2^2$ $\Rightarrow \frac{20 - 8x}{5 -2x}$ $\Rightarrow \frac{20 - 8x}{5 -2x}$ $\Rightarrow \frac{4(5 -2x)}{5 -2x}$ therefore $\frac{\|z-4\|}{\|z-1\|} = 4$ on the circle. $Part (3)$ $\|z - 1\| = \frac{1}{4} \|z - 4\|$ Subbing into $\|z - 1\| + \|z - 4\| = 5<br />$ $\frac{5}{4} \|z - 4\| = 5$ $\|z - 4\| = 4$ so we have a circle $(x-4)^2 +y^2 = 2^2$ we also get $\|z-1\| = 1$ giving the circle $(x-1)^2 +y^2 = 1$ So the system is $(x-4)^2 +y^2 = 2^2 (1)$ $(x-1)^2 +y^2 = 1 (2)$ $(1) - (2)$ $(x-4)^2 - (x-1)^2 = 3$ $-3(2x-5) = 3$ $x = 2$ using equation 2 $(2-1)^2 +y^2 = 1$ $y = 0$ so $z = 2$ which is wrong! 6. Originally Posted by bobak Thank you a lot for your response Mr Fantastic. I think it is only appropriate that I post all my working. For the first part. $Part (1)$ $\arg (z-2) - \arg (z-2i) = \frac{3 \pi}{4}<br />$ rewriting arg(Z) in arctan from where $z = x +iy$ $\arctan \left ( \frac{y}{x-2} \right ) - \arctan \left ( \frac{y-2}{x} \right ) = \frac{3 \pi}{4}$ .... $\tan \left (\arctan \left ( \frac{y}{x-2} \right ) - \arctan \left ( \frac{y-2}{x} \right ) \right )= \tan \left ( \frac{3 \pi}{4} \right )$ then using compound angle formula. $\frac{ \frac{y}{x-2} - \frac{y-2}{x} }{1 + \frac{y}{x-2} \cdot \frac{y-2}{x}} = -1$ $\frac{y-2}{x} - \frac{y}{x-2} = 1 + \frac{y}{x-2} \cdot \frac{y-2}{x}$ .... skipping some step giving the equation of he circular arc as $x^2 + y^2 = 2^2$ Now I am not too sure about how I should restrict the domain of the circle to give the equation of the arc. [snip] To get the restriction, note that: $\arg (z-2) - \arg (z-2i) = \frac{3 \pi}{4} \Rightarrow \arg \left( \frac{z - 2}{z - 2i} \right) = \frac{3 \pi}{4}$. But $\frac{z - 2}{z - 2i}$ cannot have an argument of $\frac{3 \pi}{4}$ unless its real part is less than zero AND its imaginary part is greater than zero ..... The latter restriction is the one that is easiest to apply. The former restriction doesn't affect things (as you can check). Substitute z = x + iy and simplify to get cartesian form: $\frac{[(x - 1)^2 + (y - 1)^2 - 2] + (2y + 2x - 4)i}{x^2 + (y - 2)^2} = \frac{3 \pi}{4}$. So the restriction $Im \left( \frac{z - 2}{z - 2i} \right) > 0$ is equivalent to $2y + 2x - 4 > 0 \Rightarrow y + x - 2 > 0 \Rightarrow y > -x + 2$. So you want the part of the circle $x^2 + y^2 = 2^2$ that lies above the line $y = - x + 2$. The endpoints do NOT get included since arg(0) is not defined. 7. Originally Posted by bobak [snip] $Part (2)$ $\frac{\|z-4\|}{\|z-1\|} \Rightarrow \frac{\|x-4 + iy\|}{\|x-1 + iy\|}$ $\Rightarrow \frac{(x-4)^2 + y^2}{(x-1)^2 + y^2}$ $\Rightarrow \frac{x^2 + y^2 + 16 - 8x}{x^2 + y^2 +1 -2x}$ using $x^2 + y^2 = 2^2$ $\Rightarrow \frac{20 - 8x}{5 -2x}$ $\Rightarrow \frac{20 - 8x}{5 -2x}$ $\Rightarrow \frac{4(5 -2x)}{5 -2x}$ therefore $\frac{\|z-4\|}{\|z-1\|} = 4$ on the circle. * [snip] Great work on this part (and part 1), bobak. You really nailed the algebra. Except for the line marked with an * (see post #9) 8. Originally Posted by bobak [snip] $Part (3)$ $\|z - 1\| = \frac{1}{4} \|z - 4\|$ Subbing into $\|z - 1\| + \|z - 4\| = 5<br />$ $\frac{5}{4} \|z - 4\| = 5$ $\|z - 4\| = 4$ so we have a circle $(x-4)^2 +y^2 = 2^2$ we also get $\|z-1\| = 1$ giving the circle $(x-1)^2 +y^2 = 1$ So the system is $(x-4)^2 +y^2 = 2^2 (1)$ $(x-1)^2 +y^2 = 1 (2)$ $(1) - (2)$ $(x-4)^2 - (x-1)^2 = 3$ $-3(2x-5) = 3$ $x = 2$ using equation 2 $(2-1)^2 +y^2 = 1$ $y = 0$ so $z = 2$ which is wrong! Hmmmmm ... That's a shame. It's a nice approach and should work ...... there's a mistake somewhere and I will find it. In the meantime, you can get brutal and find the intersection of the arc of $x^2 + y^2 = 2^2$ with the ellipse. Can you get the cartesian equation of the ellipse ..... The equation is of the form $\frac{(x - h)^2}{a^2} + \frac{y^2}{b^2} = 1$ where h = 5/2, giving centre at (5/2, 0), a = 5/2, giving major axis of length 5, and b = 2 (which can be got from geometric considerations). I've used the locus definition of the ellipse to get this ...... 9. I've found the problem! Originally Posted by bobak [snip] $Part (2)$ $\frac{\|z-4\|}{\|z-1\|} \Rightarrow \frac{\|x-4 + iy\|}{\|x-1 + iy\|}$ $\Rightarrow \frac{(x-4)^2 + y^2}{(x-1)^2 + y^2}$ $\Rightarrow \frac{x^2 + y^2 + 16 - 8x}{x^2 + y^2 +1 -2x}$ using $x^2 + y^2 = 2^2$ $\Rightarrow \frac{20 - 8x}{5 -2x}$ $\Rightarrow \frac{20 - 8x}{5 -2x}$ $\Rightarrow \frac{4(5 -2x)}{5 -2x}$ therefore $\frac{\|z-4\|}{\|z-1\|} = 4$ on the circle. Mr F says: This is the mistake that causes the trouble with part 3 ...... What you've actually found is that $\frac{\|z-4\|^2}{\|z-1\|^2} = 4\,$ ! Therefore $\frac{\|z-4\|}{\|z-1\|} = 2$ ...... So $\frac{1}{\lambda} = 2$, NOT 4 .... [snip] Part 3 looks OK now using the original method. I get $x = \frac{10}{9}$ and $y^2 = +\frac{\sqrt{224}}{9}$. I haven't checked whether this point lies on the ellipse (I leave that little chore for you), but it does lie on the arc - good news. And you could use the brutal method as a check - good practice. I've used the wonders of hindsight to edit post #8. 10. By the way ..... did you notice that the locus defined by $\|z - 4\| = 2 \|z - 1\|$ is the circle $x^2 + y^2 = 2^2$ ....? A circle defined by $\|z - z_1\| = \lambda \|z - z_2\|$, $\lambda > 0 \,$ and $\, \lambda \neq 1 \,$, is called an Apollonius Circle. 11. Okay picking up where i left off. using $\frac{\|z-4\|}{\|z-1\|} = 2$ and the same method I applied earlier for part 3 I get $x= \frac{10}{9}$ and $y = \pm \frac{ 4 \sqrt{14}}{9}$ But we must neglect the negative solution of y as it lies underneath, the line $y = - x + 2$ so answering the initial question $z = \frac{10}{9} + \frac{ 4 \sqrt{14}}{9}i$ The book however wrote $z = \frac{10}{9} \pm \frac{ 4 \sqrt{14}}{9}i$ But I guess they are paying less attention than we are. I'll use the brute force method to check as you suggested. I need to transform this $\|z-1\| + \|z-4\| = 5<br />$ into an ellipse equation first, From the definition of the elpise i know the the sum distance of a point on the loci to each focus is 2a so $2a = 5$ so $a= \frac{5}{2}$ and the mid point of the two foci is clearly $\frac{5}{2}$ as we know the position of the foci we can use either foci to find e I did $ae+h = 4$ giving e = $\frac{3}{5}$ then using $b^2 =a^2(1-e^2)$ I got $b=2$ so the equation of our elipse is $\frac{(x - \frac{5}{2})^2}{\frac{5}{2}^2}+ \frac{y^2}{2^2} = 1<br />$ messing around a bit i get $x^2 - 5x + \frac{25}{16} y^2 = 0 <br />$ then subbing $y^2 = 4 - x^2$ i get a quadratic in x $9x^2 +80x -100 = 0$ $(9x - 10)(x+10) = 0$ x = -10 can clearly be rejected by geometric considerations, we get $x = \frac{10}{9}$ which is very reassuring. Mr F, i didn't notice that last part about the Apollonius Circle. never come across that before, shall do some reading. Thanks for help, i am very surprised that you thanked me for asking a question. 12. Originally Posted by bobak [snip] i am very surprised that you thanked me for asking a question. Ha ha. Don't be. Sometimes questions can be useful too .....	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 141, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9363727569580078, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/208314-partial-fractions.html	1Thanks • 1 Post By Plato # Thread: 1. ## Partial Fractions Hi everyone! The problem I need help with is: Preform partial fraction decomposition: x / 4x2 + 4x + 1 In the solution to this exercise the author first factorizes the denominator, which factorizes into (2x + 1)2, but than he writes that x / (2x + 1)2 = A / (2x + 1) + B / (2x + 1)2, I am unsure why this is the case? I understand that it can't be A / (2x + 1) + B / (2x + 1) either, but the authors approach seems wrong to me (don't worry I know I am wrong, hehe), because (2x + 1)2 is the original factorization of 4x2 + 4x + 1, so adding (2x + 1) to it, would in my eyes give a bigger number than 4x2 + 4x + 1 . Also, my teacher in Calculus 1 at university said that for the exam it is most important that we learn how to draw graphs and than the rest will come on its own. I assume that was meant as a joke, but he never jokes, so now I am trying to read into his statement to see if there is a underlying hidden message in it. Thanks to anyone for reading this and helping out 2. ## Re: Partial Fractions Originally Posted by Nora314 Preform partial fraction decomposition: x / 4x2 + 4x + 1 x / (2x + 1)2 = A / (2x + 1) + B / (2x + 1)2, I am unsure why this is the case? I understand that it can't be A / (2x + 1) + B / (2x + 1) either, but the authors approach seems wrong to me (don't worry I know I am wrong, hehe), because (2x + 1)2 is the original factorization of 4x2 + 4x + 1, so adding (2x + 1) to it, would in my eyes give a bigger number than 4x2 + 4x + 1 . When we add $\frac{x}{(x-2)}+\frac{x+1}{(x-2)^2}$ what is the LCD? If you see that correctly then you answer your own question. 3. ## Re: Partial Fractions Thank you so much, , I understand now!	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9389340281486511, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/25110/why-are-orbits-elliptical/25116	# Why are orbits elliptical? Almost all of the orbits of planets and other celestial bodies are elliptical, not circular. Is this due to gravitational pull by other nearby massive bodies? If this was the case a two body system should always have a circular orbit. Is that true? - ## 9 Answers No, any ellipse is a stable orbit, as shown by Johannes Kepler. A circle happens to be one kind of ellipse, and it's not any more likely or preferable than any other ellipse. And since there are so many more non-circular ellipses (infinitely many), it's simply highly unlikely for two bodies to orbit each other in a perfect circle. - 8 However, tidal and frictional forces will tend to dissipate energy while maintaining angular momentum, eventually resulting in an ellipse getting more and more rounded out into a circle. All of the planets are on very nearly circular elliptical orbits. – Andrew Aug 25 '11 at 18:58 @Andrew Except for many extrasolar planets, that's not true. – Mark Eichenlaub May 5 '12 at 6:11 ## Did you find this question interesting? Try our newsletter email address A circle is a very difficult shape to maintain. Even the slightest deviation, and a circle is bypassed. Orbits are elliptical when any of the following things happen: • Another object strikes the planet in such a way to change its orbit. It would have to be massive compared to the primary object, at least a sizable fraction. • Gravitational interaction with other nearby objects, especially if resonance occurs. This is why Pluto has such an elliptical orbit. • Differing albedo can cause differences to occur over a long period of time. I'm sure I could come up with other reasons as well. In order to have a perfectly circular orbit, one must achieve the perfect speed for one's distance from the body around which they are orbiting. The lecture Astronomy 106, Orbital Velocity gives the formula, which is: # $$V_c = \sqrt {GM \over r}$$ Any deviation from this results in an elliptical orbit. - In addition to the other answers I want to remark that the calculations from the conversation and force laws give you conic sections for the two body systems, parabolas, hyperbolas and ellipses (including circles). Ellipses are the only paths for orbits because the other paths never come near the starting point again. - Extending this idea, can almost-orbits be modelled with hyperbolas or parabolas? That is, the path of 2 celestial bodies that near each other but don't act on each other enough to create an orbit? – Eric Hu Feb 28 at 1:47 On the most fundamental level, ellipticity comes from the conservation of energy, angular momentum, and the $1/r^2$ gravitational force law. Any freshman taking classical mechanics should be able to take these three constraints and get ellipses. The impressive thing is how Newton took Keplers laws and worked backwards to get the laws of gravity and conservation. - Um, no, freshmen can't do it. It requires a tricky change of variables the way it is usually done. If you think it's easy--- try it. The problem is usually presented to 3rd year students, but you can do it with freshmen techiques if you don't want deep understanding. Newton did it with deep understanding. It was Hooke who derived the inverse square law from Kepler, and this was easier. Hooke didn't get the ellipse. – Ron Maimon May 5 '12 at 1:54 If this was the case a two body system should always have a circular orbit. There are a number of highly-eccentric orbits that aren't anywhere close to circular, but they're more difficult to maintain over time, as they're more likely to be biased by other objects. For some spacecraft (eg, STEREO; watch the first movie), they actually use this behaviour so that the spacecraft orbit the earth such that the moon will effectively 'throw' the spacecraft where they're trying to get it. The Voyager spacecraft used this multi-planet Gravitational Assist known as Planetary Grand Tour to achieve solar escape velocity with very modest fuel requirements. It is not uncommon to use such assists as it can radically reduce the fuel costs of a mission (to the point of making it even possible), in exchange for increased mission durations. - Many objects presently in orbit around larger objects were originally 'captured' by the gravity of the larger one as the smaller one happened to pass by closely enough. Of all the possible combinations of speed and direction of the two objects relative to each other at the time of gravitational capture, only a very special subset will result in a circular orbit; all others are elliptical with varying degrees of eccentricity. - Could you explain why "all others are elliptical"? – Manishearth♦ Nov 22 '12 at 7:00 A circle is only a special case of an ellipse and obeys all the same mathematical rules. So even circular orbits are elliptical. Origin chance and third+ body perturbations will have random effects on the ellipticity, so no surprise purely circular orbits are rare to non-existent being only one of an infinite number of possible outcomes. What I find much more interesting is how close to circular our planetary orbits actually are. Excluding Pluto as being not-a-planet any more, that is. - Perhaps the elliptic orbits arise because of an expanding universe. If the universe were stationary, the orbits might have been circular. - 2 The derivation of elliptical orbits by Kepler was carried out well before any modern notion of an expanding universe was held. – Nic Jan 13 '12 at 12:25 2 It would be better if the answer was based on facts with some evidence rather than a supposition. – Stuart Woodward Jan 22 '12 at 11:56 Every circle is elliptical since no answer for PI has been or can fully be achieved. The elliptical model fits as well as any. When taking into consideration Higgs Boson and Dark matter and there possible inflection on string theory in the fabric of space with multiple universes, it is very possible that the distortion of space-time make it impossible to achieve the theoretical circle. Once these distortions have been fully understood, we may find that all elliptic's are in actuality the eternal, just perceived now as a different form. The problem exists within our capability to measure what requires eternal rules or current rule have built in ambiguities. We have not achieved the eternal, therefore perfection is out of reach. You can always say; "For all practical purposes" but, that does not lead to the perfect or sacred knowledge in the quest for certainty. - This really doesn't make much sense, could you elaborate? – Manishearth♦ Dec 10 '12 at 23:05 ## protected by Qmechanic♦Jan 3 at 18:21 This question is protected to prevent "thanks!", "me too!", or spam answers by new users. To answer it, you must have earned at least 10 reputation on this site.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9518690705299377, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/80721/vertex-cover-reduction?answertab=oldest	# Vertex Cover Reduction I'm having a hard time figuring out how to reduce a problem to prove that it is NP-complete. Basically, my problem is this: Given an undirected graph, find a set $S$ such that for all nodes in $G \setminus S$ (i.e. all other nodes), there exists at least one edge from each node in $G \setminus S$ to a node in $S$. Expanding on this, I want to know if I can find this particular set with size at most equal to some integer $k$. - I've discovered that this is the problem of a dominating set. Which does help – CCSab Nov 10 '11 at 3:00 – joriki Nov 10 '11 at 9:24 Answered, but since my rep is low, couldn't properly answer it. I edited it into my original question. – CCSab Nov 10 '11 at 9:40 Thanks. I moved it to a community wiki answer, and I see you've already accepted it. The effect is the same, reputation-wise, since neither community wiki answers nor self-answers generate points. – joriki Nov 10 '11 at 9:49 ## 1 Answer A dominating set is simply a subset $S$ of $V$ such that each member of $V \setminus S$ contains at least one edge to a member of $S$. We reduce vertex-cover to dominating set in the following way: Create a new graph such that there exists a vertex for each vertex in G. Furthermore, add a new vertex for each edge in $G$, such that there now exists a triangle of edges. That is to say, for an edge, $e$ = ($u, v$) there now exists a new vertex, $z$, and the edges ($u, v$), ($u, z$), ($v, z$). We have to prove that our graph $G$ has a vertex cover of size $k$ iff the new graph has a dominating set of size $k$. If we have a set, $S$, that is a vertex cover of $G$, then we know that the same set is a dominating set on the new graph. By definition of vertex cover, each edge in $G$ is incident to at least one vertex in $S$. Therefore, each triangle of vertices in the new graph has at least one member of which belongs to $S$. As a result, we know that each edge in the new graph is either in $S$ or adjacent to $S$, which is the definition of a dominating set. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 26, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9576338529586792, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/281175/complex-number-vector-prove/281182	# Complex number vector prove Prove that, where $z$ denotes complex numbers: a.) $\|z_1 - z_2\| \le \|z_1\| +\|z_2\|$ b.) $\|\|z_1\| - \|z_2\|\| \le \|z_1 +z_2\|$ For a. the book answer gives: $\|z_1 -z_2\| = \|z_1 + (-z_2)\| \le \|z_1\| +\|-z_2\| =\|z_1\| +\|z_2\|$, but that proof isn't helpful at all, to me, because it doesn't explain anything. How can I prove this using a clear explaination instead of the way the book did? - ## 2 Answers Using Mejrdad's hint, for example: $$z_k:=x_k+y_ki\,\,\,,\,,k=1,2\,\,,\,x_k,y_k\in\Bbb R\Longrightarrow$$ $$\begin{align}(1)&\;\;\;\;\;\;\|z_1-z_2\|=\|(x_1-x_2)+(y_1-y_2)i\|=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\ (2)&\;\;\;\;\;\;\|z_1\|+\|z_2\|=\sqrt{x_1^2+y_1^2}+\sqrt{x_2^2+y_2^2}\end{align}$$ So squaring both equations above: $$x_1^2+y_1^2+x_2^2+y_2^2-2(x_1x_2+y_1y_2)\leq x_1^2+y_1^2+x_2^2+y_2^2+2\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}\Longleftrightarrow$$ $$-x_1x_2-y_1y_2\leq\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}\stackrel{\text{squaring}}\Longleftrightarrow$$ $$x_1^2x_2^2+y_1^2y_2^2+2x_1x_2y_1y_2\leq x_1^2x_2^2+y_1^2y_2^2+x_1^2y_2^2+x_2^2y_1^2\Longleftrightarrow$$ $$(x_1y_2-x_2y_1)^2\geq 0$$ And since the last inequality is trivial we get what we want going backwards( Why is it possible to argue that way?) - 1 +1 It looks fine and nice. I don't understand why some tutors want the students not to use some well-known facts when they solve the routine problems. In fact, the triangle inequality is a very basic tool when we face these kinds of inequalities. Any way, Nice illustration, Don. – Babak S. Jan 18 at 4:01 Thanks a lot Don! – Q.matin Jan 18 at 6:20 I think you might be able to do it by letting z = a + bi and then using the triangle inequality to prove the given relationships. - 1 But, the OP, was suggested to use the same method. Do you think your approach would change the way? – Babak S. Jan 18 at 3:22 @BabakSorouh: It's the same thing, I just explained that it's the triangle inequality since the book seemed to have given no explanation of why it's true. – Mehrdad Jan 18 at 3:24	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9399594068527222, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/43274/list	Return to Answer 3 added 1087 characters in body Hi Hailong, I would add one more observation to the other comments. Let me not worry about (2) vs (3) as the difference is only about the zero module so this is more of a philosophical question than a mathematical one. More specifically, a module could never satisfy $S_n$ for any $n$ that's larger than the dimension of the module, but not larger than $\dim X$. Kind of along the same lines, let $A\to B$ be a surjective morphism of rings (commutative with an identity) and $M$ a $B$-module. I.e., ${\rm Spec}\, B$ is a closed subset of ${\rm Spec}\, A$. Now both ${\rm depth}\, M$ and ${\rm supp}\, M$ are independent of the fact whether one views $M$ as a $B$-module or an $A$-module. It is reasonable that then whether it is $S_n$ would be also independent. The main difference between (1) and (2) is whether one wants to compare to the support of the module (i.e., view it over ring/annihilator) or the whole ring. To me, the second seems more natural. This way a sheaf/module that is $S_n$ on a subscheme remains $S_n$ when viewed on an ambient scheme. The definition (1) seems to prefer to compare to the fixed ring. One way some people try to bridge the gap between the two definitions is to say "$M$ is $S_n$ over its support", meaning that one should mod out by annihilator first before applying (either of the) definition(s). Then the two definitions are equivalent. As for (3), some people go the distance to say "a non-zero module is $S_n$ if..." 2 typo fixed Hi Hailong, I would add one more observation to the other comments. Let me not worry about (2) vs (3) as the difference is only about the zero module so this is more of a philosophical question than a mathematical one. I would just like to point out that there is a very useful characterization of depth and dimension of a module, namely Grothendieck's vanishing theorem which says that at any $x\in X$, the local cohomology of $M$ vanishes for $i$ strictly between the depth and the dimension of the module and does not vanish for $i$ equal either the depth or the dimension. In my eyes this suggest that one out to should use the dimension of the module in the definition, i.e., use (2). Another argument to support the use of (2) is that we like to say that CM is equivalent to "$S_n$ for all $n$". Now if you use definition (1) then only modules supported on the entire $X$ have even a chance to be CM, but I don't see how one would gain from assuming that. More specifically, a module could never satisfy $S_n$ for any $n$ that's larger than the dimension of the module, but not larger than $\dim X$. 1 Hi Hailong, I would add one more observation to the other comments. Let me not worry about (2) vs (3) as the difference is only about the zero module so this is more of a philosophical question than a mathematical one. I would just like to point out that there is a very useful characterization of depth and dimension of a module, namely Grothendieck's vanishing theorem which says that at any $x\in X$, the local cohomology of $M$ vanishes for $i$ strictly between the depth and the dimension of the module and does not vanish for $i$ equal either the depth or the dimension. In my eyes this suggest that one out to use the dimension of the module in the definition, i.e., use (2). Another argument to support the use of (2) is that we like to say that CM is equivalent to "$S_n$ for all $n$". Now if you use definition (1) then only modules supported on the entire $X$ have even a chance to be CM, but I don't see how one would gain from assuming that. More specifically, a module could never satisfy $S_n$ for any $n$ that's larger than the dimension of the module, but not larger than $\dim X$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 39, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9408872127532959, "perplexity_flag": "head"}
http://crypto.stackexchange.com/questions/2405/how-are-the-primes-used-to-generate-rsa-keys	# How are the primes used to generate RSA keys? I am confused about how keys in RSA asymmetric encryption are generated and what the implications for open communications are. Textbooks say the one-way function is merely two primes (with some critical constraints) -- so how do the two primes generated get turned into keys? Does this mean that having a public key is simply having one of the primes in a translated or equivalent form? For example, if I were to encrypt a message with someone else's public key and the one-way function could be broken by a third party, then could they decrypt the message with a key generated by the other prime? Does the other prime have to go through some special process to become a key? To go further, if the encrypted message was posted in a public forum or sniffed off of a wireless network, etc. (and again, the one-way function was broken by a mysterious third party), does that mean the message could be decrypted by the third party? Would it matter which prime was used? What i am wondering is what is the process whereby the primes behind the one-way function are converted into keys and is that process the same regardless of encryption used? - Welcome to Cryptography Stack Exchange. Your question was migrated here because of being not directly related to software development (the topic of Stack Overflow), and being fully on-topic here. Please register your account here, too, to be able to comment and accept an answer. – Paŭlo Ebermann♦ Apr 20 '12 at 7:40 I edited your title (and the text) to make clear this is about RSA - there are other asymmetric encryption schemes which don't use primes this way. – Paŭlo Ebermann♦ Apr 20 '12 at 7:43 ## 1 Answer Textbooks say the one-way function is merely two primes... Yes, that's the key part. It turns out that if you have a number that's a product of two large primes, deducing which primes factor that number is quite difficult. So, what we understand from this is that if you have $n = pq$, and you know $p$ and $q$, you can generate $n$, but if you have only $n$, it is hard to generate $p$ and $q$. These are not keys by and of themselves - they're parts of a process used to find a key. Specifically, a number $e$ is chosen and then you need to find $de = 1 \mod (p-1)(q-1)$. I've covered the mathematics in detail here, but the essential part to understand is that it is difficult to compute $p-1$ or $q-1$ without $p$ and $q$, which you don't have if you have only $n$. Thus, if you know $e$, you have no way to get to $d$ (it's important to understand it isn't theoretically impossible. It's just very, very hard practically). $e$ and $d$ form the key - $e$ is chosen and $d$ is usually computed, although not always by the mechanism I described. ...if ... the one-way function could be broken by a third party, then could they decrypt the message with a key generated by the other prime? Yes. The message wouldn't be generated by another prime in the case of RSA, but if they could somehow undo the one-way function, then the cryptosystem is broken and all your messages would be readable. Would it matter which prime was used? There are two cases here: • In the case of which primes are used for RSA - it does matter. They need to be big! At one time, the choice of prime did matter, for resistance against Pollard's p-1. However, it is now thought that for more advanced methods this is not the case. • In the case of the keys (which may or may not be prime) - for RSA you can actually interchange the keys, which makes RSA a trapdoor permutation. However, this is unusual and many other one-way functions are just trapdoor functions - which means they don't have this property. What i am wondering is what is the process whereby the primes behind the one-way function are converted into keys and is that process the same regardless of encryption used? The $de = 1 \mod (p-1)(q-1)$ process is not the only way to relate $d$ to $e$. You can also use $\lambda(n)$, the Carmichael function, which is what happens in PKCS#1. Other public key systems have different conversion functions, not all of which rely on the use of prime number factorisation. Common examples are the discrete logarithm problem and point multiplication over elliptic curves. (You can use this script to render the Mathjax in this answer; it'll automatically render if this Q gets migrated). -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 24, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9622408151626587, "perplexity_flag": "middle"}
http://mathhelpforum.com/number-theory/50631-elem-num-theo-help-w-proof.html	# Thread: 1. ## elem.num.theo - help w/proof So - I know all the steps, I just need help with a gap. Prove $n! > n^2 \exists n \geq4$ Ok, proof by Induction: Since 4! = 24 which is $>(4)^2 = 16$, $n!>n^2$ for n=4. Let k be a positive integer [math\geq4[/tex] and suppose $k!>k^2$. Then, $k!(k+1) > k^2(k+1)$ $(k+1)! > k^2(k+1) = k^3+k^2$ Here's where I'm stuck. I know $k\geq4$ but how can I show that $(k+1)^2 < k^3+k^2$? From then, I can show $(k+1)! > k^3+k^2 > (k+1)^2$ Therefore, $(k+1)! > (k+1)^2$, so $n! > n^2 for n \geq4$. Please help me out with this way of solving the proof, I'm sure there are other ways. 2. Originally Posted by cassiopeia1289 So - I know all the steps, I just need help with a gap. Prove $n! > n^2 \exists n \geq4$ Ok, proof by Induction: Since 4! = 24 which is $>(4)^2 = 16$, $n!>n^2$ for n=4. Let k be a positive integer [math\geq4[/tex] and suppose $k!>k^2$. Then, $k!(k+1) > k^2(k+1)$ $(k+1)! > k^2(k+1) = k^3+k^2$ Here's where I'm stuck. I know $k\geq4$ but how can I show that $(k+1)^2 < k^3+k^2$? From then, I can show $(k+1)! > k^3+k^2 > (k+1)^2$ Therefore, $(k+1)! > (k+1)^2$, so $n! > n^2 for n \geq4$. Please help me out with this way of solving the proof, I'm sure there are other ways. you will have that $k^3 + k^2 > (k + 1)^2$ if you can show $k^3 > 2k + 1$ for $k \ge 4$ 3. so how do I go from $(k+1)!>k^3+k^2$ to $k^3>2k+1$ - where did the 2k+1 come from? 4. Originally Posted by cassiopeia1289 so how do I go from $(k+1)!>k^3+k^2$ to $k^3>2k+1$ well, if i told you that, i'd be doing the problem for you, wouldn't i? - where did the 2k+1 come from? we want to show that $k^3 + k^2 > (k + 1)^2 = k^2 + {\color{red}2k + 1}$ thus we need to show that $k^3 > 2k + 1$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 31, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.943143367767334, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/75655/is-there-a-log-space-algorithm-for-divisibility?answertab=active	# Is there a log-space algorithm for divisibility? Is there an algorithm to test divisibility in space $O(\log n)$, or even in space $O(\log(n)^k)$ for some $k$? Given a pair of integers $(a, b)$, the algorithm should return TRUE if $b$ is divisible by $a$, and FALSE otherwise. I understand that there is no proof that divisibility is not in $L$ since that would imply $P \ne L$ which is open. Also I understand that if there is a nondeterministic algorithm in $O(\log(n)^k)$ then there is a deterministic algorithm in $O(\log(n)^{2k})$, by Savitch's theorem. I believe I've figured out an $L$ algorithm to verify $a*b=c$, and also an $FL$ algorithm to compute $c$ (essentially the method I was taught in grade school, reusing ink when possible), but I haven't been able to adapt it to divisibility. Is such an algorithm for divisibility known? - Isn't the euclidean algorithm log space and with the identity $a \| b$ iff $\gcd(a,b)=a$, wouldn't this give you a logspace algorithm? – JSchlather Oct 25 '11 at 6:08 1 If so, yes, but I don't know how to make the Euclidean algorithm log-space either. The way I would ordinarily implement it is O(n) space. Note that n ~ log(b) here, so log-space means O(log(log(b))) space. – Dan Brumleve Oct 25 '11 at 6:13 1 The euclidean algorithm is most certainly an overkill; if b divides a, it will be discovered in the first step of the algorithm (where we divide a by b and take the remainder)... – Gadi A Oct 25 '11 at 7:53 1 – Dan Brumleve Oct 25 '11 at 8:40 1 Integer division is known to be in L. Page 22 of the slides “My Favorite Ten Complexity Theorems of the Past Decade II” by Lance Fortnow refers to Chiu 1995, although I cannot locate this reference right now. If you can access Chiu, Davida, and Litow 2001, I am sure that it contains the reference to it. – Tsuyoshi Ito Oct 25 '11 at 13:46 show 5 more comments ## 2 Answers (This is an updated version of my comment on the question.) Beame, Cook, and Hoover [BCH86] showed that integer divisibility is in L. More recently, Chiu, Davida, and Litow [CDL01] showed that integer division is also in L. ### References [BCH86] Paul W. Beame, Stephen A. Cook, and H. James Hoover. Log depth circuits for division and related problems. SIAM Journal on Computing, 15(4):994–1003, Nov. 1986. DOI: 10.1137/0215070 [CDL01] Andrew Chiu, George Davida, and Bruce Litow. Division in logspace-uniform NC1. Theoretical Informatics and Applications, 35(3):259–275, May 2001. DOI: 10.1051/ita:2001119. - 1 By the way, according to the introduction of [BCH86], there seems to be a folklore O((log n)^2)-space algorithm. – Tsuyoshi Ito Oct 26 '11 at 23:13 2 – Tsuyoshi Ito Oct 26 '11 at 23:41 1 – Kaveh Nov 2 '11 at 16:59 1 @Kaveh: That is true. I did not mention that result in the answer because I wanted to keep the answer simple by focusing on space complexity by a Turing machine. – Tsuyoshi Ito Nov 10 '11 at 21:40 Edited to add: As Gadi points out in a comment, this answer is wrong. If you have a logspace algorithm to verify $x \times y = z$, then since you're not concerned with running time, you can simply check, for all $c$ with $1 < c \le b$, whether $a \times c = b$. - 2 I believe he meant that the algorithm is logspace in the sum of lengths of x,y,z. What you suggest is not logspace since writing c takes up space on the work tape. – Gadi A Oct 25 '11 at 12:32	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 20, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9184472560882568, "perplexity_flag": "middle"}
http://rjlipton.wordpress.com/2009/11/04/on-mathematical-diseases/	## a personal view of the theory of computation Mathematical diseases: symptoms and examples Underwood Dudley is a number theorist, who is perhaps best known for his popular books on mathematics. The most famous one is A Budget of Trisections, which studies the many failed attempts at the ancient problem of trisecting an angle with only a ruler and a compass. This problem is impossible, yet that has not stopped some people from working day and night looking for a solution. Trying to find such a solution is an obsession for some; it’s almost like they have a malady that forces them to work on the problem. Today I plan on talking about other mathematical obsessions. They are like diseases that affect some, and make them feel they have to work on certain mathematical problems. Perhaps P=NP is one? Dudley’s book is quite funny, in my opinion, although it does border on being a little bit unkind. As the title suggests, in “A Budget of Trisections,” he presents one attempt after another at a general method for trisecting any angle. For most he points out that when the angle is equal to some value what the exact error is. For others he adds a comment like: This construction almost worked, if only the points ${A}$ and ${B}$ and ${C}$ had really been co-linear it would have worked. Perhaps the author could move ${\dots}$ His book is about the kind of mathematical problems that I will discuss today: problems that act almost like a real disease. I cannot resist a quote from Underwood that attacks bloggers. Note he uses “he” to refer to himself in this quote: He has done quite a bit of editing in his time–the College Mathematics Journal for five years, the Pi Mu Epsilon Journal for three, the Dolciani Mathematical Expositions book series (six years), and the New Mathematical Library book series (three years). As a result he has a complete grasp of the distinction between “that” and “which” (very rare) and the conviction that no writing, including this, should appear before the public before passing through the hands, eyes, and brain of an editor. Take that, bloggers! (Bold added by me.) Oh well. What Is a Mathematical Disease? This is the flu season in Atlanta, and many are getting it. I hope you either miss the bug, or if you are unfortunate enough to get it, get a mild case. There is another type of “bug” that affects mathematicians—the attempt to solve certain problems. These problems have been called “diseases,” which is a term coined by the great graph theorist Frank Harary. They include many famous problems from graph theory, some from algebra, some from number theory, some from complexity theory, and so on. The symptoms of the flu are well known—I hope again you stay away from fever, chills, and the aches—but the symptoms for a mathematical disease (MD) are less well established. There are some signs however that a problem is a MD. 1. A problem must be easy to state to be a MD. This is not sufficient, but is required. Thus, the Hodge-Conjecture will never be a disease. I have no clue what it is about. 2. A problem must seem to be accessible, even to an amateur. This is a key requirement. When you first hear the problem your reaction should be: that is open? The problem must seem to be easy. 3. A problem must also have been repeatedly “solved” to be a true MD. A good MD usually has been “proved” many times—often by the same person. If you see a paper in arXiv.org with many “updates” that’s a good sign that the problem is a MD. Unlike real diseases, MD’s have no known cure. Even the solution of the problem will not stop attempts by some to continue working on it. If the proof shows that something is impossible—like the situation with angle trisection—those with the MD will often still work hard on trying to get around the proof. Even when there is a fine proof, those with the disease may continue trying to find a simple proof. For example, Andrew Wiles’ proof of Fermat’s Last Theorem has not stopped some from trying to find Pierre de Fermat’s “the truly marvellous proof.” Some Mathematical Diseases Here are some of the best known MD’s along with a couple of lesser known ones. I would like to hear from you with additional suggestions. As I stated earlier Harary was probably the first to call certain problems MD’s. His original list was restricted to graph problems, however. ${\bullet }$ Graph Isomorphism: This is the classic question of whether or not there is a polynomial time algorithm that can tell if two graphs are isomorphic. The problem seems so easy, but it has resisted all attempts so far. I admit to being mildly infected by this MD: in the 1970′s I worked on GI for special classes of graphs using a method I called the beacon set method. There are some beautiful partial results: for example, the work of László Babai, Yu Grigoryev, and David Mount on the case where the graphs have bounded multiplicity of eigenvalues is one of my favorites. Also the solution by Eugene Luks of the bounded degree case is one of the major milestones. I would like to raise one question that I believe is open: Is there a polynomial time algorithm for the GI problem for expander graphs? I asked several people at the recent Theory Day and no one seem to know the answer. Perhaps you do. ${\bullet }$ Group Isomorphism: This problem is not as well known as the GI problem. The question is given two finite groups of size ${n}$ are they isomorphic? The key is that the groups are presented by their multiplication tables. The best known result is that isomorphism can be done in time ${n^{\log n +O(1)}}$. This result is due to Zeke Zalcstein and myself and independently Bob Tarjan. It is quite a simple observation based on the fact that groups always have generator sets of cardinality at most ${\log n}$. I have been affected with this MD for decades. Like some kind of real diseases I get “bouts” where I think that I have a new idea, and I then work hard on the problem. It seems so easy, but is also like GI—very elusive. I would be personally excited by any improvement over the above bound. Note, the hard case seems to be the problem of deciding isomorphism for ${p}$-groups. If you can make progress on such groups, I believe that the general case might yield. In any event ${p}$-groups seem to be quite hard. ${\bullet }$ Graph Reconstruction: This is a famous problem due to Stanislaw Ulam. The conjecture is that the vertex deleted subgraphs of a graph determine the graph up to isomorphism, provided it has at least ${3}$ vertices. It is one of the best known problems in graph theory, and is one of the original diseases that Harary discussed. I somehow have been immune to this disease—I have never thought about it at all. The problem does seem to be solvable; how can all the subgraphs not determine a graph? My only thought has been that this problem somehow seems to be related to GI. But, I have no idea why I believe that is true. ${\bullet }$ Jacobian Conjecture: This is a famous problem about when a polynomial map has an inverse. Suppose that we consider the map that sends a pair of complex numbers ${(x,y)}$ to ${(p(x,y),q(x,y))}$ where ${p(x,y)}$ and ${q(x,y)}$ are both integer polynomials. The conjecture is that the mapping is 1-1 if and only if the mapping is locally 1-1. The reason it is called the Jacobian Conjecture is that the map is locally 1-1 if and only if the determinant of the matrix $\displaystyle \left( {\begin{array}{cc} p_{x}(x,y) & q_{x}(x,y) \\ p_{y}(x,y) & q_{y}(x,y) \\ \end{array} } \right)$ is a non-zero constant. Note, ${p_{x}(x,y)}$ is the partial derivative of the polynomial with respect to ${x}$. The above matrix is called the Jacobian of the map. This is a perfect example of a MD. I have worked some on it with one of the experts in the area—we proved a small result about the problem. During the time we started to work together, within a few months the full result was claimed twice. One of the claims was by a faculty member of a well known mathematics department. They even went as far to schedule a series of “special” talks to present the great proof. Another expert in the area had looked at their proof and announced that it was “correct.” Eventually, the talks were cancelled, since the proof fell apart. ${\bullet }$ Crypto-Systems: This is the quest to create new public key crypto-systems. While factoring, discrete logarithm, and elliptic curves seem to be fine existing public key systems, there is a constant interest in creating new ones that are based on other assumptions. Some of this work is quite technical, but it seems a bit like an MD to me. There are amateurs and professionals who both seem to always want to create a new system. Many times these systems are broken quite quickly—it is really hard to design a crypto-system. A recent example of this was the work of Sarah Flannery and David Flannery in creating a new system detailed in their book In Code. The book gives the story of her discovery of her system, and its eventual collapse. ${\bullet }$ P=NP: You all know this problem. See this for a nice list of attempts over the years to resolve the problem. Thanks to Gerhard Woeginger for maintaining the list. Open Problems What are other MD’s? What is your favorite? Why do some problems become diseases? While others do not? I would love to see some progress made on group isomorphism—I guess I have a bad case of this disease. I promise that if you solve it I will stop thinking about it. Really. ### Like this: from → History, P=NP, People, Proofs 54 Comments leave one → 1. November 4, 2009 7:40 pm What a great subject! To lead off, a wholly benign, utterly useless, and wonderfully enjoyable MD is the study of self-replicating structures in Conway’s Life. The accomplishments of the Life community over the last 39 years are so amazing, that all one can say is … Golly! 2. November 4, 2009 9:02 pm A nice list Professor! The following 2 problems spring up to my head; they both have a number theoretic flavor though. The first one is the Goldbach conjecture. I mean, I find it really amazing that this problem is accessible to people like me with a modest mathematical problem and is so profound that it has avoided attacks by best brains. The next one is the 3x+1 problem. This “seemingly” toy problem has much to offer as shown by its resistance to attempts at solving it. 3. November 4, 2009 11:59 pm I think my favorite bug is the 4-color conjecture. 633 configurations is still too many! 4. subruk November 5, 2009 1:32 am Although not so easy to state, I think Riemann Hypothesis has attracted quite a lot of attention. See here for a list of attempted proofs. • rjlipton * November 5, 2009 8:02 am Yes. It also has been claimed a few times. One claim made the front page of the New York Times. 5. November 5, 2009 4:32 am As someone who’s been bitten by the reconstruction bug, I’ll tell you that it is indeed related to graph isomorphism on a fundamental level. Here’s a brief sketch of why: If we label the vertices of a graph, then reconstruction is trivial — we just glue the induced subgraphs together so that the labels match up, and in fact we only need three induced subgraphs to reconstruct our original graph. The reconstruction conjecture is that, if we forget the labels, then this “gluing” process is still unique (up to graph isomorphism, of course). This feels like it should be intuitively true, but there’s a crucial problem: namely, if the graph has a large automorphism group, then its induced subgraphs can get “put into” the original graph in many different ways, and it’s not clear that the global properties of the graph don’t change. So much of the work on graph reconstruction centers around understanding just how automorphism groups of graphs behave, and how they relate to combinatorial properties. (From the other direction, by the way, it’s an old result of Bollobas that almost all graphs are reconstructible, since random graphs don’t allow us to embed large subgraphs into them in more than one way.) 6. November 5, 2009 5:30 am Here is my favourite mathematical disease. In a recent talk entitled “A DISCOURSE ON THREE COMBINATORIAL DISEASES” Alexander Rosa has ranked number one the graceful tree conjecture (GTC) as a combinatorial disease [1]. Definitions of the three related conjectures have been listed below: 1. Ringel’s conjecture (RC,1963). The complete graph K2n+1 with 2n + 1 vertices can be decomposed into 2n + 1 subgraphs, each isomorphic to a given tree with n edges. 2. Kotzig’s conjecture (KC,1967). The complete graph K2n+1 can be cyclically decomposed into 2n + 1 subgraphs isomorphic to a given tree with n edges. 3. Graceful tree conjecture (GTC,1967). Every tree has a graceful labelling. Despite of huge efforts very little known for graceful trees e.g., any tree with <34 vertices has a graceful labeling [2] and all trees of diameter up to 5 are graceful [3]. There is even no a "zero-knowledge proof" for the graceful tree conjecture. [1] http://www.math.ilstu.edu/cve/speakers/Rosa-CVE-Talk.pdf [2] http://www.projectgtv.cn/index_en.html (Graceful Tree Verification Project) [3] http://www.combinatorics.org/Surveys/ds6.pdf • rjlipton * November 5, 2009 7:53 am Your right. How can this be open? It seems so simple. Yet it is very hard. Thanks for the details. • November 5, 2009 11:25 am Graceful Tree Conjecture is my favorite too. Read my earlier post here. • November 5, 2009 1:04 pm Let me give short explanation why GTC is a mathematical disease. Rosa has identifi ed essentially three reasons why a graph fails to be graceful: (1) G has “too many vertices and not enough edges,” (2) G “has too many edges,” and (3) G “has the wrong parity.” If for any graceful tree an arbitrary vertex can be assigned the label 0 (rotatable tree) then the proof of the GTC would be piece of cake. Unfortunately not all trees are rotatable. Similarly if all trees are alpha-valuable (a kind of balanced labeling stronger than graceful (beta-valuable) labeling) then the proof of GTC follows easily. What remains is an algorithmic proof attempt based on the induction on the diameter d (the length of the longest path in a tree) of a tree. Unfortunately most of the trees with diameter greater than 4 have no alpha-labeling. In the past I have attempted twice (once in 1975, settled a class of symmetric trees and again 1980, settled all trees of diameter 4). I didn’t give up, it is a disease after all. See a letter from Gerhard Ringel at: http://www.flickr.com/photos/49058045@N00/3947426742/ • November 7, 2009 12:41 am Additional note: If one day the graceful tree conjecture will be settled, there is even stronger one! See a letter from Donald E. Knuth at: http://www.flickr.com/photos/49058045@N00/2094508909/ 7. November 5, 2009 8:47 am There are also some really super simple problems that function like diseases such as the Goldbach Conjecture: http://en.wikipedia.org/wiki/Goldbach%27s_conjecture Everyone has dreamed about solving these at least a little bit at some stage in their life. There’s even a (fictional) book written about this problem about a man obsessed with solving it: 8. Visitor November 5, 2009 10:38 am I think there is a confusion here between two different types of “diseases”. There is the “disease” where an amateur mathematician tries to prove a big conjecture or problem. This indeed entails the problem to be formulated simple enough, so that the amateur can at least convince him/herself that he/she has the solution. I claim that this “mathematical disease” is in many cases a symptom of a real mental problem — it usually appears in eccentric figures, with a tendency to obsessive and compulsive behavior. These people are usually referred to as “cranks”. Most P vs NP papers fall into this category. On the other hand, there are “healthy”, or fruitful, mathematical obsessions, that is, an obsession of a professional mathematician to solve certain important (and even not so important) problem. And in this case, it’s not necessary at all that the problem has a simple formulation . • rjlipton * November 5, 2009 4:26 pm For a long time the irrational/rational nature of zeta(3) was open. It was solved by a non-mathematician, Roger Apéry. I believe that he was not a professional mathematician and was almost dismissed asa “crank”. But he had a proof. • ninguem November 5, 2009 5:21 pm http://en.wikipedia.org/wiki/Roger_Apéry He was a professor at a small French university. He was past the age most mathematicians prove big theorems, he had a history of bad proofs, not a big research output and, I was told, also an alcoholic. But he was a professional mathematician and not a crank. We still don’t know whether zeta(3) is transcendental. • rjlipton * November 6, 2009 8:10 am Never said was a crank. I did understand that he was in another area. Sorry if I got it wrong. • Anonymous November 5, 2009 11:16 pm Actually, Apery was a math professor at Caen with a fairly prestigious background (he was a prize-winning student at ENS) and a number of research publications; although he wasn’t a terribly high-profile researcher, he was by no means an outsider to the mathematical research community. The reason people were initially skeptical was that he gave a very weird talk presenting the proof. In the talk, he stated some implausible-looking identities and recurrences, with no hint of how to prove them, and he showed that they implied the irrationality of zeta(3). He apparently didn’t respond well to questions, and this led people to wonder whether he actually had a proof of the identities. Maybe he had just conjectured them based on numerical experiments, and perhaps they weren’t even true. He eventually came out with a paper that proved everything, and he probably had the proofs at the time he gave the talk, but he certainly didn’t explain it at all clearly at that time. I think it wasn’t until Don Zagier came up with his own proofs of Apery’s assertions that everyone became convinced it definitely worked (although everyone got very excited even before that, once they checked everything numerically and found that it all seemed to work). 9. Ted Carroll November 5, 2009 10:55 am The Palendromic Number Conjecture. I suck non-mathematicians in with that one all the time — especially computer people because there’s a proof that the conjecture holds for binary. 10. November 5, 2009 11:43 am Conway’s Thrackle conjecture bites me about once every two years… 11. ryanw November 5, 2009 6:34 pm Allow me to get a little bit infected… If we assume the graph reconstruction conjecture, does this imply anything interesting about the complexity of graph isomorphism? If you want to tell whether two n-node graphs are isomorphic, and you know that this “reduces” to checking whether there is an “isomorphism matching” between two sets of n graphs on n-1 nodes each, can you get a recursive algorithm? 12. November 5, 2009 8:34 pm As one of the many who search for new cryptosystems, allow me to defend the affliction (before someone develops a vaccine)! Sure, factoring- and discrete-log-based systems are great for Alice and Bob’s everyday secret messages, but… What if you don’t believe that these problems are truly hard? What if someone builds a quantum computer? (What if it happens sooner rather than later?) What if your device is constrained and can’t handle 2048-bit exponentiations? What if you need ‘extra features,’ like delegation, revocation, or homomorphisms? What if you want to be guaranteed security even if some/most of your secret key leaks out via a side channel? Your standard-issue cryptosystems don’t admit very satisfactory answers to these questions. That’s why we need to look for more… 13. November 6, 2009 5:13 am Let me adopt the MD term under a slight protest and mention (in two installements, alas without serious editing,) some of my favourites MDs that I spent most time studying. 1) The rate of error-correcting binary codes (and spherical codes). Asymptotically.(Infected by Nati Linial) A very easy to describe problem. You want an error correcting binary code on n bits with minimal distance delta n. What is the largest possible rate as a function of delta.(A closely related problem is about the densest sphere packing in high dimensional spaces.) Is the Gilbert-Varshamov lower bound the correct one? Can the MRRW upper bound (the best known one) be (even slightly) improved? A (somewhat) related (easier) problem: Can you find an elementary construction (not based on algebraic geometry) for large-alphabets codes with better rate than Gilbert-Varshamov. (The zig-zag success for expanders give some little hope.) 2) The g-conjecture for spheres; This is probably the first problem on my list. I started working on it the earliest and probably spent more time on it than on any other. It is a little hard to explain and motivate. But there are quite a few people who thought about the problem. (There are a few posts about it on my blog). 3) The Hirsch conjecture (and strongly polynomial LP) I wrote about it amply on my blog. 4) The Erdos-Rado Delta system-conjecture This is on Gowers’s possible future polymath projects so let me not elaborate further here. 5) The Cap set conjecture It is about the largest size of a subset A of (Z_3)^n not having three elements that sum to zero.Closely related to Roth’s and Szemeredi’s theorems. 14. November 6, 2009 5:21 am 6) Borsuk’s conjecture. I dont think Jeff Kahn and I were “obsessed” about the problem while working on it for quite a few years. It is not clear if an obsession mode is a good sign. Our approach to the problem is somewhat related to a famous open problem which is still open and is on Alexander Rosa’s list and was always high on Jeff’s list: The Erdos-Faber-Lovasz conjecture. (Jeff settled the EFL cnjecture asymprotically and Jeff and Paul Seymour solved it fractionally.) 7) Bible codes This represents an applied topic that I intensively (and obsessively) spent much time in the late 90s. At the end I was a coauthor of a 4-authors paper containing a thorough refutation of the scientific evidence for the existence of bible codes. It was a good (while strange) introduction for me on various issues regarding statistics, science, scepticism, even philosophy of science. 8) Learnability vs rationality One tempting “cure” for various deseases, especially of conceptual nature, is “learnability” via VC-dimension. I was very optimistic at some time about the usefulness of replacing “rational” by “learnable” in the foundations of theoretical economics. 9) Infeasibility of quantum computers This represents a current main research interest. Let me mention some flirting with famous problems 10) 4CT Once I had some idea about 4CT (which asserts that every planar cubic graph is 3-edge colorable) and relating it to Tverberg’s theorem, and I remember Laci Lovasz asked me: “Can you use your approach to prove that a bipartite cubic graph is 3-edge colorable?” (Which is an easy graph theory result.) Dealing with bipartite cubic or even with bipartite planar cubic graphs looks like a good test-case for various hypothetical approaches. 11) NP ne P and related issues It is probably a good instinct whenever you study some new notion about Boolean functions (or simplicial complexes which are just monotone Boolean functions) to spend a little (let me repeat: a little) time on thinking: does this new notion has baring on NP=!P or other questions in computational complexity? Most often you can easily realize that the answer is no, and sometimes you realize that the answer is no after some more effort. 12) A little flirt with Poincare I was interested in triangulation of manifolds for which the links of vertices are of the simplest possible kind: stacked spheres. (They are the boundaries of a set of simplices glued together along facets.) For dimension greater than 3 I proved that such a simply connected manifold is a sphere. For dimension 3 I could not prove it and it is a very very very special case of the Poincare conjecture. (I still cannot prove this special case directly.) If you drop the assumption that the manifolds are simply connected then for d>3 you are left with very simple handle body manifolds. I do not know (and am curious to know) which 3-manifolds have a triangulation where are links are stacked spheres. • November 6, 2009 12:15 pm Gil Kalai, would you mind commenting further (either here or on your own blog) about “infeasibility of quantum computers?” I have finally started learning some basic quantum techniques, because I think they might be useful to prove non-quantum statements that interest me, but I have serious reservations about the physical realization of quantum computers. Are you saying a research direction exists that attempts to prove their physical impossibility, for example through a better understanding of decoherence? • November 7, 2009 11:05 am Hmm, probably I should have written: ” 9) feasibility of quantum comuters.” This is the area of understanding fault tolerant quantum computation, e.g. how to handle decoherence. Since the concept of quantum computers is quite revolutionary it is natural to have some reservations like you have. Nevertheless, there are good responses to various concerns that were raised over the years, and it is fair to say that, at this stage, optimism is not only instrumental but also fairly justified. As you correctly understood my interests are in the negative direction. Learning quantum information/computation is certainly a very good thing to do. And the idea that quantum techniques will have applications in other areas (similar to applications of probability) is a nice idea that was also discussed in this blog. (So far, there are handful of examples). In any case, even in the likely event that quantum tricks will not help the non quantum statements that interest you, you may certainly enjoy learning the quantum tricks. (Here is the post on the “quantum method”. http://rjlipton.wordpress.com/2009/03/28/erds-and-the-quantum-method/) • December 22, 2009 1:42 am I forgot one (that I took from Itai Benjamini, Yuval Peres and Oded Schramm). It is the “Dying prtcolation conjecture” which asserts that for three dimensional percolation at the critical probability p_c, the probability for an infinite open cluster is zero. 15. proaonuiq November 6, 2009 12:12 pm In the Harary paper they cite the Rudrata disease, an old and widespread one. I like specially the RCD-variant (Rudrata problem for Cayley Digraphs), also old and widespread. Be aware: the later infection is harder to cure than the more general Rudrata disease, since the problem seems simpler ! Afaik the complexity of this RCD-disease is still unkown even for very restricted cases and imo this problem is a good candidate for the class of intermediate problems (NPI ? PI ?). I ignore if some cases of known polynomial sequential complexity (i.e. 2-circulants: Woeginger is co-author of a paper about this)can be parallelized efficiently. The point is that to my knowledge the complexity of the GI problem for CD is also unknown even for general circulants. On the other hand it is known that CD are tools for describing groups succintly so i wonder how the Group isomorphism problem could be related to GI via CD isomorphism. 16. November 6, 2009 1:51 pm The Splay Conjecture and the Sunflower Lemma are major ones. 17. November 8, 2009 8:47 pm People have been seeking (and are still seeking) the Philosopher’s Stone for hundreds of years. Physics even gives hope that transformations of base metals into gold can be economically achieved. While not impossible, a cost-effective solution is likely improbable and success would undermine the cost of gold so much, that markets would make any gains from the innovation very short-lived. People still have MD over Fermat’s Last Theorem. Proved or no, they are still seeking Fermat’s “elegant” proof. Einstein went to his grave denying the Heisenberg Uncertainty Principle and people (even physicists) still confuse it with the Observer Effect. Arrow’s Impossibility Theorem hasn’t halted popular faith in the institution of democracy, nor should it. In the absence of a first-best solution, second-best solutions are not only tolerable but desirable. Time travel is axiomatically impossible, yet even Stephen Hawking has retreated in the face of popular zeal for this science fiction myth and justified it with pseudo-scientific conjecture. Gamblers and mathematicians seek fool-proof winning systems consisting of linear combinations of wagers with negative expected value and still think they can succeed. When climate scientists regard their theories as “settled”, it flies in the face of so many previous notions which were considered “settled” and subsequently disproven. “Settled science” is an oxymoron. There is a converse to futile efforts toward discovery. We know, for example, that chess has either a winning strategy for White or a non-losing strategy for Black. The strategy is likely unimplementable by a human being. Yet the search for this difficult strategy continues and, if found, would hardly kill the game as an enjoyable past-time. After all, we still teach our children Tic-Tac-Toe. The ultimate question which is both unprovable and undeniable is the existence of God. Mankind’s finite intellect and limited sensory capabilities make either pursuit entirely futile. An extradimensional entity is beyond anyone’s ability to comprehend or observe. Continued discussion, though, is fruitful but continued argument is a waste of time. Scientists never could reconcile matters of faith with the shortcomings of their profession. Indeed, they refuse to admit shortcomings of their profession. The bottom line is that stubborn people, (the most stubborn of which are scientists) don’t like being told anything is impossible. There is value in people trying to attain that which is merely believed to be impossible (like heavier than air flight). Attempting to prove the known impossible is folly, but some useful insights might be gained in the attempt. Many an innovation was discovered by accident while searching for something else. “There are more things in heaven and earth…than are dreamt of in your philosophy.” 18. JamesD November 10, 2009 3:12 am “While factoring, discrete logarithm, and elliptic curves seem to be fine existing public key systems,…” I can’t see how this is true. They are only fine if you are satisfied with very weak security guarantees. You have security until somebody builds a quantum computer. Maybe that won’t be for a while. However, there has been a lot of momentum building on several experimental implementations. (Also, at least as much work is now being done in secret as in public, so don’t count on someone telling you when your security blanket has been pulled away.) Therefore, given how heavily reliant we currently are on public-key cryptography, finding new schemes has to be a pretty urgent problem. I can’t see how it can be called a “mathematical disease.” I can still see the other problems on your list afflicting computer scientists and mathematicians twenty years from now. If we don’t have better public-key cryptosystems twenty years from now, then we will be in trouble. (On the other hand, maybe it is possible to run the modern Internet and other networks without public-key cryptosystems? Instead, perhaps some infrastructure could be deployed. Given the ubiquity of public-key crypto, I think this would be a challenge, but I don’t know enough about it.) 19. JamesD November 10, 2009 3:21 am I wrote, “I can still see the other problems on your list afflicting computer scientists and mathematicians twenty years from now.” But what will happen to the P versus NP question when we have quantum computers? Will this question still be seen as important, or will the research just move on to BQP versus QMA? Honest question: How important is P versus NP in a quantum world? 20. DC November 10, 2009 7:46 pm Just a note, Underwood Dudley’s “Budget of Trisections” has been reprinted by the MAA as “The Trisectors”, so if difficulty is encountered finding an old used edition you can just go over to the MAA (or wherever you buy books) and get it new. If memory serves— I no longer have the older book, published by Springer— the new “Trisectors” includes expanded commentary about trisections, in the same vein as Dudley’s other books on cranks and numerology, before the material that is identical to what appeared in the original “Budget of Trisections.” It is a great book. (I am not Underwood Dudley.) • rjlipton * November 10, 2009 9:51 pm Even if you are him, it is still a great book. Thanks for pointer to new name. 21. proaonuiq November 12, 2009 12:21 pm Just to complete my previous comment on Rudrata disease on this thread: as well as Rudrata problems on CD are good candidates for intermediate problems in P and NP, the undirected version, wich is a restricted version of a well known loved conjecture (wich will probably be solved positivelly, hopefully soon) is a good candidate for a problem similar to factoring. 22. Charles Wells November 12, 2009 4:52 pm Didn’t the Collatz conjecture cause a lot of overloading on many many mainframes when computer scientists first became aware of it in the 1970′s? Surely that counts as a disease. 23. November 16, 2009 12:46 am The Frey-Mazur conjecture is a personal MD (see conjecture 3 of this paper: http://www.springerlink.com/content/k251nu0703j75441/fulltext.pdf). 24. November 16, 2009 4:19 am Although it’s more recreational in flavour, I think the Collatz conjecture (aka the Hailstone conjecture) fits the first two criteria of an MD. I’m not sure how often it has been claimed that it has been solved but, as Charles notes, it would have certainly chewed up a lot of processing power creating visualisations of the hailstone function. My favourite “surely that’s not open” problem is the Frankl conjecture (aka the Union-closed sets conjecture): “for any finite union-closed family of finite sets, other than the family consisting only of the empty set, there exists an element that belongs to at least half of the sets in the family.” I’ve mentioned this to several people who have all said something along the lines of “that should be easy”, wrestled with it for several hours and given up, exasperated. • rjlipton * November 16, 2009 9:01 am Great example. What is the best known on this conjecture? Do you know? • November 16, 2009 6:13 pm I’ve only really engaged with the Union-closed set conjecture recreationally. That is to say, I haven’t looked up any “serious” work that has been done on it. The Wikipedia entry notes that the conjecture has been proven for several special cases. • retiritter December 9, 2009 5:43 pm When talking about the best known result(s), we should use the more abstract lattice theoretic formulation of the conjecture: Every nontrivial finite lattice L has a join-irreducible element that is at or below at most half the elements of L. The conjecture is known to be true for lattices having some additional structures, e.g. for modular lattices. Furthermore there are various results for lattices fulfilling some bound-condition, e.g. for lattices having a limited number of elements or having a limited number of join-irreducible elements. Links can be found at Wikipedia and elsewhere. None of these results gives any clue, how the prove the conjecture in general. The “best” known result from my understanding is a tautology: For every integer n >= 2 the conjecture is true for more than half of the isomorphism classes of lattices having n elements. It is tautologic, since for any lattice not fulfilling the conjecture the conjecture is obviously fulfilled by the dual lattice – having the reverse partial ordering -, and since for any n there are lattices isomorphic with their dual lattice. 25. Jurjen December 29, 2009 9:51 am There’s one mathematical disease that I am missing here: (video) compression! Just Google “Jan Sloot” to see what I mean. 26. Jerome JEAN-CHARLES November 2, 2010 10:02 pm Here is another MD : The Rota Basis conjecture : in a vector space of dimension at n. Given n basis : write one per line. Can you reorder each line ( in itself) so that in each of the n column you have a basis too? Case n = 3 is ok but messy . n = 4 too messy for me. There is also some matroid formulation with some partial reduction to equivalent conjecture ( Timothy Chow ..) • Josh May 23, 2012 12:46 am Have you seen a direct proof of n=4 or 5? Do you have a reference? • Jerome May 29, 2012 5:07 am No I do not have seen proof, it seems doable by computer using brute force for n=4 ? ### Trackbacks %d bloggers like this:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 23, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9541533589363098, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2008/05/28/matrices-iv/?like=1&_wpnonce=68d26d17fa	# The Unapologetic Mathematician ## Matrices IV Like we saw with the tensor product of vector spaces, the dual space construction turns out to be a functor. In fact, it’s a contravariant functor. That is, if we have a linear transformation $T:U\rightarrow V$ we get a linear transformation $T^:V^\rightarrow U^$. As usual, we ask what this looks like for matrices. First, how do we define the dual transformation? It turns out this is the contravariant functor represented by $\mathbb{F}$. That is, if $\mu:V\rightarrow\mathbb{F}$ is a linear functional, we define $T^(\mu)=\mu\circ T:U\rightarrow\mathbb{F}$. In terms of the action on vectors, $\left[T^(\mu)\right](v)=\mu(T(v))$ Now let’s assume that $U$ and $V$ are finite-dimensional, and pick bases $\left\{e_i\right\}$ and $\left\{f_k\right\}$ for $U$ and $V$, respectively. Then the linear transformation $T$ has matrix coefficients $t_i^k$. We also get the dual bases $\left\{\epsilon^j\right\}$ of $U^$ and $\left\{\phi^l\right\}$ of $V^$. Given a basic linear functional $\phi^l$ on $V$, we want to write $T^(\phi^l)$ in terms of the $\epsilon^j$. So let’s evaluate it on a generic basis vector $e_i$ and see what we get. The formula above shows us that $\left[T^(\phi^l)\right](e_i)=\phi^l(T(e_i))=\phi^l(t_i^kf_k)=t_i^k\delta_k^l=t_i^l$ In other words, we can write $T^(\phi^l)=t_j^l\epsilon^j$. The same matrix works, but we use its indices differently. In general, given a linear functional $\mu$ with coefficients $\mu_l$ we find the coefficients of $T^(\mu)$ as $t_j^l\mu_l$. The value $\left[T^(\mu)\right](v)=\mu(T(u))$ becomes $\mu_lt_i^lu^i$. Notice that the summation convention tells us this must be a scalar (as we expect) because there are no unpaired indices. Also notice that because we can use the same matrix for two different transformations we seem to have an ambiguity: is the lower index running over a basis for $U$ or one for $U^*$? Luckily, since every basis gives rise to a dual basis, we don’t need to care. Both spaces have the same dimension anyhow. ### Like this: Posted by John Armstrong \| Algebra, Linear Algebra ## 2 Comments » 1. [...] the linear functional on into one on . But this is just the dual transformation ! Then we can evaluate this on the column vector to get the same result: [...] Pingback by \| May 30, 2008 \| Reply 2. [...] the way we defined the dual transformation was such that we can instead apply the dual to the linear functional , and [...] Pingback by \| May 22, 2009 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 33, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9022909998893738, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/108180/in-class-today-we-had-to-find-a-closed-generating-function-for-a-n-2n1-the	# In class today, we had to find a closed generating function for $A_n=2n+1$. The sequence of the odd natural numbers. Anyone with an idea? In class today, we had to find a closed generating function for $A_n=2n+1$, where $n\in \mathbb{N}$ and $A_n$ is the sequence of odd natural numbers. Anyone with an idea? I tried several thing but could not get it. - 8 What is the derivative of a geometric series? – deinst Feb 11 '12 at 17:19 ## 3 Answers Consider $\displaystyle f(x)= x+x^3+x^5+x^7+\cdots= \frac x{1-x^2}$ Then $f'(x)=1+3x^2+5x^4+\cdots$ and : $$\left(\frac {x}{1-x^2}\right)'=\frac{2x^2}{(x^2-1)^2}-\frac{1}{x^2-1}=1+3x^2+5x^4+\cdots$$ and your generating function is : $$1+3t^1+5t^2+\cdots= \frac{2t}{(t-1)^2}-\frac{1}{t-1}=\frac{t+1}{(t-1)^2}$$ - An alternative approach is to write down a recurrence for the sequence and produce the generating function from the recurrence. Clearly the recurrence here is $A_n=A_{n-1}+2$, with initial condition $A_0=1$. If we assume that $A_n=0$ for all integers $n<0$, we can use the Iverson bracket to do away with the initial condition and write the recurrence simply as $$A_n=A_{n-1}+2-[n=0]\;.$$ Now multiply both sides by $x^n$ and sum over $n$ to get the generating function $f$: $$\begin{align}f(x)=\sum_nA_nx^n&=\sum_n\left(A_{n-1}x^n+2x^n-[n=0]x^n\right)\\ &=\sum_nA_{n-1}x^n+2\sum_nx^n-\sum_n[n=0]x^n\\ &=x\sum_nA_{n-1}x^{n-1}+\frac2{1-x}-1\\ &=x\sum_nA_nx^n+\frac2{1-x}-1\\ &=xf(x)+\frac{1+x}{1-x}\;, \end{align}$$ so $$(1-x)f(x)=\frac{1+x}{1-x}\;,$$ and $$f(x)=\frac{1+x}{(1-x)^2}\;.$$ - Great explanation - a general way to proceed instead of a single insight about the properties of a particular infinite series! – Rex Kerr Feb 11 '12 at 22:36 $$\sum_{n=0}^\infty (2n+1)x^n=\sum_{n=0}^\infty\big( 2(n+1)-1\big) x^n =2\left(\sum_{n=0}^\infty \frac{d}{dx}x^{n+1}\right)-\left(\sum_{n=0}^\infty x^n\right)$$ $$=2\frac{d}{dx}\left(\frac{1}{1-x}-1\right)-\frac{1}{1-x}=\frac{2}{(1-x)^2}-\frac{1}{1-x}=\frac{1+x}{(1-x)^2}.$$ -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 8, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9198065400123596, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/47951/calculate-partial-autocorrelation	# calculate partial autocorrelation [duplicate] Possible Duplicate: PACF manual calculation I am trying to find a formula for how to calculate partial autocorrelation between variables. We know that aucorrealtion between variables at different lags are given by: $$\hat\rho_h=\frac{\sum^T_{t=h+1}(y_t-\bar y)(y_{t-h}-\bar y)}{\sum^T_{t=1}(y_t-\bar y)^2}$$ I know also that partial autocorrelation is the autocorrelation between `y[t]` and `y[t–h]` after removing any linear dependence on `y[1], y[2], ..., y[t–h+1]`. But how do you remove any linear dependence on `y[1], y[2], ..., y[t–h+1]`? Does there exist some formula for this? - ## marked as duplicate by whuber♦Jan 17 at 16:05 This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question. ## 1 Answer Googling for "partial autocorrelation" will lead you to Wikipedia: The partial correlation between $X$ and $Y$ given a set of $n$ controlling variables $Z = \{Z_1, Z_2, \dots, Z_n\}$, written $\rho_{XY\cdot Z}$, is the correlation between the residuals $R_X$ and $R_Y$ resulting from the linear regression of $X$ with $Z$ and of $Y$ with $Z$, respectively. - but there is written method very unclear,i can't undertsand how to use it for practically,let's take some example of time series,otherwise i can't guess it – user466441 Jan 17 at 15:18 any help?please – user466441 Jan 17 at 15:25 Are you unclear on the concept of residuals from a regression? If so, I suggest you read the first few chapters from any textbook on regression. This will help you much more than any cookbook recipe we could possibly give you here. – Stephan Kolassa Jan 17 at 15:28 no i am not unclear,just i need to know practically how to remove linear dependence between variables – user466441 Jan 17 at 15:34 generally more help is when you give them direct hint,not something different unclear method,let us suppose that i have data from 1981 year to 1995 like this.23 45 12 10 9 8 6 4 5 33 7 66 54 89 76 how to calculate partial auto correlation between time 1982 and 1989? – user466441 Jan 17 at 15:44 show 5 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9241197109222412, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/152104-iisomorphic.html	# Thread: 1. ## iisomorphic I have an example of 2sylow subgroup of S6. I want to prove that this subgroup isomorphous to Z2XD4 2. Originally Posted by rtrt1 I have an example of 2sylow subgroup of S6. I want to prove that this subgroup isomorphous to Z2XD4 The Sylow 2-sbgps. of $S_4$ are easily seen to be isomorphic with $D_4$ ; if now you observe that in $S_6$ you have two numbers more as part of the permutations that don't appear in $S_4$ then you're easily done. Tonio	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9516043663024902, "perplexity_flag": "middle"}
http://www.nag.com/numeric/CL/nagdoc_cl23/html/S/s21bhc.html	NAG Library Function Documentnag_elliptic_integral_complete_K (s21bhc) 1 Purpose nag_elliptic_integral_complete_K (s21bhc) returns a value of the classical (Legendre) form of the complete elliptic integral of the first kind. 2 Specification #include <nag.h> #include <nags.h> double nag_elliptic_integral_complete_K (double dm, NagError fail) 3 Description nag_elliptic_integral_complete_K (s21bhc) calculates an approximation to the integral $Km = ∫0 π2 1-m sin2⁡θ -12 dθ ,$ where $m<1$. The integral is computed using the symmetrised elliptic integrals of Carlson (Carlson (1979) and Carlson (1988)). The relevant identity is $Km = RF 0,1-m,1 ,$ where ${R}_{F}$ is the Carlson symmetrised incomplete elliptic integral of the first kind (see nag_elliptic_integral_rf (s21bbc)). 4 References Abramowitz M and Stegun I A (1972) Handbook of Mathematical Functions (3rd Edition) Dover Publications Carlson B C (1979) Computing elliptic integrals by duplication Numerische Mathematik 33 1–16 Carlson B C (1988) A table of elliptic integrals of the third kind Math. Comput. 51 267–280 5 Arguments 1: dm – doubleInput On entry: the argument $m$ of the function. Constraint: ${\mathbf{dm}}<1.0$. 2: fail – NagError Input/Output The NAG error argument (see Section 3.6 in the Essential Introduction). 6 Error Indicators and Warnings NE_INTERNAL_ERROR An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance. NE_REAL On entry, ${\mathbf{dm}}=〈\mathit{\text{value}}〉$; the integral is undefined. Constraint: ${\mathbf{dm}}<1.0$. On failure, the function returns zero. NW_INTEGRAL_INFINITE On entry, ${\mathbf{dm}}=1.0$; the integral is infinite. On failure, the function returns the largest machine number (see nag_real_largest_number (X02ALC)). 7 Accuracy In principle nag_elliptic_integral_complete_K (s21bhc) is capable of producing full machine precision. However round-off errors in internal arithmetic will result in slight loss of accuracy. This loss should never be excessive as the algorithm does not involve any significant amplification of round-off error. It is reasonable to assume that the result is accurate to within a small multiple of the machine precision. 8 Further Comments You should consult the s Chapter Introduction, which shows the relationship between this function and the Carlson definitions of the elliptic integrals. In particular, the relationship between the argument-constraints for both forms becomes clear. For more information on the algorithm used to compute ${R}_{F}$, see the function document for nag_elliptic_integral_rf (s21bbc). 9 Example This example simply generates a small set of nonextreme arguments that are used with the function to produce the table of results. 9.1 Program Text Program Text (s21bhce.c) None. 9.3 Program Results Program Results (s21bhce.r)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 10, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.6676052808761597, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/72684/quick-algorithm-for-finding-real-solutions-for-a-system-polynomial-equations	## Quick algorithm for finding real solutions for a system polynomial equations ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Would you please recommend a computer program that could give quick answer Yes/No for the question: does there exist a real solution of a given system of polynomial equations with integer coefficients. I will need it to solve a huge list of such systems, each of them is over $\mathbb{R}^6$, has $6$ equations of degree $4$ or less. Coefficients are also quite small. Maple's Triangularize procedure for most of the cases works too long, so applying it for big list is almost impossible. - 1 How dense are your polynomials? – Igor Rivin Aug 11 2011 at 15:29 Also, what do the monomials look like (that is, do variables actually appear with degree 4, or is the degree of any given variable bounded by 1 or 2?) – Igor Rivin Aug 11 2011 at 15:38 Take a look at the implementation of Tarski-Seidenberg Theorem explained at xorshammer.com/2009/05/14/… I do not know whether this particular implementation is efficient, but it seems worth trying (easy to install, ...). – boumol Aug 11 2011 at 16:43 1 I asked a similar question at mathoverflow.net/questions/1493/…, have a look at the answers. QEPCAD somewhat faster than Maple/Mathematica, but less pleasant to use. If you problem allows, try solving over complex numbers, and then restrict. – Boris Bukh Aug 11 2011 at 18:32 If the OP's problem is "generic", then, since the number of equations equals the number of unknowns, the solution is a zero-dimensional variety, the number of complex points of which is bounded by the product of the degrees, or 4096, so the problem should not be so bad by sub-resultants. – Igor Rivin Aug 11 2011 at 21:47 show 1 more comment ## 1 Answer Have you thought obtain a Groebner basis of the set of defining polynomials of your system? Maybe, it could simplify considerably the aspect of the system, and you would need a little time to calculate the real solutions of the system. For that, you could use the free software wxMaxima. If you don't have it, you can download in http://maxima.sourceforge.net/download.html. I would like post this comment just as a comment and not as an answer, but I don't know how to do it. - Mathematica and Maple also can find Groebner basis quite effectively. In Mathematica there is also `FindInstance[]`, which finds an instance of solution or gives an empty list if it can't find any. I don't know if it work always. May be a combination of both could help. – Andrew Aug 11 2011 at 18:03 I need the process to be very automatical, as there is a huge list I of systems I want to check. Dealing with each of the systems is not a hard problem, but it's a "handmade" work, I won't be able to check all the systems by hand. Most of software can compute Groebner basis, but is there a fast algorithm which can say Yes/No to real roots problem, having a Groebner basis of the ideal? – Al Tal Aug 12 2011 at 19:52	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9385828375816345, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/102090-first-calculus-assignment-year-print.html	# first calculus assignment of the year Printable View • September 13th 2009, 12:47 PM OCD first calculus assignment of the year Hi, first post here. Also first calculus assignment; it's called preparation for calculus so I feel like it's supposed to be easy, and I feel a bit of shame that I'm asking for help, but here goes. 1. Find the intercepts (if any). $y=(x-1)/(x-2)$ 2. Check for symmetry with respect to both axes and to the origin. Show work. $x^2y-x^2+4y=0$ I hope this is in the right place (it may really be more of pre-calculus work) and thanks! • September 13th 2009, 01:15 PM VonNemo19 Quote: Originally Posted by OCD Hi, first post here. Also first calculus assignment; it's called preparation for calculus so I feel like it's supposed to be easy, and I feel a bit of shame that I'm asking for help, but here goes. 1. Find the intercepts (if any). $y=(x-1)/(x-2)$ 2. Check for symmetry with respect to both axes and to the origin. Show work. $x^2y-x^2+4y=0$ I hope this is in the right place (it may really be more of pre-calculus work) and thanks! 1. when y is zero, what is x? Conversly, when x is zero, what is y? 2. Begin by solving for y. a)if your insert any -x and x which both yield a positive y, then there is symetry about y. b)If x gives y, and -x gives -y, thenthere is symetry about the origin. c) assuming that y is not a function, then if x gives two values of y (+/-y), then there is symetry about the x axis. • September 13th 2009, 01:30 PM OCD thanks! I think I've got 1 now. The problem with # two is splitting up the x and the y at the beginning. That's where I keep getting stuck because I subtract $-x^2$ from the left side and then try to divide the other $x^2$ to separate it from the y, but I know this isn't right. All times are GMT -8. The time now is 08:12 AM.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9404033422470093, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-statistics/122767-correlation-coefficient-help.html	# Thread: 1. ## correlation coefficient help Hey, im a first year in leeds uni doing maths and there are some things im starting to get quiet stressed over and they arnt even hard so you'll probably see quite a few of my posts pop up with my exams coming in a few days. The question im stuck on is: Let X1 , X2 and X3 be independent random variables with zero mean and variance 1. Find the correlation coefficient between Y1 = 2(X1) + (X2) and Y2 = (X2) ¡ 2(X3) . I really dont know where to start and its quite conserning because i dont think its even hard . I beleive that cov(x,y) = E(XY) -E(X)E(Y) but with a zero mean of the variables im unsure how this works. Thanks for all help. 2. Originally Posted by dhowlett Hey, im a first year in leeds uni doing maths and there are some things im starting to get quiet stressed over and they arnt even hard so you'll probably see quite a few of my posts pop up with my exams coming in a few days. The question im stuck on is: Let X1 , X2 and X3 be independent random variables with zero mean and variance 1. Find the correlation coefficient between Y1 = 2(X1) + (X2) and Y2 = (X2) ¡ 2(X3) . I really dont know where to start and its quite conserning because i dont think its even hard . I beleive that cov(x,y) = E(XY) -E(X)E(Y) but with a zero mean of the variables im unsure how this works. Thanks for all help. You want to find the population correlation coefficient $\rho_{Y_1,Y_2}$ In you book, $\rho{X,Y}=\frac{cov(X,Y)}{\sigma_X \sigma_Y}=\frac{E[(X-\mu_X)(Y-\mu_Y)]}{\sigma_X \sigma_Y}$ You can simply substitute $Y_1$ into $X,$ and $Y_2$ into $Y$. Since $\sigma_{Y_1}=1, \sigma_{Y_2}=1, \mu_{Y1}=0, \mu_{Y2}=0$, The result becomes $\rho_{Y_1,Y_2}= cov(Y_1,Y_2)=E[Y_1Y_2]$ 3. Thankyou. Though i am confused. The question says the random variables are independent, doesn't that mean the correlation coefficient is 0? the answer to the question is 0.2. E(xy) = E(x)E(y) doesn't it when they are independent? This gives E(y1) = 2E(x1) + E(x2) and E(y2) = E(x2) - 2E(x3) but the question states the mean of all these random variables is 0 thus that gives 0 for both of them. But the answer is 0.2 :S. 4. Originally Posted by dhowlett Thankyou. Though i am confused. The question says the random variables are independent, doesn't that mean the correlation coefficient is 0? the answer to the question is 0.2. E(xy) = E(x)E(y) doesn't it when they are independent? This gives E(y1) = 2E(x1) + E(x2) and E(y2) = E(x2) - 2E(x3) but the question states the mean of all these random variables is 0 thus that gives 0 for both of them. But the answer is 0.2 :S. Are you looking for the linear correlation or the population correlation? 5. i dont know unfortunetly, i copyed out the whole question. Im really getting lost on these "introduction to probability" exams. 6. Originally Posted by dhowlett i dont know unfortunetly, i copyed out the whole question. Im really getting lost on these "introduction to probability" exams. I think you are looking for the population correlation coefficient, $\rho_{Y_1,Y_2}$ When $Y_1$ and $Y_2$ are dependent random variables, after algebraic reduction, it became $\rho_{Y_1,Y_2}=Cov(Y_1,Y_2)=E[XY].$ In the case where $Y_1$ and $Y_2$ are independent, $E[XY]$ becomes $E[X]E[Y].$ The linear correlation coefficient is $r=\frac{\Sigma Y_1Y_2}{\sqrt{(\Sigma Y_1)^2(\Sigma Y_2)^2}}$. If this does not look familiar, the population correlation coefficient must be what you are looking for. 7. Originally Posted by dhowlett Thankyou. Though i am confused. The question says the random variables are independent, doesn't that mean the correlation coefficient is 0? the answer to the question is 0.2. If the two random variables are independent, the coveriance =0. It follows that the population correlation coefficient must equal to zero. It cannot be anything but zero; however, it's not true for the linear regression coefficient of correlation. For independent variable, the linear correlation need not be zero. 0.2 can be true. I have a hunch that you are to find the linear correlation coefficient.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 17, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9312329292297363, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/17797/list	## Return to Answer 2 Slightly expanded. I agree that the correspondence between representations of the fundamental group(oid) and locally constant sheaves is not very well documented in the basic literature. Whenever it comes up with my students, I end up having to sketch it out on the blackboard. However, my recollection is that Spanier's Algebraic Topology gives the correspondence as a set of exercises with hints. In any case, one direction is easy to describe as follows. Suppose that $X$ is a good connected space X (e.g. a manifold). Let $\tilde X\to X$ denote its universal cover. Given a representation of its fundamental $\rho:\pi_1(X)\to GL(V)$, one can form the sheaf of sections of the bundle $(\tilde X\times V)/\pi_1(X)\to X$. More explicitly, the sections of the sheaf over U can be identified with the continuous functions $f:\tilde U\to V$ satisfying $$f(\gamma x) = \rho(\gamma) f(x)$$ for $\gamma\in \pi_1(X)$. This sheaf can be checked to be locally constant. Essentially the same procedure produces a flat vector bundle, i.e. a vector bundle with locally constant transition functions. This is yet another object equivalent to a representation of the fundamental group. With regard to your other comments, perhaps I should add emphasize that the Narasimhan-Seshadri correspondence is between stable vector bundles of degree 0 and irreducible unitary representations of the fundamental group. The bundle is constructed as indicated above. Anyway, this sounds like a good Diplom thesis problem. Have fun. 1 I agree that the correspondence between the fundamental group(oid) and locally constant sheaves is not very well documented in the basic literature. Whenever it comes up with my students, I end up having to sketch it out on the blackboard. However, my recollection is that Spanier's Algebraic Topology gives the correspondence as a set of exercises with hints. In any case, one direction is easy to describe as follows. Suppose that $X$ is a good connected space X (e.g. a manifold). Let $\tilde X\to X$ denote its universal cover. Given a representation of its fundamental $\rho:\pi_1(X)\to GL(V)$, one can form the sheaf of sections of the bundle $(\tilde X\times V)/\pi_1(X)\to X$. More explicitly, the sections of the sheaf over U can be identified with the continuous functions $f:\tilde U\to V$ satisfying $$f(\gamma x) = \rho(\gamma) f(x)$$ for $\gamma\in \pi_1(X)$. This sheaf can be checked to be locally constant. With regard to your other comments, perhaps I should add that the Narasimhan-Seshadri correspondence is between stable vector bundles of degree 0 and irreducible unitary representations of the fundamental group. Anyway, this sounds like a good Diplom thesis problem. Have fun.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9373617768287659, "perplexity_flag": "head"}
http://mathhelpforum.com/discrete-math/206356-help-mathematical-induction-problem-print.html	# Help with mathematical induction problem Printable View • October 29th 2012, 06:18 PM Walshy Help with mathematical induction problem Suppose we want to prove that: 1/2 * 3/4 ... 2n-1/2n < 1/sqrt(3n) for all positive integers. (a) Show that if we try to prove this inequality using mathematical induction, the basis step works, but the inductive step fails. (b) Show that mathematical induction can be used to prove the stronger inequality: 1/2 * 3/4 ... 2n-1/2n < 1/sqrt(3n+1) So far I have proven the basis step works in part a by plugging in 1, however, I do not know how to say the inductive step fails. I have no clue on part b. Thanks. • October 30th 2012, 01:12 PM emakarov Re: Help with mathematical induction problem Let P(n) be 1/2 * 3/4 ... (2n-1)/2n < 1/sqrt(3n) (note that parentheses around 2n-1 are mandatory). You verified that P(1) is true. The induction step consists of proving that P(k) implies P(k + 1) for all k >= 1. Strictly speaking, this is a true statement just because the conclusion P(k + 1) is true. After all, the overall problem is to prove P(n) for all n. What fails is a particular, most natural way of proving the step. Indeed, the natural way is to say $\frac{1}{2} \cdot \frac{3}{4} \cdot \dots\cdot\frac{2n-1}{2n}\cdot \frac{2n+1}{2n+2} < \frac{1}{\sqrt{3n}}\cdot \frac{2n+1}{2n+2}$ by the induction hypothesis and then to try proving that the right-hand side is < $\frac{1}{\sqrt{3(n+1)}}$. However, $\frac{1}{\sqrt{3n}}\cdot\frac{2n+1}{2n+2} < \frac{1}{\sqrt{3(n+1)}}$ has no solutions. When we strengthen the induction hypothesis, the method above works. Note that the new strict inequality fails for n = 1, so you should either prove a non-strict inequality $\frac{1}{2}\cdot\frac{3}{4}\cdot\dots\cdot\frac{2n-1}{2n}\le\frac{1}{\sqrt{3n+1}}$ for all positive integers n and then say that $\frac{1}{\sqrt{3n+1}}<\frac{1}{\sqrt{3n}}$, or consider n = 2 in the base step. All times are GMT -8. The time now is 09:18 AM.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9003940224647522, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/12611/why-can-we-treat-quantum-scattering-problems-as-time-independent/12628	# Why can we treat quantum scattering problems as time-independent? From what I remember in my undergraduate quantum mechanics class, we treated scattering of non-relativistic particles from a static potential like this: 1. Solve the time-independent Schrodinger equation to find the energy eigenstates. There will be a continuous spectrum of energy eigenvalues. 2. In the region to the left of the potential, identify a piece of the wavefunction that looks like $Ae^{i(kx - \omega t)}$ as the incoming wave. 3. Ensure that to the right of the potential, there is not piece of the wavefunction that looks like $Be^{-i(kx + \omega t)}$, because we only want to have a wave coming in from the left. 4. Identify a piece of the wavefunction to the left of the potential that looks like $R e^{-i(kx + \omega t)}$ as a reflected wave. 5. Identify a piece of the wavefunction to the right of the potential that looks like $T e^{i(kx - \omega t)}$ as a transmitted wave. 6. Show that $\|R\|^2 + \|T\|^2 = \|A\|^2$. Interpret $\frac{\|R\|^2}{\|A\|^2}$ as the probability for reflection and $\frac{\|T\|^2}{\|A\|^2}$ as the probability for transmission. This entire process doesn't seem to have anything to do with a real scattering event - where a real particle is scattered by a scattering potential - we do all our analysis on a stationary waves. Why should such a naive procedure produce reasonable results for something like Rutherford's foil experiment, in which alpha particles are in motion as they collide with nuclei, and in which the wavefunction of the alpha particle is typically localized in a (moving) volume much smaller than the scattering region? - 3 – Marek Jul 22 '11 at 11:50 3 This really bothered me too when I first took quantum mechanics. – Ted Bunn Jul 22 '11 at 14:11 ## 6 Answers This is fundamentally no more difficult than understanding how quantum mechanics describes particle motion using plane waves. If you have a delocalized wavefunction $\exp(ipx)$ it describes a particle moving to the right with velocity p/m. But such a particle is already everywhere at once, and only superpositions of such states are actually moving in time. Consider $$\int \psi_k(p) e^{ipx - iE(p) t} dp$$ where $\psi_k(p)$ is a sharp bump at $p=k$, not a delta-function, but narrow. The superposition using this bump gives a wide spatial waveform centered about at x=0 at t=0. At large negative times, the fast phase oscillation kills the bump at x=0, but it creates a new bump at those x's where the phase is stationary, that is where $${\partial\over\partial p}( p x - E(p)t ) = 0$$ or, since the superposition is sharp near k, where $$x = E'(k)t$$ which means that the bump is moving with a steady speed as determined by Hamilton's laws. The total probability is conserved, so that the integral of psi squared on the bump is conserved. The actual time-dependent scattering event is a superposition of stationary states in the same way. Each stationary state describes a completely coherent process, where a particle in a perfect sinusoidal wave hits the target, and scatters outward, but because it is an energy eigenstate, the scattering is completely delocalized in time. If you want a collision which is localized, you need to superpose, and the superposition produces a natural scattering event, where a wave-packet comes in, reflects and transmits, and goes out again. If the incoming wavepacked has an energy which is relatively sharply defined, all the properties of the scattering process can be extracted from the corresponding energy eigenstate. Given the solutions to the stationary eigenstate problem $\psi_p(x)$ for each incoming momentum $p$, so that at large negative x, $\psi_p(x) = exp(ipx) + A \exp(-ipx)$ and $\psi_p(x) = B\exp(ipx)$ at large positive x, superpose these waves in the same way as for a free particle $$\int dp \psi_k(p) \psi_p(x) e^{-iE(p)t}$$ At large negative times, the phase is stationary only for the incoming part, not for the outgoing or reflected part. This is because each of the three parts describes a free-particle motion, so if you understand where free particle with that momentum would classically be at that time, this is where the wavepacket is nonzero. So at negative times, the wavepacket is centered at $$x = E'(k)t$$ For large positive t, there are two places where the phase is stationary--- those x where $$x = - E'(k) t$$ $$x = E_2'(k) t$$ Where $E_2'(k)$ is the change in phase of the transmitted k-wave in time (it can be different than the energy if the potential has an asymptotically different value at $+\infty$ than at $-\infty$). These two stationary phase regions are where the reflected and transmitted packet are located. The coefficient of the reflected and transmitted packets are A and B. If A and B were of unit magnitude, the superposition would conserve probability. So the actual transmission and reflection probability for a wavepacket is the square of the magnitude of A and of B, as expected. - First suppose that the Hamiltonian $H(t) = H_0 + H_I(t)$ can be decomposed into free and interaction parts. It can be shown (I won't derive this equation here) that the retarded Green function for $H(t)$ obeys the equation $$G^{(+)}(t, t_0) = G_0^{(+)}(t, t_0) - {i \over \hbar} \int_{-\infty}^{\infty} {\rm d} t' G_0^{(+)}(t,t') H_I(t') G^{(+)}(t', t_0)$$ where $G_0^{(+)}$ is the retarded Green function for $H_0$. Letting this equation act on a state $\left\| \psi(t_0) \right>$ this becomes $$\left\| \psi(t) \right> = \left\| \varphi(t) \right> - {i \over \hbar} \int_{-\infty}^{\infty} {\rm d} t' G_0^{(+)}(t,t') H_I(t')\left\| \psi(t') \right>$$ where $\varphi(t) = G_0^{(+)}(t,t') \left\| \psi(t_0) \right>$. Now, we suppose that until $t_0$ there is no interaction and so we can write $\left \|\psi(t_0) \right>$ as superposition of momentum eigenstates $$\left\| \psi(t_0) \right> = \int {\rm d}^3 \mathbf p a(\mathbf p) e^{-{i \over \hbar} E t_0} \left\| \mathbf p \right>.$$ A similar decomposition will also hold for $\left\| \phi(t) \right>$. This should inspire us in writing $\left\| \psi(t) \right >$ as $$\left\| \psi(t) \right> = \int {\rm d}^3 \mathbf p a(\mathbf p) e^{-{i \over \hbar} E t} \left\| \psi^{(+)}_{\mathbf p} \right>$$ where the states $\left\| \psi^{(+)}_{\mathbf p} \right>$ are to be determined from the equation for $\left\|\psi(t) \right>$. Now, the amazing thing (which I again won't derive due to the lack of space) is that these states are actually eigenstates of $H$: $$H \left\| \psi^{(+)}_{\mathbf p} \right> = E \left\| \psi^{(+)}_{\mathbf p} \right>$$ for $E = {\mathbf p^2 \over 2m}$ (here we assumed that the free part is simply $H_0 = {{\mathbf p}^2 \over 2m}$ and that $H_I(t)$ is independent of time). Similarly, one can derive advanced eigenstates from advanced Green function $$H \left\| \psi^{(-)}_{\mathbf p} \right> = E \left\| \psi^{(-)}_{\mathbf p} \right>.$$ Now, in one dimension and for an interaction Hamiltonian of the form $\left< \mathbf x \right\| H_I \left\| \mathbf x' \right> = \delta(\mathbf x - \mathbf x') U(\mathbf x)$ it can be further shown that $$\psi^{(+)}_p \sim \begin{cases} e^{{i \over \hbar}px} + A(p) e^{-{i \over \hbar}px} \quad x< -a \cr B(p)e^{{i \over \hbar}px} \quad x> a \end{cases}$$ where $a$ is such that the potential vanishes for $\|x\| > a$ and $A(p)$ and $B(p)$ are coefficients fully determined by the potential $U(x)$. Similar discussion again applies for wavefunctions $\psi^{(-)}_p$. Thus we have succeeded in reducing the dynamical problem into a stationary problem by writing the non-stationary states $\psi(t, x)$ in the form of stationary $\psi^{(+)}_p(x)$. - -1 This answer is no good. You are turning off the scattering potential at $t=-\infty$ for no reason, the Hamiltonian in a scattering problem of the sort the OP is asking about is time independent. The answer is ridiculously formal, and all the interesting things are in the "it can be shown...". – Ron Maimon Aug 17 '11 at 2:32 @Ron: I don't quite understand your objection. Physically, the $t = -\infty$ part of the potential never matters in a scattering problem since particles are infinitely away from the potential (that is usually generated by their being close anyway). So this is only technicallity that I prefer to work with that doesn't change anything (rather, it's very convenient in more general situations). As for the "it can be shown" parts... well, I can show them but the answer would be twice as long. Will you remove the downvote if I include the derivations? And as for being formal... so what? – Marek Aug 17 '11 at 6:26 The answer to this is the same as the answer to why you solve the Time-Independent-Schrodinger-Equation to find the time evolution of a bound particle. First you solve the TISE to find the stationary states $\psi_n$, then you write the particle's wavefunction $\Psi(t=0)$ in terms of a superposition of the $\psi_n$. Since you know how the stationary states evolve in time, you now know (at least in principle) how ANY wavefunction evolves in time. It's the same thing for scattering. You figure out what happens for the energy eigenstates, and now you know what will happen for any wavepacket (which you would write as a superposition of energy eigenstates, of course). And here it's even easier than the bound states: if all you care about is R and T, and your wavepacket has a narrow range of energies (for which T is nearly constant), then the value of T for your wavepacket is the same as what you just calculated for the energy eigenstate. Huzzah! If your wavepacket involves a superpostion of a wide range of energies, with a wide range of T's, then your life will be more complicated, of course. But in scattering experiments, folks usually try to employ nearly monoenergetic beams. Because quantum mechanics classes spend so much time mired in the details of solving the TISE (either for scattering or bound states), they often lose sight of one of the motivations for solving the TISE: it's a tool for finding the time behavior of any initial condition. - 2 I'm baffled by @Marek's statement that the Hamiltonian is explicitly time-dependent. It certainly doesn't need to be and often isn't. For instance, Rutherford scattering: $H=p^2/(2m)+q_1q_2/(4\pi\epsilon_0r)$. Note the absence of time dependence. In a scattering situation, the wavefunction is time-dependent, not generally the Hamiltonian. In any situation in which the Hamiltonian is explicitly time-dependent, the procedure described in the original question wouldn't work, so in the context of this question we're certainly assuming time-independent Hamiltonians. – Ted Bunn Jul 22 '11 at 18:49 1 @Ted: also note that the process Mark describes is not what AC describes in his answer. We don't evolve solutions in time at all. To give complete justification one needs to proceed as in the usual scattering theory (which is best dealt with in the Dirac picture and not Schrodinger picture). This is a huge subject and it certainly is not about simple solving of TISE (even though it can be reduced to this sometimes)... – Marek Jul 22 '11 at 19:36 1 I don't dispute any of this, but I don't think any of it is relevant to the question at hand. Note that it's explicitly about scattering from a static potential. One should be able to understand why the "usual" undergraduate quantum mechanics procedure for treating, e.g., Rutherford scattering, or scattering from a delta-function potential, or a square barrier gives the right answer. (Continued ...) – Ted Bunn Jul 22 '11 at 19:47 3 There's no need to introduce time-dependence in any of those cases: you could solve the time-dependent equation numerically for a wave packet, or you can solve the time-dependent Schrodinger equation analytically. As I understand it, Mark's question is why those two ways of treating the problem give the same answer. – Ted Bunn Jul 22 '11 at 19:49 1 @Ted: well, I was just trying to describe why the problem is about something else than simple solving of TISE. As for the real justification, I hinted at it in my comment under the question: it follows from the L-S equation. What AC describes is either another way of solving the scattering problem (and so irrelevant to the question) or a (wrong) justification of why the "usual" way works. Either way, I find this answer unsatisfactory. – Marek Jul 22 '11 at 20:12 show 8 more comments There is already a detailed and correct derivation, in my answer I can try to address the qualitative side of "why". In a scattering problem, there is always a hierarchy of well-separated scales. In your example of an alpha particle in Rutherford experiment, you refer to localization in space which means a certain spread in the momentum/energy. However, as long as this spread is smaller than the characteristic energy scale on which the scattering amplitudes changes, the time-independent at well-defined energy should give correct results. In terms of lengths this scale separation required for the time-independent picture to work is that the wave-packet of the alpha particle should be larger that the neighbourhood of the nucleus where the scattering happens. Typically, this is the case -- if it is not, the alpha particle is likely to have very uncertain (in Heisenberg sense) energy/momentum. - Here I would like to expand some of the arguments given in Ron Maimon's nice answer. i) Let us divide the 1D $x$-axis into three regions $I$, $II$, and $III$, with a localized potential $V(x)$ in the middle region $II$ having a compact support. (Clearly, there are physically relevant potentials that hasn't compact support, e.g. the Coulomb potential, but this assumption simplifies the following discussion.) ii) Time-independent and monochromatic. The particle is free in the regions $I$ and $III$, so we can solve the time-independent Schrödinger equation $$\hat{H}\psi(x) ~=~E \psi(x), \qquad\qquad \hat{H}~=~ \frac{\hat{p}^2}{2m}+V(x),\qquad\qquad E> 0, \qquad\qquad (1)$$ exactly there. We know that the 2nd order linear ODE has two linearly independent solutions, which in the free regions $I$ and $III$ are plane waves $$\psi_{I}(x) ~=~ a^{+}_{I}(k)e^{ikx} + a^{-}_{I}(k)e^{-ikx}, \qquad\qquad k> 0, \qquad\qquad (2)$$ $$\psi_{III}(x) ~=~ a^{+}_{III}(k)e^{ikx} + a^{-}_{III}(k)e^{-ikx}, \qquad\qquad (3)$$ Just from linearity of the Schrödinger equation, even without solving the middle region $II$, we know that the four coefficients $a^{\pm}_{I/III}(k)$ are constrained by two linear conditions. This observation leads, by the way, to the time-independent notion of the scattering $S$-matrix and the transfer $M$-matrix $$\begin{pmatrix} a^{-}_{I}(k) \\ a^{+}_{III}(k) \end{pmatrix}~=~ S(k) \begin{pmatrix} a^{+}_{I}(k) \\ a^{-}_{III}(k) \end{pmatrix}. \qquad\qquad (4)$$ $$\begin{pmatrix} a^{+}_{III}(k) \\ a^{-}_{III}(k) \end{pmatrix}~=~ M(k) \begin{pmatrix} a^{+}_{I}(k) \\ a^{-}_{I}(k) \end{pmatrix}. \qquad\qquad (5)$$ see e.g. Griffiths' book, Introduction to Quantum Mechanics, Section 2.7, and this answer. iii) Time-dependence of monochromatic wave. The dispersion relation reads $$\frac{E(k)}{\hbar} ~\equiv~\omega(k)~=~\frac{\hbar k^2}{2m}, \qquad\qquad (6)$$ The specific form $(6)$ of the dispersion relation will not matter in what follows. The full time-dependent monochromatic solution in the free regions I and III becomes $$\Psi_r(x,t) ~=~ \sum_{\sigma=\pm}a^{\sigma}_r(k)e^{\sigma ikx-i\omega(k)t} ~=~\underbrace{e^{-i\omega(k)t}}_{\text{phase factor}} \Psi_r(x,0), \qquad r ~\in~ \{I, III\}. \qquad (7)$$ The solution $(7)$ is a sum of a right mover ($\sigma=+$) and a left mover ($\sigma=-$). For now the words right and left mover may be taken as semantic names without physical content. The solution $(7)$ is fully delocalized in the free regions I and III with the probability density $\|\Psi_r(x,t)\|^2$ independent of time $t$, so naively, it does not make sense to say that the waves are right or left moving, or even scatter. However, it turns out, we may view the monochromatic wave $(7)$ as a limit of a wave packet, and obtain a physical interpretation in that way, see next section. iv) Wave packet. We now take a wave packet $$A^{\sigma}_r(k)~=~0 \qquad \text{for} \qquad \|k-k_0\| ~\geq~ \frac{1}{L}, \qquad\sigma~\in~\{\pm\}, \qquad r ~\in~ \{I, III\},\qquad (8)$$ narrowly peaked about some particular value $k_0$ in $k$-space, $$\|k-k_0\| ~\leq~ K, \qquad\qquad (9)$$ where $K$ is some wave number scale, so that we may Taylor expand the dispersion relation $$\omega(k)~=~ \omega(k_0) + v_g(k_0)(k-k_0) + {\cal O}\left((k-k_0)^2\right), \qquad\qquad (10)$$ and drop higher-order terms ${\cal O}\left((k-k_0)^2\right)$. Here $$v_g(k)~:=~\frac{d\omega(k)}{dk}\qquad\qquad (11)$$ is the group velocity. The wave packet (in the free regions I and III) is a sum of a right and a left mover, $$\Psi_r(x,t)~=~ \Psi^{+}_r(x,t)+\Psi^{-}_r(x,t), \qquad\qquad r ~\in~ \{I, III\},\qquad\qquad (12)$$ where $$\Psi^{\sigma}_r(x,t)~:=~ \int dk~A^{\sigma}_r(k)e^{\sigma ikx-i\omega(k)t}, \qquad\qquad\sigma~\in~\{\pm\}, \qquad\qquad r ~\in~ \{I, III\},$$ $$~\approx~ e^{i(k_0 v_g(k_0)-\omega(k_0))t} \int dk~A^{\sigma}_r(k)e^{ ik(\sigma x- v_g(k_0)t)}$$ $$~=~\underbrace{e^{i(k_0 v_g(k_0)-\omega(k_0))t}}_{\text{phase factor}} ~\Psi^{\sigma}_r(x-\sigma v_g(k_0)t,0).\qquad\qquad (13)$$ The right and left movers $\Psi^{\sigma}$ will be very long spread-out wave trains of sizes $\geq \frac{1}{K}$, but we are still able to identity via eq. $(13)$ their time evolution as just 1. a collective motion with group velocity $\sigma v_g(k_0)$, and 2. an overall time-dependent phase factor of modulus $1$, which is the same for the right and the left mover. In the limit $K \to 0$, with $K >0$, the approximation $(10)$ becomes better and better, and we recover the time-independent monochromatic wave, $$A^{\sigma}_r(k) ~\longrightarrow ~a^{\sigma}_r(k_0)~\delta(k-k_0)\qquad \text{for} \qquad K\to 0. \qquad\qquad (14)$$ It thus makes sense to assign a group velocity to each of the $\pm$ parts of the monochromatic wave $(7)$, because it can understood as an appropriate limit of the wave packet $(13)$. - I also struggled to understand it myself. Why I think this confuses many people is that they try to interpret the time-independent scattering wavefunction as describing one single collision of a particle from the target and it is this interpretation which is not correct and leads to the confusion! I think that the easiest way of seeing why the time-independent approach works lies in the definition of the scattering process which the wavefunction describes. The time-independent scattering solution describes the situation in which the target is being continuously bombarded by a flux of non-interacting projectiles approaching with different impact parameters (this is how most of the scattering experiments work). Therefore the process you are trying to describe is stationary. This is the actual reason why the time-independent formulation works. You can see that from e.g. the classical book on scattering (Taylor: Scattering theory), where the scattering process is defined (Chapter 3, section d) very clearly in terms of the continuous flux of the incoming particles. You can convince yourself that this interpretation of the time-independent scattering solution is indeed correct by simply noting that the probability flux (either incoming or outgoing) that you can calculate from the scattering wavefunction has the units of probability per unit time per unit area, i.e. it describes a stationary scattering process. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 85, "mathjax_display_tex": 30, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9400739669799805, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/113327-improved-euler-s-method-help.html	# Thread: 1. ## Improved Euler's Method Help I am trying to use the Improved Euler's Method to solve a homework problem, but I am having trouble with the notation and my book is extremely unhelpful. Where the lower formula is the Improved Euler's method. I am given: y = 5e^(x(x+2)) y' = 2y(x+1) y(0) = 5 dx = .3 (I'm guessing this is step size...) My main difficulty is notation and conception of the problem: the notation is confusing me, and I'm not sure what is actually happening in the formula. Does f(x,y) mean the function with x and y substituted, or something else? The problem asks for the first approximation. If someone could explain the problem in steps that would be great. 2. Usually they tell you that $y'=f(x,y)$. So in your case $f(x,y)=2y(x+1)$ and $y_0=y(x_0)=5$ and $x_0 = 0$. So find $z_1$ and then $y_1, x_1$, then $z_2, y_2, x_2$, etc. They should probably mention that $x_n = x_{n-1}+\Delta x$ (Usually $h$ is used for the step size though.)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9291970729827881, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/154022-odd-even-functions.html	# Thread: 1. ## Odd and Even functions Prove that: a) if f(x) is an even function, $\int^a_{-a}f(x)dx=2\int^a_0 f(x)dx$ b)if f(x) is an odd function, $\int^a_{-a} f(x)dx=0$ I tried using proof by contradiction: f(x) is even, then $\int^a_{-a}f(x)dx\neq 2\int^a_0 f(x)dx$ My problem is I don't know how to continue. I figure if I get the first one right I can do the other. Thanks! 2. You have to use the conditions on even and odd functions. I would then break up both LHS integrals into two pieces: one on either side of the origin.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.921088695526123, "perplexity_flag": "head"}
http://mathoverflow.net/questions/54926?sort=votes	## Longest element of Weyl groups. ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) What is a reduced expression of the longest element of each type of Weyl group. For type $A_n$ it is just $s_n(s_ns_{n-1})...(s_n...s_1)$. I know for type $B_n,C_n,E_7,E_8$,$G_2$ and $D_n$ (n even) it is just $-id$, although I don't have an explicit reduced expression for them. For type $D_n$ (n odd) and type $E_6$ I don't know what are the longest elements. Any reference where it is written explicitely ? - Just for the action of the longest element on the root system, it will always act as $-\sigma$ for a diagram automorphism $\sigma$. $\sigma$ always has order 2, and in types $A_n, E_6$ and $D_n$ for $n$ odd it is the unique non-trivial such diagram automorphism. – Peter Tingley Feb 10 2011 at 20:31 1 Take a look at arxiv.org/pdf/1108.1048v2.pdf table 1. – B. Bischof Apr 17 2012 at 16:44 ## 4 Answers 2-color your Dynkin diagram, black and white. Let $w$ be the product of the white simple reflections, $b$ the product of the black. Note that $w$ and $b$ are well-defined, as the reflections you're multiplying to make them, commute. You'll have to pick the order if you want an actual word, in what follows. If $G$ is not $A_{even}$: the affinization of the diagram is also 2-colorable, so you can choose the affine vertex to be white. Let $\chi = w b$, a Coxeter element. The Coxeter number $h$ is even, and $\chi^{h/2} = w_0$. So you get a reduced word $wbwbwb\ldots wb$, where the total number of letters is $h$ (and each letter is a product of commuting reflections). If $G$ is, unfortunately, $A_{even}$: you have to pick $w$ vs. $b$, and the diagram automorphism shows that the choice is unavoidable. The Coxeter number is odd. But you still get a reduced word, $wbwb\ldots bw$, again with $h$ letters. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. EDIT: This is a belated attempt (motivated by a question from Yongjun Xu) to answer the question more precisely than I did at first, with more emphasis on careful choice of a Coxeter element when the Coxeter number `$h$` is even (as happens for all irreducible Weyl groups except those of type `$A_\ell$` with `$\ell$` even). As Allen points out, for even type `$A$` the algorithm needs to be modified somewhat. I think the best conceptual approach (leaving even type `$A$` aside) is based on the Coxeter element, treated in detail in Bourbaki's Chapter V and later in Chapter 3 of my 1990 book Reflection Groups and Coxeter Groups. My exercise 2 at the end of 3.19 deals with the longest element explicitly, but requires a special choice of the Coxeter element `$w$` as in the treatment of the Coxeter plane in 3.17: here you start by dividing the simple reflections into two sets, each consisting of mutually orthogonal reflections. For instance, with the usual numbering in type `$A_3$` you can take `$w = s_1 s_3 s_2$` but not `$w =s_1 s_2 s_3$` (whose square is non-reduced). Then you automatically get a reduced expression `$w_\circ = w^{h/2}$` for the longest element. Bourbaki discusses all of this in section 6.2: see Proposition 2, where it is understood that `$W$` is irreducible and the Coxeter element (denoted `$c$`) has the special format fixed earlier in the section. (The wording of my exercise was corrected a long time ago in the list of revisions which I keep on my homepage, but my original answer here overlooked that correction. After the 1992 reprinting of my book, publishers began using print-on-demand technology involving photocopies, so no further corrections can be made.) As in Bourbaki's Chapter V, the treatment actually involves arbitrary irreducible finite reflection groups, not just Weyl groups. Besides even rank `$A_\ell$`, the odd rank dihedral groups have to be omitted because their Coxeter numbers are odd. - You can get some reduced expressions using LiE: http://www-math.univ-poitiers.fr/~maavl/LiE/ via the command long_word(Xn) where Xn is the Dynkin diagram. - There are lots of reduced expressions, which combinatorists have studied. John Stembridge has also developed good software for such explorations. – Jim Humphreys Feb 9 2011 at 23:11 A good reference to try is Bourbaki "Lie groups and Lie Algebras, Chapters 4-6" Look at the plates at the end of the book, which contain all kinds of useful information about each one of the types. - 2 But, alas, not reduced expressions for the longest elements :) – Mariano Suárez-Alvarez Feb 9 2011 at 22:57	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 31, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9206359386444092, "perplexity_flag": "head"}
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http://physics.stackexchange.com/questions/33916/what-is-the-escape-velocity-of-a-black-hole/33948	# What is the escape velocity of a Black Hole? The escape velocity of Earth is $v=\sqrt{\frac {GM}{R}}$, where $M$ is the mass of the Earth and $R$ it's radius (approximating it as a sphere), and is much less than light speed $c$. What is the escape velocity of a Black hole? Is it much more than light speed? - The escape velocity formula is $v_e=\sqrt{\frac {2GM}{R}}$ – voix Aug 10 '12 at 21:05 – Chris White Aug 29 '12 at 4:42 ## 2 Answers In General relativity, energy formula of a body thrown straight up to the infinity is $\large {E=\frac{mc^2}{\sqrt{1-R_S/R}}}$ As we know relativistic energy formula is $\large {E=\frac{mc^2}{\sqrt{1-v^2/c^2}}}$ So $\large {\frac{mc^2}{\sqrt{1-v_e^2/c^2}}=\frac{mc^2}{\sqrt{1-R_S/R}}}$ hence escape velocity equation in General relativity is ${\large {v_e^2=c^2\frac{R_S}{R}}}$ where $R_S=2GM/c^2$ - Schwarzschild radius of a black hole, and $R>R_S$ It's easy to derive that ${\large {v_e=c\sqrt{\frac{R_S}{R}}}=\sqrt{\frac {2GM}{R}}}$ So escape velocity formula in General relativity and Newton gravity is the same. - So the escape velocity at the event horizon would go to the speed of light as the ratio in the radical goes to 1? Am I reading that right? That would still quite simply imply infinite energy, so it seems consistent. – AlanSE Aug 11 '12 at 12:36 @AlanSE - Yes, for a point-like object. – voix Aug 11 '12 at 20:24 The escape velocity from the surface (i.e., the event horizon) of a Black Hole is exactly $c$, the speed of light. Actually the very prediction of the existence of black holes was based on the idea that there could be objects with escape velocity equal to $c$. - Where the "surface" is the event horizon. There isn't necessarily any material surface there. – Keith Thompson Aug 10 '12 at 21:38 Yes, that means event horizon. – Anixx Aug 10 '12 at 21:38 But if the escape velocity is $c$, why can't light escape? – seriousdev Aug 11 '12 at 14:18 It can if it comes from a place just above the horizon. But it looses much of its energy and becomes redshifted. – Anixx Aug 11 '12 at 22:20	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 15, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9208444356918335, "perplexity_flag": "middle"}
http://mathematica.stackexchange.com/questions/16207/find-asymptotics-of-a-function?answertab=oldest	# Find asymptotics of a function Can Mathematica find the asymptotics of a function in the following sense? I have ````Log[1/n^2]/Log[4^(-Sqrt[Log[n]]) + (2^(-Sqrt[Log[n]]) - 1)^2] ```` and I would like to know an asymptotic approximation when $n$ is large. That is a simple function that is within a constant multiplicative value in the limit when $n \rightarrow \infty$. If instead it was ````Log[1/n^2]/Log[4^(-Sqrt[Log[n]]) + (2^(-Sqrt[Log[n]]))^2] ```` then I know that ````Log[1/n^2]/Log[4^(-Sqrt[Log[n]]) + (2^(-Sqrt[Log[n]]))^2]/Sqrt[Log[n]] ```` tends to a constant value. My question is how could you use Mathematica to discover that $\sqrt{\log{n}}$ is the right answer in the second case and to find whatever the right solution is in the first case? By trial and error I happen to know that the right answer in the first case is somewhere between $2^{\log^{1/2}{n}}$ and $2^{\log^{1/2+\epsilon}{n}}$. - Could you please put parentheses around `2^-...` and `4^-...` ? – b.gatessucks Dec 12 '12 at 20:52 Done. I hope that is clearer. – lip1 Dec 12 '12 at 20:55 What is connection between 2nd and 3rd formulas? – Vitaliy Kaurov Dec 12 '12 at 21:00 You ought to help Mathematica ought with some simple analysis first. The solution in the first case looks like $O(2^{\sqrt{\text{Log}[n]}}\text{Log}\left[n\right])$; recognizing this, you can ask MMA to take the limit of the ratio for you (demonstrating it is correct and obtaining the constant in the process). – whuber Dec 12 '12 at 21:02 @VitaliyKaurov, The third is just the second divided by $\sqrt{\log{n}}$. This is a constant in the limit so $\sqrt{\log{n}}$ is asymptotically equal to the second formula under my definition. – lip1 Dec 12 '12 at 21:04 show 6 more comments ## 1 Answer For the first equation, the substitution $z= 2^{\sqrt{\log n}}$ with the inverse relation $n = \exp[ (\log_2 z)^2]$ seems to be worth to try. Note that with $n\to\infty$ also $z\to\infty$. So we try ```sub=Simplify[Log[1/n^2]/ Log[4^(-Sqrt[Log[n]]) + (2^(-Sqrt[Log[n]]) - 1)^2] /. n -> Exp[Log[2, z]^2], z > 1]``` with the result `-((2 Log[z]^2)/(Log[2]^2 Log[(2 - 2 z + z^2)/z^2])).` And next `PowerExpand[Series[sub, {z, \[Infinity], 4}] /. {z -> 2^Sqrt[Log[n]]}]` which yields the asymptotic expansion $$\log (n) 2^{\sqrt{\log (n)}}+\frac{2 \log (n)}{3\ 2^{\sqrt{\log (n)}}}+\frac{\log (n)}{\left(2^{\sqrt{\log (n)}}\right)^2}+\frac{56 \log (n)}{45 \left(2^{\sqrt{\log (n)}}\right)^3}+\frac{4 \log (n)}{3 \left(2^{\sqrt{\log (n)}}\right)^4}+O\left(\left(\frac{1}{2^{\sqrt{\log (n)}}}\right)^5\right).$$ I believe for the second problem another substitution might do the trick. - That's very nice, thanks. – lip1 Dec 12 '12 at 21:24 lang-mma	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9129230380058289, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/68734/list	## Return to Answer 3 added 4 characters in body One way to see that there is a vector bundle $E$ over $J(C)$ with $C^{(n)}\cong \mathbb{P}(E)$ is using semi continuity. Consider the closed immersion $C^{(n-1)}\hookrightarrow C^{(n)}$. This is a divisor on a smooth variety and so corresponds to a line bundle $L$. We take the push forward $u_(L)$ where $u:C^{(n)}\to J(C)$. Now using semi-continuity and Riemann-Roch, for $n$ large this is a vector bundle . $E$. In order to give a morphism $\phi:C^{(n)}\to \mathbb{P}(E)$, it suffices to check that the natural map $u^u_L\to L$ is surjective. This is easy to see by checking it over fibers of $u$. Also it is easy to check that $\phi$ is an isomorphism and that the pullback of $\mathscr{O}(1)$ is $C^{(n-1)}$, by looking at the fibers of $u$. But it is not clear to me why $E=\text{some line bundle}\otimes P_n$. 2 deleted 3 characters in body One way to see that there is a vector bundle $E$ over $J(C)$ with $C^{(n)}\cong \mathbb{P}(E)$ is using semi continuity. Consider the closed immersion $C^{(n-1)}\hookrightarrow C^{(n)}$. This is a divisor on a smooth variety and so corresponds to a line bundle $L$. We take the push forward $u_(L)$ where $u:C^{(n)}\to J(C)$. Now using semi-continuity and Riemann-Roch, for $n$ large this is a vector bundleby the Riemann Roch. In order to give a morphism $\phi:C^{(n)}\to \mathbb{P}(E)$, it suffices to check that the natural map $u^u_L\to L$ is surjective. This is easy to see by checking it over fibers of $u$. Also it is easy to check that $\phi$ is an isomorphism and that the pullback of $\mathscr{O}(1)$ is $C^{(n-1)}$, by looking at the fibers of $u$. But it is not clear to me why $E=\text{some line bundle}\otimes P_n$. 1 One way to see that there is a vector bundle $E$ over $J(C)$ with $C^{(n)}\cong \mathbb{P}(E)$ is using semi continuity. Consider the closed immersion $C^{(n-1)}\hookrightarrow C^{(n)}$. This is a divisor on a smooth variety and so corresponds to a line bundle $L$. We take the push forward $u_(L)$ where $u:C^{(n)}\to J(C)$. Now using semi-continuity, for $n$ large this is a vector bundle by the Riemann Roch. In order to give a morphism $\phi:C^{(n)}\to \mathbb{P}(E)$, it suffices to check that the natural map $u^u_*L\to L$ is surjective. This is easy to see by checking it over fibers of $u$. Also it is easy to check that $\phi$ is an isomorphism and that the pullback of $\mathscr{O}(1)$ is $C^{(n-1)}$, by looking at the fibers of $u$. But it is not clear to me why $E=\text{some line bundle}\otimes P_n$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 49, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9554325342178345, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?t=578592	Physics Forums ## Covariant derivative of connection coefficients? The connection $$\nabla$$ is defined in terms of its action on tensor fields. For example, acting on a vector field Y with respect to another vector field X we get [tex]\nabla_X Y = X^\mu ({Y^\alpha}_{,\mu} + Y^\nu {\Gamma^\alpha}_{\mu\nu})e_\alpha = X^\mu {Y^\alpha}_{;\mu}e_\alpha[/tex] and we call $${Y^\alpha}_{;\mu}={Y^\alpha}_{,\mu} + Y^\nu {\Gamma^\alpha}_{\mu\nu}$$ the covariant derivative of the components of Y. We can similarly form the covariant derivative of the components of any rank tensor, by including other appropriate terms with the connection coefficients. So what does it mean to take the covariant derivative of the connection coefficients themselves? They are not components of a tensor? I have just come across a reference to $${\Gamma^\alpha}_{\mu\nu;\lambda}$$ and don't know what to do with it. PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug I figured this out. Apparently, its covariant derivative does have the same form as the covariant derivative of the components of a (1,2) tensor. But if someone can confirm this result is correct, I would appreciate it. Recognitions: Science Advisor That's a very bastard notation, and whoever wrote it down should explain what they mean. As you say, the connection coefficients are not a covariant object, so it is not sensible to talk about their covariant derivatives. My guess is someone probably noticed they could write down the formula for the Riemann tensor in a kind of shorthand. It is technically incorrect. Recognitions: Science Advisor ## Covariant derivative of connection coefficients? By the way, I'm not sure of your level of knowledge, but if you're still learning this stuff, I would say to avoid getting in the habit of using "comma, semicolon" notation, for two reasons: 1. Since covariant derivatives do not commute, it is unclear what is meant by objects such as $$A^\mu{}_{;\nu\rho} = \nabla_\nu \nabla_\rho A^\mu \quad \text{or} \quad \nabla_\rho \nabla_\nu A^\mu \; \text{?}$$ 2. On the printed page, little marks like commas and semicolons can be hard to see, especially in photocopies. Whoever invented the notation thought they were being clever by saving space, but seems to have forgotten that the main purpose of scientific papers is to communicate... Thanks, guys. Yeah, I never liked the semi-colon notation either. Quote by pellman I figured this out. Apparently, its covariant derivative does have the same form as the covariant derivative of the components of a (1,2) tensor. But if someone can confirm this result is correct, I would appreciate it. Sorry to drag this up, but in trying to verify the formula for the components of the Riemann tensor in a non--coordinate basis, I need to know how to take the covariant derivative of the connection coefficients. Pellman, can you let me know the resource that confirmed that [tex]\nabla_a \Gamma^b{}_{cd} = \partial_a \Gamma^b{}_{cd} + \Gamma^b{}_{ma}\Gamma^m{}_{cd} - \Gamma^m{}_{ca}\Gamma^b{}_{md}- \Gamma^m{}_{da}\Gamma^b{}_{cm} [/tex] Cheers Ah, working backward from the definition of the Riemann tensor, it would appear that $$\nabla_d \Gamma^a{}_{bc} = \partial_d \Gamma^a{}_{bc}$$ ...? Recognitions: Gold Member Since the connection coefficients aren't a tensor, taking a covariant derivative of them doesn't really make sense. Quote by elfmotat Since the connection coefficients aren't a tensor, taking a covariant derivative of them doesn't really make sense. OK, but $\Gamma^a{}_{bc}e^b$ is a vector, so it makes sense to take its covariant derivative. A covariant derivative is the covariant analogue of a regular derivative. But if you use the affine connection as the thing to operate on, even if it has a form looking like a covariant derivative, it still will not be--it will not be a tensor. You can do the operation anyway, and if it has physical usefulness it will still have physical usefulness even though it is not covariant. Quote by ianhoolihan Sorry to drag this up, but in trying to verify the formula for the components of the Riemann tensor in a non--coordinate basis, I need to know how to take the covariant derivative of the connection coefficients. Pellman, can you let me know the resource that confirmed that $$\nabla_a \Gamma^b{}_{cd} = \partial_a \Gamma^b{}_{cd} + \Gamma^b{}_{ma}\Gamma^m{}_{cd} - \Gamma^m{}_{ca}\Gamma^b{}_{md}- \Gamma^m{}_{da}\Gamma^b{}_{cm}$$ Cheers Sorry. I never found an independent confirmation. I just proved it to my own satisfaction. I don't recall the details now either. Quote by ApplePion A covariant derivative is the covariant analogue of a regular derivative. But if you use the affine connection as the thing to operate on, even if it has a form looking like a covariant derivative, it still will not be--it will not be a tensor. You can do the operation anyway, and if it has physical usefulness it will still have physical usefulness even though it is not covariant. I realise the connection coefficients are not the components of a tensor. However, in the case of the vector $\nabla_c e_b = \Gamma^a{}_{bc}e_a$ I'm pretty sure you can just treat the connection coefficient as the component of the vector: $\Gamma^a{}_{bc} = [\nabla_c e_b]^a$. Hence $$\nabla_d (\nabla_c e_b) = \partial_d \Gamma^a{}_{bc} + \Gamma^a{}_{fd}\Gamma^f{}_{bc}$$. Quote by pellman Sorry. I never found an independent confirmation. I just proved it to my own satisfaction. I don't recall the details now either. I guess my case is different to yours. Quote by ianhoolihan OK, but $\Gamma^a{}_{bc}e^b$ is a vector, so it makes sense to take its covariant derivative. Sorry, I meant $\Gamma^a{}_{bc}e_a$ in this case, as in the post above. Blog Entries: 9 Recognitions: Homework Help Science Advisor The connection coefficients are not the components of any tensor. The covariant derivative, if applied onto this set of components, would lose their meaning and purpose as a derivative. I haven't seen any source in geometry defining a covariant derivative to the connection coefficients. Quote by dextercioby The connection coefficients are not the components of any tensor. The covariant derivative, if applied onto this set of components, would lose their meaning and purpose as a derivative. I haven't seen any source in geometry defining a covariant derivative to the connection coefficients. Yes, as before, I understand this. However, as in the previous post, $\nabla_d (\nabla_c e_b) = \nabla_d(\Gamma^a{}_{bc}e_a)$ is a valid equation. Is my previous result correct? Thread Tools \| \| \| \| \|-----------------------------------------------------------------------\|------------------------------\|---------\| \| Similar Threads for: Covariant derivative of connection coefficients? \| \| \| \| Thread \| Forum \| Replies \| \| \| Special & General Relativity \| 6 \| \| \| Differential Geometry \| 3 \| \| \| General Math \| 0 \| \| \| Advanced Physics Homework \| 2 \| \| \| Differential Geometry \| 3 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 7, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9471127390861511, "perplexity_flag": "middle"}
http://www.nag.co.uk/numeric/cl/nagdoc_cl23/html/E05/e05sbc.html	# NAG Library Function Documentnag_glopt_nlp_pso (e05sbc) Note: this function uses optional arguments to define choices in the problem specification and in the details of the algorithm. If you wish to use default settings for all of the optional arguments, you need only read Sections 1 to 9 of this document. If, however, you wish to reset some or all of the settings please refer to Section 10 for a detailed description of the algorithm and to Section 11 for a detailed description of the specification of the optional arguments. ## 1 Purpose nag_glopt_nlp_pso (e05sbc) is designed to search for the global minimum or maximum of an arbitrary function, subject to general nonlinear constraints, using Particle Swarm Optimization (PSO). Derivatives are not required, although these may be used by an accompanying local minimization function if desired. nag_glopt_nlp_pso (e05sbc) is essentially identical to nag_glopt_bnd_pso (e05sac), with an expert interface and various additional arguments added; otherwise most arguments are identical. In particular, nag_glopt_bnd_pso (e05sac) does not handle general constraints. ## 2 Specification #include <nag.h> #include <nage05.h> void nag_glopt_nlp_pso (Integer ndim, Integer ncon, Integer npar, double xb[], double fb, double cb[], const double bl[], const double bu[], double xbest[], double fbest[], double cbest[], void (objfun)(Integer mode, Integer ndim, const double x[], double objf, double vecout[], Integer nstate, Nag_Comm comm), void (confun)(Integer mode, Integer ncon, Integer ndim, Integer tdcj, const Integer needc[], const double x[], double c[], double cjac[], Integer nstate, Nag_Comm comm), void (monmod)(Integer ndim, Integer ncon, Integer npar, double x[], const double xb[], double fb, const double cb[], const double xbest[], const double fbest[], const double cbest[], const Integer itt[], Nag_Comm comm, Integer inform), Integer iopts[], double opts[], Nag_Comm comm, Integer itt[], Integer inform, NagError fail) Before calling nag_glopt_nlp_pso (e05sbc), nag_glopt_opt_set (e05zkc) must be called with optstr set to ‘Initialize = e05sbc’. Optional arguments may also be specified by calling nag_glopt_opt_set (e05zkc) before the call to nag_glopt_nlp_pso (e05sbc). ## 3 Description nag_glopt_nlp_pso (e05sbc) uses a stochastic method based on Particle Swarm Optimization (PSO) to search for the global optimum of a nonlinear function $F$, subject to a set of bound constraints on the variables, and optionally a set of general nonlinear constraints. In the PSO algorithm (see Section 10), a set of particles is generated in the search space, and advances each iteration to (hopefully) better positions using a heuristic velocity based upon inertia, cognitive memory and global memory. The inertia is provided by a decreasingly weighted contribution from a particles current velocity, the cognitive memory refers to the best candidate found by an individual particle and the global memory refers to the best candidate found by all the particles. This allows for a global search of the domain in question. Further, this may be coupled with a selection of local minimization functions, which may be called during the iterations of the heuristic algorithm, the interior phase, to hasten the discovery of locally optimal points, and after the heuristic phase has completed to attempt to refine the final solution, the exterior phase. Different options may be set for the local optimizer in each phase. Without loss of generality, the problem is assumed to be stated in the following form: $minimize x∈Rndim Fx subject to ℓ ≤ x cx ≤ u ,$ where the objective $F\left(\mathbf{x}\right)$ is a scalar function, $\mathbf{c}\left(\mathbf{x}\right)$ is a vector of scalar constraint functions, $\mathbf{x}$ is a vector in ${R}^{\mathit{ndim}}$ and the vectors $\mathbf{\ell }\le \mathbf{u}$ are lower and upper bounds respectively for the $\mathit{ndim}$ variables and $\mathit{ncon}$ constraints. Both the objective function and the $\mathit{ncon}$ constraints may be nonlinear. Continuity of $F$, and the functions $\mathbf{c}\left(\mathbf{x}\right)$, is not essential. For functions which are smooth and primarily unimodal, faster solutions will almost certainly be achieved by using Chapter e04 functions directly. For functions which are smooth and multi-modal, gradient dependent local minimization functions may be coupled with nag_glopt_nlp_pso (e05sbc). For multi-modal functions for which derivatives cannot be provided, particularly functions with a significant level of noise in their evaluation, nag_glopt_nlp_pso (e05sbc) should be used either alone, or coupled with nag_opt_simplex_easy (e04cbc). For heavily constrained problems, nag_glopt_nlp_pso (e05sbc) should either be used alone, or coupled with nag_opt_nlp (e04ucc) provided the function and the constraints are sufficiently smooth. The $\mathit{ndim}$ lower and upper box bounds on the variable $\mathbf{x}$ are included to initialize the particle swarm into a finite hypervolume, although their subsequent influence on the algorithm is user determinable (see the option ${\mathbf{Boundary}}$ in Section 11). It is strongly recommended that sensible bounds are provided for all variables and constraints. nag_glopt_nlp_pso (e05sbc) may also be used to maximize the objective function, or to search for a feasible point satisfying the simple bounds and general constraints (see the option ${\mathbf{Optimize}}$). Due to the nature of global optimization, unless a predefined target is provided, there is no definitive way of knowing when to end a computation. As such several stopping heuristics have been implemented into the algorithm. If any of these is achieved, nag_glopt_nlp_pso (e05sbc) will exit with NW_SOLUTION_NOT_GUARANTEED, and the parameter inform will indicate which criteria was reached. See inform for more information. In addition, you may provide your own stopping criteria through monmod, objfun and confun. nag_glopt_bnd_pso (e05sac) provides a simpler interface, without the inclusion of general nonlinear constraints. ## 4 References Gill P E, Murray W and Wright M H (1981) Practical Optimization Academic Press Kennedy J and Eberhart R C (1995) Particle Swarm Optimization Proceedings of the 1995 IEEE International Conference on Neural Networks 1942–1948 Koh B, George A D, Haftka R T and Fregly B J (2006) Parallel Asynchronous Particle Swarm Optimization International Journal for Numerical Methods in Engineering 67(4) 578–595 Vaz A I and Vicente L N (2007) A Particle Swarm Pattern Search Method for Bound Constrained Global Optimization Journal of Global Optimization 39(2) 197–219 Kluwer Academic Publishers ## 5 Arguments Note: for descriptions of the symbolic variables, see Section 10. 1: ndim – IntegerInput On entry: $\mathit{ndim}$, the number of dimensions. Constraint: ${\mathbf{ndim}}\ge 1$. 2: ncon – IntegerInput On entry: $\mathit{ncon}$, the number of constraints, not including box constraints. Constraint: ${\mathbf{ncon}}\ge 0$. 3: npar – IntegerInput On entry: $\mathit{npar}$, the number of particles to be used in the swarm. Assuming all particles remain within constraints, each complete iteration will perform at least npar function evaluations. Otherwise, significantly fewer objective function evaluations may be performed. Suggested value: ${\mathbf{npar}}=10×{\mathbf{ndim}}$. Constraint: ${\mathbf{npar}}\ge 5$. 4: xb[ndim] – doubleOutput On exit: the location of the best solution found, $\stackrel{~}{\mathbf{x}}$, in ${R}^{\mathit{ndim}}$. 5: fb – double Output On exit: the objective value of the best solution, $\stackrel{~}{f}=F\left(\stackrel{~}{\mathbf{x}}\right)$. 6: cb[ncon] – doubleOutput On exit: the constraint violations of the best solution found, $\stackrel{~}{\mathbf{e}}=\mathbf{e}\left(\stackrel{~}{\mathbf{x}}\right)$. These may have been deemed to be acceptable given the tolerance and scaling of the constraints. See Sections 10 and 11. 7: bl[${\mathbf{ndim}}+{\mathbf{ncon}}$] – const doubleInput 8: bu[${\mathbf{ndim}}+{\mathbf{ncon}}$] – const doubleInput On entry: ${\mathbf{bl}}$ is $\mathbf{\ell }$, the array of lower bounds, bu is $\mathbf{u}$, the array of upper bounds. The first ndim entries in bl and bu must contain the lower and upper simple (box) bounds of the variables respectively. These must be provided to initialize the sample population into a finite hypervolume, although their subsequent influence on the algorithm is user determinable (see the option ${\mathbf{Boundary}}$ in Section 11). The next ncon entries must contain the lower and upper bounds for any general constraints respectively. If ${\mathbf{bl}}\left[i-1\right]={\mathbf{bu}}\left[i-1\right]$ for any $i\in \left\{1,\dots ,{\mathbf{ndim}}\right\}$, variable $i$ will remain locked to ${\mathbf{bl}}\left[i-1\right]$ regardless of the ${\mathbf{Boundary}}$ option selected. It is strongly advised that you place sensible lower and upper bounds on all variables and constraints, even if your model allows for unbounded variables or constraints. Constraints: • ${\mathbf{bl}}\left[\mathit{i}-1\right]\le {\mathbf{bu}}\left[\mathit{i}-1\right]$, for $\mathit{i}=1,2,\dots ,{\mathbf{ndim}}+{\mathbf{ncon}}$; • ${\mathbf{bl}}\left[i-1\right]\ne {\mathbf{bu}}\left[i-1\right]$ for at least one $i\in \left\{1,\dots ,{\mathbf{ndim}}\right\}$. 9: xbest[${\mathbf{ndim}}×{\mathbf{npar}}$] – doubleInput/Output Note: the $i$th component of the best position of the $j$th particle, ${\stackrel{^}{x}}_{j}\left(i\right)$, is stored in ${\mathbf{xbest}}\left[\left(j-1\right)×{\mathbf{ndim}}+i-1\right]$. On entry: if using ${\mathbf{Start}}=\mathrm{WARM}$, the initial particle positions, ${\stackrel{^}{\mathbf{x}}}_{j}^{0}$. On exit: the best positions found, ${\stackrel{^}{\mathbf{x}}}_{j}$, by the npar particles in the swarm. 10: fbest[npar] – doubleInput/Output On entry: if using ${\mathbf{Start}}=\mathrm{WARM}$, objective function values, ${\stackrel{^}{f}}_{j}^{0}=F\left({\stackrel{^}{\mathbf{x}}}_{j}^{0}\right)$, corresponding to the npar particle locations stored in xbest. On exit: objective function values, ${\stackrel{^}{f}}_{j}=F\left({\stackrel{^}{\mathbf{x}}}_{j}\right)$, corresponding to the locations returned in xbest. 11: cbest[${\mathbf{ncon}}×{\mathbf{npar}}$] – doubleInput/Output Note: the $k$th constraint violation of the $j$th particle is stored in ${\mathbf{cbest}}\left[\left(j-1\right)×{\mathbf{ncon}}+k-1\right]$. On entry: if using ${\mathbf{Start}}=\mathrm{WARM}$, the initial constraint violations, ${\stackrel{^}{\mathbf{e}}}_{j}^{0}=\mathbf{e}\left({\stackrel{^}{\mathbf{x}}}_{j}^{0}\right)$, corresponding to the npar particle locations. On exit: the final constraint violations, ${\stackrel{^}{\mathbf{e}}}_{j}$, corresponding to the locations returned in xbest. 12: objfun – function, supplied by the userExternal Function objfun must, depending on the value of mode, calculate the objective function and/or calculate the gradient of the objective function for a $\mathit{ndim}$-variable vector $\mathbf{x}$. Gradients are only required if a local minimizer has been chosen which requires gradients. See the option ${\mathbf{Local Minimizer}}$ for more information. The specification of objfun is: void objfun (Integer mode, Integer ndim, const double x[], double objf, double vecout[], Integer nstate, Nag_Comm comm) 1: mode – Integer Input/Output On entry: indicates which functionality is required. ${\mathbf{mode}}=0$ $F\left(\mathbf{x}\right)$ should be returned in objf. The value of objf on entry may be used as an upper bound for the calculation. Any expected value of $F\left(\mathbf{x}\right)$ that is greater than objf may be approximated by this upper bound; that is objf can remain unaltered. ${\mathbf{mode}}=1$ ${\mathbf{Local Minimizer}}='\mathrm{e04ucc}'$ only First derivatives can be evaluated and returned in vecout. Any unaltered elements of vecout will be approximated using finite differences. ${\mathbf{mode}}=2$ ${\mathbf{Local Minimizer}}='\mathrm{e04ucc}'$ only $F\left(\mathbf{x}\right)$ must be calculated and returned in objf, and available first derivatives can be evaluated and returned in vecout. Any unaltered elements of vecout will be approximated using finite differences. ${\mathbf{mode}}=5$ $F\left(\mathbf{x}\right)$ must be calculated and returned in objf. The value of objf on entry may not be used as an upper bound. ${\mathbf{mode}}=6$ ${\mathbf{Local Minimizer}}='\mathrm{e04dgc}'$ only All first derivatives must be evaluated and returned in vecout. ${\mathbf{mode}}=7$ ${\mathbf{Local Minimizer}}='\mathrm{e04dgc}'$ only $F\left(\mathbf{x}\right)$ must be calculated and returned in objf, and all first derivatives must be evaluated and returned in vecout. On exit: if the value of mode is set to be negative, then nag_glopt_nlp_pso (e05sbc) will exit as soon as possible with NE_USER_STOP and ${\mathbf{inform}}={\mathbf{mode}}$. 2: ndim – IntegerInput On entry: the number of dimensions. 3: x[ndim] – const doubleInput On entry: $\mathbf{x}$, the point at which the objective function and/or its gradient are to be evaluated. 4: objf – double Input/Output On entry: the value of objf passed to objfun varies with the argument mode. ${\mathbf{mode}}=0$ objf is an upper bound for the value of $F\left(\mathbf{x}\right)$, often equal to the best constraint penalised value of $F\left(\mathbf{x}\right)$ found so far by a given particle if the objective function is strictly positive (see Section 10). Only objective function values less than the value of objf on entry will be used further. As such this upper bound may be used to stop further evaluation when this will only increase the objective function value above the upper bound. ${\mathbf{mode}}=1$, $2$, $5$, $6$ or $7$ objf is meaningless on entry. On exit: the value of objf returned varies with the argument mode. ${\mathbf{mode}}=0$ objf must be the value of $F\left(\mathbf{x}\right)$. Only values of $F\left(\mathbf{x}\right)$ strictly less than objf on entry need be accurate. ${\mathbf{mode}}=1$ or $6$ Need not be set. ${\mathbf{mode}}=2$, $5$ or $7$ $F\left(\mathbf{x}\right)$ must be calculated and returned in objf. The entry value of objf may not be used as an upper bound. 5: vecout[ndim] – doubleInput/Output On entry: if ${\mathbf{Local Minimizer}}=\mathrm{e04ucc}$, the values of vecout are used internally to indicate whether a finite difference approximation is required. See nag_opt_nlp (e04ucc). On exit: the required values of vecout returned to the calling function depend on the value of mode. ${\mathbf{mode}}=0$ or $5$ The value of vecout need not be set. ${\mathbf{mode}}=1$ or $2$ vecout can contain components of the gradient of the objective function $\frac{\partial F}{\partial {x}_{i}}$ for some $i=1,2,\dots {\mathbf{ndim}}$, or acceptable approximations. Any unaltered elements of vecout will be approximated using finite differences. ${\mathbf{mode}}=6$ or $7$ vecout must contain the gradient of the objective function $\frac{\partial F}{\partial {x}_{i}}$ for all $i=1,2,\dots {\mathbf{ndim}}$. Approximation of the gradient is strongly discouraged, and no finite difference approximations will be performed internally (see nag_opt_conj_grad (e04dgc)). 6: nstate – IntegerInput On entry: nstate indicates various stages of initialization throughout the function. This allows for permanent global arguments to be initialized the least number of times. For example, you may initialize a random number generator seed. ${\mathbf{nstate}}=2$ objfun is called for the very first time. You may save computational time if certain data must be read or calculated only once. ${\mathbf{nstate}}=1$ objfun is called for the first time by a NAG local minimization function. You may save computational time if certain data required for the local minimizer need only be calculated at the initial point of the local minimization. ${\mathbf{nstate}}=0$ Used in all other cases. 7: comm – Nag_Comm * Pointer to structure of type Nag_Comm; the following members are relevant to objfun. user – double * iuser – Integer * p – Pointer The type Pointer will be void . Before calling nag_glopt_nlp_pso (e05sbc) you may allocate memory and initialize these pointers with various quantities for use by objfun when called from nag_glopt_nlp_pso (e05sbc) (see Section 3.2.1 in the Essential Introduction). 13: confun – function, supplied by the userExternal Function confun must calculate any constraints other than the box constraints. If no constraints are required, confun may be NULL. For information on how a NAG local minimizer will use confun see the documentation for nag_opt_nlp (e04ucc). The specification of confun is: void confun (Integer mode, Integer ncon, Integer ndim, Integer tdcj, const Integer needc[], const double x[], double c[], double cjac[], Integer nstate, Nag_Comm comm) 1: mode – Integer Input/Output On entry: indicates which values must be assigned during each call of confun. Only the following values need be assigned, for each value of $k\in \left\{1,\dots ,{\mathbf{ncon}}\right\}$ such that ${\mathbf{needc}}\left[k-1\right]>0$: ${\mathbf{mode}}=0$ the constraint values ${c}_{k}\left(\mathbf{x}\right)$. ${\mathbf{mode}}=1$ rows of the constraint jacobian, $\frac{\partial {c}_{k}}{\partial {x}_{\mathit{i}}}\left(\mathbf{x}\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{ndim}}$. ${\mathbf{mode}}=2$ the constraint values ${c}_{k}\left(\mathbf{x}\right)$ and the corresponding rows of the constraint jacobian, $\frac{\partial {c}_{k}}{\partial {x}_{\mathit{i}}}\left(\mathbf{x}\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{ndim}}$. On exit: may be set to a negative value if you wish to terminate the solution to the current problem. In this case nag_glopt_nlp_pso (e05sbc) will terminate with NE_USER_STOP and ${\mathbf{inform}}={\mathbf{mode}}$ as soon as possible. 2: ncon – IntegerInput On entry: the number of constraints, not including box bounds. 3: ndim – IntegerInput On entry: the number of variables. 4: tdcj – IntegerInput On entry: the stride separating matrix column elements in the array cjac. 5: needc[ncon] – const IntegerInput On entry: the indices of the elements of c and/or cjac that must be evaluated by confun. If ${\mathbf{needc}}\left[k-1\right]>0$, the $k-1$th element of c, corresponding to the values of the $k$th constraint, and/or the available elements of the $k-1$th row of cjac, corresponding to the derivatives of the $k$th constraint, must be evaluated at $\mathbf{x}$ (see argument mode). 6: x[ndim] – const doubleInput On entry: $\mathbf{x}$, the vector of variables at which the constraint functions and/or the available elements of the constraint Jacobian are to be evaluated. 7: c[ncon] – doubleOutput On exit: if ${\mathbf{needc}}\left[k-1\right]>0$ and ${\mathbf{mode}}=0$ or $2$, ${\mathbf{c}}\left[k-1\right]$ must contain the value of ${c}_{k}\left(\mathbf{x}\right)$. The remaining elements of c, corresponding to the non-positive elements of needc, need not be set. 8: cjac[${\mathbf{ncon}}×{\mathbf{tdcj}}$] – doubleInput/Output Note: the derivative of the $k$th constraint with respect to the $i$th component, $\frac{\partial {c}_{k}}{\partial {x}_{i}}$, is stored in ${\mathbf{cjac}}\left[\left(k-1\right)×{\mathbf{tdcj}}+i-1\right]$, which we denote as ${\mathbf{CJAC}}\left(k,i\right)$. On entry: the elements of cjac are set to special values which enable nag_glopt_nlp_pso (e05sbc) to detect whether they are changed by confun. On exit: if ${\mathbf{needc}}\left[k-1\right]>0$ and ${\mathbf{mode}}=1$ or $2$, the elements of cjac corresponding to the $k$th row of the constraint jacobian should contain the available elements of the vector $\nabla {c}_{k}$ given by $∇ck= ∂ck ∂x1 , ∂ck ∂x2 ,…, ∂ck ∂xn ,$ where $\frac{\partial {c}_{k}}{\partial {x}_{i}}$ is the partial derivative of the $k$th constraint with respect to the $i$th variable, evaluated at the point $\mathbf{x}$; elements of cjac that remain unaltered will be approximated internally using finite differences. The remaining rows of cjac, corresponding to non-positive elements of needc, need not be set. It must be emphasized that unassigned elements of cjac are not treated as constant; they are estimated by finite differences, at nontrivial expense. An interval for each element of $\mathbf{x}$ is computed automatically at the start of the optimization. The automatic procedure can usually identify constant elements of cjac, which are then computed once only by finite differences. 9: nstate – IntegerInput On entry: nstate indicates various stages of initialization throughout the function. This allows for permanent global arguments to be initialized a minimum number of times. For example, you may initialize a random number generator seed. Note that unless the option ${\mathbf{Optimize}}=\mathrm{CONSTRAINTS}$ has been set, objfun will be called before confun. ${\mathbf{nstate}}=2$ confun is called for the very first time. This argument setting allows you to save computational time if certain data must be read or calculated only once. ${\mathbf{nstate}}=1$ confun is called for the first time during a NAG local minimization function. This argument setting allows you to save computational time if certain data required for the local minimizer need only be calculated at the initial point of the local minimization. ${\mathbf{nstate}}=0$ Used in all other cases. 10: comm – Nag_Comm * Pointer to structure of type Nag_Comm; the following members are relevant to confun. user – double * iuser – Integer * p – Pointer The type Pointer will be void . Before calling nag_glopt_nlp_pso (e05sbc) you may allocate memory and initialize these pointers with various quantities for use by confun when called from nag_glopt_nlp_pso (e05sbc) (see Section 3.2.1 in the Essential Introduction). confun should be tested separately before being used in conjunction with nag_glopt_nlp_pso (e05sbc). 14: monmod – function, supplied by the userExternal Function A user-specified monitoring and modification function. monmod is called once every complete iteration after a finalization check. It may be used to modify the particle locations that will be evaluated at the next iteration. This permits the incorporation of algorithmic modifications such as including additional advection heuristics and genetic mutations. monmod is only called during the main loop of the algorithm, and as such will be unaware of any further improvement from the final local minimization. If no monitoring and/or modification is required, monmod may be NULLFN. The specification of monmod is: void monmod (Integer ndim, Integer ncon, Integer npar, double x[], const double xb[], double fb, const double cb[], const double xbest[], const double fbest[], const double cbest[], const Integer itt[], Nag_Comm comm, Integer inform) 1: ndim – IntegerInput On entry: the number of dimensions. 2: ncon – IntegerInput On entry: the number of constraints. 3: npar – IntegerInput On entry: the number of particles. 4: x[${\mathbf{ndim}}×{\mathbf{npar}}$] – doubleInput/Output Note: the $i$th component of the $j$th particle, ${x}_{j}\left(i\right)$, is stored in ${\mathbf{x}}\left[\left(j-1\right)×{\mathbf{ndim}}+i-1\right]$. On entry: the npar particle locations, ${\mathbf{x}}_{j}$, which will currently be used during the next iteration unless altered in monmod. On exit: the particle locations to be used during the next iteration. 5: xb[ndim] – const doubleInput On entry: the location, $\stackrel{~}{\mathbf{x}}$, of the best solution yet found. 6: fb – doubleInput On entry: the objective value, $\stackrel{~}{f}=F\left(\stackrel{~}{\mathbf{x}}\right)$, of the best solution yet found. 7: cb[ncon] – const doubleInput On entry: the constraint violations, $\stackrel{~}{\mathbf{e}}=\mathbf{e}\left(\stackrel{~}{\mathbf{x}}\right)$, of the best solution yet found. 8: xbest[${\mathbf{ndim}}×{\mathbf{npar}}$] – const doubleInput Note: the $i$th component of the position of the $j$th particle's cognitive memory, ${\stackrel{^}{x}}_{j}\left(i\right)$, is stored in ${\mathbf{xbest}}\left[\left(j-1\right)×{\mathbf{ndim}}+i-1\right]$. On entry: the locations currently in the cognitive memory, ${\stackrel{^}{\mathbf{x}}}_{\mathit{j}}$, for $\mathit{j}=1,2,\dots ,{\mathbf{npar}}$ (see Section 10). 9: fbest[npar] – const doubleInput On entry: the objective values currently in the cognitive memory, $F\left({\stackrel{^}{\mathbf{x}}}_{\mathit{j}}\right)$, for $\mathit{j}=1,2,\dots ,{\mathbf{npar}}$. 10: cbest[${\mathbf{ncon}}×{\mathbf{npar}}$] – const doubleInput Note: the $k$th constraint violation of the $j$th particle's cognitive memory is stored in ${\mathbf{cbest}}\left[\left(j-1\right)×{\mathbf{ncon}}+k-1\right]$. On entry: the constraint violations currently in the cognitive memory, $\stackrel{^}{\mathbf{e}}=\mathbf{e}\left({\stackrel{^}{\mathbf{x}}}_{\mathit{j}}\right)$, for $\mathit{j}=1,2,\dots ,{\mathbf{npar}}$, evaluated at ${\stackrel{^}{\mathbf{x}}}_{j}$. 11: itt[$7$] – const IntegerInput On entry: iteration and function evaluation counters (see description of itt below). 12: comm – Nag_Comm Pointer to structure of type Nag_Comm; the following members are relevant to monmod. user – double * iuser – Integer * p – Pointer The type Pointer will be void . Before calling nag_glopt_nlp_pso (e05sbc) you may allocate memory and initialize these pointers with various quantities for use by monmod when called from nag_glopt_nlp_pso (e05sbc) (see Section 3.2.1 in the Essential Introduction). 13: inform – Integer Input/Output On entry: ${\mathbf{inform}}=0$ On exit: setting ${\mathbf{inform}}<0$ will cause near immediate exit from nag_glopt_nlp_pso (e05sbc). This value will be returned as inform with NE_USER_STOP. You need not set inform unless you wish to force an exit. 15: iopts[$\mathit{dim}$] – IntegerCommunication Array Note: the dimension of the array iopts, corresponding to the array length liopts (see e05zkc), must be at least $\mathrm{100}$. On entry: optional parameter array as generated and possibly modified by calls to nag_glopt_opt_set (e05zkc). The contents of iopts MUST NOT be modified directly between calls to nag_glopt_nlp_pso (e05sbc), nag_glopt_opt_set (e05zkc) or nag_glopt_opt_get (e05zlc). 16: opts[$\mathit{dim}$] – doubleCommunication Array Note: the dimension of the array opts, corresponding to the array length lopts (see e05zkc), must be at least $\mathrm{100}$. On entry: optional parameter array as generated and possibly modified by calls to nag_glopt_opt_set (e05zkc). The contents of opts MUST NOT be modified directly between calls to nag_glopt_nlp_pso (e05sbc), nag_glopt_opt_set (e05zkc) or nag_glopt_opt_get (e05zlc). 17: comm – Nag_Comm Communication Structure The NAG communication argument (see Section 3.2.1.1 in the Essential Introduction). 18: itt[$7$] – IntegerOutput On exit: integer iteration counters for nag_glopt_nlp_pso (e05sbc). ${\mathbf{itt}}\left[0\right]$ Number of complete iterations. ${\mathbf{itt}}\left[1\right]$ Number of complete iterations without improvement to the current optimum. ${\mathbf{itt}}\left[2\right]$ Number of particles converged to the current optimum. ${\mathbf{itt}}\left[3\right]$ Number of improvements to the optimum. ${\mathbf{itt}}\left[4\right]$ Number of function evaluations performed. ${\mathbf{itt}}\left[5\right]$ Number of particles reset. ${\mathbf{itt}}\left[6\right]$ Number of violated constraints at completion. Note this is always calculated using the ${L}^{1}$ norm and a nonzero result does not necessarily mean that the algorithm did not find a suitably constrained point with respect to the single norm used. 19: inform – Integer Output On exit: indicates which finalization criterion was reached. The possible values of inform are: inform Meaning $<0$ Exit from a user-supplied subroutine. 0 nag_glopt_nlp_pso (e05sbc) has detected an error and terminated. 1 The provided objective target has been achieved. (${\mathbf{Target Objective Value}}$). 2 The standard deviation of the location of all the particles is below the set threshold (${\mathbf{Swarm Standard Deviation}}$). If the solution returned is not satisfactory, you may try setting a smaller value of ${\mathbf{Swarm Standard Deviation}}$, or try adjusting the options governing the repulsive phase (${\mathbf{Repulsion Initialize}}$, ${\mathbf{Repulsion Finalize}}$). 3 The total number of particles converged (${\mathbf{Maximum Particles Converged}}$) to the current global optimum has reached the set limit. This is the number of particles which have moved to a distance less than ${\mathbf{Distance Tolerance}}$ from the optimum with regard to the ${L}^{2}$ norm. If the solution is not satisfactory, you may consider lowering the ${\mathbf{Distance Tolerance}}$. However, this may hinder the global search capability of the algorithm. 4 The maximum number of iterations without improvement (${\mathbf{Maximum Iterations Static}}$) has been reached, and the required number of particles (${\mathbf{Maximum Iterations Static Particles}}$) have converged to the current optimum. Increasing either of these options will allow the algorithm to continue searching for longer. Alternatively if the solution is not satisfactory, re-starting the application several times with ${\mathbf{Repeatability}}=\mathrm{OFF}$ may lead to an improved solution. 5 The maximum number of iterations (${\mathbf{Maximum Iterations Completed}}$) has been reached. If the number of iterations since improvement is small, then a better solution may be found by increasing this limit, or by using the option ${\mathbf{Local Minimizer}}$ with corresponding exterior options. Otherwise if the solution is not satisfactory, you may try re-running the application several times with ${\mathbf{Repeatability}}=\mathrm{OFF}$ and a lower iteration limit, or adjusting the options governing the repulsive phase (${\mathbf{Repulsion Initialize}}$, ${\mathbf{Repulsion Finalize}}$). 6 The maximum allowed number of function evaluations (${\mathbf{Maximum Function Evaluations}}$) has been reached. As with ${\mathbf{inform}}=5$, increasing this limit if the number of iterations without improvement is small, or decreasing this limit and running the algorithm multiple times with ${\mathbf{Repeatability}}=\mathrm{OFF}$, may provide a superior result. 7 A feasible point has been found. The objective has not been minimized, although it has been evaluated at the final solutions given in xb and xbest (${\mathbf{Optimize}}=\mathrm{CONSTRAINTS}$). If you wish to continue from the final position gained from a previous simulation with adjusted options, you may set the option ${\mathbf{Start}}=\mathrm{WARM}$, and pass back in the returned arrays xbest, fbest, and cbest. You should either record the returned values of xb, fb and cb for comparison, as these will not be re-used by the algorithm, or include them in xbest, fbest and cbest respectively by overwriting the entries corresponding to one particle with the relevant information. 20: fail – NagError *Input/Output The NAG error argument (see Section 3.6 in the Essential Introduction). nag_glopt_nlp_pso (e05sbc) will return NW_SOLUTION_NOT_GUARANTEED if no error has been detected, and a finalization criterion has been achieved which cannot guarantee success. This does not indicate that the function has failed, merely that the returned solution cannot be guaranteed to be the true global optimum. The value of inform should be examined to determine which finalization criterion was reached. ## 6 Error Indicators and Warnings NE_ALLOC_FAIL Dynamic memory allocation failed. NE_BAD_PARAM On entry, argument $⟨\mathit{\text{value}}⟩$ had an illegal value. NE_BOUND On entry, ${\mathbf{bl}}\left[i\right]={\mathbf{bu}}\left[i\right]$ for all box bounds $i$. Constraint: ${\mathbf{bu}}\left[i\right]>{\mathbf{bl}}\left[i\right]$ for at least one box bound $i$. On entry, ${\mathbf{bl}}\left[⟨\mathit{\text{value}}⟩\right]=⟨\mathit{\text{value}}⟩$ and ${\mathbf{bu}}\left[⟨\mathit{\text{value}}⟩\right]=⟨\mathit{\text{value}}⟩$. Constraint: ${\mathbf{bu}}\left[i\right]\ge {\mathbf{bl}}\left[i\right]$ for all $i$. NE_DERIV_ERRORS Derivative checks indicate possible errors in the supplied derivatives. NE_ILLEGAL_CALLBACK nag_glopt_nlp_pso (e05sbc) has been called with ${\mathbf{ncon}}>0$ and . Only use NULL with ${\mathbf{ncon}}=0$. NE_INT On entry, ${\mathbf{ncon}}=⟨\mathit{\text{value}}⟩$. Constraint: ${\mathbf{ncon}}\ge 0$. On entry, ${\mathbf{ndim}}=⟨\mathit{\text{value}}⟩$. Constraint: ${\mathbf{ndim}}\ge 1$. On entry, ${\mathbf{npar}}=⟨\mathit{\text{value}}⟩$. Constraint: ${\mathbf{npar}}\ge 5$. NE_INTERNAL_ERROR An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance. NE_INVALID_OPTION Either the option arrays have not been initialized for nag_glopt_nlp_pso (e05sbc), or they have become corrupted. The option arrays are not consistent. The option arrays have not been initialized for nag_glopt_nlp_pso (e05sbc). The option ${\mathbf{Optimize}}=\mathrm{CONSTRAINTS}$ is active, however ${\mathbf{ncon}}=0$. NE_USER_STOP User requested exit $⟨\mathit{\text{value}}⟩$ during call to confun. User requested exit $⟨\mathit{\text{value}}⟩$ during call to monmod. User requested exit $⟨\mathit{\text{value}}⟩$ during call to objfun. NW_FAST_SOLUTION After initialization ${\mathbf{fb}}=⟨\mathit{\text{value}}⟩$ but ${\mathbf{Target Objective Value}}=⟨\mathit{\text{value}}⟩$. On entry, ${\mathbf{fbest}}\left[⟨\mathit{\text{value}}⟩\right]=⟨\mathit{\text{value}}⟩$ but ${\mathbf{Target Objective Value}}=⟨\mathit{\text{value}}⟩$. Target $⟨\mathit{\text{value}}⟩$ achieved after the first iteration. ${\mathbf{fb}}=⟨\mathit{\text{value}}⟩$. The option ${\mathbf{Optimize}}=\mathrm{CONSTRAINTS}$ has been set and the first point sampled satisfied all constraints. NW_NOT_FEASIBLE Unable to locate strictly feasible point. $⟨\mathit{\text{value}}⟩$ constraints remain violated. NW_SOLUTION_NOT_GUARANTEED A finalization criterion was reached that cannot guarantee success. On exit, ${\mathbf{inform}}=⟨\mathit{\text{value}}⟩$. ## 7 Accuracy If NE_NOERROR (or NW_FAST_SOLUTION) or NW_SOLUTION_NOT_GUARANTEED on exit, a criterion will have been reached depending on user selected options. As with all global optimization software, the solution achieved may not be the true global optimum. Various options allow for either greater search diversity or faster convergence to a (local) optimum (See Sections 10 and 11). Provided the objective function and constraints are sufficiently well behaved, if a local minimizer is used in conjunction with nag_glopt_nlp_pso (e05sbc), then it is more likely that the final result will at least be in the near vicinity of a local optimum, and due to the global search characteristics of the particle swarm, this solution should be superior to many other local optima. Caution should be used in accelerating the rate of convergence, as with faster convergence, less of the domain will remain searchable by the swarm, making it increasingly difficult for the algorithm to detect the basins of attraction of superior local optima. Using the options ${\mathbf{Repulsion Initialize}}$ and ${\mathbf{Repulsion Finalize}}$ described in Section 11 will help to overcome this, by causing the swarm to diverge away from the current optimum once no more local improvement is likely. On successful exit with guaranteed success, NE_NOERROR (or NW_FAST_SOLUTION). This may happen if a ${\mathbf{Target Objective Value}}$ is assigned and is reached by the algorithm at a satisfactorily constrained point. It will also occur if a constrained point is found when ${\mathbf{Optimize}}=\mathrm{CONSTRAINTS}$ is set. On successful exit without guaranteed success, NW_SOLUTION_NOT_GUARANTEED is returned. This will happen if another finalization criterion is achieved without the detection of an error. In both cases, the value of inform provides further information as to the cause of the exit. ## 8 Further Comments The memory used by nag_glopt_nlp_pso (e05sbc) is relatively static throughout. Indeed, most of the memory required is used to store the current particle locations, the cognitive particle memories, the particle velocities and the particle weights. As such, nag_glopt_nlp_pso (e05sbc) may be used in problems with high dimension number (${\mathbf{ndim}}>100$) without the concern of computational resource exhaustion, although the probability of successfully locating the global optimum will decrease dramatically with the increase in dimensionality. Due to the stochastic nature of the algorithm, the result will vary over multiple runs. This is particularly true if arguments and options are chosen to accelerate convergence at the expense of the global search. However, the option ${\mathbf{Repeatability}}=\mathrm{ON}$ may be set to initialize the internal random number generator using a preset seed, which will result in identical solutions being obtained. ## 9 Example Note: a modified example is supplied with the NAG Library for SMP & Multicore and links have been supplied in the following subsections. This example uses a particle swarm to find the global minimum of the $2$ dimensional Schwefel function: $minimize x∈R2 f = ∑ j=1 2 xj sinxj$ subject to the constraints: $3.0 x1 - 2.0 x2 <10.0 , -1.0 < x12 - x22 + 3.0 x1 x2 < 50000.0 , -0.9 < cos x1 / 200 2 + x2 / 100 < 0.9 , -500 ≤ x1 ≤ 500 , -500 ≤ x2 ≤ 500 .$ The global optimum has an objective value of ${f}_{\mathrm{min}}=-731.707$, located at $\mathbf{x}=\left(-394.15,-433.48\right)$. Only the third constraint is active at this point. The example demonstrates how to initialize and set the options arrays using nag_glopt_opt_set (e05zkc), how to query options using nag_glopt_opt_get (e05zlc), and finally how to search for the global optimum using nag_glopt_nlp_pso (e05sbc). The problem is solved twice, first using nag_glopt_nlp_pso (e05sbc) alone, and secondly by coupling nag_glopt_nlp_pso (e05sbc) with nag_opt_nlp (e04ucc) as a dedicated local minimizer. In both cases the default option ${\mathbf{Repeatability}}=\mathrm{ON}$ is used to produce repeatable solutions. ### 9.1 Program Text Program Text (e05sbce.c) None. ### 9.3 Program Results Program Results (e05sbce.r) ## 10 Algorithmic Details The following pseudo-code describes the algorithm used with the repulsion mechanism. $INITIALIZE for j=1, npar xj = R ∈ Uℓbox,ubox x^ j = R ∈ Uℓbox,ubox Start=COLD x^ j 0 Start=WARM vj = R ∈ U -V max ,Vmax f^j = F x^ j Start=COLD f^ j 0 Start=WARM e^ j = e x^ j Start=COLD e^ j 0 Start=WARM wj = Wmax Weight Initialize=MAXIMUM Wini Weight Initialize=INITIAL R ∈ U W min , W max Weight Initialize=RANDOMIZED end for x~ = 12 ℓbox + ubox f~ = F x~ e~ = e x~ Ic = Is = 0 SWARM while (not finalized), Ic = Ic + 1 for j = 1 , npar xj = BOUNDARYxj,ℓbox,ubox fj = F xj ej = e xj if fj / fscale + ϕ wj ej < f^j / fscale + ϕwj e^j f^j = fj , x^ j = xj if e j < e~ or e j ≈ e~ and fj < f~ f~ = fj , x~ = xj end for if new f~ LOCMINx~,f~,e~,Oi , Is=0 [see note on repulsion below for code insertion] else Is = Is + 1 for j = 1 , npar vj = wj vj + Cs D1 x^j - xj + Cg D2 x~ - xj xj = xj + vj if xj - x~ < dtol reset xj, vj, wj; x^j = xj else update wj end for if (target achieved or termination criterion satisfied) finalized=true monmod xj end LOCMINx~,f~,e~,Oe$ The definition of terms used in the above pseudo-code are as follows. $\mathit{npar}$ the number of particles, npar ${\mathbf{\ell }}_{\mathrm{box}}$ array of ndim lower box bounds ${\mathbf{u}}_{\mathrm{box}}$ array of ndim upper box bounds ${\mathbf{x}}_{j}$ position of particle $j$ ${\stackrel{^}{\mathbf{x}}}_{j}$ best position found by particle $j$ $\stackrel{~}{\mathbf{x}}$ best position found by any particle ${f}_{j}$ $F\left({\mathbf{x}}_{j}\right)$ ${\stackrel{^}{f}}_{j}$ $F\left({\stackrel{^}{\mathbf{x}}}_{j}\right)$, best value found by particle $j$ $\stackrel{~}{f}$ $F\left(\stackrel{~}{\mathbf{x}}\right)$, best value found by any particle ${e}_{k}\left(\mathbf{x}\right)$ $k$th (scaled) constraint violation at $\mathbf{x}$, evaluated as $\mathrm{min}\phantom{\rule{0.125em}{0ex}}\left({c}_{k}\left(\mathbf{x}\right)-{l}_{{\mathbf{ndim}}+k},0.0\right)+\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left({c}_{k}\left(\mathbf{x}\right)-{u}_{{\mathbf{ndim}}+k},0.0\right)$; this may be scaled by the maximum $k$th constraint found thus far $\mathbf{e}\left(\mathbf{x}\right)$ the array of ncon constraint violations, ${e}_{\mathit{k}}\left(\mathbf{x}\right)$, for $\mathit{k}=1,2,\dots ,{\mathbf{ncon}}$, at a point $\mathbf{x}$ ${\mathbf{e}}_{j}$ $\mathbf{e}\left({\mathbf{x}}_{j}\right)$, the array of constraint violations evaluated at ${\mathbf{x}}_{j}$ ${\stackrel{^}{\mathbf{e}}}_{j}$ $\mathbf{e}\left({\stackrel{^}{\mathbf{x}}}_{j}\right)$, the array of constraint violations evaluated at ${\stackrel{^}{\mathbf{x}}}_{j}$ $\stackrel{~}{\mathbf{e}}$ $\mathbf{e}\left(\stackrel{~}{\mathbf{x}}\right)$, the array of constraint violations evaluated at $\stackrel{~}{\mathbf{x}}$ ${\mathbf{v}}_{j}$ velocity of particle $j$ ${w}_{j}$ weight on ${\mathbf{v}}_{j}$ for velocity update, decreasing according to ${\mathbf{Weight Decrease}}$ ${\mathbf{V}}_{\mathrm{max}}$ maximum absolute velocity, dependent upon ${\mathbf{Maximum Variable Velocity}}$ ${I}_{c}$ swarm iteration counter ${I}_{s}$ iterations since $\stackrel{~}{\mathbf{x}}$ was updated ${f}_{\mathrm{scale}}$ objective function scaling defined by the options ${\mathbf{Constraint Scaling}}$, ${\mathbf{Objective Scaling}}$ and ${\mathbf{Objective Scale}}$. ${\mathbf{D}}_{1}$,${\mathbf{D}}_{2}$ diagonal matrices with random elements in range $\left(0,1\right)$ ${C}_{s}$ the cognitive advance coefficient which weights velocity towards ${\stackrel{^}{\mathbf{x}}}_{j}$, adjusted using ${\mathbf{Advance Cognitive}}$ ${C}_{g}$ the global advance coefficient which weights velocity towards $\stackrel{~}{\mathbf{x}}$, adjusted using ${\mathbf{Advance Global}}$ $\mathit{dtol}$ the ${\mathbf{Distance Tolerance}}$ for resetting a converged particle $\mathbf{R}\in U\left({\mathbf{\ell }}_{\mathrm{box}},{\mathbf{u}}_{\mathrm{box}}\right)$ an array of random numbers whose $i$-th element is drawn from a uniform distribution in the range $\left({{\mathbf{\ell }}_{\mathrm{box}}}_{\mathit{i}},{{\mathbf{u}}_{\mathrm{box}}}_{\mathit{i}}\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{ndim}}$ ${O}_{i}$ local optimizer interior options ${O}_{e}$ local optimizer exterior options $\varphi \left({w}_{j}\right)$ a function of ${w}_{j}$ designed to increasingly weight towards minimizing constraint violations as ${w}_{j}$ decreases $\mathrm{LOCMIN}\left(\mathbf{x},f,\mathbf{e},O\right)$ apply local optimizer using the set of options $O$ using the solution $\left(\mathbf{x},f,\mathbf{e}\right)$ as the starting point, if used (not default) monmod monitor progress and possibly modify ${\mathbf{x}}_{j}$ BOUNDARY apply required behaviour for ${\mathbf{x}}_{j}$ outside bounding box, (see ${\mathbf{Boundary}}$) new ($\stackrel{~}{f}$) true if $\stackrel{~}{\mathbf{x}}$, $\stackrel{~}{\mathbf{c}}$, $\stackrel{~}{f}$ were updated at this iteration Additionally a repulsion phase can be introduced by changing from the default values of options ${\mathbf{Repulsion Finalize}}$ (${r}_{f}$), ${\mathbf{Repulsion Initialize}}$ (${r}_{i}$) and ${\mathbf{Repulsion Particles}}$ (${r}_{p}$). If the number of static particles is denoted ${n}_{s}$ then the following can be inserted after the new($\stackrel{~}{f}$) check in the pseudo-code above. $else if ( ns ≥ rp and ri ≤ Is ≤ ri + rf ) LOCMINx~,f~,e~,Oi use -Cg instead of Cg in velocity updates if Is = ri + rf Is = 0$ ## 11 Optional Arguments This section can be skipped if you wish to use the default values for all optional arguments, otherwise, the following is a list of the optional arguments available and a full description of each optional argument is provided in Section 11.1. ### 11.1 Description of the Optional Arguments For each option, we give a summary line, a description of the optional argument and details of constraints. The summary line contains: • the keywords; • a parameter value, where the letters $a$, $i\text{ and }r$ denote options that take character, integer and real values respectively; • the default value, where the symbol $\epsilon $ is a generic notation for machine precision (see nag_machine_precision (X02AJC)), and $\mathit{Imax}$ represents the largest representable integer value (see nag_max_integer (X02BBC)). All options accept the value ‘DEFAULT’ in order to return single options to their default states. Keywords and character values are case insensitive, however they must be separated by at least one space. For nag_glopt_nlp_pso (e05sbc) the maximum length of the argument cvalue used by nag_glopt_opt_get (e05zlc) is $15$. Advance Cognitive $r$ Default$\text{}=2.0$ The cognitive advance coefficient, ${C}_{s}$. When larger than the global advance coefficient, this will cause particles to be attracted toward their previous best positions. Setting $r=0.0$ will cause nag_glopt_nlp_pso (e05sbc) to act predominantly as a local optimizer. Setting $r>2.0$ may cause the swarm to diverge, and is generally inadvisable. At least one of the global and cognitive coefficients must be nonzero. Advance Global $r$ Default$\text{}=2.0$ The global advance coefficient, ${C}_{g}$. When larger than the cognitive coefficient this will encourage convergence toward the best solution yet found. Values $r\in \left(0,1\right)$ will inhibit particles overshooting the optimum. Values $r\in \left[1,2\right)$ cause particles to fly over the optimum some of the time. Larger values can prohibit convergence. Setting $r=0.0$ will remove any attraction to the current optimum, effectively generating a Monte–Carlo multi-start optimization algorithm. At least one of the global and cognitive coefficients must be nonzero. Boundary $a$ Default$\text{}=\text{FLOATING}$ Determines the behaviour if particles leave the domain described by the box bounds. This only affects the general PSO algorithm, and will not pass down to any NAG local minimizers chosen. This option is only effective in those dimensions for which ${\mathbf{bl}}\left[i-1\right]\ne {\mathbf{bu}}\left[i-1\right]$, $i=1,2,\dots ,{\mathbf{ndim}}$. IGNORE The box bounds are ignored. The objective function is still evaluated at the new particle position. RESET The particle is re-initialized inside the domain. ${\stackrel{^}{\mathbf{x}}}_{j}$, ${\stackrel{^}{f}}_{j}$ and ${\stackrel{^}{\mathbf{e}}}_{j}$ are not affected. FLOATING The particle position remains the same, however the objective function will not be evaluated at the next iteration. The particle will probably be advected back into the domain at the next advance due to attraction by the cognitive and global memory. HYPERSPHERICAL The box bounds are wrapped around an $\mathit{ndim}$-dimensional hypersphere. As such a particle leaving through a lower bound will immediately re-enter through the corresponding upper bound and vice versa. The standard distance between particles is also modified accordingly. FIXED The particle rests on the boundary, with the corresponding dimensional velocity set to $0.0$. Constraint Norm $a$ Default$\text{}=\mathrm{L1}$ Determines with respect to which norm the constraint residuals should be constructed. These are automatically scaled with respect to ncon as stated. For the set of (scaled) violations $\mathbf{e}$, these may be, L1 The ${L}^{1}$ norm will be used, ${‖\mathbf{e}‖}_{1}=\frac{1}{{\mathbf{ncon}}}\sum _{1}^{{\mathbf{ncon}}}\left\|{e}_{k}\right\|$ L2 The ${L}^{2}$ norm will be used, ${‖\mathbf{e}‖}_{2}=\frac{1}{{\mathbf{ncon}}}\sqrt{\sum _{1}^{{\mathbf{ncon}}}{e}_{k}^{2}}$ L2SQ The square of the ${L}^{2}$ norm will be used, ${‖\mathbf{e}‖}_{{2}^{2}}=\frac{1}{{\mathbf{ncon}}}\sum _{1}^{{\mathbf{ncon}}}{e}_{k}^{2}$ LMAX The ${L}^{\infty }$ norm will be used, ${‖\mathbf{e}‖}_{\infty }=\underset{0<k\le {\mathbf{ncon}}}{\mathrm{max}}\phantom{\rule{0.25em}{0ex}}\left(\left\|{e}_{k}\right\|\right)$ Constraint Scale Maximum $r$ Default$\text{}=\text{1.0e6}$ Internally, each constraint violation is scaled with respect to the maximum violation yet achieved for that constraint. This option acts as a ceiling for this scale. Constraint: $r>1.0$. Constraint Scaling $a$ Default$\text{}=\mathrm{INITIAL}$ Determines whether to scale the constraints and objective function when constructing the penalty function. OFF Neither the constraint violations nor the objective will be scaled automatically. This should only be used if the constraints and objective are similarly scaled everywhere throughout the domain. INITIAL The maximum of the initial cognitive memories, ${\stackrel{^}{f}}_{j}$ and ${\stackrel{^}{\mathbf{e}}}_{j}$, will be used to scale the objective function and constraint violations respectively. ADAPTIVE Initially, the maximum of the initial cognitive memories, ${\stackrel{^}{f}}_{j}$ and ${\stackrel{^}{\mathbf{e}}}_{j}$, will be used to scale the objective function and constraint violations respectively. If a significant change is detected in the behaviour of the constraints or the objective, these will be rescaled with respect to the current state of the cognitive memory. Constraint Superiority $r$ Default$\text{}=0.01$ The minimum scaled improvement in the constraint violation for a location to be immediately superior to that in memory, regardless of the objective value. Constraint: $r>0.0$. Constraint Tolerance $r$ Default$\text{}={10}^{-4}$ The maximum scaled violation of the constraints for which a sample particle is considered comparable to the current global optimum. Should this not be exceeded, then the current global optimum will be updated if the value of the objective function of the sample particle is superior. Constraint Warning $a$ Default$\text{}=\mathrm{ON}$ Activates or deactivates the error exit associated with the inability to completely satisfy all constraints, NW_NOT_FEASIBLE. It is advisable to deactivate this option if the exit NW_NOT_FEASIBLE is preferred in such cases. OFF NW_NOT_FEASIBLE will not be returned. ON NW_NOT_FEASIBLE will be returned if any constraints are sufficiently violated at the end of the simulation. Distance Scaling $a$ Default$\text{}=\mathrm{ON}$ Determines whether distances should be scaled by box widths. ON When a distance is calculated between $\mathbf{x}$ and $\mathbf{y}$, a scaled ${L}^{2}$ norm is used. $L2 x,y = ∑ i \| ui ≠ ℓi , i≤ndim xi - yi ui - ℓi 2 1 2 .$ OFF Distances are calculated as the standard ${L}^{2}$ norm without any rescaling. $L2 x,y = ∑ i=1 ndim xi - yi 2 1 2 .$ Distance Tolerance $r$ Default$\text{}={10}^{-4}$ This is the distance, $\mathit{dtol}$ between particles and the global optimum which must be reached for the particle to be considered converged, i.e., that any subsequent movement of such a particle cannot significantly alter the global optimum. Once achieved the particle is reset into the box bounds to continue searching. Constraint: $r>0.0$. Function Precision $r$ Default$\text{}={\epsilon }^{0.9}$ The argument defines ${\epsilon }_{r}$, which is intended to be a measure of the accuracy with which the problem function $F\left(\mathbf{x}\right)$ can be computed. If $r<\epsilon $ or $r\ge 1$, the default value is used. The value of ${\epsilon }_{r}$ should reflect the relative precision of $1+\left\|F\left(\mathbf{x}\right)\right\|$; i.e., ${\epsilon }_{r}$ acts as a relative precision when $\left\|F\right\|$ is large, and as an absolute precision when $\left\|F\right\|$ is small. For example, if $F\left(\mathbf{x}\right)$ is typically of order $1000$ and the first six significant digits are known to be correct, an appropriate value for ${\epsilon }_{r}$ would be ${10}^{-6}$. In contrast, if $F\left(\mathbf{x}\right)$ is typically of order ${10}^{-4}$ and the first six significant digits are known to be correct, an appropriate value for ${\epsilon }_{r}$ would be ${10}^{-10}$. The choice of ${\epsilon }_{r}$ can be quite complicated for badly scaled problems; see Chapter 8 of Gill et al. (1981) for a discussion of scaling techniques. The default value is appropriate for most simple functions that are computed with full accuracy. However when the accuracy of the computed function values is known to be significantly worse than full precision, the value of ${\epsilon }_{r}$ should be large enough so that no attempt will be made to distinguish between function values that differ by less than the error inherent in the calculation. Local Boundary Restriction $r$ Default$\text{}=0.5$ Contracts the box boundaries used by a box constrained local minimizer to, $\left[{\beta }_{l},{\beta }_{u}\right]$, containing the start point $x$, where $∂i = r × ui - ℓi βli = maxℓi, xi - ∂i2 βui = minui, xi + ∂i2 , i=1,…,ndim .$ Smaller values of $r$ thereby restrict the size of the domain exposed to the local minimizer, possibly reducing the amount of work done by the local minimizer. Constraint: $0.0\le r\le 1.0$. Local Interior Iterations ${i}_{1}$ Local Interior Major Iterations ${i}_{1}$ Local Exterior Iterations ${i}_{2}$ Local Exterior Major Iterations ${i}_{2}$ The maximum number of iterations or function evaluations the chosen local minimizer will perform inside (outside) the main loop if applicable. For the NAG minimizers these correspond to: \| \| \| \| \| \|-------------------------------\|------------------------------------------------------------------------------\|------------------------------------------------------------------------------\|------------------------------------------------------------------------------\| \| Minimizer \| Argument/option \| Default Interior \| Default Exterior \| \| nag_opt_simplex_easy (e04cbc) \| maxcal \| ${\mathbf{ndim}}+10$ \| $2×{\mathbf{ndim}}+15$ \| \| nag_opt_conj_grad (e04dgc) \| ${\mathbf{max_iter}}$ \| $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(30,3×{\mathbf{ndim}}\right)$ \| $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(50,5×{\mathbf{ndim}}\right)$ \| \| nag_opt_nlp (e04ucc) \| ${\mathbf{max_iter}}$ \| $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(10,2×{\mathbf{ndim}}\right)$ \| $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(30,3×{\mathbf{ndim}}\right)$ \| Unless set, these are functions of the arguments passed to nag_glopt_nlp_pso (e05sbc). Setting $i=0$ will disable the local minimizer in the corresponding algorithmic region. For example, setting ${\mathbf{Local Interior Iterations}}=0$ and ${\mathbf{Local Exterior Iterations}}=30$ will cause the algorithm to perform no local minimizations inside the main loop of the algorithm, and a local minimization with upto $30$ iterations after the main loop has been exited. Constraint: ${i}_{1}\ge 0$, ${i}_{2}\ge 0$. Local Interior Tolerance ${r}_{1}$ Default$\text{}={10}^{-4}$ Local Exterior Tolerance ${r}_{2}$ Default$\text{}={10}^{-4}$ This is the tolerance provided to a local minimizer in the interior (exterior) of the main loop of the algorithm. Constraint: ${r}_{1}>0.0$, ${r}_{2}>0.0$. Local Interior Minor Iterations ${i}_{1}$ Local Exterior Minor Iterations ${i}_{2}$ Where applicable, the secondary number of iterations the chosen local minimizer will use inside (outside) the main loop. Currently the relevant default values are: \| \| \| \| \| \|----------------------\|------------------------------------------------------------------------------\|------------------------------------------------------------------------------\|------------------------------------------------------------------------------\| \| Minimizer \| Argument/option \| Default Interior \| Default Exterior \| \| nag_opt_nlp (e04ucc) \| ${\mathbf{minor_max_iter}}$ \| $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(10,2×{\mathbf{ndim}}\right)$ \| $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(30,3×{\mathbf{ndim}}\right)$ \| Constraint: ${i}_{1}\ge 0$, ${i}_{2}\ge 0$. Local Minimizer $a$ Default$\text{}=\mathrm{OFF}$ Allows for a choice of Chapter e04 functions to be used as a coupled, dedicated local minimizer. OFF No local minimization will be performed in either the INTERIOR or EXTERIOR sections of the algorithm. e04cbc Use nag_opt_simplex_easy (e04cbc) as the local minimizer. This does not require the calculation of derivatives. On a call to objfun during a local minimization, ${\mathbf{mode}}=5$. e04dgc Use nag_opt_conj_grad (e04dgc) as the local minimizer. Accurate derivatives must be provided, and will not be approximated internally. Additionally, each call to objfun during a local minimization will require either the objective to be evaluated alone, or both the objective and its gradient to be evaluated. Hence on a call to objfun, ${\mathbf{mode}}=5$ or $7$. e04ucc Use nag_opt_nlp (e04ucc) as the local minimizer. This operates such that any derivatives of either the objective function or the constraint Jacobian, which you cannot supply, will be approximated internally using finite differences. Either, the objective, objective gradient, or both may be requested during a local minimization, and as such on a call to objfun, ${\mathbf{mode}}=1$, $2$ or $5$. The box bounds forwarded to this function from nag_glopt_nlp_pso (e05sbc) will have been acted upon by ${\mathbf{Local Boundary Restriction}}$. As such, the domain exposed may be greatly smaller than that provided to nag_glopt_nlp_pso (e05sbc). Maximum Function Evaluations $i$ Default $=\mathit{Imax}$ The maximum number of evaluations of the objective function. When reached this will return NW_SOLUTION_NOT_GUARANTEED and ${\mathbf{inform}}=6$. Constraint: $i>0$. Maximum Iterations Completed $i$ Default$\text{}=1000×{\mathbf{ndim}}$ The maximum number of complete iterations that may be performed. Once exceeded nag_glopt_nlp_pso (e05sbc) will exit with NW_SOLUTION_NOT_GUARANTEED and ${\mathbf{inform}}=5$. Unless set, this adapts to the parameters passed to nag_glopt_nlp_pso (e05sbc). Constraint: $i\ge 1$. Maximum Iterations Static $i$ Default$\text{}=100$ The maximum number of iterations without any improvement to the current global optimum. If exceeded nag_glopt_nlp_pso (e05sbc) will exit with NW_SOLUTION_NOT_GUARANTEED and ${\mathbf{inform}}=4$. This exit will be hindered by setting ${\mathbf{Maximum Iterations Static Particles}}$ to larger values. Constraint: $i\ge 1$. Maximum Iterations Static Particles $i$ Default$\text{}=0$ The minimum number of particles that must have converged to the current optimum before the function may exit due to ${\mathbf{Maximum Iterations Static}}$ with NW_SOLUTION_NOT_GUARANTEED and ${\mathbf{inform}}=4$. Constraint: $i\ge 0$. Maximum Particles Converged $i$ Default $=\mathit{Imax}$ The maximum number of particles that may converge to the current optimum. When achieved, nag_glopt_nlp_pso (e05sbc) will exit with NW_SOLUTION_NOT_GUARANTEED and ${\mathbf{inform}}=3$. This exit will be hindered by setting ‘Repulsion’ options, as these cause the swarm to re-expand. Constraint: $i>0$. Maximum Particles Reset $i$ Default $=\mathit{Imax}$ The maximum number of particles that may be reset after converging to the current optimum. Once achieved no further particles will be reset, and any particles within ${\mathbf{Distance Tolerance}}$ of the global optimum will continue to evolve as normal. Constraint: $i>0$. Maximum Variable Velocity $r$ Default$\text{}=0.25$ Along any dimension $j$, the absolute velocity is bounded above by $\left\|{\mathbf{v}}_{j}\right\|\le r×\left({\mathbf{u}}_{j}-{\mathbf{\ell }}_{j}\right)={\mathbf{V}}_{\mathrm{max}}$. Very low values will greatly increase convergence time. There is no upper limit, although larger values will allow more particles to be advected out of the box bounds, and values greater than $4.0$ may cause significant and potentially unrecoverable swarm divergence. Constraint: $r>0.0$. Objective Scale $r$ Default$\text{}=1.0$ The initial scale for the objective function. This will remain fixed if ${\mathbf{Objective Scaling}}=\mathrm{USER}$ is selected. Objective Scaling $a$ Default$\text{}=\mathrm{MAXIMUM}$ The method of (re)scaling applied to the objective function when the function detects a significant difference between the scale and the global and cognitive memory ($\stackrel{~}{f}$ and ${\stackrel{^}{f}}_{j}$). This only has an effect when ${\mathbf{ncon}}>0$ and ${\mathbf{Constraint Scaling}}$ is active. MAXIMUM The objective is rescaled with respect to the maximum absolute value of the objective in the cognitive and global memory. MEAN The objective is rescaled with respect to the mean absolute value of the objective in the cognitive and global memory. USER The scale remains fixed at the value set using ${\mathbf{Objective Scale}}$. Optimize $a$ Default$\text{}=\mathrm{MINIMIZE}$ Determines whether to maximize or minimize the objective function, or ignore the objective and search for a constrained point. MINIMIZE The objective function will be minimized. MAXIMIZE The objective function will be maximized. This is accomplished by minimizing the negative of the objective. CONSTRAINTS The objective function will be ignored, and the algorithm will attempt to find a feasible point given the provided constraints. The objective function will be evaluated at the best point found with regards to constraint violations, and the final positions returned in xbest. The objective will be calculated at the best point found in terms of constraints only. Should a constrained point be found, nag_glopt_nlp_pso (e05sbc) will exit with NE_NOERROR and ${\mathbf{inform}}=6$. Constraint: if ${\mathbf{Optimize}}=\mathrm{CONSTRAINTS}$, ${\mathbf{ncon}}>0$ is required. Repeatability $a$ Default$\text{}=\mathrm{OFF}$ Allows for the same random number generator seed to be used for every call to nag_glopt_nlp_pso (e05sbc). ${\mathbf{Repeatability}}=\mathrm{OFF}$ is recommended in general. OFF The internal generation of random numbers will be nonrepeatable. ON The same seed will be used. Repulsion Finalize $i$ Default $=\mathit{Imax}$ The number of iterations performed in a repulsive phase before re-contraction. This allows a re-diversified swarm to contract back toward the current optimum, allowing for a finer search of the near optimum space. Constraint: $i\ge 2$. Repulsion Initialize $i$ Default $=\mathit{Imax}$ The number of iterations without any improvement to the global optimum before the algorithm begins a repulsive phase. This phase allows the particle swarm to re-expand away from the current optimum, allowing more of the domain to be investigated. The repulsive phase is automatically ended if a superior optimum is found. Constraint: $i\ge 2$. Repulsion Particles $i$ Default$\text{}=0$ The number of particles required to have converged to the current optimum before any repulsive phase may be initialized. This will prevent repulsion before a satisfactory search of the near optimum area has been performed, which may happen for large dimensional problems. Constraint: $i\ge 0$. Start $a$ Default$\text{}=\mathrm{COLD}$ Used to affect the initialization of the function. COLD The random number generators and all initialization data will be generated internally. The variables xbest, fbest and cbest need not be set. WARM You must supply the initial best location, function and constraint violation values xbest, fbest and cbest. This option is recommended if you already have a data set you wish to improve upon. Swarm Standard Deviation $r$ Default$\text{}=0.1$ The target standard deviation of the particle distances from the current optimum. Once the standard deviation is below this level, nag_glopt_nlp_pso (e05sbc) will exit with NW_SOLUTION_NOT_GUARANTEED and ${\mathbf{inform}}=2$. This criterion will be penalized by the use of ‘Repulsion’ options, as these cause the swarm to re-expand, increasing the standard deviation of the particle distances from the best point. Constraint: $r\ge 0.0$. Target Objective $a$ Default$\text{}=\mathrm{OFF}$ Target Objective Value $r$ Default$\text{}=0.0$ Activate or deactivate the use of a target value as a finalization criterion. If active, then once the supplied target value for the objective function is found (beyond the first iteration if ${\mathbf{Target Warning}}$ is active) nag_glopt_nlp_pso (e05sbc) will exit with NE_NOERROR and ${\mathbf{inform}}=1$. Other than checking for feasibility only (${\mathbf{Optimize}}=\mathrm{CONSTRAINTS}$), this is the only finalization criterion that guarantees that the algorithm has been successful. If the target value was achieved at the initialization phase or first iteration and ${\mathbf{Target Warning}}$ is active, nag_glopt_nlp_pso (e05sbc) will exit with NW_FAST_SOLUTION. This option may take any real value $r$, or the character ON/OFF as well as DEFAULT. If this option is queried using nag_glopt_opt_get (e05zlc), the current value of $r$ will be returned in rvalue, and cvalue will indicate whether this option is ON or OFF. The behaviour of the option is as follows: $r$ Once a point is found with an objective value within the ${\mathbf{Target Objective Tolerance}}$ of $r$, nag_glopt_nlp_pso (e05sbc) will exit successfully with NE_NOERROR and ${\mathbf{inform}}=1$. OFF The current value of $r$ will remain stored, however it will not be used as a finalization criterion. ON The current value of $r$ stored will be used as a finalization criterion. DEFAULT The stored value of $r$ will be reset to its default value ($0.0$), and this finalization criterion will be deactivated. Target Objective Safeguard $r$ Default$\text{}=10.0\epsilon $ If the magnitude of ${\mathbf{Target Objective Value}}$ is below $r$, then the absolute error between the optimum and the target will be used. Otherwise the relative error will be used. If a value entered is below $100.0\epsilon $, the default value will be used. Target Objective Tolerance $r$ Default$\text{}=0.0$ The optional tolerance to a user-specified target value. Constraint: $r\ge 0.0$. Target Warning $a$ Default$\text{}=\mathrm{OFF}$ Activates or deactivates the error exit associated with the target value being achieved before entry into the main loop of the algorithm, NW_FAST_SOLUTION. OFF No error will be returned, and the function will exit normally. ON An error will be returned if the target objective is reached prematurely, and the function will exit with NW_FAST_SOLUTION. Verify Gradients $a$ Default$\text{}=\mathrm{ON}$ Adjusts the level of gradient checking performed when gradients are required. Gradient checks are only performed on the first call to the chosen local minimizer if it requires gradients. There is no guarantee that the gradient check will be correct, as the finite differences used in the gradient check are themselves subject to inaccuracies. OFF No gradient checking will be performed. ON A cheap gradient check will be performed on both the gradients corresponding to the objective through objfun and those provided via the constraint Jacobian through confun. OBJECTIVE A more expensive gradient check will be performed on the gradients corresponding to the objective objfun. The gradients of the constraints will not be checked. CONSTRAINTS A more expensive check will be performed on the elements of cjac provided via confun. The objective gradient will not be checked. FULL A more expensive check will be performed on both the gradient of the objective and the constraint Jacobian. Weight Decrease $a$ Default$\text{}=\mathrm{INTEREST}$ Determines how particle weights decrease. OFF Weights do not decrease. INTEREST Weights decrease through compound interest as ${w}_{\mathit{IT}+1}={w}_{\mathit{IT}}\left(1-{W}_{\mathit{val}}\right)$, where ${W}_{\mathit{val}}$ is the ${\mathbf{Weight Value}}$ and $\mathit{IT}$ is the current number of iterations. LINEAR Weights decrease linearly following ${w}_{\mathit{IT}+1}={w}_{\mathit{IT}}-\mathit{IT}×\left({W}_{\mathit{max}}-{W}_{\mathit{min}}\right)/{\mathit{IT}}_{\mathit{max}}$, where $\mathit{IT}$ is the iteration number and ${\mathit{IT}}_{\mathit{max}}$ is the maximum number of iterations as set by ${\mathbf{Maximum Iterations Completed}}$. Weight Initial $r$ Default$\text{}={W}_{\mathit{max}}$ The initial value of any particle's inertial weight, ${W}_{\mathit{ini}}$, or the minimum possible initial value if initial weights are randomized. When set, this will override ${\mathbf{Weight Initialize}}=\mathrm{RANDOMIZED}$ or $\mathrm{MAXIMUM}$, and as such these must be set afterwards if so desired. Constraint: ${W}_{\mathit{min}}\le r\le {W}_{\mathit{max}}$. Weight Initialize $a$ Default$\text{}=\mathrm{MAXIMUM}$ Determines how the initial weights are distributed. INITIAL All weights are initialized at the initial weight, ${W}_{\mathit{ini}}$, if set. If ${\mathbf{Weight Initial}}$ has not been set, this will be the maximum weight, ${W}_{\mathit{max}}$. MAXIMUM All weights are initialized at the maximum weight, ${W}_{\mathit{max}}$. RANDOMIZED Weights are uniformly distributed in $\left({W}_{\mathit{min}},{W}_{\mathit{max}}\right)$ or $\left({W}_{\mathit{ini}},{W}_{\mathit{max}}\right)$ if ${\mathbf{Weight Initial}}$ has been set. Weight Maximum $r$ Default$\text{}=1.0$ The maximum particle weight, ${W}_{\mathit{max}}$. Constraint: $1.0\ge r\ge {W}_{\mathit{min}}$ (If ${W}_{\mathit{ini}}$ has been set then $1.0\ge r\ge {W}_{\mathit{ini}}$.) Weight Minimum $r$ Default$\text{}=0.1$ The minimum achievable weight of any particle, ${W}_{\mathit{min}}$. Once achieved, no further weight reduction is possible. Constraint: $0.0\le r\le {W}_{\mathit{max}}$ (If ${W}_{\mathit{ini}}$ has been set then $0.0\le r\le {W}_{\mathit{ini}}$.) Weight Reset $a$ Default$\text{}=\mathrm{MAXIMUM}$ Determines how particle weights are re-initialized. INITIAL Weights are re-initialized at the initial weight if set. If ${\mathbf{Weight Initial}}$ has not been set, this will be the maximum weight. MAXIMUM Weights are re-initialized at the maximum weight. RANDOMIZED Weights are uniformly distributed in $\left({W}_{\mathit{min}},{W}_{\mathit{max}}\right)$ or $\left({W}_{\mathit{ini}},{W}_{\mathit{max}}\right)$ if ${\mathbf{Weight Initial}}$ has been set. Weight Value $r$ Default$\text{}=0.01$ The constant ${W}_{\mathit{val}}$ used with ${\mathbf{Weight Decrease}}=\mathrm{INTEREST}$. Constraint: $0.0\le r\le \frac{1}{3}$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 625, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.7694838047027588, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/127945-abstract-several-questions-print.html	# Abstract-several questions Printable View • February 9th 2010, 01:33 AM WannaBe Abstract-several questions Well, I'm trying to solve some old exams in abstract algebra, and I need your help in the following: Question 1: Let $U(Z/15Z)$ be the multiplicative group of the invertible elemenets in $Z/15Z$ . Write $U(Z/15Z)$ as a product of cyclic groups. My try: It's easy to show that x is in U iff gcd(x,15)=1... Hence: U={1,2,8,4,7,13,11,14} ... If we'll take the element 2 to be a generator we'll get the cyclic group {1,2,4,8} ... If we'll take <7> we'll get {1,7,4,13} etc... I can't figure out how to write U as a product of cyclic groups when the intersection of each two cyclic subgroups of U is "bigger" than {1}... Question 2: Prove that $Z/nZ$ has only one maximal ideal (which isn't trivial) iff n is a power of a prime number. Question 3: Find a 3-sylow subgroup of $S_{8}$ and find a group that is isomorphic to it. My try: We know $o(S_{8}) = 8!$ . Hence, a 3-sylow subgroup H is from order $o(H)= 3^{2}=9$. If we'll be able to find an element in $S_{8}$ from that order- we're done...But is there any element of that order? How should I solve this one? Thanks a lot! • February 9th 2010, 11:42 PM aliceinwonderland Quote: Originally Posted by WannaBe Well, I'm trying to solve some old exams in abstract algebra, and I need your help in the following: Question 1: Let $U(Z/15Z)$ be the multiplicative group of the invertible elemenets in $Z/15Z$ . Write $U(Z/15Z)$ as a product of cyclic groups. My try: It's easy to show that x is in U iff gcd(x,15)=1... Hence: U={1,2,8,4,7,13,11,14} ... If we'll take the element 2 to be a generator we'll get the cyclic group {1,2,4,8} ... If we'll take <7> we'll get {1,7,4,13} etc... I can't figure out how to write U as a product of cyclic groups when the intersection of each two cyclic subgroups of U is "bigger" than {1}... If $\mathbb{Z}/m\mathbb{Z} \cong \mathbb{Z}/m_1\mathbb{Z} \oplus \mathbb{Z}/m_2\mathbb{Z} \oplus \cdots \oplus \mathbb{Z}/m_i\mathbb{Z}$, then $U(\mathbb{Z}/m\mathbb{Z}) \cong U(\mathbb{Z}/m_1\mathbb{Z}) \times U(\mathbb{Z}/m_2\mathbb{Z}) \times \cdots \times U(\mathbb{Z}/m_i\mathbb{Z})$. We see that if each $m_k$ is a prime number, then each $U(\mathbb{Z}/m_k\mathbb{Z})$ is a cyclic group. Thus, $U(\mathbb{Z}/15\mathbb{Z}) \cong U(\mathbb{Z}/3\mathbb{Z}) \times U(\mathbb{Z}/5\mathbb{Z})$. Quote: Question 2: Prove that $Z/nZ$ has only one maximal ideal (which isn't trivial) iff n is a power of a prime number. If n is a power of prime number, then $\mathbb{Z}/n = \mathbb{Z}/p^i\mathbb{Z}$. The maximal ideal of $\mathbb{Z}/p^i\mathbb{Z}$ is $\mathbb{Z}/p^{i-1}\mathbb{Z}$ for i>=2, which is unique for $n=p^i$. Quote: Question 3: Find a 3-sylow subgroup of $S_{8}$ and find a group that is isomorphic to it. My try: We know $o(S_{8}) = 8!$ . Hence, a 3-sylow subgroup H is from order $o(H)= 3^{2}=9$. If we'll be able to find an element in $S_{8}$ from that order- we're done...But is there any element of that order? How should I solve this one? Thanks a lot! H={e, (1,2,3), (1,3,2), (4,5,6), (4,6,5), (1,2,3)(4,5,6), (1,2,3)(4,6,5), (1,3,2)(4,5,6), (1,3,2)(4,6,5)}. Other subgroups of order 9 in S_8 can be found similarly. • February 10th 2010, 12:39 AM WannaBe Thanks a lot man! • February 10th 2010, 12:47 AM aliceinwonderland Quote: Originally Posted by WannaBe Thanks a lot man! Please check a modification of Q3. • February 10th 2010, 01:10 AM WannaBe Quote: Originally Posted by aliceinwonderland Please check a modification of Q3. modification of Q3? • February 10th 2010, 01:27 AM aliceinwonderland Quote: Originally Posted by WannaBe modification of Q3? Yeah, I had edited my answer for Q3. Even though 9 \| 8! , I was not able to find an order 9 subgroup in S_8. Sorry for any confusion. My English is no good. • February 10th 2010, 01:53 AM WannaBe Oh , you meant Question 3 LOL ... I wasn't sure about it.... Yep, I saw your "modification"... But we know that there are 2 groups of order 9 up to isomorphism: C_3 x C_3 and C_9... We don't have an element of order 9 in S8 hence the only subgroup of order 9 in S8 is C_3xC_3 ... (I think so...) Thanks a lot anyway ! All times are GMT -8. The time now is 02:57 AM.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 25, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9331120252609253, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/62312/is-every-p-point-ultrafilter-ramsey/62342	## Is every p-point ultrafilter Ramsey? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) A non-principal ultrafilter $\mathcal{U}$ on $\omega$ is a p-point (or weakly selective) iff for every partition $\omega = \bigsqcup _{n < \omega} Z_n$ into null sets, i.e each $Z_n \not \in \mathcal{U}$, there exists a measure one set $S \in \mathcal{U}$ such that $S \cap Z_n$ is finite for each $n$. A non-principal ultrafilter $\mathcal{U}$ on $\omega$ is Ramsey (or selective) iff for every partition as above, there exists a measure one set $S$ such that $\|S \cap Z_n\| = 1$ for each $n$. Clearly, every Ramsey ultrafilter is a p-point. What is known about the converse? I couldn't find anything, not even a consistency result, in any searches I've done or sources I've checked. Is very little known/published about the converse? - ## 4 Answers Amit: The converse is not true, not even under MA. This is a result of Kunen, and the paper you want to look at is "Some points in $\beta{\mathbb N}$", Math. Proc. Cambridge Philos. Soc. 80 (1976), no. 3, 385–398. There is a related notion, called $q$-point. These are ultrafilters such that any finite-to-one $f:\omega\to\omega$ is injective on a set in the ultrafilter. A Ramsey ultrafilter is one that is simultaneously a $p$-point, and a $q$-point. Miller proved ("There are no $Q$-points in Laver's model for the Borel conjecture", Proc. Amer. Math. Soc. 78 (1980), no. 1, 103–106) that it is consistent that there are no $q$-points. The consistency of the non-existence of $p$-points is significantly harder, and due to Shelah (see for example Chapter VI of his "Proper and improper forcing"). There is a fairly extensive literature on related results. You may want to start by looking at Blass' article in the Handbook of Set Theory, "Combinatorial Cardinal Characteristics of the Continuum". - Fantastic, thanks! – Amit Kumar Gupta Apr 19 2011 at 18:58 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. A few addenda to Andres Caicedo's answer: It was proved around 1970 by several people (Adrian Mathias was one of them) that the continuum hypothesis (CH) implies the existence of P-points that are not selective. (CH also implies the existence of selective ultrafilters and the existence of Q-points that are not selective. ZFC alone suffices to prove the existence of ultrafilters that are neither P-points nor Q-points.) The more difficult task of producing a model of set theory in which P-points exist but selective ultrafilters don't was achieved in Kunen's paper cited by Andres. It is a famous open problem whether there are models of set theory in which neither P-points nor Q-points exist; it is known that in such a model the cardinal of the continuum must be at least $\aleph_3$. (In contrast, there are models with `$2^{\aleph_0}=\aleph_2$` with no P-points and others with no Q-points.) - Another small addendum to Andres's and Andreas's answers. It is also consistent that the answer to your question is yes. Shelah has constructed a model of ZFC in which there exists (up to isomorphism) exactly one p-point -- and that p-point is, in fact, selective. This construction is Section XVIII.4 in Shelah, Proper and Improper Forcing . - Oh, yes. If there is a unique $p$-point, it follows that it must be Ramsey. By the way, the consistency of this uniqueness result is significantly more elaborate than the consistency of "there is a unique Ramsey ultrafilter" (also due to Shelah). – Andres Caicedo Apr 20 2011 at 1:53 Andres, is there an easy argument why a unique p-point must be selective? Also, it could be interesting to the OP to point out that you can pick any Ramsey ultrafilter in a CH model and have it survive to become the unique one in an extension. But then again, selective ultrafilters all look the same (in the sense of what Andreas calls 'complete combinatorics'). I would agree that the model is more involved though I have heard good arguments that this is mostly due to the presentation. – Peter Krautzberger Apr 20 2011 at 3:59 3 Peter, isn't it the case that any non-principal ultrafilter Rudin-Keisler below a p-point must itself be a p-point? In which case, if you have only one p-point, it must be minimal under the Rudin-Keisler ordering, and hence Ramsey. – Tanmay Inamdar Apr 20 2011 at 13:36 Doh! Thanks Tanmay! I shouldn't ask questions after midnight... – Peter Krautzberger Apr 20 2011 at 18:20 I just looked at Shelah's construction and I think it's worth pointing out that Shelah really needs (or at least thinks he needs) a Ramsey ultrafilter to start with, i.e., the preservation lemma requires the "surviving" P-point to be selective. – Peter Krautzberger Apr 26 2011 at 21:38 show 1 more comment Another small and slightly trivial addendum: If there are no p-points, then every p-point is a Ramsey ultrafilter. (Duh!) As Andreas Blass remarked above, this situation is consistent, which is easier to prove than the consistency of a unique p-point. ("It is usually significantly harder to prove there is a unique object than to prove there is none". See Shelah's Proper and improper forcing VI.5) -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 22, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9439817070960999, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/94634/how-many-smooth-functions-are-non-analytic	# How many smooth functions are non-analytic? We know from example that not all smooth (infinitely differentiable) functions are analytic (equal to their Taylor expansion at all points). However, the examples on the linked page seem rather contrived, and most smooth functions that I've encountered in math and physics are analytic. How many smooth functions are not analytic (in terms of measure or cardinality)? In what situations are such functions encountered? Are they ever encountered outside of real analysis (e.g. in physics)? - 2 Note that the cardinality of the set of analytic functions on an open subset is at most countable (since either $\mathbb{R}$ or $\mathbb{C}$ is second countable, and on each sufficiently small neighborhood an analytic function is determined by countable set of coefficients), while the set of smooth functions obviously has larger cardinality. – sos440 Dec 28 '11 at 7:34 3 @sos440: What you say about analytic functions is incorrect. Note that there are uncountably many choices for each coefficient. Also, even if you restrict to a countable selection of coefficients, there can be uncountably many choices for the sequences of coefficients, for example the set of functions $\displaystyle{\sum \frac{a_n}{n!}x^n}$ where $a_n\in\{-1,1\}$. – Jonas Meyer Dec 28 '11 at 7:47 @jonasMeyer : Oh, that's my mistake. What I meant was the dimensional sense, though it se – sos440 Dec 28 '11 at 12:19 1 ...seems still wrong. – sos440 Dec 28 '11 at 12:25 @sos440: Yes, $\{x\mapsto e^{ax}:a\in\mathbb R\}$ is linearly independent. – Jonas Meyer Dec 29 '11 at 2:44 ## 4 Answers It is not difficult to see that the collection of $C^{\infty}$ functions that fail to be analytic at each point is $c$-dense in the space of continuous functions defined on a compact interval (sup metric). To see this, let $f$ be such a continuous function and choose $\epsilon > 0.$ Next, pick any $C^{\infty}$ and nowhere analytic function $\phi$ that is bounded between $-1$ and $1.$ (Take $\frac{2}{\pi}$ times the arctangent of an unbounded example, if an example bounded between $-1$ and $1$ isn't handy.) Let $P$ be a polynomial whose sup-metric distance from $f$ is less than $\frac{\epsilon}{3}$ (Weierstrass's Approximation Theorem). Now let $g = \left(\frac{\epsilon}{3}\right)\phi + P.$ Then, for each of the $c$-many real numbers $\delta$ such that $0 < \delta < \frac{\epsilon}{3},$ the function $g + \delta$ is: (a) $C^{\infty}$, (b) nowhere analytic, (c) belongs to the $\epsilon$-ball centered at $f.$ This last part involves the triangle inequality, and earlier we need the fact that if $\phi$ is $C^{\infty}$ and nowhere analytic, then the composition $\arctan \circ \phi$ is $C^{\infty}$ and nowhere analytic and $\left(\frac{\epsilon}{3}\right)\phi + P + \delta$ is $C^{\infty}$ and nowhere analytic. Note that we can also easily get $c$-many such functions arbitrarily close (sup metric) to any continuous function defined on $\mathbb R$ by appropriately splicing together functions on the intervals $...\; [-2,-1],$ $[-1,0],$ $[0,1],$ $[1,2],\; ...$ Any type of cardinality result is pretty much maxed out by this result, but by considering stronger forms of "largeness" we can do better. The results I know about involve Baire category and the idea of prevalance (complement of a Haar null set), and each implies the $c$-dense result above (and much more). Back in 2002 I posted a couple of lengthy essays in sci.math about $C^{\infty}$ and nowhere analytic functions. For some reason they were never archived by google's sci.math site, but they can be found at the Math Forum sci.math site. One day I might LaTeX these essays for posting in this group, but I doubt I'll have time in the near future. ESSAY ON NOWHERE ANALYTIC C-INFINITY FUNCTIONS [parts 1 and 2; May 9 & 19, 2002] http://mathforum.org/kb/message.jspa?messageID=387148 http://mathforum.org/kb/message.jspa?messageID=387149 - I'm having trouble reading your posts. For example, my browser doesn't recognize Cellérier (which is displayed as CellÃÂÃÂ©rier) and other accented letters. – t.b. Dec 29 '11 at 4:26 – Dave L. Renfro Dec 29 '11 at 21:26 In terms of cardinality, there are the same number of smooth and analytic functions, $2^{\aleph_0}$. The constant functions are enough to see that there are at least $2^{\aleph_0}$ analytic functions. The fact that a continuous function is determined by its values on a dense subspace, along with my presumption that you are referring to smooth functions on a separable space, imply that there are at most $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$ smooth functions. Added: In light of the question edit, I should mention that the cardinality of the set of smooth nonanalytic functions is also $2^{\aleph_0}$. This can be seen by taking the constant multiples of some bump function. I don't know about measures, but analytic functions are a very special subclass of smooth functions (something which I'm sorry to leave vague at the moment, but hopefully someone will give a better answer here (Added: Now Dave L. Renfro has)). They are also important, useful, and relatively easy to work with, which is part of why they are so prevalent in the math and physics you have seen. Where are they encountered? Bump functions are important in differential equations and manifolds, so I would guess they're important in physics. Bump functions are smooth and not analytic. - In response to your first question, it's not hard to see there are continuum many continuous functions, and hence continuum many analytic functions (consider the values of said functions at rational points; also, the constant functions provide at least continuum many. On the other hand, there are 2^c many Lebesgue measurable functions of the reals: If S is any subset of the cantor set, the indicator function of S is Lebesgue measurable, since it has measure 0. There are, however, only continuum many Borel measurable functions. Maybe you meant to ask, "How many smooth functions are not analytic?" I can sort of see a picture proof in my had that in the topology of continuous functions of the reals, non-analytic functions are a dense subset of the set of smooth functions. - What topology are you referring to in your last paragraph? – Jonas Meyer Dec 28 '11 at 7:49 The uniform topology. – Elan B. Dec 28 '11 at 8:08 'Maybe you meant to ask, "How many smooth functions are not analytic?"': Thanks, I did. I edited the OP – tba010 Dec 28 '11 at 9:19 I came upon this quirky example, find smooth $f(x)$ such that $f(f(x)) = \sin x$ with $f(0)=0$ and $f'(0) = 1.$ The fact that this can be solved is quite nontrivial. It turns out that it is analytic between each $(k \pi, (k+1) \pi)$ but only $C^\infty$ at $0-$ and all multiples of $\pi.$ Go figure. The answer is periodic, resembles the sine wave but with slightly larger amplitude. This can be seen since, for $0 < x \leq \frac{\pi}{2},$ the function is between $\sin x$ and $x.$ -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 56, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9428719282150269, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/115533-solved-difference-harmonic-series-1-n-comparsion.html	# Thread: 1. ## [SOLVED] The difference of the harmonic series and 1/n comparsion I recently had this classic example to determine if it converged or diverged [MATH\frac{sin(n)}{n}][/tex] I learned that you can use a comparison test using the squeeze theorem. $\frac{-1}{n}\le\frac{sin(n)}{n}\le\frac{1}{n}$ Then you would take the limit of $\lim_{n\to\infty}\frac{1}{n}$ which was go to 0. Because the function was greater than the original you can say the original converged as well to a limit of 0. However, how would the harmonic series, diverge if you are basically determining the limit of the same thing Edit: Poor question, trying to mark as solved, please disregard, if misused the concept of series and sequence 2. In... http://www.mathhelpforum.com/math-he...eries-4-a.html ... it has been demonstrated that the series... $\sum_{n=1}^{\infty} \frac{\sin n}{n}$ , $\sum_{n=1}^{\infty} \frac{\cos n}{n}$ (1) ... are both convergent and also their sum has been computed. I don't know if it exists an easier way to arrive at the same results... probably it exists ... Kind regards $\chi$ $\sigma$ 3. Thank you very much, it seems to me I am confusing the terms series and sequence perhaps, if there is a real difference	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9513891339302063, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/36017/complete-set-of-observables-in-classical-mechanics?answertab=oldest	# Complete set of observables in classical mechanics I'm reading "Symplectic geometry and geometric quantization" by Matthias Blau and he introduces a complete set of observables for the classical case: The functions $q^k$ and $p_l$ form a complete set of observables in the sense that any function which Poisson commutes (has vanishing Poisson brackets) with all of them is a constant. I wonder why is it so? That is why do we call it a complete set of observables? As I understand it means the functions satisfying the condition above form coordinates on a symplectic manifold, but I don't see how. - – Yrogirg Sep 11 '12 at 12:22 ## 4 Answers Any observable $H$ in classical mechanics defines a flow of states by regarding it as a Hamiltonian. This flow acts on observables $f$ by $df/dt = \{H,f\}$ (this is Hamilton's equation). The idea of a complete set of observables is that it is a set for which any observable with constant flow for all members of the set (ie. Poisson commute with the set) is constant. Intuitively, these flows move all over the phase space, so if $f$ is nonconstant, the flow of $f$ along one of the observables in the complete set can detect this. These functions don't have to form coordinates. EDIT: To complement QMechanic's counterexample, here is a compact counterexample: consider the 2-sphere with its ordinary symplectic form and the functions $\mathrm{cos} 2\theta$, $\mathrm{sin}\phi$, and $cos \phi$, where $\theta$ is the polar angle, and $\phi$ is the azimuthal angle. These are symmetric accross the equator, so they don't distinguish points, but it is pretty clear that they are a complete set. - "I don't think that this means the set of observables can uniquely determine states" The one (but strong) reason I can think of to support this is that the contrary would mean the manifold could be covered with only one map. – Yrogirg Sep 11 '12 at 9:39 @Yrogirg That's a good point. It is of course possible to have a minimal complete set which is larger than the dimension of the phase space. Then that the functions separate points just means they define an embedding into some higher dimensional Euclidean space. For example, the ps and qs are really only defined in their coordinate charts, so to turn them into globally defined functions, we can use a partition of unity so that they are zero outside of their charts. Then it takes several coordinate charts to cover the phase space in general. – user404153 Sep 11 '12 at 17:45 Ok, final clarifications, so you don't see any classical physical meaning in a classical complete set of observables? And in QM $\hat q^2$ and $\hat p$ still form a complete set for a 1D particle? – Yrogirg Sep 12 '12 at 8:14 1 I have tried to give a description free of any mention of operators, Hilbert spaces, or anything else from the quantum picture. And yes, those do form a complete set for a 1D particle. – user404153 Sep 12 '12 at 8:41 I) Here is my interpretation of OP's question(v1). The mentioned quote is from p.11 in Section 2.2 just below eq.(2.22). Section 2.2 is devoted to the case where phase space is a $2n$ dimensional real vector space $V$ with $2n$ global coordinates $$(x^1, \ldots, x^{2n})~=~(q^1, \ldots, q^n; p_1 \ldots, p_n),$$ and canonical Poisson bracket, cf. eq. (2.22). This is a special case of a general symplectic manifold. An observable $f$ is by definition a smooth function on $V$, i.e., $f\in C^{\infty}(V)$. Or in plain words, $f$ is a smooth function of $x^1, \ldots, x^{2n}$. On the other hand, the $2n$ coordinates $x^1, \ldots, x^{2n},$ form a complete set of generators for the algebra $(C^{\infty}(V),+,\cdot)$. Let us assume that the function $f$ Poisson commutes (has vanishing Poisson brackets) with all of the $2n$ variables, $$\forall x\in V \forall I\in\{1,\ldots, 2n\}~:~\{x^I, f(x)\}~=~0.$$ By the definition (2.21) of the canonical Poisson bracket, we deduce that $f$ has vanishing derivatives wrt. all the positions and momenta. Hence $f$ is just a constant function. II) On the other hand, let us imagine that we have $2n$ differential functions $g^1, \ldots, g^{2n},$ such that $$\forall f\in C^{\infty}(V): [ (\forall x\in V\forall I\in\{1,\ldots, 2n\}~:~\{g^I(x), f(x)\}~=~0)~ \Rightarrow ~ f ~\text{is constant} ].$$ OP essentially asks in a comment: Do $g^1, \ldots, g^{2n},$ locally form a coordinate system? By the word locally we mean: Given a fixed point $x_{(0)}\in V$, does there exist a sufficiently small neighborhood of $x_{(0)}$, such that $g^1, \ldots, g^{2n},$ could serve as coordinate functions there? Answer: In general the answer is No, but if we e.g. additionally assume that the Jacobian matrix $$\left(\frac{\partial g^I}{\partial x^J}\right)_{1\leq I,J \leq 2n}$$ is invertible in the fixed point $x_{(0)}$, then the answer is Yes by the inverse function theorem. Counterexample: Let $n=1$, i.e. the phase space $V$ is $2$-dimensional with coordinates $(x^1,x^2)=(q,p)$. Let the fixed point be $x_{(0)}=(0,0)$. Let $g^1(q,p)=q^2$ and $g^2(q,p)=p$. The map $x\mapsto$ $g(x)$ is not invertible in $x_{(0)}=(0,0)$, so that $g^1$ and $g^2$ cannot serve as coordinate functions. On the other hand, clearly only a constant function $f$ would have (identically) vanishing Poisson brackets with $g^1$ and $g^2$. - It's nice, but I meant why do we call it "a complete set of observables"? – Yrogirg Sep 10 '12 at 4:10 oh, what is left in your answer is to show the things backwards, from commuting to the fact that they are coordinates. – Yrogirg Sep 10 '12 at 4:15 or if you agree with my interpretation of what "a complete set of observables" should mean, then it can be moved to Math.SE – Yrogirg Sep 10 '12 at 4:37 I updated the answer. – Qmechanic♦ Sep 11 '12 at 14:47 Nice counterexample. :) – user404153 Sep 11 '12 at 17:50 show 1 more comment Blau called them a 'complete set' in analogy to the quantum mechanical picture, where a observable commuting (read Poission-commuting in the classical case) with a complete set of commuting observables is proportional to the unit, i.e. a 'constant'. This is called (first) Schur's lemma. - @Yrogirg indeed another way to say 'complete set of observerables' is to say that these ps and qs form an irreducible representation of the Heisenberg algebra (which is equivalent by Schur's lemma to saying that the only observables which Poisson commute with all the ps and qs are constant). – user404153 Sep 10 '12 at 21:33 @user404153 so specifying the values of all the classical observables from the complete set won't specify a unique classical state? – Yrogirg Sep 11 '12 at 4:23 @Yrogirg That's not quite what I mean. Certainly if one specifies the position and the momentum of a state, one specifies the classical state completely (after all, positions and momenta are supposed to be the coordinates of the space of classical states). What I mean is that if you specify an observable's Poisson brackets with all the ps and qs, then one has specified the observable up to the addition of a constant. Equivalently, an observable which Poisson commutes with all the ps and qs is a constant. – user404153 Sep 11 '12 at 4:59 @user404153 ok, but what has it to do with classical physics alone, without referencing to QM? That's just math interpretation, if that was what I was looking for, I'd asked the question at Math.SE. – Yrogirg Sep 11 '12 at 5:13 1 @Yrogirg I am confused by your question then. Are you asking why the ps and qs form a complete set of observables? You can show {p,f} = df/dq and {q,f} = df/dp for any function f(p,q), so certainly if {p,f} = {q,f} = 0, f is constant. – user404153 Sep 11 '12 at 8:27 show 2 more comments Blau's definition is a classical analog of the Schur's lemma. The reasoning behind this definition is the requirement that under a faithful quantization map which carries functions on the phase space to operators on some Hilbert space, the representation of the algebra of the quantum observables is irreducible. The irreducibility requirement has a physical origin as irreducible representations correspond to "single" quantum systems and if a representation is reducible, then it can be reduced to independent subsystems. Of course due to the Groenwold-Van Hove theorem, in general, no such quantization map exists. We usually give up faithfulness for the sake of irreducibility. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 51, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9052290916442871, "perplexity_flag": "head"}
http://mathoverflow.net/questions/40876/second-order-taylor-expansion-to-solve-system-of-equations/50168	## Second order Taylor expansion to solve system of equations ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Suppose you need to solve $f(\mathbf{x})=\mathbf{0}$ where $f:\mathbb{R}^n \to \mathbb{R}^m$, $m,n>1$. Newton's method relies on first order Taylor expansion of f. Where can I find details of analogous method using second order Taylor expansion? I found at least a dozen numerical analysis books which mention this method, but give no details or applications - 2 I'd be interested in learning about when using a second order technique might be useful and better than a first order method (Newton or Picard). It seems to me that higher methods are more prone to instabilities. – Deane Yang Oct 2 2010 at 23:17 @Deane: yes, methods with higher-order convergence tend to be finicky in that if you're a teeny bit far away from the solution you want to converge to, there is a great chance of converging to a different solution, or worse, diverge. – J. M. Oct 3 2010 at 0:30 @Yaroslav: Again, as I mentioned in your m.SE question, you need to figure out first what it means to invert a rank-3 tensor before you can figure out how to generalize Halley to multivariate nonlinear systems. Otherwise, there's Householder's method, but since multiplication of vectors, matrices and rank-3 tensors is noncommutative, you need to look at how to multiply the terms to get to your multivariate generalization. – J. M. Oct 3 2010 at 0:39 Halley's method is just one particular form of second order method. Alternative would be to use quadratic formula. Are you sure I just need to take Halley's formula and "plug it in"? – Yaroslav Bulatov Oct 3 2010 at 1:07 1 Since "order" is so overloaded a word :P , that's the reason I prefer "quadratically/cubically convergent" and "first/second order Taylor expansion" when discussing such iterative methods. :) – J. M. Oct 3 2010 at 14:22 show 2 more comments ## 2 Answers The single-variable version of what you are looking for is called Halley's method. (See, for example, MathWorld's article on Halley's method.) Maybe there's a fairly straightforward way to generalize it to multivariable functions. Or, if nothing else, this gives you another search term. - It seems like Halley's method is a hybrid between 1st order Taylor expansion and second order Taylor expansion books.google.com/…'s+method&hl=en&ei=jrenTNDRFJG-sAP1hr2HDQ&sa=X&oi=book_result&ct=result&resnum=4&ved=0CDUQ6AEwAw#v=onepage&q=Halley's%20method&f=false – Yaroslav Bulatov Oct 2 2010 at 22:57 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. I haven't gotten around to downloading and reading it (and I'm wondering how I missed this when I was searching for results related to Halley's method), but apparently a multivariate version of the Halley iteration has already been developed decades ago. Maybe this might be of use. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9117596745491028, "perplexity_flag": "middle"}
http://mathforum.org/mathimages/index.php/Talk:Summation_Notation	# Response to checklist Original response to checklist, in black, written by Kate 14:05, 14 July 2011 (UTC) You'll find my comments in red AnnaP AnnaP 22:33, 14 July 2011 (UTC) ## References and footnotes • Images are all cited. The one issue with this is that, based on a conversation we had last night, I went back to the blog where I found the cupcake liner picture to figure out if he minded his images being used, and it looks like he does. But I'm planning on baking cupcakes with friends this weekend, so I figure I'll just take a picture then and replace the current one on Monday. • All mathematical content is from my general knowledge (with corrections by Prof. Maurer), and so no references are provided ## Good writing ### Quality of prose and page structuring • Subject headings and/or first sentences make the purpose of each section clear • Each section is relevant to the topic • Page builds up from a one-sentence definition to examples of more complicated sums ### Integration of images and text • Math writing is used far more often than images, as the point of the page is to explain a notational convention and few images are relevant ### Links to other pages • The HelperPage template links back to Matrix and Markus-Lyapunov fractals ### Examples, Calculations, Applications, Proofs • Many numerical examples and calculations are included, effort has been made to integrate them well into the text ### Mathematical Accuracy and precision of language • To the best of my knowledge, all statements are accurate and error-free • I discussed this page with Prof. Maurer, and the only major change since then is the addition of the cupcake example • Definitions have been provided in green bubbles for sequence (& series), and dummy variable (which also has a small hidden section elaborating) • When you say this: Indexing with two variables can also be written using only one sigma:$\sum_{i,j=1}^n {a_{ij}} = \sum_{i=1}^n {\sum_{j=1}^n a_{ij}}$, you are summing to n in both cases. Since that isn't what you're doing above, you should offer a bit more explanation of that. Kate 13:54, 15 July 2011 (UTC): Fixed. • In your polynomials section, I'd take your example one step forward and say $p_3 (x) = a_0 x^0 + a_1 x^1 + a_2 x^2 + a_3 x^3 =a_0 + a_1 x^1 + a_2 x^2 + a_3 x^3$. Kate 13:54, 15 July 2011 (UTC): Fixed. ### Layout • Text is in short paragraphs, but is not broken up by images, as there are very few images relevant to this topic. • Hide/Shows have been used to: • Hide more challenging mathematical content at the bottom • Hide an additional explanation that not all readers may need closer to the top • Hide the answers to example problems that the reader is asked to think about • Hide the example problems, in case the reader isn't interested in them • Page has been viewed in a variety of window sizes, nothing weird happened • Did you think about making all of the subsections under "more on summation notation" simply their own sections. In think that might look/flow better. Kate 13:54, 15 July 2011 (UTC): I hadn't thought about it, but now that you've mentioned it, I changed it. • In the summation and distribution section, add a bit more space between the two equations, and maybe label one "multiplication" and the other "division" Kate 13:54, 15 July 2011 (UTC): done. # Older comments Diana 18:35, 6/17: A lot of students are really confused by i when they first see this notation. You address it briefly in the "indexing" section, but I think it's worth saying more about it. For the layman, especially someone who's never done programming or logic, the idea of using i as a placeholder is often foreign. Kate 13:40, 20 June 2011 (UTC): Can you tell me more about how it's confusing? I guess I'm so used to it that I'm not really sure what parts need to be expanded on or how to expand them. The problem that Diana is talking about is sort of a more sophisticated version of how some kids in their pre-algebra class really struggle that x can mean any number. Then, some kids will struggle that you can make any variable any number--if it's unknown, they think it must be "x"! It's a very, very common problem with understanding the abstraction. So, this issue with indices is a similar problem with really getting the abstraction. So, that's my explanation of the problem... now in terms of a solution, it may be best to have an entire paragraph on "dummy variable" instead of just your mouse over. I'd also suggest pointing this out with a sum that goes to something smaller than 20, as you do in the later sections. If you want me to, I can spend some more time working on the wording AnnaP 7/5 Okay, so I think I understand the problem better now. I'll try and figure out what should go in a paragraph about dummy variables. Do you think cupcake liners could be a good analogy? Like, when you make cupcakes, you set out all the cupcake liners where you intend to make cupcakes, but you don't actually have any cupcakes until you fill them with batter. So the cupcake liners are like indices, because they help define the pattern, but you don't actually have anything until you "fill" them with actual numbers or expressions. I'm also thinking that no matter how good an explanation I give, the best way to actually understand it is to do a lot of example sums, which isn't really feasible in a website. Cupcake liners might work--I'd suggest putting in a picture just to make sure everyone knows what you're talking about. Also, I'd specific that putting down the liners doesn't mean that you have to put the exact same thing in each. In fact, all of your cupcakes will be at least a little bit different! I added this example in - is the way I did it ok? I didn't want to put it in there the first time I mention "dummy variable", because that's in my table, and then I wasn't sure where to put it, so I made a combination balloon-link and put the paragraph in a hidden section before the example problems. (Kate 18:57, 11 July 2011 (UTC)) Chris 7/13/11 The cupcake idea is excellent is terms of place-holding but problematic in terms of the fact that a cook wants each cupcake to be similar. I brainstormed for a while to come up with a better idea but couldn't find one. I think the value of the analogy outweighs the problems. I like the way Kate integrated it into the page and included the cupcake liners image. As a minor point, your section on the matrix is specifically about a square (n by n) matrix. I'd suggest making it an n by m matrix, just to make your example a bit more generic. AnnaP 7/5 Changed. xd 20:36, 14 June 2011 (UTC) I think it has all the general information it has. It is not possible for this page to be comprehensive and listing all the sums in closed forms. Maybe it is useful to put on some common sums like $\sum_{i=1}^{n} i =\frac{(1+n)n}{2}$ done!	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 3, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9713024497032166, "perplexity_flag": "middle"}
http://mathhelpforum.com/geometry/212437-intersection-known-line-two-lines-unknown-variable-each.html	# Thread: 1. ## Intersection of a known line and two lines with an unknown variable in each. I'm trying to do number 7 on page 10 of this document: http://www.edexcel.com/migrationdocu...athematics.pdf I have done parts a and b. The answers give x=-4 but I don't know how that is obtained. The method used for part b is questionable as I would use implicit differentiation. You cannot simply convert to an expression starting with y= when that type of equation isn't suitable. It is also questionable why the value of a is not used in b but that is a minor point. 2. ## Re: Intersection of a known line and two lines with an unknown variable in each. This is in the wrong sub-forum. You should post this in the university math board. 3. ## Re: Intersection of a known line and two lines with an unknown variable in each. It is pre-university mathematics since it is a question from A-level. 4. ## Re: Intersection of a known line and two lines with an unknown variable in each. I thought you meant that you needed help with implicit differentiation. 5. ## Re: Intersection of a known line and two lines with an unknown variable in each. Originally Posted by Stuck Man It is pre-university mathematics since it is a question from A-level. Frankly, I don't know what problem you asking about. All of this could have been avoided if you would just take the time to past the question. It is so easy to use very basic LaTeX. On the toolbar you will see $\boxed{\Sigma}$ clicking on that give the LaTeX wraps, [tex] [/tex]. The code goes between them. Here are some examples. [tex]y^{2}=4ax [/tex] gives $y^2=4ax$ [tex] \frac{2x+1}{x^{2}+x+1} [/tex] gives $\frac{2x+1}{x^2+x+1}$ [tex] \sqrt{x+1} [/tex] gives $\sqrt{x+1}$ OR You can also click on Reply with Quote. That will allow you to ‘steal’ any code that you see that you like. 6. ## Re: Intersection of a known line and two lines with an unknown variable in each. I am trying to do part c. I have figured there is an infinite number of points on the line y=15 where x<0 that could be the point of intersection. The answers give x=-4 which is presumably the nearest point that allows t in both of the tangents to be an integer or a rational number. The question does not give enough information to solve the coordinates of A and B. 7. ## Re: Intersection of a known line and two lines with an unknown variable in each. Hi Stuck Man! I'm afraid your assumption about x=-4 is incorrect. There is a piece of information in the problem that you have not used yet. It says that both tangent lines meet on the directrix of C where y=15. The parabola $y^2=4ax$ has a directrix at x=-a. In other words, both tangents go through the point (-4,15). For the meaning of the word directrix, see for instance: Conic section - Wikipedia, the free encyclopedia.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9454153180122375, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/2317/including-air-resistance-what-is-the-escape-velocity-from-earth/2321	# Including air resistance, what is the escape velocity from Earth? Including air resistance, what is the escape velocity from the surface of the earth for a free-flying trajectile? - ## 3 Answers For the lower atmosphere, where most of the air is, the temp, pressure, and density of air is given by: $T=T_0-Lh$ $p=p_0\left(\frac{T}{T_0}\right)^{\frac{gM}{RL}}$ $\rho = \frac{pM}{RT}$ using the following constants: • sea level standard atmospheric pressure p0 = 101325 Pa • sea level standard temperature T0 = 288.15 K • Earth-surface gravitational acceleration g = 9.80665 m/s2. • temperature lapse rate L = 0.0065 K/m • universal gas constant R = 8.31447 J/(mol·K) • molar mass of dry air M = 0.0289644 kg/mol The force due to air resistance can be written: $F = -\rho v^2 C_d A$ where $C_d$ is the coefficient of drag, $v$ is the velocity, and $A$ is the surface area of the projectile. The goal is to get out of the atmosphere (where force of gravity is roughly constante) with the Earth's escape velocity, $11.2$ km/s. For a bullet-shaped 1-kg projectile of steel, $C_d \approx 0.04$ and $A \approx 4\times 10^{-4}$ m$^2$. This leads to a initial velocity of $13.5$ km/s. While not much higher than the vaccuum value, it is still high enough that the bullet would probably vaporize in the atmosphere. Interestingly, if one used a sphere instead of a bullet, then $C_d=0.4$, $A=4\times 10^{-3}$ m$^2$, the air resistance is 100 times higher, and thus a much greater velocity is needed. But since the drag scales like $v^2$, this leads to much higher drag. So much so, that even if one could launch the ball at the speed of light (Newtonians only, please!), it still could not make it out of the atmosphere! - 1 I doubt that a $v^2$ dependence is the whole story for objects with speeds on order of km/s. It is certainly not the whole story for objects at or above Mach 1, which means that achieving those kinds of velocities in the lower atmosphere renders these kinds of estimates very rough indeed. – dmckee♦ Dec 28 '10 at 16:01 Is the drag force is still the same when the bullet speed is much higher than the sound speed? – hwlau Dec 28 '10 at 16:04 That is what I thought too. I'm no expert, but from looking around the web, it seems the $v^2$ scaling is pretty universal, even for supersonic flight. But the drag coefficient may change, as "wave drag" becomes important around the sound barrier. In hypersonic flight, the problem actually reverts to a simpler one, where the air molecules are treated as elastic scattering events, which explains the $\rho v^2$ scaling. – Jeremy Dec 28 '10 at 16:27 4 why you are asking and answering the same question at the same time? – Graviton Dec 29 '10 at 15:56 It depends on the altitude, but let's assume sea level. The base escape velocity, assuming a vacuum, is 11.2 km/s. From there, it depends on the aerodynamical properties of the object. From Hitchhiker's Guide to Model Rocketry, we get this formula: $$rag = \frac{1}{2}V ^ 2 \times (Air Density) \times CD \times (Projected Area)$$ The CD, or Coefficient of Drag, is a dimensionless number dependent on the shape of an object. A typical model rocket has a CD of about 0.75, while a high performance, highly streamlined rocket may be as low as 0.4. Smaller fins will reduce the CD, but also cause the stability to go down. So I'm going to assume for simplicities sake that we've got a 0.4 CD rocket. I'll assume a 1 m diameter circular projected area, or pi/4 projected area. That leaves us still needing an adequate Air Density model. The air density model I'm using is pulled from Wikipedia, and is: I might try and model this sometime, but let me just give you the force at sea level, given a normal escape velocity. The drag is 2414 kN. Assuming this object was roughly cylndrical, having a density of 1, and is 10 m long, that would give it a weight of roughly 8000 kg. That would be somewhat less then the affect of gravity on the same object. I'd guess from these numbers that for this object, if you were going around 14-15 km/s, you'd be able to escape the Earth in one shot, but I'd have to crunch the numbers more carefully. - – Jeremy Dec 28 '10 at 15:52 Only the Streamlined body is lower, but, well, thanks for that source. Very interesting. – PearsonArtPhoto Dec 28 '10 at 15:54 Anything from 13.5 to15 km/s is what we reach after tedious calculations. - 5 Welcome to Physics.SE! The users of this site run the from interested amateurs just starting out to working professionals. While we don't have a formal requirement for citing sources or exhibiting calculations, you may find that such unsupported answers are not always well received. – dmckee♦ Nov 29 '11 at 4:20 1 A hint of what factors you have or havn't included would be nice. – Nic Nov 6 '12 at 13:45 This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. – Sklivvz♦ Dec 9 '12 at 12:02	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 19, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9324119687080383, "perplexity_flag": "middle"}
http://mathhelpforum.com/statistics/81406-cd-random-order-problem.html	# Thread: 1. ## CD random order problem Hi, I was listening to a CD a week ago in random playback. At one point three of the tracks played in the correct order (24 tracks on the CD). For example: 8, 4, 17, 10, 11, 12, ... (Can't remember the exact track numbers.) How do I calculate the probability of this type of thing happening by chance? I tried to work it out, but got nowhere (other than I suppose model it using random numbers). Thanks. 2. 20 tracks can arranged among themselves in 20! ways. hence n(s) = 20! Considering the 3 tracks which played in a order as 1. Now, total number of tracks = 18 and these can be arranged among themselves in 18! ways. n(e) = 18!. Probability of the event is 18!/20! = 1/380. Please correct me If I am wrong. Thanks, Ashish 3. Thanks Ashish, Something that was puzzling me was how to allow for the fact it might be any three tracks that could have been played in order. With just four tracks in total, if I take 123 as the group, I get 2! ways of arranging them, and the same if I assume 234, so a total of 4 ways out of 24. However, 1234, 2341, and 4123 are the only arrangements (I've double-counted 1234). Any pointers? With more than 5 or 6 tracks I got totally stumped! Steve 4. Anyone else? I still can't see how to use the hint given by Ashish to the case where any three tracks on the CD appear in the correct order. 5. Originally Posted by steve23144 I was listening to a CD a week ago in random playback. At one point three of the tracks played in the correct order (24 tracks on the CD). For example: 8, 4, 17, 10, 11, 12, ... How do I calculate the probability of this type of thing happening by chance? I will assume the no track is ever repeated until the complete permutation of all twenty four tracks is complete. There are $24!=6200448401733239439360000$ ways to play the CD is random mode. To get a particular sequence of three say 11, 12, & 13 the probability of that happing is $\frac{22!}{24!}= 0.00181$. The find the probability of at least one sequence of three is a bit more complicated. 6. Originally Posted by Plato I will assume the no track is ever repeated until the complete permutation of all twenty four tracks is complete. There are $24!=6200448401733239439360000$ ways to play the CD is random mode. To get a particular sequence of three say 11, 12, & 13 the probability of that happing is $\frac{22!}{24!}= 0.00181$. The find the probability of at least one sequence of three is a bit more complicated. Thanks Plato, Does that mean the quickest way to find the answer is to do a computer simulation after all then, or can I break the problem down in some way to make it easier? 7. Originally Posted by steve23144 Thanks Plato, Does that mean the quickest way to find the answer is to do a computer simulation after all then, or can I break the problem down in some way to make it easier? Are you asking for "The find the probability of at least one sequence of three being played in a row"? 8. Originally Posted by Plato Are you asking for "The find the probability of at least one sequence of three being played in a row"? Yes, that's what I'd like, or even better the probability of at least one sequence of at least three tracks on the CD play in their correct order. Thanks. 9. I would like show you the extent of the difficulty involved in your question. Consider the sequence $1,\,2,\, 3,\cdots,\, n-1,\, n$. How many ways can we rearrange the string so that none of the patterns $(12),\,(2,3),\, (3,4),\cdots,\, (n-1,n)$ will appear. The answer is rather hard to understand: $\left( {n - 1} \right)!\left[ {\sum\limits_{k = 1}^{n - 1} {\left( { - 1} \right)^k \frac{{n - k}}{{k!}}} } \right]$. Now subtracting that number from $n!$ will give the number of ways of having a string in which at least one of those patterns wii appear.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9596109390258789, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/203626/is-this-composition-of-polynomial-correct	# Is this composition of polynomial correct? Suppose $p(x)=x^2+5x+3$ and $q(x)=3x^3-3x+7$. Write the expression $(q \circ p)(x)$ as a sum of terms, each which is a constant times the power of $x$. Here is my work for the problem: $(q\circ p)(x)=3(x^2+5x+3)^3-3(x^2+5x+3)+7$ $(q\circ p)(x)=3(x^6+125x^3+27)-3x^2-15x-9+7$ $(q\circ p)(x)=3x^6+375x^3+81-3x^2-15x-2$ $(q\circ p)(x)=3x^6+375x^3-3x^2-15x+79$ Did I do something wrong while working with this problem? The final answer I got wasn't in the multiple choice answers. I went over this problem 3 times and cannot find what I am doing wrong. - Your understanding of composition is fine; it’s just your algebra that went astray. See @Jasper’s answer. – Brian M. Scott Sep 27 '12 at 20:42 ## 2 Answers The mistake you made is assuming that $(a+b+c)^3=a^3+b^3+c^3$. Note that $(a+b+c)(a+b+c)=a^2+ab+ac+ba+b^2+bc+ca+cb+c^2$. Simplify and multiply by $a+b+c$ again to get the cube. Alternatively you can rewrite $q(x)=3x(x^2-1)+7$ and work from there. - Is there a shortcut to multiply this out? I did $(a+b+c)^2$ first then multiplied that by $(a+b+c)$ again. It was very long and messy. – Steven N Sep 27 '12 at 20:45 @Steven This is a common (and very important) fallacy to overcome! Powers do not "distribute" over + or -... try to unlearn this misconception! – rschwieb Sep 27 '12 at 20:45 As Jasper noted you did wrong here: $$...=3(x^2+5x+3)^3...$$ $$...=3(x^6+125x^3+27)-...$$ In fact $$(a+b+c)^3=a^3 + 3a^2b + 3a^2c + 3ab^2 + 6abc + 3ac^2 + b^3 + 3b^2c + 3bc^2 + c^3$$ we have $q(p(x))=3x^6+45x^5+252x^4+645x^3+753x^2+390x+79$ -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 15, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.956359326839447, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/B-spline	# B-spline In the mathematical subfield of numerical analysis, a B-spline is a spline function that has minimal support with respect to a given degree, smoothness, and domain partition. B-splines were investigated as early as the nineteenth century by Nikolai Lobachevsky. A fundamental theorem states that every spline function of a given degree, smoothness, and domain partition, can be uniquely represented as a linear combination of B-splines of that same degree and smoothness, and over that same partition.[1] The term "B-spline" was coined by Isaac Jacob Schoenberg and is short for basis spline.[2][3] B-splines can be evaluated in a numerically stable way by the de Boor algorithm. Simplified, potentially faster variants of the de Boor algorithm have been created but they suffer from comparatively lower stability.[4][5] In the computer science subfields of computer-aided design and computer graphics, the term B-spline frequently refers to a spline curve parametrized by spline functions that are expressed as linear combinations of B-splines (in the mathematical sense above). A B-spline is simply a generalisation of a Bézier curve, and it can avoid the Runge phenomenon without increasing the degree of the B-spline. ## Definition Given m real valued ti, called knots, with $t_0 \le t_1 \le \cdots \le t_{m-1}$ a B-spline of degree n is a parametric curve $\mathbf{S}:[t_{n}, t_{m-n-1}] \to \mathbb{R}$ composed of a linear combination of basis B-splines bi,n of degree n $\mathbf{S}(t)= \sum_{i=0}^{m-n-2} \mathbf{P}_{i} b_{i,n}(t) \mbox{ , } t \in [t_{n},t_{m-n-1}]$. The points $\mathbf{P}_{i} \in \mathbb{R}$ are called control points or de Boor points. There are m−n-1 control points, and the convex hull of the control points is a bounding volume of the curve. The m-n-1 basis B-splines of degree n can be defined, for n=0,1,...,m-2, using the Cox-de Boor recursion formula $b_{j,0}(t) := \left\{ \begin{matrix} 1 & \mathrm{if} \quad t_j \leq t < t_{j+1} \\ 0 & \mathrm{otherwise} \end{matrix} \right.,\qquad j=0,\ldots, m{-}2$ $b_{j,n}(t) := \frac{t - t_j}{t_{j+n} - t_j} b_{j,n-1}(t) + \frac{t_{j+n+1} - t}{t_{j+n+1} - t_{j+1}} b_{j+1,n-1}(t) ,\qquad j=0,\ldots, m{-}n{-}2.$ Note that j+n+1 can not exceed m-1, which limits both j and n. When the knots are equidistant the B-spline is said to be uniform, otherwise non-uniform. If two knots tj are identical, any resulting indeterminate forms 0/0 are deemed to be 0. Note that when one sums a run of adjacent n-degree basis B-splines one obtains, from this recursion $\sum_{j=j'}^{j''} b_{j,n}(t) = \frac{t - t_{j'}}{t_{j'+n} - t_{j'}} b_{j',n-1}(t) \quad + \sum_{j=j'{+}1}^{j''} b_{j,n{-}1}(t) \quad + \; \frac{t_{j''+n+1} - t}{t_{j''+n+1} - t_{j''+1}} b_{j''+1,n-1}(t)$ for any sum with $0\le j' < j''\le m{-}n{-}2.$ When $j''\ge j'+n-1$ here, then this sum is, by this recursion, identically equal to 1, within the limited subrange $t_{j'{+}n} \le t \le t_{j''{+}1}$, (since this interval excludes the supports of the two basis B-splines in the separate terms at the ends of this sum). ### Uniform B-spline When the B-spline is uniform, the basis B-splines for a given degree n are just shifted copies of each other. An alternative non-recursive definition for the m−n-1 basis B-splines is $b_{j,n}(t) = b_n(t - t_j), \qquad\; j = 0, \ldots, m-n-2$ with $b_{n}(t) := (t_{n+1}-t_{0}) \sum_{i=0}^{n+1} \omega_{i,n}(t - t_i)_+^{n} \,\;$ and $\omega_{i,n} := \prod_{j=0, j \neq i}^{n+1} \frac{1}{t_j - t_i} \,\;$ where $(t - t_i)_+^n := \left\{\begin{matrix} (t - t_i)^n &\mbox{if}\ t \ge t_i \\ 0 &\mbox{if}\ t < t_i \end{matrix}\right.$ is the truncated power function. ### Cardinal B-spline Define B0 as the indicator function (or characteristic) function of $[-\tfrac{1}{2}, \tfrac{1}{2}]$, and Bk recursively as the convolution product $B_k := B_{k-1} * B_0, ~k =1, 2, \dots$ then Bk are called (centered) cardinal B-splines. This definition goes back to Schoenberg. Bk has compact support $[-\tfrac{k+1}{2}, \tfrac{k+1}{2}]$ and is an even function. As $k \rightarrow \infty$ the normalized cardinal B-splines tend to the Gaussian function.[6] ## Notes When the number of de Boor control points is one more than the degree and $t_0 = \dotsb = t_n = 0$ and $t_{n+1} = \dotsb = t_{2n} = 1$ (thus $t \in [0, 1]$), the B-Spline degenerates into a Bézier curve. In particular, the B-Spline basis function $b_{i,n}(t)$ coincides with the n-th degree univariate Bernstein polynomial.[7] The shape of the basis functions is determined by the position of the knots. Scaling or translating the knot vector does not alter the basis functions. The spline is contained in the convex hull of its control points. A basis B-spline of degree n $b_{i,n}(t)\,\;$ is non-zero only in the interval [ti, ti+n+1] that is $b_{i,n}(t) = \left\{\begin{matrix} >0 & \mathrm{if} \quad t_{i} \le t < t_{i+n+1} \\ 0 & \mathrm{otherwise} \end{matrix} \right.$ In other words if we manipulate one control point we only change the local behaviour of the curve and not the global behaviour as with Bézier curves. Also see Bernstein polynomial for further details. ## Examples ### Constant B-spline The constant B-spline is the simplest spline. It is defined on only one knot span and is not even continuous on the knots. It is just the indicator function for the different knot spans. $b_{j,0}(t) = 1_{[t_j,t_{j+1}]} = \left\{\begin{matrix} 1 & \mathrm{if} \quad t_j \le t < t_{j+1} \\ 0 & \mathrm{otherwise} \end{matrix} \right.$ ### Linear B-spline The linear B-spline is defined on two consecutive knot spans and is continuous on the knots, but not differentiable. $b_{j,1}(t) = \left\{\begin{matrix} \frac{t - t_j}{t_{j+1} - t_j} & \mathrm{if} \quad t_j \le t < t_{j+1} \\ \frac{t_{j+2} - t}{t_{j+2} - t_{j+1}} & \mathrm{if} \quad t_{j+1} \le t < t_{j+2} \\ 0 & \mathrm{otherwise} \end{matrix} \right.$ ### Uniform quadratic B-spline Quadratic B-splines with uniform knot-vector is a commonly used form of B-spline. The blending function can easily be precalculated, and is equal for each segment in this case. Relatively simple formula exists only for uniform grid $t_{j+k}=t_{j}+h*k, k=0,...,3$ $b_{j,2}(t) = \begin{cases} (t-t_j)^2/(2h^{2}), & t_j\le t \le t_{j+1} \\ ((t-t_{j})^2 - 3(t-t_{j+1})^{2})/(2h^{2}), & t_{j+1} \le t \le t_{j+2}\\ ((t-t_{j})^2-3(t-t_{j+1})^{2}+3(t-t_{j+2})^{2})/(2h^{2}), & t_{j+2} \le t \le t_{j+3}\\ 0 & \mbox{otherwise}. \end{cases}$ Put in matrix-form, it is:[8] $\mathbf{S}_i(t) = \begin{bmatrix} t^2 & t & 1 \end{bmatrix} \frac{1}{2} \begin{bmatrix} 1 & -2 & 1 \\ -2 & 2 & 0 \\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} \mathbf{p}_{i-1} \\ \mathbf{p}_{i} \\ \mathbf{p}_{i+1} \end{bmatrix}$ for $t \in [0,1], i = 1,2 \ldots m-2$ ### Cubic B-Spline A B-spline formulation for a single segment can be written as: $\mathbf{S}_{i} (t) = \sum_{k=0}^3 \mathbf{P}_{i-3+k} b_{i-3+k,3} (t) \mbox{ ; }\ t \in [0,1]$ where Si is the ith B-spline segment and P is the set of control points, segment i and k is the local control point index. A set of control points would be $P_i^w = ( w_i x_i, w_i y_i, w_i z_i, w_i)$ where the $w_i$ is weight, pulling the curve towards control point $P_i$ as it increases or moving the curve away as it decreases. An entire set of segments, m-2 curves ($S_3,S_4,\dotsc,S_m$) defined by m+1 control points ($P_0,P_1,\dotsc,P_m, m \ge 3$), as one B-spline in t would be defined as: $\mathbf{S}(t) = \sum_{i=0}^{m-1} \mathbf{P}_{i} b_{i,3} (t)$ where i is the control point number and t is a global parameter giving knot values. This formulation expresses a B-spline curve as a linear combination of B-spline basis functions, hence the name. There are two types of B-spline - uniform and non-uniform. A non-uniform B-spline is a curve where the intervals between successive control points are not necessarily equal (the knot vector of interior knot spans are not equal). A common form is where intervals are successively reduced to zero, interpolating control points. Comparison between a uniform cubic B-spline (yellow) and a cubic Hermite spline (dark red). ### Uniform cubic B-splines Cubic B-splines with uniform knot-vector is the most commonly used form of B-spline. The blending function can easily be precalculated, and is equal for each segment in this case. Put in matrix-form, it is: $\mathbf{S}_i(t) = \begin{bmatrix} t^3 & t^2 & t & 1 \end{bmatrix} \frac{1}{6} \begin{bmatrix} -1 & 3 & -3 & 1 \\ 3 & -6 & 3 & 0 \\ -3 & 0 & 3 & 0 \\ 1 & 4 & 1 & 0 \end{bmatrix} \begin{bmatrix} \mathbf{p}_{i-1} \\ \mathbf{p}_{i} \\ \mathbf{p}_{i+1} \\ \mathbf{p}_{i+2} \end{bmatrix}$ for $t \in [0,1].$ ## P-spline The term P-spline stands for "penalized B-spline". It refers to using the B-spline representation where the coefficients are determined partly by the data to be fitted, and partly by an additional penalty function that aims to impose smoothness to avoid overfitting.[9] ## References 1. Carl de Boor (1978). A Practical Guide to Splines. Springer-Verlag. pp. 113–114. 2. Carl de Boor (1978). A Practical Guide to Splines. Springer-Verlag. pp. 114–115. 3. Gary D. Knott (2000), . Springer. p. 151 4. Lee, E. T. Y. (December 1982). "A Simplified B-Spline Computation Routine". Computing (Springer-Verlag) 29 (4): 365–371. doi:10.1007/BF02246763. 5. Lee, E. T. Y. (1986). "Comments on some B-spline algorithms". Computing (Springer-Verlag) 36 (3): 229–238. doi:10.1007/BF02240069. 6. Brinks R: On the convergence of derivatives of B-splines to derivatives of the Gaussian function, Comp. Appl. Math., 27, 1, 2008 7. Prautzsch et al., Hartmut (2002). Bezier and B-Spline Techniques. Springer-Verlag. pp. 60–66. ISBN 3-540-43761-4. 8. Eilers, P.H.C. and Marx, B.D. (1996). Flexible smoothing with B-splines and penalties (with comments and rejoinder). Statistical Science 11(2): 89-121.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 39, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8825377821922302, "perplexity_flag": "middle"}
http://mathhelpforum.com/trigonometry/4770-angles-degrees-arcs.html	# Thread: 1. ## Angles, Degrees, and Arcs Hey there, I've got a few questions. I have no idea how to figure out these kind of word problems. I'm taking an online class and I've got no source of help but here. So, is there some sort of formula I must follow with these? Draw a diagram? What does the "bearing" of 42.26 do for me? A fire is sighted due west of lookout A. The bearing of the fire from lookout B, 8.6 miles due south of A, is N 42°26'W. How far is the fire from B (to the nearest tenth of a mile)? Oh, and this one is probably easy, just simple algebra and such, but I dont know what to do... sin A = cos 3A Thanks in advance, Juan. 2. Originally Posted by juangohan9 Hey there, I've got a few questions. I have no idea how to figure out these kind of word problems. I'm taking an online class and I've got no source of help but here. So, is there some sort of formula I must follow with these? Draw a diagram? What does the "bearing" of 42.26 do for me? A fire is sighted due west of lookout A. The bearing of the fire from lookout B, 8.6 miles due south of A, is N 42°26'W. How far is the fire from B (to the nearest tenth of a mile)? $42^{\circ}26'=42^{\circ}+\frac{26}{60}^{\circ}=<br /> \frac{<br /> 1260+13}{30}$ a right triangle is formed when you draw a line from A to B to the fire, the distance from the Fire to base B is the hypotenuse, the distances from A to B and A to the Fire are the bases of the right triangle. do you know secant? $\sec\left(\frac{1273}{30}\right)=\frac{\overline{B F}}{\overline{AB}}$ therefore: $\sec\left(\frac{1273}{30}\right)=\frac{\overline{B F}}{8.6}$ thus: $8.6\sec\left(\frac{1273}{30}\right)=\overline{BF}$ $11.7\approx\overline{BF}$ so the distance is 11.7 miles. note: $\sec A=\cos^{-1}A=\frac{hypotenuse}{adjacent}$ there is also another way of finding it, using tan and the pythagoreum theorum, would you like to see it? Attached Thumbnails 3. Originally Posted by juangohan9 sin A = cos 3A Thanks in advance, Juan. Thus, $\sin x - \cos 3x=0$ Use co-function identity, $\sin x - \sin\left( \frac{\pi}{2} - 3x \right)=0$ --- For reference use formula, $\sin x-\sin y=2\cos \frac{x+y}{2} \sin \frac{x-y}{2}$ --- Thus, $2\cos \left(\frac{\pi}{4}-x\right)\sin \left( \frac{\pi}{4} +2x \right) =0$ Divide by two, thus, $\cos \left(\frac{\pi}{4}-x\right)\sin \left( \frac{\pi}{4} +2x \right) =0$ Set each factor equal to zero thus, $\left\{ \begin{array}{c} \cos \left(\frac{\pi}{4}-x\right)=0\\\sin \left( \frac{\pi}{4} +2x \right) =0$ From the first one you have, $\frac{\pi}{4}-x=\frac{\pi}{2}+\pi k$ Thus, $x=-\frac{\pi}{4}-\pi k$ To make your answer look cooler sine $-k$ is an integer too is does not matter if you write, $x=-\frac{\pi}{4}+\pi k$ Since the number being constantly added and subtracted is $\pi$ it does not matter if you write, $x=-\frac{\pi}{4}+\pi+\pi k$ Simplify, $x=\frac{3\pi}{4}+\pi k$ For the second one you have, $\sin \left(\frac{\pi}{4}+2x \right)=0$ Thus, $\frac{\pi}{4}+2x=\pi k$ Thus, $2x=-\frac{\pi}{4}+\pi k$ Since the number being constantly subtracted and added is $\pi$ it does not matter if, $2x=\pi-\frac{\pi}{4}+\pi k$ Thus, simplify, $2x=\frac{3\pi}{4}+\pi k$ Thus, $x=\frac{3\pi}{8}+\frac{\pi}{2} k$ 4. Hello, Juan! I'll explain $N\,42^o26'\,W$ . . The $N$ means we start by facing North. . . Then $42^o26'\,W$ means we turn $42^o26'$ to the west. And that is the direction indicated by $N\,42^o26'\,W$. Similarly, $S30^oE$ means: face South, turn 30° to the east. A fire is sighted due west of lookout $A$. The bearing of the fire from lookout $B$, 8.6 miles due south of $A$, is $N\,42^o26'\,W$ How far is the fire from $B$ to the nearest tenth of a mile? Code: ``` F * - - - - - - - * A \ \| \ \| \ \| d \ \| 8.6 \ \| \ θ \| \ \| * B``` The fire is at $F$. .Angle $\theta \,= \,42^o26' \,\approx \,42.4333^o$. .Let $d = BF.$ In right triangle $FAB$, we have: . $\cos\theta\,=\,\frac{8.6}{d}\quad\Rightarrow\quad d\,=\,\frac{8.6}{\cos\theta}$ Therefore: . $d \:= \:\frac{8.6}{\cos42.4333^o} \:= \:11.6521...\:\approx\:11.7$ miles. Solve for $A:\;\;\sin A \:= \:\cos 3A$ Here's a back-door approach to this problem . . . If $\sin A \,= \,\cos B$, then $A$ and $B$ are complementary: . $A + B \,=\,90^o$ Look at this right triangle. Code: ``` * B c * * * * a * * A * * * * * C b``` We have: . $\sin A = \frac{a}{c}$ .and . $\cos B = \frac{a}{c}$ See? . . . Angles $A$ and $B$ are from the same right triangle. ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ Back to the problem: . $\sin A \:= \:\cos3A$ Since $A$ and $3A$ are complementary: . $A + 3A \:= \:90^o\quad\Rightarrow\quad 4A \:=\:90^o$ Therefore: . $\boxed{A\:=\:22.5^o}$ Note: TPHacker's solution is complete.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 57, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9405075311660767, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/100266?sort=oldest	## A covering problem for the Hamming cube ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Consider the set of all $k$-subsets of $\{1,\dots,n\}$, naturally identified with a subset $A$ of $\{0,1\}^n$ where each element has exactly $k$ ones. Is there a sharp bound known for $\epsilon$-covering of this set in the Hamming distance? More specifically, suppose that $k = \gamma n$ where $\gamma \in(0,1/2)$ is fixed. The cardinality of $A$ is asymptotically $\|A\| \sim e^{n h(\gamma)}$ where $h(\cdot)$ is the binary entropy function. Is there an $\epsilon$-covering of $A$ in Hamming distance with $e^{ \frac{C n}{\log n}}$ elements and say $\epsilon \le \frac{n}{(\log n)^{2}}$? In other words, how large $\epsilon$ needs to be to be able to go from cardinality being exponential in $n$ to it being exponential in $n / \log n$. - ## 1 Answer If you're not concerned with constant factors on the covering radius (and it looks like you're not), then you should be able to get the right answer on volume arguments alone. On one hand, in order to cover all of $A$ with "balls" of radius $\epsilon$, you need to take at least $\|A\|/\|B(\epsilon)\|$ points, where $B(r)$ denotes the number of points in $A$ within Hamming distance $r$ of a fixed point in $A$. On the other hand, suppose you greedily choose balls of radius $\epsilon/2$, so long as these balls are completely disjoint. Certainly you will not choose more than $\|A\|/\|B(\epsilon/2)\|$ balls. But having done so, if you double the radius of each ball, you will have covered all of $A$ (by the maximality of the set of balls you started with). So you see you get the right answer up to not worrying about a factor of $2$ on the radius. For more details, I think the keywords to search for are "covering codes" as well as "Johnson scheme" (the latter being the coding theory terminology for considering all binary strings of a fixed Hamming weight). - @Ryan, thanks for your response. I agree that a volume argument is pretty tight. I had tried it and I guess you end up with a bound of the form $\binom{n}{k} / [ \sum_{i=0}^r \binom{k}{i} \binom{n-k}{i} ]$ on the number of points required for an $r$-covering. I am having some difficulty, evaluating this asymptotically. Some rather rough calculations seems to suggest that you cannot get reduce the number from being exponential in $n$ (say $e^{cn(1+o(1)}$) if you require $k = \gamma n$ and do want $r$ to grow sublienar in $n$. – passerby51 Jun 21 at 23:02 ... what I mean is I always seem to get the denominator to grow at most polynomially in $n$, hence the total number to be $e^{c n (1 - o(1)}$. Any tricks to get a $\log n$ drop in the exponent is appreciated. – passerby51 Jun 21 at 23:09	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 39, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9517689347267151, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2008/02/25/integration-by-parts/?like=1&source=post_flair&_wpnonce=134bd2d71b	# The Unapologetic Mathematician ## Integration by Parts Now we can use the FToC as a mirror to work out other methods of finding antiderivatives. The linear properties of differentiation were straightforward to reflect into the linear properties of integration. This time we’ll reflect the product rule through the FToC to get a method called “integration by parts”. The product rule tells us that the derivative of the product of two functions $\left[fg\right](x)=f(x)g(x)$ is given by the “Leibniz rule”: $\left[fg\right]'(x)=f'(x)g(x)+f(x)g'(x)$. Now we take the antiderivative of both sides: $\displaystyle f(x)g(x)=\int\left[fg\right]'(x)dx=\int f'(x)g(x)dx+\int f(x)g'(x)dx$ Adding specific limits of integration and rearranging a bit we find the usual formula for integration by parts: $\displaystyle \int\limits_a^bf(x)g'(x)dx=f(b)g(b)-f(a)g(a)-\int\limits_a^bf'(x)g(x)dx$ So if we can recognize our integrand as the product of a function $f(x)$ that’s easy to differentiate and a function $g'(x)$ that’s easy to integrate, then we might be able to simplify things, though we have to be careful about the new terms that crop up from evaluating $f$ and $g$ at the boundary points $a$ and $b$. As a side note, physicists love to use this technique (and more general analogues) by waving their hands hard enough to push the boundaries far enough away that they can be ignored. There are some — like my departmental colleague Frank Tipler — who think this is the source of most problems modern physics seems to have. Myself, I take no position on the matter. I’ve upset enough people for this month already. ### Like this: Posted by John Armstrong \| Analysis, Calculus ## 6 Comments » 1. I’ve had almost more integration by parts than I can stand during the last few weeks, with all the calculus tutoring I do. (Perhaps I should send my students to this site.) This is one of my favorite techniques, because I love how it can help solve some seemingly impossible integration problems, especially some particularly horrible ones that you can’t imagine could possibly have a “nice” integral. Comment by \| February 25, 2008 \| Reply 2. Integration by parts is one of my least favourite techniques. Unless I specifically need some reduction formula, I don’t use it if I can avoid it. (IME, any integrand which would succumb to integration by parts succumbs more easily to a Risch-Norman attack.) Comment by Pseudonym \| February 26, 2008 \| Reply 3. Pseudonym: I actually don’t compute many antiderivatives at all, myself. You’re right that there are often other methods. On the other hand, I know I’m going to use it at certain points in the future. Besides, it’s a great illustration of the use of FToC as a mirror. Comment by \| February 26, 2008 \| Reply 4. [...] of Variables Just like we did for integration by parts we’re going to use the FToC as a mirror, but this time we’ll reflect the chain [...] Pingback by \| February 27, 2008 \| Reply 5. [...] where else have we seen derivatives as factors in integrands? Right! integration by parts! Here our formula says [...] Pingback by \| March 6, 2008 \| Reply 6. [...] where again we throw away negligible terms. Now we can handle the first term here using integration by parts: [...] Pingback by \| July 16, 2012 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 10, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9447800517082214, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/tagged/bounded-variation+complex-analysis	# Tagged Questions 4answers 189 views ### $\{z\in C:\|z\| = \|\operatorname{re}(z)\| +\|\operatorname{im}(z)\|\}$ open or closed [duplicate] Possible Duplicate: Proving that a complex set in open/closed/neither and bounded/not bounded I think $\{z\in C:\|z\| = \|\operatorname{re}(z)\| +\|\operatorname{im}(z)\|\}$ is closed. But I have ... 1answer 81 views ### Boundedness on strips in the complex plane for functional equations [closed] We know that the recurrence for $b>0$ (1) $f(0)=1$ (2) $f(z+1)=b{f(z)}$ has $f(z)=b^z$ as the only entire solution that is bounded on the strip $S=\{z: 0<\Re(z)\le 1\}$. The image of $S$ ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 8, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9528095722198486, "perplexity_flag": "middle"}
http://michaelnielsen.org/polymath1/index.php?title=Overlapping_Schwarz	# Overlapping Schwarz ### From Polymath1Wiki Our goal is to either analytically solve, or numerically approximate, the 2nd Neumann eigenfunction of the Laplacian on a generic acute triangle $\Omega \equiv$ABC (see figure). Since we don't have an analytic solution on this domain, but can solve problems on sectors or the circle, we propose a domain decomposition strategy in the spirit of the overlapping Schwarz iteration. Let 0 be the incenter of the triangle, and denote by Ω0 the interior of the incircle. This is tangent to the segments AB,BC and CA at F,D,E respectively. Denote by Γ10 the segment of the circle connecting E,F. Denote by Ω1 the sector AEF, and by Γ01 the arc connecting E,F. Note that Ω1 and Ω2 overlap. Repeat this process for the other three vertices. Let us denote the exact second Neumann eigenvalue as λ. Now we'll proceed by iteration. At step n, suppose $u_i^n$ for i = 1,2,3 satisfies $-\Delta u_i^n = \lambda^n_i u_i, x\in \Omega_i, \frac{\partial u_i^n}{\partial \nu}=0, x\in \partial \Omega_i \setminus{\Gamma_{0i}}$ and the Robin condition $u_0^{n-1} \frac{\partial u_i^{n}}{\partial \nu} - \frac{\partial u_0^{n-1}}{\partial \nu} u_i^n =0, x\in \Gamma_{0i}$ The function $u_0^n$ solves $-\Delta u_0^n = \lambda^n_0 u_0^n, x\in \Omega_0, u_i^{n} \frac{\partial u_0^{n}}{\partial \nu} - \frac{\partial u_i^{n}}{\partial \nu} u_0^n =0, x\in \Gamma_{i0},$ Now $(u_{i}^n$, λi) solve generalized eigenfunction problems on sectors of circles. If one knows $u_0^{n-1}, \frac{\partial u_0^{n-1}}{\partial \nu}$ on the curves Γ0i, then one can use local Fourier-Bessel expansions to get $u_i^n$. Then one uses their traces onto Γ10, and has the correct data to solve the eigenvalue problem for $u_0^n$ on the disk. One can use Fourier-Bessel expansions to do this as well. The claim is that as $n\rightarrow \infty$, the sequences $u_i^n$ converge to the restriction of the actual eigenfunction u on the sub-domains. Clearly we have to prescribe a starting guess for the iteration. ## Solving on the wedge Ωi We are at step n of the iteration. Suppose i = 1 is fixed for concreteness, and the traces of $u_0^{n-1}$ and $\frac{\partial u_0^{n-1}}{\partial \nu}$ on Γ01 are known. $-\Delta u_1^n = \lambda^n_1 u_1, x\in \Omega_1, \frac{\partial u_1^n}{\partial \nu}=0 x\in \partial \Omega_1\setminus{\Gamma_{01}}, u_0^{n-1} \frac{\partial u_1^{n}}{\partial \nu} - \frac{\partial u_0^{n-1}}{\partial \nu} u_1^n =0, x\in \Gamma_{01}$ We will try the method of particular solutions of Fox and Henrici, adapted to this problem. We know that since the opening angle is α, in Ω1 the functions $w_k(r,\theta):= J_{\frac{\pi k}{\alpha}} \sqrt{\lambda} r) \cos(\frac{\pi k}{\alpha} \theta)$ will satisfy the Neumann conditions on the line segments AE,AF < math > ,aswellassatisfytheequation < math > − Δwk = λwk. So, we suppose $u_1^n = \sum_{k=1}^M c_k w_k (r,\theta)$. We want to find the coefficients ck so as to satisfy the boundary condition on Γ01. Now, Γ01 is an arc of radius ρ1 = AE. Let (ρ1,θj) be 2M collocation points along this curve. At each point, we want to enforce $0= u_0^{n-1}(\rho_1,\theta_j) \frac{\partial u_1^{n}}{\partial \nu} (\rho_1,\theta_j)- \frac{\partial u_0^{n-1}}{\partial \nu}(\rho_1,\theta_j) u_1^n(\rho_1,\theta_j)$ $= u_0^{n-1}(\rho_1,\theta_j) \sum_{k=1}^M c_k \frac{\partial}{\partial r} J_{\frac{\pi k}{\alpha}} (\sqrt{\lambda} \rho_1) \cos(\frac{\pi k}{\alpha} \theta_j) - \frac{\partial u_0^{n-1}}{\partial \nu}(\rho_1,\theta_j) \sum_{k=1}^M c_k J_{\frac{\pi k}{\alpha}} (\sqrt{\lambda} \rho_1) \cos(\frac{\pi k}{\alpha} \theta_j) .$ This is equivalent to solving the rectangular nonlinear system $A(\lambda) \vec{c}=0$ where $a_{jk}(\lambda) = u_0^{n-1}(\rho_1,\theta_j) \frac{\partial}{\partial r} J_{\frac{\pi k}{\alpha}} (\sqrt{\lambda} \rho_1) \cos(\frac{\pi k}{\alpha} \theta_j) - \frac{\partial u_0^{n-1}}{\partial \nu}(\rho_1,\theta_j)J_{\frac{\pi k}{\alpha}} (\sqrt{\lambda} \rho_1) \cos(\frac{\pi k}{\alpha} \theta_j)$ We find the solutions by looking for values of λ so that the smallest singular value of A(λ) approaches 0. This is the Moler approach to the original Fox-Henrici-Moler paper. Once we locate the solution $\vec{c}$, we have the iterate $u_1^n$. We do this same process for the other wedges as well. ### Solving on the disk Ω0 We are at step n of the iteration, and have solved for the functions $u_i^n$. We therefore have their traces on the arcs Γi0. The function $u_0^n$ solves $-\Delta u_0^n = \lambda^n_0 u_0^n, x\in \Omega_0, u_i^{n} \frac{\partial u_0^{n}}{\partial \nu} - \frac{\partial u_i^{n}}{\partial \nu} u_0^n =0, x\in \Gamma_{i0}$ We shall again use a Fourier-Bessel ansatz: let zm(r,θ) = Jm(λr)eimθ, and assume $u_0^n(r,\theta) = \sum_{k=0}^M d_m z_m(r,\theta)$. Repeat the process above of enforcing the (non-standard) boundary conditions at collocation points along Γi0. ### {Convergence?} At the end of the nth step, we have 4 functions: $u_{i}^n$, i=0,1,2,3 and 4 eigenvalues. We repeat the process until the eigenvalues are all the same number.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 28, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8872346878051758, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?t=129395	Physics Forums ## Boson gas... Let's suppose we have a Boson Non-interacting gas under an Harmonic potential so $$V(x)= \omega (k) x^{2}$$ the question is if we know what the Partition function is $$Z= Z (\beta )$$ we could obtain the specific Heat, and other important Thermodinamical entities...but could we know what the "dispersion relation" w(k) for k real is? , i have looked several books about "Solid State" but i don't find any info about how to get dispersion relations using partition functions or similar..or if we can find an Integral or differential equation for the w(k)..thanks PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> Nanocrystals grow from liquid interface>> New insights into how materials transfer heat could lead to improved electronics We can really only know the partition function if we know w(k). For a given Hamiltonian for which we can find the eigenspectrum, the partition function for n non-interacting, once we know the partition function for a single particle in this potential we are able to formulate the many-body partition function. I this what you are asking is to calculate the response function for the system, which is different from w(k) Sorry "Epicurus" i'm not Brittish or American so my english sometimes sounds ambigous..my problem is.. -Let's suppose we know the TOTAL partition function for the system $$Z(\beta )$$ - If we have a Non-interacting Boson gas we have that: $$Z(\beta)= \prod _k Z_k (\beta)$$ - I wish to calculate fro this...the "dispersion relation" $$\omega (k)$$ using the functions i know (Total partition function and Specific Heat, Gibss function and similar that can be obtained from the Total partition function )... for example getting a differential equation or other type of equation for $$\omega (k)$$ so it can be solved by numerical methods to obtain the "frecuencies"..Hope it's clearer (my question) now... ## Boson gas... 1-The expression you have written down in the third line is incorrect. - Are you refering that for a Non-interacting gas the "total partition function" (Harmonic approach) isn't equal to the product of all the partition function for all the particles taking N=1 ?.... Yes that correct. You talking about the distinguishable case, not the bosonic case. - Well in any case...is there any form to obtain the "structure" (unit cell) of the gas or the dispersion relation, speed of sound $$c(k)= \frac{d \omega }{dk}$$ or any quantity related to the "frecuencies"...? I know that from the partition function you could calculate "Entropy" , "Energy" (U) and other Thermodinamical functions but not the "frecuencies"..perhaps you could use X-ray scattering or other method but if you don't know the "shape" (unit cell) of the gas i think you can't do anything. Thread Tools \| \| \| \| \|-----------------------------------\|----------------------------------------\|---------\| \| Similar Threads for: Boson gas... \| \| \| \| Thread \| Forum \| Replies \| \| \| High Energy, Nuclear, Particle Physics \| 7 \| \| \| High Energy, Nuclear, Particle Physics \| 7 \| \| \| Beyond the Standard Model \| 2 \| \| \| Astrophysics \| 0 \| \| \| High Energy, Nuclear, Particle Physics \| 2 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 7, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8840452432632446, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/93983/convergence-in-distribution-and-convergence-of-quantile	# convergence in distribution and convergence of quantile Suppose real-valued random variables $\{X_{n}\}$ converges to $X$ in distribution. Then, will the quantile of the distribution of $\{X_n\}$ converge to the quantile of $X$? . - ## 1 Answer Yes. If $X$ is a random variable with distribution function $F$, then for $0<p<1$ define the quantile function as $Q(p)=\inf(x: F(x)\geq p)$. Then $X_n\to X$ in distribution if and only if $Q_n(p)\to Q(p)$ at all continuity points $p$ of $Q$. Added: It's a nice exercise to prove this result from the definition. On the other hand, it is Proposition 5 (page 250) in A Modern Approach to Probability Theory by Bert Fristedt and Lawrence Gray, and is also proved in Chapter 21 of Asymptotic Statistics by A. W. van der Vaart. - But what if $Q$ is not continuous at $p$? – webster Dec 25 '11 at 11:09 Suppose we know that $F$ is strictly increasing. Will this additional condition help? – webster Dec 25 '11 at 14:08 The discontinuity points of $Q$ correspond to the "flat spots" on the graph of $F$ (Draw a picture!). So if $F$ is strictly increasing, then $Q$ is continuous at all $0<p<1$. – Byron Schmuland Dec 25 '11 at 15:26 Yes. But what will be the formal proof? – webster Dec 25 '11 at 16:25	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 20, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8690173029899597, "perplexity_flag": "head"}
http://mathoverflow.net/questions/18392/signed-minimum	## Signed minimum? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I am looking for references to papers which might have defined a 'signed minimum' equivalent to $$smin(x,y) ::= \left(\frac{\textit{signum}(x)+\textit{signum}(y)}{2}\right)\cdot \min(\|x\|,\|y\|)$$ where $\textit(signum)(x)$ is $-1$ for $x\lt 0$, $1$ for $x\gt 0$ and an arbitrary (finite) value for $x=0$. Also, any simpler expression for $smin$ would be appreciated. The above definition works for $\mathbb{R}, \mathbb{Q}$, and $\mathbb{Z}$. Intuition: the 'signed minimum' between x and y is the one closest to $0$ if they both have the same sign, otherwise it's 0. - Why doesn't this work for $\mathbb{Z}$? You can just do case-by case analysis and verify that the result always going to be an integer. – Mikola Mar 16 2010 at 18:13 It does work, I guess my phrasing is a tad awkward. – Jacques Carette Mar 16 2010 at 19:04 ## 1 Answer $smin(x,y)$ can also be described as the number of smallest absolute value in the closed interval between $x$ and $y$. When $x \le y$ are integers, this is also the value of the game $\{x-1 \mid y+1\}$ in the sense of combinatorial game theory (Conway, On Numbers And Games). -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 16, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9271743893623352, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/218051/math-contest-find-number-of-roots-of-fx-fracn2-involving-a-strange-int?answertab=active	# Math contest: Find number of roots of $F(x)=\frac{n}{2}$ involving a strange integral. Edit summary: A good answer appeared. CW full answer added, based on given answers. Removing my ugly-looking attempts, as they still remain in the rev. history. Here's a final-round calculus contest problem in our district in 2009. Test paper isn't available online. I originally proved $F(n/2)<n/2$ but failed to prove $F(n)>n/2$ and the monotonicity of $F(n)$. Define $$F(x)=\int\limits_0^x \mathrm e^{-t} \left(1+t+\frac{t^2}{2!}+\cdots+\frac{t^n}{n!}\right) \mathrm d t$$ where $n>1, n \in \mathbb N^+$; and an equation: $\displaystyle F(x)=\frac{n}{2}$. Find the number of roots of this equation within the interval $\displaystyle I=\left(\frac{n}{2},n\right)$. Tedious Previous Attempts Removed. - 2 People, please take your time, I regret to say that I should've fallen asleep by now. Thanks again! – FrenzY DT. Oct 21 '12 at 15:00 2 Is the fact that $F'(x)=e^{-x} (1+x+\cdots +\frac{x^{n}}{n!})$ of any use? – dado Oct 21 '12 at 15:15 @dado I just understood what you meant. Exactly! (I thought you were asking if $F(x)$ were of any use. – FrenzY DT. Oct 26 '12 at 2:18 ## 3 Answers ### Proving $\displaystyle F(n)>\frac{n}{2}$: Looks like you've made the life far more complicated than needed. For $t>0$, write $$(1-\frac tn)e^t=\sum_{k\ge 0}\frac 1{k!}(1-\frac kn)t^k<\sum_{k=0}^n \frac{t^k}{k!}$$ (drop the negative coefficients and increase the rest). Rewrite this as $$e^{-t}\sum_{k=0}^n \frac{t^k}{k!}> 1-\frac tn.$$ Now, I hope even a dumb calculus student can integrate the right hand side from $0$ to $n$. - How did you foresee this? I never have envisaged that infinite series. Please offer some advice. – FrenzY DT. Oct 24 '12 at 2:28 1 I just know only a few decent functions whose integral from $0$ to $n$ is $n/2$. The constant $1/2$ did not seem to work, so I took the next simplest one, which is linear. Of course, to multiply $e^{-t}$ and the partial sum is a nightmare but we can put the exponent on the other side and, since we want to compare with a (truncated) power series, it makes sense to expand the product $(1-\frac tn)e^t$ into a power series as well, which is not a big deal. The rest is pure luck. If it had failed, I would try something else, but since it worked, I stopped here. – fedja Oct 24 '12 at 2:36 (+1) and checkmarked, because this answer exploits series which is my weak point. – FrenzY DT. Oct 24 '12 at 3:04 Great answer. This gives a nice inequality for the incomplete gamma function. – Albert Steppi Oct 24 '12 at 15:01 ### Proof that $\displaystyle F(\frac{n}{2})<\frac{n}{2}$ $$F(x)= \int\limits_0^x \mathrm e^{-t} \left(1+t+\frac{t^2}{2!}+\cdots+\frac{t^n}{n!}\right) \mathrm dt <\int\limits_0^x \mathrm e^{-t}\mathrm e^t\mathrm dt=x \,\,\Rightarrow\,\, F(\frac{n}{2})<\frac{n}{2}.$$ ### Proof that $\displaystyle F(n)>\frac{n}{2}$ See accepted answer and comment. $$(1-\frac tn)e^t = \sum_{k=0}^{+\infty}\frac 1{k!}\underbrace{(1-\frac kn)}_{\in(-\infty,1]}t^k<\sum_{k=0}^n \frac{t^k}{k!} \,\,\Rightarrow\,\, e^{-t}\sum_{k=0}^n \frac{t^k}{k!}> 1-\frac tn \,\,\Rightarrow\,\, F(n)> n-1 \ge \frac{n}{2}.$$ ### Proof that $F(n)$ is monotonically increasing on $\displaystyle(\frac{n}{2},n)$ $$F^\prime(x) = \mathrm e^{-x} \sum\limits_{i=0}^n \frac{x^i}{i!} = \frac{\text{Power series with positive coefficients and $x>0$}}{\text{Exponentiation}} > 0.$$ Summing up the previous three proofs, OP's equation has only $1$ root on the interval $\displaystyle \left(\frac{n}{2},n\right)$. $\square$ - You had the right idea, but this seems to be a fairly difficult problem. My solution depends upon a non-trivial result about the median of the gamma distribution. My answer is probably not that satisfying, since it seems unlikely that participants in a math contest would know this stuff off hand; It took me a fair amount of research to come up with this solution. Hopefully someone else will come up with a more elementary argument. Edit: I've decided to offer a bounty on this question since I really want to see a simpler solution. I offer 100 points to anyone who is able to prove that $$\int_{0}^{n}e^{-t}\left(1 + t + {t^2\over 2!} + \ldots + {t^{n} \over n!}\right)dt > {n\over 2}$$ for all positive integers $n$, in a simpler manner more consistent with the spirit of a math competition. I know this is a somewhat subjective requirement, what I'm looking for is an argument a bright first year calculus student could come up with without access to the internet or other research materials. My Solution: It was a good idea to define $G(x)=F(x)-{n \over 2}$. We can use the Fundamental Theorem of Calculus to compute the derivative of $G(x)$. This yields $$G^{'}(x)=e^{-x}(1+x+{x^2 \over 2!}+{x^3 \over 3!}+\ldots + {x^n \over n!})$$ Claim: $G^{'}(x) > 0$ $\forall x \in [{n \over 2}, n]$ Proof Since $e^{-x}$ is positive for all $x$, this depends on whether or not $$p_n(x) = 1 + x +\ldots + {x^n \over n!} > 0$$. Since we are dealing with the interval $[{n\over 2}, n]$ for positive $n$, all terms of this sum are positive for $x$ in our interval. Thus $p_n(x)>0$ for $x \in [{n \over 2}, n]$. Thus we have shown that $G^{'}(x) > 0$ for $x \in [{n \over 2}, n]$. $\square$ This shows that $G(x)$ is monotonically increasing. Now we look at $G({n \over 2})$. Claim: $G({n \over 2}) < 0$. Proof. Since $1 + x + \ldots + {x^n \over n!} < e^x$ for all positive $x \in \mathbb{R}$. $$G({n \over 2}) = \int_{0}^{{n \over 2}}e^{-t}(1 + t +\ldots + {t^n \over n!})dt - {n \over 2}$$ $$< \int_{0}^{{n \over 2}}e^{-t}e^{t}dt - {n \over 2} < {n \over 2} - {n \over 2} = 0$$ $\square$ Next we consider $G(n)$. Claim: $G(n)>0$. Proof. This is the hard part $$G(n) = \int_{0}^{n}e^{-t}p_n(t)dt - {n \over 2}$$ $$p_n(x) = e^{x} - R_n(x) = e^{x} - \int_{0}^{x}{e^{t} \over n!}(x-t)^ndt$$ where we have applied Taylor's formula with the Cauchy form for the remainder. Hence $$G(n) = \int_{0}^n\left(1 - e^{-t}\int_{0}^{t}{e^{u}\over n!}(t - u)^ndu\right) dt - {n\over 2}$$ $$= \int_{0}^n\left(1 - \int_{0}^{t}{e^{u-t}\over n!}(t - u)^ndu\right) dt - {n\over 2}$$ Making the change of variables $v = t - u$ in the inner integral we get $$G(n) = \int_{0}^n\left(1 - {1\over n!}\int_{0}^{t}e^{-v}v^ndv\right) dt - {n\over 2} = \int_{0}^{n}1 - {\gamma(n+1,t) \over n!}dt - {n \over 2}$$ Where $\gamma(s,x)$ is the lower incomplete gamma function. $\gamma(s,x)$ is complementary with $\Gamma(s,x)$, the upper incomplete gamma function. $\gamma(s,x) + \Gamma(s,x) = \Gamma(s)$, the standard gamma function. Applying this fact yields $$G(n) = {1\over n!}\int_{0}^{n}\Gamma(n+1,t)dt - {n \over 2}$$ Since $\Gamma(n+1,t)$ is a decreasing function in $t$ for fixed $n$ we have $$G(n) > n{\Gamma(n+1,n)\over n!} - {n \over 2}$$ The integral is greater than the minimum value of the integrand in the interval in question times the length of the interval. Now let's take a look at the Gamma Distribution. In particular we want to consider the distribution $Gamma(n+1,1)$ which has cumulative distribution function $$f(x) = {\gamma(n+1,x) \over n!}.$$ Where $x \in (0,\infty)$. Thus if $X$ is a random variable drawn from this distribution then $$P(X < x) = {\gamma(n+1,x) \over n!}$$ We also have $$P(X > x) = {\Gamma(n+1,x) \over n!}$$ Which follows from the complementary nature of the incomplete gamma functions. According to this paper, the median $\nu$ of $Gamma(n+1,1)$ satisfies $$n + {2 \over 3} < \nu < n + \ln (2)$$. In particular this tells us that $n$ is less than the median of the distribution. Thus $${\Gamma(n+1,n) \over n!} = P(X > n) > {1 \over 2}$$ which tells us that $$G(n) > n{\Gamma(n+1,n) \over n!} - {n \over 2} > {n\over 2} - {n\over 2} = 0$$ $\square$ We have proven that $G(x)$ increases strictly monotonically on $[{n\over 2},n]$, that $G({n\over 2})$ < 0, and that $G(n) > 0$. Hence it follows that $G(x)$ has one and only root in this interval. Thus the equation $$F(x) = {n\over 2}$$ has one and only one root in $[{n\over 2}, n]$. - (+1) Thank you for showing me how to finish off the incomplete gamma function. I had also tried this path but stopped at the incomplete $\gamma$/$\Gamma$ function. I also found out that $G'(x)>0$ is extremely easy to prove because $\sum_{i=0}^n \frac{n^i}{i!}>0$ and $\exp\{\text{blah}\cdots\}>0$ when $x>0$. – FrenzY DT. Oct 22 '12 at 5:22 Let's try to find a more elementary solution w/o prob & stats because I can't rely on the median of gamma distribution during the alloted contest time. Hope you will understand. – FrenzY DT. Oct 22 '12 at 5:25 I appreciate your effort. Also that there is another good solution. – FrenzY DT. Oct 24 '12 at 3:05	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 76, "mathjax_display_tex": 25, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9465304017066956, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?t=38964&page=3	Physics Forums Thread Closed Page 3 of 4 < 1 2 3 4 > ## mass gap in Yang-Mills theories Quote by nrqed I am not sure I understand what you mean. If I follow your reasoning, you seem to be saying that short range of strong force -> massive carriers -> mass generated by SSB But that's not the explanation at all for the short range of the strong force. The gluons are massless, they don't have a mass generated by the SSB. The short range is explained by the nonlinear nature of QCD. Pat First of all it is true that gluons are generally considered to be massless, altough mass-values of a few MeV can also be possible !!! After spontaneous breakdown of symmetry gluons DO acquire mass. The process responsible for this is dynamical mass generation. The best example (of a massive-gluon-state...) are the glueball-condensates (constructed solely out of gluons) which give rise to an effective-gluon-mass without breaking gauge-invariance. Ofcourse, i have to be honest and say that the gluons themselves are massless and we are talking about an EFFECTIVE mass here. That was my point. I agree on the non lineair nature though... regards marlon Recognitions: Gold Member Staff Emeritus Gluon confinement without having mass is still not fully explained in QCD, although a lot of progress on the answer has been made. It is BELIEVED that a reverse effect like the asymptotic freedom, which has just won three physicists the Nobel proze, is responsible. Quote by selfAdjoint Gluon confinement without having mass is still not fully explained in QCD, although a lot of progress on the answer has been made. It is BELIEVED that a reverse effect like the asymptotic freedom, which has just won three physicists the Nobel proze, is responsible. Correct, The best model up till now that describes the confinement phenomenon is the dual abelian higgs model. In this model gluons with both colour and NO colour are predicted. So not all gluons undergo confinement since not all gluons contain colour. These last gluons are called the abelian gluons. If somebody wanna know more, please consult the last link i provided in the "elementary particles presented thread" regards marlon Quote by humanino Hey Marlon, what's up old dude ! I am not sure that the mass gap is accountable by Higgs field. I heard stuff like "10% of the mass of the proton is due to the higgs field. The 90% remaining is the weight of the glue". But I don't undestand it. Hi, Humanino... You are right on this statement although the 10/90 comparison in weight is something i have never heard of... Your post is a very good one in my opinion since it insists on making clear the different kinds of mass we need to look at in these subjects. For starters, we have the real physical mass that we measure in experiments. this mass is gained by the Higgsfield in the case of real matter-particles. Now ofcourse massless gauge-bosons can also acquire mass (yes, i am primarily referring to GLUONS here via interaction with the Higgsfield) This mass is of a different kind though since it is called effective mass. This is mass generated by the self-interactions of such particles. Just look at how gluon-condensates are formed out of dynamical mass-generation. When you are talking about the glue, you are basically referring to this kind of mass. I am sure you know these things like effective-mass in solid state physics and the quasi-particles in many-body-problems. These particles reduce one many-body-problem that we cannot solve, by many one body problems that we CAN solve by lumping together all the interactions of one particle with many surrounding particles and putting this into the self-energy of one particle and "forgetting about all the other surrounding particles". This final particle (the quasi-particle) is then considered to be free at first extent.... regards marlon Do recall that the massive gauge bosons of the weak force have a real physical mass.... marlon Recognitions: Homework Help Science Advisor Quote by nrqed I am not sure I understand what you mean. If I follow your reasoning, you seem to be saying that short range of strong force -> massive carriers -> mass generated by SSB But that's not the explanation at all for the short range of the strong force. The gluons are massless, they don't have a mass generated by the SSB. The short range is explained by the nonlinear nature of QCD. Pat Quote by marlon First of all it is true that gluons are generally considered to be massless, altough mass-values of a few MeV can also be possible !!! After spontaneous breakdown of symmetry gluons DO acquire mass. The process responsible for this is dynamical mass generation. The best example (of a massive-gluon-state...) are the glueball-condensates (constructed solely out of gluons) which give rise to an effective-gluon-mass without breaking gauge-invariance. Ofcourse, i have to be honest and say that the gluons themselves are massless and we are talking about an EFFECTIVE mass here. That was my point. I agree on the non lineair nature though... regards marlon Hi Marlon, I agree that it is dynamical mass generation. My point was that this has nothing to do with the Higgs mechanism (which seemed to be what you were hinting at in your previous post). That was the only point I wanted to make. Best regards, Pat Recognitions: Homework Help Science Advisor Quote by marlon Hi, Humanino... You are right on this statement although the 10/90 comparison in weight is something i have never heard of... Your post is a very good one in my opinion since it insists on making clear the different kinds of mass we need to look at in these subjects. For starters, we have the real physical mass that we measure in experiments. this mass is gained by the Higgsfield in the case of real matter-particles. Now ofcourse massless gauge-bosons can also acquire mass (yes, i am primarily referring to GLUONS here via interaction with the Higgsfield) ..... Now I am quite confused. Could you point to me the interaction term between the gluons and the Higgs in the Standard Model? As far as I know, the only interaction between gauge bosons and the Higgs is through the covariant derivative $\vec D_\mu \Phi = (\partial_\mu - {i\over 2} g \vec \tau \cdot \vec A_\mu -{i \over 2} g' B_\mu) \Phi$ (see Cheng and Li, page 349). Where the A's and the B are the fields which will become the W^+, W^-, Z^0 and the photon after SSB. There is no coupling between the Higgs and the gluons. So what am I missing? Regards Pat I never said that, sorry. Gluonmass is EFFECTIVE mass, that is my point. Your formula is the standard one in QFT but not complete for the needs of QCD. This is a QED-thing. There are models (like the dual Landau Ginzburg-model) that predict massive gluons via the Higgs-field. The criticism on this model is the fact that this mass value is quite low and the Higgs-particles themselves have low mass. The big question then is ofcourse : if this mass is so low, how come we did not see these Higgs-particles yet. I am sure you will agree this is a very powerful counter-statement. Problem is that this model does the BEST job in describing the nature of meson and baryon-configurations. Probably (this is just my thought, so it sure ain't no FACT) the idea of magnetic monopoles forming the flux-tube is the correct way to look at confinement (because of elegance and more over SYMMETRY) yet problems arise with how to construct the linear interquark-potential. regards marlon So : here are quotations from "Gauge Fields and strings". Even the negative results are interesting, remember that it is an excellent book to read. Quote by M. Polyakov, excerpt from §6 (from intro §6) We have seen in the previous chapters that in Abelian systems the problem of charge confinement is solved by instantons. In Non-Abelian theories instantons are also present. However, due to the large perturbative fluctuations, dicussed in Chapter 2, it is difficult to judge whether they play a decisive role in forming a mass gap and a confining regime. In such theories we had a kind of instanton liquid which is difficult to treat. It is possible that due to some hidden symmetries, present in these systems, instantons may form a useful set of variables for an exact description of the system, but this has not yet been shown. At the same time, due to the fact that instantons carry non-trivial topology (they describe configurations of the fields which cannot be "disentangled"), some manifestations of instantons cannot be mixed up with perturbative fluctuations. [...] (from end of §6.2) As happened in the case of n-fields, the instanton contribution has an infrared divergence. This implies that in the multi-instanton picture, individual instantons tend to grow and to overlap. The vaive dilute gas approximation is certainly inapplicable then, and we should expect somethig like dissociation of dipole-like instantons to their elementary constituents, as happened in the case of n-fields. However, even one loop computations on the multi-instanton background have not yet been performed, and nothing similar to the Coulomb plasma of the previous section has been discovered. This is connected partly with the fact that multi-instanton solutions have not yet been explicitely parameterized up to now. I expect many interesting surprises await us, even on the one loop level, in this hard problem. [...] (from end of §6.2) So, our conclusion is that on the present level of understanding of instanton dynamics, we cannot obtain any exact dynamical statements concerning Non-Abelian gauge theory. In the case of n-fields the situation is slightly better, since we were able to demonstrate the apearance of the mass-gap on a qualitative level. Even in this case one would like to have much deeper understanding of the situation. There are reasons to believe that some considerable progress will be achieved in the near future. In the case of gauge fields we have to pray for luck. At the same time, the existence of fields with topological charge has a deep qualitative influence on the dynamical structure of the theory. [...] (from end of §6.3) (...) exchange of a massless fermion pair leads to long-range forces between instantons and anti-instantons. The result of this may have several alternative consequences. The first one is that since (6.87) implies quenching of large fluctuations in the presence of massless fermions, the system looses the confining property and we would end up with massless gauge fields together with fermions. This option seems highly improbable to me on the basis of some analogies and some model considerations. However, I am not aware of any strict statements permitting us to reject it. The second possibility, which in my opinion is realized in the theory, is the following. Due to the strong binding force between fermions the chiral symmetry gets spontaneously broken and as a result the fermions acquire mass. After that has happened, the long range force between instantons and anti-instantons disappears, being screened by the fermionic mass term in the effective lagrangian. The only remaining effect of anomalous non-conservation will consist of giving a mass to the corresponding Goldstone boson. There is also another improbable option, namely that instantons get confined but some type of large fluctuations, not suppressed by fermions, disorder the system. (follows a short but excellent account on the $$\theta$$-term and the (failure of the) search for axion particle) Recognitions: Science Advisor I'm a little skeptical of that mass gap paper. Probably b/c im partial to lattice QCD, and its rather apparent that the way the gap appears in that (admittedly numerical) formalism strikes me at odds with some of the claims of the paper. I'll reread it again more thoroughly later. Progress towards understanding the mass-gap in QCD : Precise Quark Mass Dependence of Instanton Determinant The fermion determinant in an instanton background for a quark field of arbitrary mass is determined exactly using an efficient numerical method to evaluate the determinant of a partial wave radial differential operator. The bare sum over partial waves is divergent but can be renormalized in the minimal subtraction scheme using the result of WKB analysis of the large partial wave contribution. Previously, only a few leading terms in the extreme small and large mass limits were known for the corresponding effective action. Our approach works for any quark mass and interpolates smoothly between the analytically known small and large mass expansions. Gerald V. Dunne, Jin Hur, Choonkyu Lee, Hyunsoo Min (hep-th/0410190) Quote by marlon Correct, The best model up till now that describes the confinement phenomenon is the dual abelian higgs model. In this model gluons with both colour and NO colour are predicted. So not all gluons undergo confinement since not all gluons contain colour. These last gluons are called the abelian gluons. regards marlon I believe the best explanation now for confinement begins with W. P. Joyce "Quark state confinement as a consequence of the extension of the Bose-Fermi recoupling to SU(3) colour" J. Phys. A: Math. Gen. 36 (2003) 12329 - 12341 This work can now be fit into a much more general framework, either via Joyce's so-called omega algebras (recent work) or equivalently from the perspective of higher categories where these algebraic structures appear naturally. Moreover the mathematics has a close tie to LQG (this is mostly unpublished) and a big motivation for it was instantons, or rather Twistor theory, because the biggest hurdle seemed to be a sufficiently rich non-abelian cohomology. More on all this elsewhere, and at a later date. Cheers Kea Penrose developed Twistor theory from a deep understanding (I believe) of GR. The correspondence is in terms of sheaf cohomology. This was extended to H2 by Hughston and Hurd in the 80s to study the Klein-Gordon equation (ie. adding mass). Ross Street, in his classic '87 paper on Oriented Simplices, explains why non-Abelian cohomology in higher dimensions is difficult. This paper lays out the structure of a 'nerve' of a strict n-category. But for reasons I won't go into here, physics seems to require much more than this: a fully higher categorical cohomology, which is still being developed by Street and others. The question is: what does this have to do with the mass gap issue? Recall that Heisenberg said that he was led to the uncertainty principle by recalling Einstein's words to the effect "the theory always dictates what is observable". In other words, the classical theory is reproduced in a very different (and complicated) way to the idea of taking 'hbar to zero'. For instance, in a topos one must be careful to define what one means by the reals, because the Cauchy reals and Dedekind reals aren't the same. Well, the crazy physical idea...... the classical limit we should be thinking about is something to do with twistors. Now it turns out that Roy Kerr discovered his solution to Einstein's equations by thinking about this sort of maths. Anyway, if there IS NO 'fixed background', which of course there isn't, then the mass gap that we have in the MORE FUNDAMENTAL unified theory goes away because the only 'proper' classical solutions concentrate the mass in things like Kerr black holes. Kea Blog Entries: 6 Recognitions: Gold Member A question: in which measure is true that to explain the mass-gap implies to explain the mass of the proton? It is sort of assuming that a proton and a glueball are almost the same thing, isn't it? Quote by arivero A question: in which measure is true that to explain the mass-gap implies to explain the mass of the proton? It is sort of assuming that a proton and a glueball are almost the same thing, isn't it? I'm saying we can't explain the mass gap without quantum gravity - and if we understand that, the mass of the proton should follow. Recognitions: Science Advisor Sorry that doesn't make much sense. Gravitational effects are completely negligable at that length scale. Even if the mass gap is seen via perturbative effects, gravity will miss it order by order in the series. However if gravity did couple to the theory in some way, it would not only lead to some complicated lagrangian, but presumably incorporate a host of gauge symmetry breaking terms to make it feasible. Moroever, we would have to introduce fine tuning terms many orders of magnitude uglier than the dual abelian higgs model. As clearly stated in the millenium problem writeup, most people expect the mass gap to appear in the quartic interaction sector of the theory (A ^ A)^2. Not only b/c of duality transitions, but also b/c it would make sense and generalize simpler toy model lagrangians, where existence of mass gaps have been rigorously shown to exist. Finally, the mass gap has been solved by computer and found to be within 1-2% of the predicted value, via lattice QCD. It means we have the right equation, solving it analytically is what now remains to be done. Adding adhoc speculation about extra non field theoretic interactions is more or less ruled out. Quote by Haelfix It means we have the right equation, solving it analytically is what now remains to be done. Solving it analytically might require quantum gravity. Thread Closed Page 3 of 4 < 1 2 3 4 > Thread Tools \| \| \| \| \|------------------------------------------------------\|----------------------------------------\|---------\| \| Similar Threads for: mass gap in Yang-Mills theories \| \| \| \| Thread \| Forum \| Replies \| \| \| High Energy, Nuclear, Particle Physics \| 4 \| \| \| Quantum Physics \| 1 \| \| \| General Physics \| 2 \| \| \| General Physics \| 2 \| \| \| General Physics \| 2 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9416568875312805, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/93218?sort=oldest	## Convergence of stochastic process ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) As we know, to prove the convergence of stochastic process, we could either show the convergence of finite dimensional distribution and tightness of the process, or use techniques of martingale problems. What about the following Markov process: $L=\frac{1}{2}p(1-p)\frac{d^{2}}{dp^{2}}-\frac{\theta}{2}p\frac{d}{dp}+\log(\theta) p(1-p)(2p-1)\frac{d}{dp}, p\in[0,1]$ We can see that the generator explodes when $\theta\rightarrow0$. How can we find the limit of this process as $\theta\rightarrow0$. Apparently, the techniques of martingale problems are not applicable here! - ## 1 Answer This drifted Wright--Fisher diffusion seems to converge to the rather degenerate process $X_0 = p$, $X_t = 1/2$ for $t>0$, which is why I would do it by hand: Show that for every $t>0$ and every $p\in(0,1)$, the process started from $p$ is with high probably near $1/2$ at time $t$, uniformly for $p\in[\epsilon,1-\epsilon]$, say (you can use standard arguments from diffusion theory for that, i.e. speed measure, Green function...). This yields convergence in finite-dimensional distributions. Convergence in Skorokhod topology will not hold, since continuous functions are closed in this topology, hence a continuous process cannot converge to a process with a jump. - I have the similar idea, but i just don't know how to verify it. Thank you so much for your answer! – John Young Apr 5 2012 at 15:05	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9323979020118713, "perplexity_flag": "head"}
http://www.ams.org/bookstore?fn=20&arg1=stmlseries&ikey=STML-37	New Titles \| FAQ \| Keep Informed \| Review Cart \| Contact Us Quick Search (Advanced Search ) Browse by Subject General Interest Logic & Foundations Number Theory Algebra & Algebraic Geometry Discrete Math & Combinatorics Analysis Differential Equations Geometry & Topology Probability & Statistics Applications Mathematical Physics Math Education Return to List $$p$$-adic Analysis Compared with Real Svetlana Katok, Pennsylvania State University, University Park, PA SEARCH THIS BOOK: Student Mathematical Library 2007; 152 pp; softcover Volume: 37 ISBN-10: 0-8218-4220-X ISBN-13: 978-0-8218-4220-1 List Price: US\$30 Member Price: US\$24 Order Code: STML/37 The book gives an introduction to $$p$$-adic numbers from the point of view of number theory, topology, and analysis. Compared to other books on the subject, its novelty is both a particularly balanced approach to these three points of view and an emphasis on topics accessible to undergraduates. In addition, several topics from real analysis and elementary topology which are not usually covered in undergraduate courses (totally disconnected spaces and Cantor sets, points of discontinuity of maps and the Baire Category Theorem, surjectivity of isometries of compact metric spaces) are also included in the book. They will enhance the reader's understanding of real analysis and intertwine the real and $$p$$-adic contexts of the book. The book is based on an advanced undergraduate course given by the author. The choice of the topic was motivated by the internal beauty of the subject of $$p$$-adic analysis, an unusual one in the undergraduate curriculum, and abundant opportunities to compare it with its much more familiar real counterpart. The book includes a large number of exercises. Answers, hints, and solutions for most of them appear at the end of the book. Well written, with obvious care for the reader, the book can be successfully used in a topic course or for self-study. This book is published in cooperation with Mathematics Advanced Study Semesters. Readership Undergraduate and graduate students interested in $$p$$-adic numbers. Reviews "...the book gives a good impetus to students to study the "p-adic worlds" more deeply. This role of the book is not only supported by carefully selected material but also by the fact that it is written in a very lively and lucid style." "I think that the reading of this book could animate some students to start to do research $$p$$-adic work. A good decision from my point of view!"	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9202556610107422, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/77816?sort=votes	## Do these matrix rings have non-zero elements that are neither units nor zero divisors? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) First, a disclaimer: This is a repost of a question I asked on stackexchange (no answer there). Let $R$ be a commutative ring (with $1$) and $R^{n \times n}$ be the ring of $n \times n$ matrices with entries in $R$. In addition, suppose that $R$ is a ring in which every non-zero element is either a zero divisor or a unit [For example: take any finite ring or any field.] My question: Is every non-zero element of $R^{n \times n}$ a zero divisor or a unit as well? We know that if $A \in R^{n \times n}$, then $AC=CA=\mathrm{det}(A)I_n$ where $C$ is the classical adjoint of $A$ and $I_n$ is the identity matrix. This means that if $\mathrm{det}(A)$ is a unit of $R$, then $A$ is a unit of $R^{n \times n}$ (since $A^{-1}=(\mathrm{det}(A))^{-1}C$). Also, the converse holds, if $A$ is a unit of $R^{n \times n}$, then $\mathrm{det}(A)$ is a unit. I would like to know if one can show $0 \not= A \in R^{n \times n}$ is a zero divisor if $\mathrm{det}(A)$ is zero or a zero divisor. Things to consider: 1) This is true when $R=\mathbb{F}$ a field. Since over a field (no zero divisors) and if $\mathrm{det}(A)=0$ then $Ax=0$ has a non-trivial solution and so $B=[x\|0\|\cdots\|0]$ gives us a right zero divisor $AB=0$. 2) You can't use the classical adjoint to construct a zero divisor since it can be zero even when $A$ is not zero. For example: $$A=\begin{pmatrix} 1 & 1 & 1 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \end{pmatrix} \qquad \mathrm{implies} \qquad \mathrm{classical\;adjoint} = 0$$ (All $2 \times 2$ sub-determinants are zero.) 3) This is true when $R$ is finite (since $R^{n \times n}$ would be finite as well). 4) Of course the assumption that every non-zero element of $R$ is either a zero divisor or unit is necessary since otherwise take a non-zero, non-zero divisor, non-unit element $r$ and construct the diagonal matrix $D = \mathrm{diag}(r,1,\dots,1)$ (this is non-zero, not a zero divisor, and is not a unit). 5) This is somewhat related to the question: http://mathoverflow.net/questions/42647/rings-in-which-every-non-unit-is-a-zero-divisor 6) This is definitely true when $n=1$ and $n=2$. It is true for $n=1$ by assumption on $R$. To see that $n=2$ is true notice that the classical adjoint contains the same same elements as that of $A$ (or negations): $$A = \begin{pmatrix} a_{11} & a_{12} \cr a_{21} & a_{22} \end{pmatrix} \qquad \Longrightarrow \qquad \mathrm{classical\;adjoint} = C = \begin{pmatrix} a_{22} & -a_{12} \ -a_{21} & a_{22} \end{pmatrix}$$ Thus if $\mathrm{det}(A)b=0$ for some $b \not=0$, then either $bC=0$ so that all of the entries of both $A$ and $C$ are annihilated by $b$ so that $A(bI_2)=0$ or $bC \not=0$ and so $A(Cb)=\mathrm{det}(A)bI_2 =0I_2=0$. Thus $A$ is a zero divisor. 7) Apparently strange behavior can occur when $R$ is non-commutative (not surprising). Like a matrix can be both a left inverse and left zero divisor. [The determinant keeps this from happening in the commutative case.] - I'm not sure why, but the "\\" are being ignored in my matrices! – Bill Cook Oct 11 2011 at 14:06 1 You must use \\\\ – Wilberd van der Kallen Oct 11 2011 at 14:14 4 Better suggestion than Wilberd's: use \cr rather than \\ in matrices. MO has re-taught me that most characters that could be considered special have alternate command names in LaTeX, which is very helpful when the MO software wants to interpret them as special characters. E.g. "<" is bad, but you can use \lt instead, you can use \lbrace for "{" etc. – Thierry Zell Oct 11 2011 at 14:30 1 Sorry for taking too long to write up the answer at math.SE -- David beat me to it by 20 minutes :-). Here it is: math.stackexchange.com/questions/71235/… – joriki Oct 11 2011 at 16:29 1 Actually, as far as I've seen definitions, $0$ is usually not called a zero divisor (so an integral domain is a non-trivial commutative ring without zero divisors). But in the current setting saying "zero divisor of zero" all the time would indeed by tiresome. – Marc van Leeuwen Mar 10 at 11:15 show 4 more comments ## 1 Answer The answer given by David Speyer can be strengthened as follows. If $A$ is a non-invertible $n\times n$ matrix with entries in $R$ as described in the problem, then the linear maps $R^n \to R^n$ defined by either left or right multiplication are non-injective. In particular, $A$ is both a left-zero-divisor and a right-zero-divisor. This is a consequence of McCoy's rank theorem. You can find a nice, brief account of this in Section 2 of this paper by Kodiyalam, Lam, and Swan. One consequence of the theorem is that for any commutative ring $R$, a square matrix $A$ over $R$ has linearly independent columns if and only if its determinant is not a zero-divisor, if and only if its rows are linearly independent. So if every element of $R$ is either invertible or a zero-divisor, this means that every square matrix over $R$ defines a linear transformation that is either invertible or non-injective. - 1 You can find a proof of McCoy's theorem here: www2.im.uj.edu.pl/actamath/PDF/30-215-218.pdf – Manny Reyes Oct 11 2011 at 16:46 2 To repeat a comment from my answer, once you know that $A:R^n \to R^n$ has a kernel, say $Av=0$, then let $V$ be the $n \times n$ matrix whose columns are all copies of $v$ and you get $AV=0$. That's a nice proof, thanks for finding it! – David Speyer Oct 11 2011 at 17:13 Right, David - thanks for being explicit for me! – Manny Reyes Oct 11 2011 at 17:22 Many thanks Manny! – Bill Cook Oct 11 2011 at 17:24	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 68, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9009101986885071, "perplexity_flag": "head"}
http://mathhelpforum.com/discrete-math/144813-combinatorial-analysis.html	# Thread: 1. ## Combinatorial analysis A set of playing cards consists of 36 cards. These 36 cars are subdivided into 9 types (A,B,C, (...)I) with four cards each. The cards within these subtypes do not differ (A1=A2=A3=A4). How many possibilites are there, to take up 9 cards? The order of taking up cards and the order of the taken up cards do not matter (AAABBCCCD=ABABACDCC). I somehow do not get a correct result. First i thought of how man possibilites there are to take 9 cards from 36, but i dont know how to get rid of all possibilites. I tried 36!/27!4!^9, but this is totally wrong... 2. Originally Posted by Schdero A set of playing cards consists of 36 cards. These 36 cars are subdivided into 9 types (A,B,C, (...)I) with four cards each. The cards within these subtypes do not differ (A1=A2=A3=A4). How many possibilites are there, to take up 9 cards? The order of taking up cards and the order of the taken up cards do not matter (AAABBCCCD=ABABACDCC). I somehow do not get a correct result. First i thought of how man possibilites there are to take 9 cards from 36, but i dont know how to get rid of all possibilites. I tried 36!/27!4!^9, but this is totally wrong... Here is a feasible but somewhat laborious way of doing it. Consider, for example, the case where you are given 4 cards from one of the 9 subtypes, 3 from another, and 1 from yet two other subtypes. How many cases of that general sort, "4+3+1+1" are there? Well, you can choose the subtype that contributes 4 cards in 9 ways. Having choosen that subtype, you can choose the subtype that contributes 3 cards in 8 ways, and the two subtypes that contribute 1 card each can be chosen in $\binom{7}{2}$ ways. In all, there are $9\cdot 8\cdot\binom{7}{2}=1512$ different cases of being dealt 9 cards that follow this general "4+3+1+1" pattern. So, I think, it comes down to listing all these possible ways of partitioning the number 9 into a sum of at most 9 non-negative integers ( $\leq 4$) and, for each of these partitionings, determining how many ways there are to get them by combining the 9 subtypes of cards accordingly. Such a list of partitionings might start like this (I am leaving out trailing 0s) $\begin{array}{lcl}9 &=&4+4+1\\<br /> &=& 4+3+2\\<br /> &=& 4+3+1+1\\<br /> &=& 4+2+2+1\\<br /> &\vdots& \vdots<br /> \end{array}$ To keept this list reasonably small, you should only list the sums that have non-increasing terms (from left to right). 3. Originally Posted by Schdero A set of playing cards consists of 36 cards. These 36 cars are subdivided into 9 types (A,B,C, (...)I) with four cards each. The cards within these subtypes do not differ (A1=A2=A3=A4). How many possibilites are there, to take up 9 cards? If we expand $(1+x+x^2+x^3+x^4)^9$ the coefficient of $x^9$ is 19855. That is the answer to your question done with a generating function. 4. Originally Posted by Plato If we expand $(1+x+x^2+x^3+x^4)^9$ the coefficient of $x^9$ is 19855. That is the answer to your question done with a generating function. That's a great way of doing it, except for two possible problems: (1) that the OP does not know about the use of generating functions in enumerative combinatorics, and, (2) that he does not have access to a CAS to determine the coefficient of $x^9$ so very easily as the above statement seems to suggest. In fact, it seems to me that if he does not have access to a CAS (or is not allowed to use a CAS) for that purpose, he might as well use the laborious procedure that I have suggested.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9537558555603027, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/35220/cohomology-analogue-for-central-series-of-length-more-than-two/35693	## Cohomology analogue for central series of length more than two ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) It is a basic result of group cohomology that the extensions with a given abelian normal subgroup A and a given quotient G acting on it via an action $\varphi$ are given by the second cohomology group $H^2_\varphi(G,A)$. In particular, when the action is trivial (so the extension is a central extension), this is the second cohomology group $H^2(G,A)$ for the trivial action. In the special case where G is also abelian, we classify all the class two groups with A inside the center and G as the quotient group. I am interested in the following: given a sequence of abelian groups $A_1, A_2, \dots, A_n$, what would classify (up to the usual notion of equivalence via commutative diagrams) the following: a group E with an ascending chain of subgroups: $$1 = K_0 \le K_1 \le K_2 \le \dots \le K_n = E$$ such that the $K_i$s form a central series (i.e., $[E,K_i] \subseteq K_{i-1}$ for all i) and $K_i/K_{i-1} \cong A_i$? The case $n = 2$ reduces to the second cohomology group as detailed in the first paragraph, so I am hoping that some suitable generalization involving cohomology would help describe these extensions. Note: As is the case with the second cohomology group, I expect the object to classify, not isomorphism classes of possibilities of the big group, but a notion of equivalence class under a congruence notion that generalizes the notion of congruence of extensions. Then, using the actions of various automorphism groups, we can use orbits under the action to classify extensions under more generous notion of equivalence. Note 2: The crude approach that I am aware of involves building the extension step by step, giving something like a group of groups of groups of groups of ... For intsance, in the case $n = 3$: $$1 = K_0 \le K_1 \le K_2 \le K_3 = G$$ with quotients $A_i \cong K_i/K_{i-1}$, I can first consider $H^2(A_3,A_2)$ as the set of possibilities for $K_3/K_1$ (up to congruence). For each of these possibilities P, there is a group $H^2(P,A_1)$ and the total set of possibilities seems to be: $$\bigsqcup_{P \in H^2(A_3,A_2)} H^2(P,A_1)$$ Here the $\in$ notation is being abused somewhat by identifying an element of a cohomology group with the corresponding extension's middle group. What I really want is some algebraic way of thinking of this unwieldy disjoint union as a single object, or some information or ideas about its properties or behavior. - ## 3 Answers This looks like a (slightly) non-additive version of Grothendieck's theory of "extensions panachées" (SGA 7/I, IX.9.3). There he considers objects (in some abelian category) $X$ together with a filtation $0\subseteq X_1\subseteq X_2\subseteq X_3=X$. In the first version he also fixes (just as one does for extensions) isomorphisms $P\rightarrow X_1$, $Q\rightarrow X_2/X_1$ and $R\rightarrow X_3/X_2$. However, in the next version he fixes the isomorphism class of the two extensions $0\rightarrow P\rightarrow X_2\rightarrow Q\rightarrow0$ and $0\rightarrow Q\rightarrow X_3/X_1\rightarrow R\rightarrow0$ so that if $E$ is an extension of $P$ by $Q$ and $F$ is an extension of $Q$ by $R$, then the category $\mathrm{EXTP}(F,E)$ has as objects filtered objects $X$ as above together with fixed isomorphisms of extensions $E\rightarrow X_2$ and $F\rightarrow X_3/X_1$ and whose morphisms are are morphisms of $X$'s preserving the given structures. The morphisms of $\mathrm{EXTP}(F,E)$ are necessarily isomorphisms so we are dealing with a groupoid. Similarly for objects $A$ and $B$ $\mathrm{EXT}(B,A)$ is the groupoid of extensions of $B$ by $A$. Grothendieck then shows that $\mathrm{EXTP}(F,E)$ is a torsor over $\mathrm{EXT}(R,P)$ (in the category of torsors, Grothendieck had previously defined this notion). The action on objects of an extension $0\rightarrow P\rightarrow G\rightarrow R\rightarrow0$ is given by first taking the pullback of it under the map $X/X_1\rightarrow R$ and then using the obtained action by addition on extensions of $P$ by $F$. To more or less complete the picture, there is an obstruction to the existence of an object of $\mathrm{EXTP}(F,E)$: We have that $E$ gives an element of $\mathrm{Ext}^1(Q,P)$ and $F$ one of $\mathrm{Ext}^1(R,Q)$ and their Yoneda product gives an obstruction in $\mathrm{Ext}^2(P,Q)$. The case at hand is similar (staying at the case of $n=3$ and with the caveat that I haven't properly checked everything): We choose fixed isomorphisms with $K_2$ and a given central extension and with $K_3/K_1$ and another given central extension (assuming that we have three groups $P$, $Q$ and $R$ as before) getting a category $\mathrm{CEXTP}(F,E)$ of central extensions. We shall shortly modify it but to motivate that modification it seems a good idea to start with this. We get as before an action of $\mathrm{CEXT}(R,P)$ on $\mathrm{CEXTP}(F,E)$ as we can pull back central extensions just as before. It turns however that the action is not transitive. In fact we can analyse both the difference between two elements of $\mathrm{CEXTP}(F,E)$ and the obstructions for the non-emptyness of it by using the Hochschild-Serre spectral sequence. To make it easier to understand I use a more generic notation. Hence we have a central extension $1\rightarrow K\rightarrow G\rightarrow G/K\rightarrow1$ and an abelian group $M$ with trivial $G$-action. There is then a succession of two obstructions for the condition that a given central extension of $M$ by $G/K$ extend to a central extension of $M$ by $G$. The first is $d_2\colon H^2(G/K,M)\rightarrow H^2(G/K,H^1(K,M))$, the $d_2$-differential of the H-S s.s. Now, we always have a map $H^2(G/K,M)\rightarrow H^2(G/K,H^1(K,M))$ given by pushout of $1\rightarrow G\rightarrow G/K\rightarrow1$ along the map $K\rightarrow \mathrm{Hom}(K,M)=H^1(K,M)$ given by the action by conjugation of $K$ on the given central extension of $M$ by $K$ (equivalently this map is given by the commutator map in that extension). It is easy to compute and identify $d_2$ but I just claim that it is equal to that map by an appeal to the What Else Can It Be-principle (which works quite well for the beginnings of spectral sequences with the usual provisio that the WECIB-principle only works up to a sign). This means that we can cut down on the number of obstructions by redefining $\mathrm{CEXTP}(F,E)$. We add as data a group homomorphism $\varphi\colon K_3/K_1\rightarrow\mathrm{Hom}(Q,P)$ that extends $Q\rightarrow \mathrm{Hom}(Q,P)$ which describes the conjugation action on $K_2$ and only look the elements of $\mathrm{CEXTP}(F,E)$ for which the action is the given $\varphi$ to form $\mathrm{CEXTP}(F,E;\varphi)$. Now the action of $\mathrm{CEXT}(R,P)$ on $\mathrm{CEXTP}(F,E;\varphi)$ should make $\mathrm{CEXTP}(F,E;\varphi)$ a $\mathrm{CEXT}(R,P)$-(pseudo)torsor. Furthermore, there is now only a single obstruction for non-emptyness which is given by $d_3\colon H^2(R,M)\rightarrow H^3(P,M)$. Going to higher lengths there are two ways of proceeding in the original Grothendieck situation: Either one can look at the the two extensions of one length lower, one ending with the next to last layer (i.e., $X_{n-1}$) and the other being $X/X_1$. This reduces the problem directly to the original case (i.e., we look at filtrations of length $n-2$ on $Q$). One could instead look at the successive two-step extensions and then look at how adjacent ones build up three-step extensions and so on. This is essentially an obstruction theory point of view and quickly becomes quite messy. An interesting thing is however the following: We saw that in the original situation the obstruction for getting a three step extension was that $ab=0$ for the Yoneda product of the two twostep filtrations. If we have a sequence of three twostep extensions whose three step extensions exist then we have $ab=bc=0$. The obstruction for the existence of the full fourstep extension is then essentially a Massey product $\langle a,b,c\rangle$ (defined up to the usual ambiguity). The messiness of such an iterated approach is well-known, it becomes more and more difficult to keep track of the ambiguities of higher Massey products. The modern way of handling that problem is to use an $A_\infty$-structure and it is quite possible (maybe even likely) that such a structure is involved. If we turn to the current situation and arbitrary $n$ then the first approach has problems in that the midlayer won't be abelian anymore and I haven't looked into what one could do. As for the second approach I haven't even looked into what the higher obstructions would look like (the definition of the first obstruction on terms of $d_3$ is very asymmetric). - @Torsten: This looks potentially very useful, but I don't have a good background in this area. Could you give some references to expositions on the background ideas here, and also perhaps point out more explicitly which of the terminology you've made up, and which of it was made up by others? Thanks a lot. Thanks for taking the time to think about this. – Vipul Naik Aug 16 2010 at 15:03 As far as I know the "extension panachée" theory is only presented in the reference to SGA that I gave. I don't know of any reference that extends this to the central extension case. As far as I can see the only terminology I made up is CEXTP (if I recall correctly Grothendieck uses CEXT for central extensions, note that he uses all capital letters to denote the category of somethings). I guess I should also take (not very much) credit for naming the very useful WECIB-principle :-) By the way I think I was too pessimistic of extending the method to general $n$, it seems to more or less extend. – Torsten Ekedahl Aug 17 2010 at 7:54 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. There is probably no easy answer to this question. Even the special case when all groups are of order 2 seems hopeless. While it is not the same as the problem of classifying all finite groups of order a power of 2, it is similar, and probably of about the same difficulty. Classifying groups of order 2n is known (or at least thought) to be a complete mess. The number of such groups grows rapidly with n: 1, 1, 2, 5, 14, 51, 267, 2328, 56092, 10494213, 49487365422 http://www.research.att.com/~njas/sequences/A000679 Moreover if one actually looks at the collection of groups one gets, there does not seem to be much obvious nice structure. And if you change the prime 2 to some other prime such as 3, the answer you gets seems to change qualitatively: there are 2-groups that do not seem to be analogous to any 3-groups, and vice versa. Looking at the smallish groups of this form does not seem to give any hints of any usable structure on the set of groups with such an ascending chain of subgroups. The people who classify 2-groups have presumably thought quite hard about this problem, and as far as I know have not come up with any easy solution: the groups are classified with a lot of hard work and computer time. - It's not completely clear to me why this problem should be as hard as the classification of groups of order 2^n. The object being constructed here contains a lot of repetitions and multiple countings of isomorphic groups -- the hard part could lie in identifying the repetitions, which is needed for a final classification up to isomorphism. The second cohomology group $H^2(G,A)$ contains a lot of elements, may of which give groups that later turn out to be isomorphic. Also, my interest is not in explicitly computing the object but in what algebraic properties it has and how to view it. – Vipul Naik Aug 15 2010 at 23:17 2 I agree with Vipul, the difficult part for classification seems to be to reconcile two different views of the same group (i.e., the same group may have two different filtrations) and then compute the orbits under the action of the appropriate automorphism groups. The first problem can be alleviated by picking some canonical filtration such as upper or lower central series but then you have to identify which filtrations are of the chosen canonical type. However, calculations also of extension groups would be difficult but even just a conceptual framework can be useful. – Torsten Ekedahl Aug 16 2010 at 9:33 You might find the material on the interpretation of $\mathrm{Ext}$ in terms of extensions at Ext and Extensions to be useful. You probably know that $H^n(G,M) = \mathrm{Ext}^n_{\mathbb{Z}[G]}(\mathbb{Z},M)$ (which is not exactly the same as the relation between $H^2$ and extensions, but it is similar). You might be able to construct a kind of Baer sum on these central series by taking pullbacks and pushouts, which makes the set of central series into a group. This makes sense, given that the method of adding group extensions which puts it in isomorphism with $H^2$ is known as Baer multiplication (c.f. Weiss, Cohomology of Groups). -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 104, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9572389125823975, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/35658/how-can-we-define-bf-theory-on-a-general-4-manifold/38356	How can we define BF theory on a general 4-manifold? (I have rewritten the question some, with new understanding) 4d BF theory is classically presented as the TFT arising from the Lagrangian $B\wedge F$, where $B$ is an abelian 2-connection (locally a real 2-form) and $F$ is the curvature of a connection $A$. People often throw about this theory on general 4-manifolds, but it seems the naive definition from the Lagrangian above is anomalous for most situations. If one restricts to the simpler case of $A$ and $B$ being globally defined forms (not considering them as connections) the phase space is $H^1(\Sigma_4,\mathbb{R}) + H^2(\Sigma_4,\mathbb{R})$ with $A$ as the first coordinate and $B$ as the second. $A$ and $B$ are conjugate from the point of view of the above Lagrangian, but this has no chance of being symplectic if the first and second Betti numbers are not equal. If $A$ and $B$ are connections, however, the only solution to the equations of motion (which not only set $dA=dB=0$ but also their holonomies) is the trivial one, and it is not a problem However, what I'd really like to consider is $nB\wedge F$. With $A$ and $B$ full-fledged connections. Let us first integrate out $B$. $B$ is a (2-)connection, so $dB$ can be any integral 3-form. We can thus write it as $d\beta + \sum_k m^k \lambda_k$, where $\beta$ is an ordinary real 2-form, and $\lambda_k$ form a basis of integral harmonic 3-forms. Then the action becomes $dA \wedge \beta + \sum_k n m^k A\wedge\lambda_k$. The first term sets $dA=0$ after integrating over $\beta$ and the second term sets the holonomies of $A$ to be nth roots of unity. If our 4-manifold doesn't have torsion, there are precisely $n^{b_1}$ such $A$s. If we do the same thing, integrating out $A$ instead, we get Dirac deltas setting $dB=0$ and the holonomies of $B$ to be nth roots of unity. Again, if we have no torsion, there are $n^{b_2}$ such $B$s. It looks like the determinants for these Dirac deltas are all 1, so from the naive calculation above, $Z_n = n^{b_1}$ but also $Z_n = n^{b_2}$. This is a problem on a general 4-manifold, but like the simpler situation above, there is no issue in quantizing on something of the form $\Sigma_3 \times \mathbb{R}$. What I think must be happening is that when I am integrating out $A$ or $B$, I am really only integrating out the non-flat parts. There should still be some more factors that correct the discrepancy above. I think I should be more careful in defining the above integral. Perhaps it is possible to define it sensibly on 4-manifolds which bound a 5-manifold a la Chern-Simons theory. How does one define this theory on general 4-manifolds? Thanks. - 2 Answers BF theory secretly has another name - $Z_n$ gauge theory in the deconfined limit. The parameter $n$ is what appears in front of the $BF$ action. $Z_n$ gauge theory can be defined on any manifold you want - just introduce a lattice approximation of the manifold and compute the lattice gauge theory partition function. By taking the extreme deconfined limit of the lattice gauge theory one can verify that this construction is independent of the way you approximated the manifold. Basic examples In the deconfined limit the $Z_n$ flux through every plaquette of the lattice gauge theory is zero. (This is like the constraint imposed by integrating out $B$.) We have a residual freedom to specify the holonomy around all non-contractible loops. Hence $Z_n(M) = Z_n(S^4) n^{b_1(M)}$ where $Z_n(S^4)$ is a normalization constant. Requiring that $Z_n(S^3 \times S^1) =1$ gives $Z_n(S^4) = 1/n$. This condition, $Z_n(S^3 \times S^1) = 1$, is the statement that the theory has one unique ground state on $S^3$. In general, $Z_n(\Sigma^3 \times S^1)$ is $\text{tr}(e^{-\beta H})$, but since $H=0$ we are simply counting ground states. As a quick check, $Z_n(S^2 \times S^1 \times S^1) = n$, which is the number of ground states on $S^2 \times S^1$, and $Z_n(T^3\times S^1) = n^3$, which is the number of ground states on $T^3 = (S^1)^3$. Further relations Another check on the value of $Z_n(S^4)$: this is the renormalized topological entanglement entropy of a ball in the $3+1$d topological phase described by deconfined $Z_n$ gauge theory. More precisely, the topological entanglement entropy of a ball is $-\ln{Z_n(S^4)}$ which gives $-\log{n}$ in agreement with explicit wavefunction calculations. We can also consider defects. The $BF$ action is $\frac{n}{2\pi} \int B \wedge d A$. Pointlike particles (spacetime worldlines) that minimally couple to $A$ carry $Z_n$ charge of $1$. Similarly, string excitations (spacetime worldsheets) that minimally couple to $B$ act like flux tubes carry $Z_n$ flux of $2\pi/n$. Now when a minimal particle encircles a minimal flux, we get a phase of $2 \pi/n$ (AB phase), but this also follows from the $BF$ action. Without getting into two many details, the term in the action like $B_{12} \partial_t A_3$ fixes the commutator of $A_3$ and $B_{12}$ to be $[A_3(x),B_{12}(y)] = \frac{2\pi i}{n} \delta^3(x-y)$ (flat space). The Wilson-line like operators given by $W_A = e^{i \int dx^3 A_3}$ and $W_B = e^{i \int dx^1 dx^2 B_{12}}$ thus satisfy $W_A W_B = W_B W_A e^{2 \pi i /n}$ which is an expression of the braiding flux above that arises since the particle worldline pierced the flux string worldsheet. Comments on the comments Conservative thoughts If I understand you correctly, what you want to do is sort of argue directly from the continuum path integral and show how the asymmetry you mentioned arises. I haven't done this calculation, so I can't be of direct help on this point right now. I'll ponder it though. That being said, it's not at all clear to me that treating the action as $\int A dB$ leads one to just counting 2-cycles. Of course, I agree that 2-cycles are how you get non-trivial configurations of $B$, but in a conservative spirit, it's not at all clear to me, after gauge fixing and adding ghosts and whatever else you need to do, that the path integral simply counts these. The way that I know the path integral counts 1-cycles is that I have an alternative formulation, the lattice gauge theory, that is easy to define and unambiguously does the counting. However, I don't know how it works for the other formulation. I suppose a naive guess would be to look at lattice 2-gauge theory for $B$, but this doesn't seem to work. Ground states One thing I can address is the issue of ground states. In fact, I favor this approach because one doesn't have to worry about so many path integral subtleties. All you really needs is the commutator. Take the example you raise, $S^2 \times S^1$. There are two non-trivial operators in this case, a $W_B$ for the $S^2$ and a $W_A$ for the $S^1$. Furthermore, these two operators don't commute, which is why there are only $n$ ground states. If we define $\|0\rangle$ by $W_A \|0 \rangle = \|0\rangle$, then the $n$ states $\{\|0\rangle, W_B \|0\rangle, ..., W_B^{n-1} \| 0 \rangle\}$ span the ground state space and are distinguished by their $W_A$ eigenvalue. Importantly, you can also label states by $W_B$ eigenvalue, but you still only get $n$ states. Poincare duality assures you that this counting of states will always work out no matter how you do it. Furthermore, the operator $W_B$ has a beautiful interpretation in terms of tunneling flux into the non-contractible loop measured by $W_A$. It's easier to visualize the analogous process in 3d, but it still works here. You can also see the difference between the theory in 4d and 4d. Since both $B$ and $A$ are 1-forms you have different possibilities. The analogue of $S^2 \times S^1$ might be $S^1 \times S^1$ and this space does have $n^2$ states. However, that is because both $A$ and $B$ can wrap both cycles. The remarkable thing is that the $Z_n$ gauge theory formulation always gets it right. You simply count the number of one-cycles and that's it. Getting at other spaces The state counting approach gets you everything of the form $Z_n(\Sigma^3 \times S^1)$, but even spaces like $S^2 \times S^2$ can be accessed. You can either do the direct euclidean gauge theory computation, or you can regard $Z_n(S^2 \times S^2) = \|Z(S^2 \times D^2)\|^2$ i.e. the inner product of a state on $S^2 \times S^1 = \partial (S^2 \times D^2)$ generated by imaginary time evolution. - Thanks, Physics Monkey. This is a good answer so far, but it does not address the inconsistency that I'm talking about. If one takes the other polarization ($B$ as coordinates, integrating out $A$), the theory is [ $U(1)$-n->$U(1)$ ] 2-gauge theory (roughly speaking, instead of holonomies around 1-cycles, we have holonomies around 2-cycles, also in $\mathbb{Z}_n$). This interpretation gives the calculation $Z_n = n^{b_2}$. – user404153 Sep 5 '12 at 19:52 Let me be more particular with my complaint. The number of ground states on $S^2 \times S^1$ in my mind should be $n^2$. Once you pick two elements of $\mathbb{Z}_n$ this uniquely defines for you a flat 2-connection $B$ (that is, a global 2-form with $dB=0$) and a flat 1-connection $A$ each with the prescribed holonomies around the sphere and circle factor, respectively. That is to say, $dB=0$ does not imply that $B$ has null periods. I guess this really comes down to what we choose our gauge transformations to be. – user404153 Sep 5 '12 at 19:56 Here's what I mean by my last comment. The theory is symmetric under $B\rightarrow B+\beta$ and $A\rightarrow A+\alpha$ for a flat 2-connection $\beta$ and a flat 1-connection $\alpha$. There is a unique ground state up to these gauge transformations. – user404153 Sep 5 '12 at 19:59 Ok, I think I understand more what you're getting at. I'll edit my answer above. – Physics Monkey Sep 7 '12 at 1:09 Thanks for the updated answer. I certainly see what you mean for $S^2 \times S^1$ and agree with you now (also, my third comment is wrong, those are not actually symmetries). The $A$ and $B$ coordinates are conjugate is another way to look at why $W_A$ and $W_B$ must not commute (indeed, the linking number prescription is like a topological Heisenberg algebra). I am continuing to think about more general 4-manifolds. I will get back to you hopefully soon with some resolution or maybe more questions. – user404153 Sep 7 '12 at 3:38 I figured this out a little while ago, so as not to leave any of you hanging, here is a (almost complete) resolution. First, there is a lattice formulation of the 2-gauge theory that agrees with the 1-gauge theory up to the topological factor $n^{\chi(\Sigma_4)}$. One has a $\mathbb{Z}_n$ variable for each 2-cell, gauge transformations on 1-cells, and importantly, 2-gauge transformations (gauge transformations between gauge transformations) on 0-cells. The number of configurations is $n^{b_2}$. Then we divide by the volume of the gauge group, which is the number of gauge transformations divided by the number of 2-gauge transformations. So the partition function of this lattice theory is $n^{b_2-b_1+b_0}$. This gives the same answer as the $A$-theory on any 4-manifold fibered over $S^1$, in particular the 4-torus @Physics Monkey and I were discussing. To get this answer from the path integral, I gauge fix $B$ using the Lorentz gauge $d*B=0$, introducing a Lagrange multiplier term $\langle \pi_1, \delta B\rangle$ as well as some ghost terms not involving $B$ or $\pi_1$. $\pi_1$ has gauge symmetries, being a 1-form, which I also fix, introducing a term $\langle E_0, \delta \pi_1\rangle$ and some more ghost terms which do not involve $B$, $\pi_1$, or $E_0$. Now I scale $B\rightarrow B/n$, $\pi_1 \rightarrow n\pi_1$, and $E_0\rightarrow E_0/n$ to reduce the integrand to the case with $n=1$ (that integral can be shown to be 1). I just need to find how the path integral measure scales under these transformations. $DB$ will scale by $n$ to the power of the dimension of the $B$s being integrated over. It's important to notice that the zero modes of the integrand are precisely the harmonic 2-forms, a space of dimension $b_2$. Then if $B_2$ is the (infinite) dimension of all 2-forms, $DB$ scales by $n^{b_2-B_2}$. Terms like $n^{B_k}$ can all be removed by some suitable regularization a la Witten's paper on abelian S-duality. What we are left with is the answer above, $n^{b_2-b_1+b_0}$. Maybe there is some way to understand the topological factor that appears? -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 162, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9449101090431213, "perplexity_flag": "head"}
http://mathoverflow.net/questions/118083/what-axioms-are-between-ac-and-countable-choice	## what axioms are between AC and Countable choice ! ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hi All. Need some information. We all know axiom of choice (AC) and countable choice. Which axioms are between these two. I mean weaker that Axiom of choice but stronger than countable choice ? - 3 What do you need this information for? Do you have a specific theorem that follows from ZFC but not from ZF+Countable choice? – Goldstern Jan 4 at 20:46 ## 2 Answers Dependent choice, for example. Or choice for well-ordered families, see http://mathoverflow.net/questions/118060 - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. There are many many many axioms. For example you have the Principle of Dependent Choice, you also have the generalized axiom of choice for $\kappa$, "For every family of size $\kappa$ of non-empty sets there is a choice function." and whenever $\kappa>\aleph_0$ this is a strictly stronger axiom. In a similar fashion the principle of dependent choice is extended to what is known as $\mathsf{DC}_\kappa$. There are many choice principles and they come in different flavours and shapes. Some good places to start would be: 1. Howard & Rubin's Consequences of the Axiom of Choice. 2. Herrlich's The Axiom of Choice. 3. G. Moore's Zermelo's Axiom of Choice. 4. Jech's The Axiom of Choice. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9268921613693237, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/2791?sort=oldest	## “Understanding” Gal(\bar Q/Q) ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I have heard people say that a major goal of number theory is to understand the absolute Galois group of the rational numbers G = Gal(Q bar/Q). What do people mean when they say this? The Kronecker-Weber theorem gives a good idea of what the abelianization of the G looks like. But in one of Richard Taylor's MSRI talks, Taylor said that he's never heard of anyone proposing a similar direct description of G and that to understand G one studies the representations of G. I know that there is a strong interest in showing the Langlands reciprocity conjecture [Edit: What I had in mind in writing this is evidently Clozel's conjecture, not the Langlands reciprocity conjecture - see Kevin Buzzard's post below] - that L-functions attached to l-adic Galois representations coincide with L-functions attached to certain automorphic representations. And I've heard people refer to the Tannakian philosophy which I understand as (roughly speaking) asserting that G is determined by all of its finite dimensional representations. Here is a representation of G understood not to be a representation of G as an abstract group but as a group together with a labeling of some of the conjugacy classes of G by rational primes (the Frobenius elements)? When people talk about "understanding G" do they mean proving [Edit: Clozel's conjecture] (in view of the Tannakian philosophy)? If not, what do they mean? If so, this conceptualization seems quite abstract to me. Is this what people mean when they say "understand G"? Can [Edit: Clozel's conjecture] be used to give more tangible statements about G? Something that I have in mind as I write this is the inverse Galois problem (does every finite group occur as a Galois group of a normal extension of Q?) and Gross' conjecture (mostly proven by now) that for each prime p there exists a nonsolvable extension of Q ramified only at p. But I am open to and interested in other senses and respects in which one might "understand" G - Thanks to the respondees so far. I'm hoping for a more complete answer and so decided to offer a bounty. – Jonah Sinick Oct 31 2009 at 5:05 Also, sorry for the nitpick, but I changed slightly the TeXing in the title -- feel free to revert. + Here's a quote from A.N.Parshin: "Langlands program sounds nice, but then I go and ask people -- if you go through with it, would you be able to get the full structure of absolute Galois group? And they say no, it won't be still completely known. And [studying the absolute Galois group] is, I think, more interesting. – Ilya Nikokoshev Oct 31 2009 at 10:57 Ilya, there are some interesting things in your post and I thank you for writing it, the respect in which I did not feel satisfied by your response is that it didn't (directly) address any of the questions that I asked above (about what people mean when they talk about understanding G, and whether the Langlands reciprocity conjecture has more tangible/explicit interpretations concerning the structure of G such as offering the possibility of a solution to the inverse Galois problem) – Jonah Sinick Oct 31 2009 at 18:22 The conjecture that Kevin attributes to Clozel, I would in fact attribute to Langlands, and call "the Langlands reciprocity conjecture", as you did originally. (Here "reciprocity" is used in distinction to "functoriality", which is roughly what Kevin is referring to when he writes about Langlands's conjectures.) – Emerton Feb 12 2010 at 6:56 ## 9 Answers What would it mean to understand this Galois group? You could mean several things. You could mean trying to give the group in terms of some smallish generators and relations. This would be nice, and help to answer questions like the inverse Galois problem that Greg Muller mentioned, and having a certain family of "generating" Galois automorphisms would allow you to study questions about e.g. the representation theory in quite explicit terms. However, the Galois group is an uncountable profinite group, and so to give any short description in terms of generators and relations leads you into subtle issues about which topology you want to impose. You could also ask for a coherent system of names for all Galois automorphisms, so that you can distinguish them and talk about them on an individual basis. One system of names comes from the dessins d'enfant that Ilya mentioned: associated to a Galois automorphism we have some associated data. • We have its image under the cyclotomic character, which tells us how it acts on roots of unity. By the Kronecker-Weber theorem this tells us about the abelianization of the Galois group. • We also have an element in the free profinite group on two generators, which (roughly speaking) tells us something about how abysmally acting on the coefficients of a power series fails to commute with analytic continuation. These two names satisfy some relations, called the 2-, 3-, and 5-cycle relation, which are conjectured to generate all relations (at least the last time I checked), but it is difficult to know whether they actually do so. If they do, then the Galois group is the so-called Grothendieck-Teichmuller group. The problem with this perspective is that the names aren't very explicit (and we don't expect them to be: we may need the axiom of choice to show they exist, and there are only two Galois automorphisms of C that are measurable functions!) and it seems to be a difficult problem to determine whether the Grothendieck-Teichmuller group really is the whole thing. (Or it was the last time I checked.) However, the cyclotomic character is a nice, and fairly canonical, name associated for Galois automorphisms. We could try to generalize this: there are Kummer characters telling us what a Galois automorphism does to the system of real positive roots of a positive rational number number (these determine a compatible system of roots of unity, or equivalent an element of the Tate module of the roots of unity). This points out one of the main difficulties, though: we had to make choices of roots of unity to act on, and if Galois theory taught us nothing else it is that different choices of roots of an irreducible polynomial should be viewed as indistinguishable. Different choices differ by conjugation in the Galois group. This brings us to the point JSE was making: if we take the "symmetry" point of view seriously, we should only be interested in conjugacy-invariant information about the Galois group. Assigning names to elements or giving a presentation doesn't really mesh with the core philosophy. So this brings us to how many people here have mentioned understanding the Galois group: you understand it by how it manifests, in terms of its representations (as permutations, or on dessins, or by representations, or by its cohomology), because this is how it's most useful. Then you can study arithmetic problems by applying knowledge about this. If I have two genus 0 curves over Q, what information distinguishes them? If I have two lifts of the same complex elliptic curve to Q, are they the same? How can I get information about a reduction of an abelian variety mod p in terms of the Galois action on its torsion points? Et cetera. - 4 The 2, 3 and 5-cycle relations of GT translate the action on the moduli space curbes of genus 0 with n marked points. I don't think it was seriously conjectured that these generate all the relations. Grothendieck's original conjecture is that you have to consider all the M_{g,n} for 3g-3+n < 3 to get a full set of relations. Lochack, Nakamura and Schneps have defined a subgroup of GT (containing G_Q) that acts on the fundamental groups of all the M_{g,n}'s. It is probably strictly smaller than GT but that hasn't been proved. – YBL Nov 25 2009 at 0:19 What is the most current status on the 2-, 3- and 5-cycle relations? – Dr Shello Dec 15 2010 at 9:45 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Perhaps some articles in this book and the articles here help, and the discussion on dessin d'enfants (more) on this site. - The book link has changed. This is the correct link: library.msri.org/books/Book41/index.html – arsmath Dec 20 2010 at 19:21 Thanks for the corrected link! – Thomas Riepe Dec 21 2010 at 5:49 One aspect of understanding the absolute Galois group I've heard of is the 'Inverse Galois Problem'. This simply askes, Is every finite group the Galois group of some extension of Q? Its known for solvable groups, but it is unsolved (and apparently very hard) for general finite groups. The inverse Galois problem is equivalent to asking whether any finite group can be the quotient of Gal(Q^bar, Q). In this way, its a more concrete question about the absolute Galois group than Langlands reciprocity. - Taylor said that he's never heard of anyone proposing a similar direct description of G and that to understand G one studies the representations of G. I remember Mazur telling me this when I was a grad student. He made this point in the following way. You shouldn't really think of Gal(Qbar/Q) as a group which has elements, but as a "group up to conjugacy" -- thus, the aspects of Galois groups that really make sense to think about are the conjugacy-invariant things: conjugacy classes (like Frobenii) and representations. To unpack this a bit more: a Galois group is a fundamental group. But to talk about a fundamental group (as opposed to a groupoid) you need to choose a basepoint. To talk about an absolute Galois group you also need to choose a basepoint, which is to say an algebraic closure Qbar/Q. (So just as one should talk about pi_1(X,) rather than pi_1(X), one should talk about Gal(Qbar/Q) rather than Gal(Q).) But a basepoint you can just draw with a pencil. A Galois closure of Q is not so easy. - I like this answer very much. But I wonder: doesn't Q come with a standard algebraic closure? Namely, the algebraic closure of Q inside C. Why doesn't this serve as a good "canonical basepoint" for Qbar/Q? (Okay, canonical up to Gal(C/R) ...) – Charles Rezk Oct 31 2009 at 16:46 Seconding Charles' question! (Which I think also covers mine, namely why Gal(Qbar/Q) is deep and magical but the absolute Galois group of F_p isn't so much.) – Harrison Brown Oct 31 2009 at 17:24 What people mean by "canonical" would be: for an equation `x^2 + 1 = 0` we have chosen canonically a solution `x`. The fact that you're in `C` doesn't help --- there are still 2 solutions, you need a choice. – Ilya Nikokoshev Nov 3 2009 at 9:33 " why Gal(Qbar/Q) is deep and magical but the absolute Galois group of F_p isn't so much." --- it's much larger, to start with and contains all `Gal F_p` in some sense. – Ilya Nikokoshev Nov 3 2009 at 9:34 I feel like this is true of most groups. E.g. the symmetric group - you need to choose an initial ordering of the letters you're permuting. – David Corwin Apr 29 at 13:16 Another aspect of "what it might mean to understand Gal(Q)" -- Shafarevich conjectured that the absolute Galois group of Q^{ab} is a free profinite group. That's pretty explicit! - does "free profinite group" mean "profinite completion of a free group"? – Ben Webster♦ Oct 27 2009 at 14:12 From one of Harbater's papers on this: "A profinite group Π is free on a generating set S that converges to 1 if every map S → G to a profinite group G that converges to 1 uniquely extends to a group homomorphism Π → G." Here "converges to 1" refers to the profinite topology. – JSE Oct 27 2009 at 14:19 A good reference for this is Ribes-Zaleski ["Profinite groups"](books.google.com/…) page 91-94. One should note that any finite set is "converging to 1" and that in the abelian case a free profinite group (free proabelian group!) on a set converging to 1 is just a product of \Z^hat's. – Lars Oct 27 2009 at 20:51 Another quite good reference is "Field Arithmetic" of Fried-Jarden. It also discusses the connection between arithmetic/geometric properties of fields and group theoretic properties of the absolute Galois groups. E.g., Ax' Theorem: if a field K is PAC ("pseudo-algebraically closed") which just means that any geometrically irreducible variety has a rational point, then its absolute Galois group is projective, and its counter part due to Lubotzky-v.d.Dries any projective profinite group can be realized as the absolute Galois group of a PAC field. – Lior Bary-Soroker Nov 25 2009 at 2:46 1 Is there any serious reason to believe (or disbelieve) this conjecture? – David Hansen Jun 4 2011 at 3:37 I'm far from expect in this topic, but here's my attempt. First, and that's something quite straightforward, people want to study `Gal Q` (this is how I will denote it; this common shortcut is defined as `Gal F := Gal \bar F/F`) because we like Galois groups. For simple fields we know their Galois groups and we know how tremendously important they have been, from solving the equations (Abel et al) to doing group cohomology in class field theory. As algebraic geometry matured, people learned that one of the deep reasons for importance of Galois group is that it's the fundamental group `pi_1(Spec F)` (depending on the definition, sometimes this is true only after a profinite completion, I'll omit this fine point further below). This allows us to use the whole apparatus of topology as well as the geometric intuition. Representations of `Gal F` thus have a natural geometric meaning as some bundles (local systems) on the coverings of `Spec F`. This alone would make their study pretty important. It's very reasonable to study a space in terms of geometric objects that live on it. The whole Galois group is very complicated. Fortunately, since the Galois groups of `F_q` and `Q_p` (`p`-adic numbers) are known, we know lots of factorgroups of it. This is the standard topic of algebraic number theory courses. Moreover, if we restrict ourselves to the Abelian part of Galois group, its structure has been completely established by the class field theory. In other words, all one-dimensional representations of `Gal Q` are known. That begs a natural question about higher-dimensional reps — indeed this goes on by the name of Langlands program. The Langlands program is such a huge topic that I don't feel able even starting to talk about it. It might be a good idea to post questions here if you'll feel brave enough to learn it :) One more topic in the discussion of the structure of `Gal F` is about the specific (conjugacy classes) of operators that live there, called Frobenius operators. The amazing thing about them is that they behave, formally, in a way similar to the knot operators in the fundamental group of a threefold without knots. This image is reinforced by the computations of the dimension of `Spec Z` that give an answer of 3, providing fruitful connections to the theory of real 3-manifolds, knots and their L-functions. Finally, here's the direction that was explored by many people and was probably made famous through Grothendieck's work. Consider a variety `Z := P^1-(0, 1, \infty)`, the projective plane without three points. It is really interesting as it makes sense over any field. Now suppose we consider it as a scheme over Q and ask for its fundamental group. Then one would have an exact sequence, if I'm not mistaken, ````0 --> pi_1(Z/\bar Q) --> pi_1(Z/Q) --> Gal Q --> 0 ```` which implies, by standard reasoning, that `Gal Q` acts on `pi_1(Z/\bar Q)`, which is a very straightforward group — it's simply a topological fundamental group of a plane without two points, so it's freely generated by two loops. Grothendieck called it a cartographic group and related it to dessin's d'enfants. The standard references are very well-written in T.'s answers and there are hopefully more questions on this topic coming to MathOverflow ;) As you see, this is a very interesting direction that we should continue to explore. One more thing I like about it is that it's also very suitable to MathOverflow format, since there are both many specific questions as well as opportunities to write reviews and to produce new results by collaboration. - Conc. primes as knots and Spec Z as 3-manifold: Fits that to the Poincare conjecture? Topologists view 3-manifolds as Kirby-equivalence classes of framed links. How would that be with Spec Z? Then, topologists have things like virtual 3-manifolds, has that analogies in arithmetics? – Thomas Riepe Nov 3 2009 at 14:01 I'm not a specialist in this topic, and it's a different from the original question here. Do you mind posting a separate question about this? – Ilya Nikokoshev Nov 3 2009 at 14:59 Khare-Larsen-Savin have used functoriality (ie the Langlands program) to help solve the inverse Galois problem, see for example link text. They show that some finite groups of Lie type occur as Galois groups. As for the Langlands program in general, solving it may not give an answer to "What is Gal(Qbar/Q)?", however it can certainly help. For example, the use of automorphic forms in Iwasawa theory has been quite fruitful such as Ribet's proof of the converse to Herbrand's theorem, the proof of the Main conjecture by Mazur-Wiles and Wiles, etc. For example, Eisenstein series, their congruences with cusp forms and the associated Galois representations are used in Ribet's paper (Inv. Math. 34 (1976)) to construct unramified p-extensions. It can be easier to hide an automorphic form up your sleeve. - $\newcommand{\bb}{\mathbb}$ This is a response to Charles' remark to JSE's answer, why doesn't $\bar{\bb Q}$ come with a standard algebraic closure inside the complex numbers $\bb C$? First, if one considers an abstract extension $K/\bb Q$, then $K$ has $d = [K:\bb Q]$ embeddings into the complex numbers, which can not, a priori, be distinguished in any way. (e.g. maps from $K = \bb Q[x]/(x^3-2)$ to $\bb C$ require a "choice" of $2^{1/3}$ in $\bb C$. Of course, this doesn't answer Charles' question, which, I imagine, is more along the following lines. Why doesn't one simply start with the complex numbers, and then consider the set of algebraic numbers inside $\bb C$? The resulting field is clearly isomorphic to $\bar{\bb Q}$, and, moreover, comes with a canonical embedding into $\bb C$. The problem arises when one wants to define Frobenius elements. Defining such elements amounts to giving a choice of embedding from $\bar{\bb Q}$ into $\bar{\bb Q}_p$. So there is a choice to be made for every $p$! Thinking of $\bar{\bb Q}$ inside $\bb C$ fixes this choice for "$p=0$" only. To make this completely explicit, consider the splitting field $K$ of $x^3 - 2$. In the "fields live inside $\bb C$" optic, $K$ is the field $\bb Q(2^{1/3},\sqrt{-3})$ where $2^(1/3)$ is real and the imaginary part of $\sqrt{-3}$ is positive. Clearly $Gal(K/\bb Q) = S_3$, where we can think of $(123)$ as sending $2^{1/3}\mapsto e^{2\pi i/3} 2^{1/3}$. If $p = 7$, then $Frob_p$ has order 3 in $S_3$. We can ask, does $Frob_p = (123)$ or $(132)$? We find that $$Frob_p = (123) \quad\mbox{if } \sqrt{-3}\equiv 2\mod 7$$ $$Frob_p = (132) \quad\mbox{if } \sqrt{-3}\equiv 5\mod 7$$ and knowing that the imaginary part of $\sqrt{-3}$ is positive does not allow us to determine $Frob_p$ without choosing an embedding of $K$ into $\bar{\bb Q}_7$. - Thanks, this is a point that I have long wondered about and your remark clarifies the matter completely. – Jonah Sinick Nov 1 2009 at 18:44 Thanks! So what I take from this is that the only canonical constructions are the completions Q_p (with Q_∞=R). So if X is the algebraic space associated to Q, then for each p we get a covering space X_p->X, associated to the field of algebraic numbers in Q_p. If X had a distinguised base point, we could name the X_p's using subgroups of π_1X. Since we can't, all we know is the conjugacy class of π_1X_p inside π_1X. For rational primes p, this is the congugacy class of the subgroup "generated" by Frob_p. – Charles Rezk Nov 1 2009 at 20:12 I hate the fact that you can't even use html characters in comments. – Charles Rezk Nov 1 2009 at 20:13 1 Charles, yes, that's correct, although literally to have a Frobenius at p one should work with extensions unramified at p, which amounts to working with pi_1(Z[1/N]) for some set N, and taking p not dividing N. (For p\|n, one has to work with the still complicated group Gal(Qbar_p/Q_p), rather than Gal(Fbar_p/F_p), which is topologically cyclic and generated by Frobenius. – Lavender Honey Nov 1 2009 at 23:29 Very very nice!!! – Filippo Alberto Edoardo Nov 6 at 23:06 I am pretty sure that when different number theorists say "one of the main goals of number theory is to understand Gal(Q-bar/Q)" they may well mean different things. One example of what someone might mean, which has been touched upon above, but which I'd like to stress (because it's what I mean whenever I start seminars with this generic-sounding statement!) is Clozel's conjectures in his Ann Arbor paper. Langlands would conjecture that for any automorphic representation of GL_n over a number field K, there should be an associated n-dimensional complex representation of what people might now call the Langlands group of K (Langlands really conjectured something a little more precise: he conjectured that a certain category that he could not quite define, but whose objects he understood, should have the structure of a Tannakian category; I've tried to make this statement "concrete" in the above). One of the problems with Langlands' conjecture was simply that it's not the sort of thing that can be checked by example, because no-one really knows an intrinsic construction of the Langlands group of K. However, there are theorems in the literature about constructing Galois representations from automorphic representations, for example class field theory (which tells us what the abelianisation of this Langlands group should look like: ((adeles of K)^* / K_^* ) ) and Deligne-et-al's theorem attaching p-adic Galois representations to holomorphic cusp forms for GL_2 over Q. Hang on a minute though---Langlands was conjecturing the existence of complex Galois representations of some group we don't have a definition of---how does Deligne's theorem fit into this? Here we have a p-adic representation (which wouldn't be continuous if you chose an isomorphism Q_p-bar = C and considered the induced complex representation) of a group that at least we have a definition of (Gal(Q-bar/Q) in Deligne's case). So in fact Deligne isn't proving a special instance of Langlands' conjecture, he's proving something else. Clozel formulated precisely what that "something else" should be: he conjectured that to any algebraic automorphic representation of GL_n(adeles of K) there should be an attached p-adic Galois representation. I think that nowadays one would also conjecture that conversely given a geometric p-adic Galois representation (i.e. a representation Gal(K-bar/K)-->GL_n(Q_p-bar) which looks reasonable in some way that can be made precise) it should arise in this way. People like Emerton and Kisin are well on the way to proving this for K=Q and n=2. So "proving Clozel's conjecture" would be one way of making precise "understanding Gal(Q-bar/Q)". What this would boil down to in this case would be understanding all the representations of Gal(Q-bar/Q) which were (a) taking values in Aut(V) with V a f.d. vector space over the p-adics, (b) unramified outside a finite set of primes (c) de Rham at p, and one would be understanding them in the sense that one would be parametrising the set of unramified conjugacy classes Frob_ell in terms of something totally different (automorphic representations, which are analytic gadgets). The book of Harris and Taylor is to a large extent the state of the art nowadays (but the theory is currently moving so fast) [EDIT: see Toby Gee's comment below]. However the base field K is essentially always either totally real or CM in this book, so in some sense the general case of Clozel's conjecture is still wide wide open. Note finally that in some sense what Langlands conjectures is still open even for modular elliptic curves. If E is a modular elliptic curve over Q then the Galois representation naturally associated to E is its Tate module. But the representation of the Langlands group attached to E would be a representation to GL2(C), whose image would land in U2(C), the trace of Frob_p would be a sub p/sqrt(p), and (in the non-CM case) the conjugacy classes of the Frob_p's would be "evenly distributed in U(2)"---a statement which turns into the Sato-Tate conjecture when you unravel it. Note that Taylor et al proved the Sato-Tate conjecture but they certainly did not prove it by constructing this representation! Indeed it's in some sense an ill-defined question to construct this representation, because we don't know what the Langlands group is. - A much better answer than mine, +1. – Ilya Nikokoshev Nov 2 2009 at 11:34 A pedantic note: Harris-Taylor isn't quite state of the art (though I guess it is to a "large extent"): in particular the hypotheses there on the existence of a discrete series place have been completely removed by Shin and Chenevier-Harris, amongst others. – TG Nov 2 2009 at 12:50 Thanks Toby. When I wrote the comment about Harris-Taylor, what I really meant was "I know things have happened since then but don't quite know who to attribute things to so I won't go there". Passing remark: I now don't quite understand the canonical thing to do. Am I supposed to edit my post to reflect your comments, which will then make your comment look ridiculous? I think I'll just leave it. – Kevin Buzzard Nov 2 2009 at 14:42 Kevin - Thanks for your interesting and informative post. I added an edit to my post (under the heading [EDIT] to avoid taking your response out of context just as you wish to avoid taking Toby's comment out of context) to account for your correction. – Jonah Sinick Nov 3 2009 at 4:21 I should perhaps clarify something: amongst the algebraic automorphic representations are those "of Artin type" (I think there must be a better name for them, but it escapes me)---I am talking about the ones which should correspond to finite image Galois representations. For these representations, Clozel's conjecture and Langlands' conjecture should "basically coincide" and should both predict finite image Galois reps to GL(n,complexes). Even proving this statement (attaching reps to pi's of Artin type on GL(n)) would be an extraordinary achievement. – Kevin Buzzard Nov 3 2009 at 16:20 show 1 more comment	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 38, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9498919248580933, "perplexity_flag": "middle"}
http://mathhelpforum.com/math-topics/69107-math-help-plz-pure-30-a.html	# Thread: 1. ## math help plz for pure 30 In another city, the schools vary in size from 200 to 2 000 students. As a result, the volleyball league is organized into three divisions. The first division is for schools with less than 500 students, the second division is for schools with 500 to 1 000 students, and the third division is for schools with more than 1 000 students. The table below shows the populations of two schools from 1995 to 2004. School A has a population that declined from 1995 to 2004. School B has a population that grew from1995 to 2004. 1. Use exponential regression equations of the form y = abt, where y is the school population and t is the number of years after 1995, to model the populations of School A and School B. Express a to the nearest whole number and b to the nearest thousandth. school A : y = -0.030x + 3.167 log b = -0.030 ----> b = 10^(-0.030) = 0.933 log a = 3.167 ----> a = 10^3.167 = 1469 school B: y = 0.015x + 2.682 log b = 0.015 ----> b = 10^(0.015) = 1.035 log a = 2.682 ----> a = 10^2.682 = 481 2. Determine, to the nearest tenth of a percent, the average annual rate of increase or decrease for each school. School A: 1353(1 - r) ^9 = 755 (1 - r) ^9 = 755 / 1353 = 0.55802 1 - r = (0.55802) ^1/9 r =- 6.2% School B: 449(1 + r) ^9 = 646 (1 + r)^9 = 646/449 = 1.4388 1 + r = (1.4388)^1/9 = 1.0412 r = 4.2% 3. Assuming the same annual rate of decrease, predict the population of School A in September 2009. Show the mathematical basis for your prediction. 4. Assuming the same annual rate of increase and using the values of a and b from the regression equation, predict the calendar year in which the population of School B reaches 1 000 for the first time. Justify your prediction graphically and algebraically. 5. Assuming the same annual rates of decrease and increase, predict the calendar years in which Schools A and B play in the same division of the league. Justify 6. • Using the same set of axes, sketch the graphs of the regression equations for the populations of School A and School B as a function of the years after 1995. • Determine the point of intersection of the two graphs, and explain the significance of the intersection point in the context of this project. i did first two but struggling with rest of those. help me out here. 2. ## urgent project help please well i am not relly gud at exp reg. i did first two but struggling with rest of those. help me out here plz. and could you plz tell me whther the first 2 are correct or not? thanks in advance 3. Originally Posted by mathproject In another city, the schools vary in size from 200 to 2 000 students. As a result, the volleyball league is organized into three divisions. The first division is for schools with less than 500 students, the second division is for schools with 500 to 1 000 students, and the third division is for schools with more than 1 000 students. The table below shows the populations of two schools from 1995 to 2004. School A has a population that declined from 1995 to 2004. School B has a population that grew from1995 to 2004. 1. Use exponential regression equations of the form y = abt, where y is the school population and t is the number of years after 1995, to model the populations of School A and School B. Express a to the nearest whole number and b to the nearest thousandth. school A : y = -0.030x + 3.167 log b = -0.030 ----> b = 10^(-0.030) = 0.933 log a = 3.167 ----> a = 10^3.167 = 1469 school B: y = 0.015x + 2.682 log b = 0.015 ----> b = 10^(0.015) = 1.035 log a = 2.682 ----> a = 10^2.682 = 481 2. Determine, to the nearest tenth of a percent, the average annual rate of increase or decrease for each school. School A: 1353(1 - r) ^9 = 755 (1 - r) ^9 = 755 / 1353 = 0.55802 1 - r = (0.55802) ^1/9 r =- 6.2% School B: 449(1 + r) ^9 = 646 (1 + r)^9 = 646/449 = 1.4388 1 + r = (1.4388)^1/9 = 1.0412 r = 4.2% 3. Assuming the same annual rate of decrease, predict the population of School A in September 2009. Show the mathematical basis for your prediction. 4. Assuming the same annual rate of increase and using the values of a and b from the regression equation, predict the calendar year in which the population of School B reaches 1 000 for the first time. Justify your prediction graphically and algebraically. 5. Assuming the same annual rates of decrease and increase, predict the calendar years in which Schools A and B play in the same division of the league. Justify 6. • Using the same set of axes, sketch the graphs of the regression equations for the populations of School A and School B as a function of the years after 1995. • Determine the point of intersection of the two graphs, and explain the significance of the intersection point in the context of this project. i did first two but struggling with rest of those. help me out here. You have two regression equations of the form $y = a (b^t)$. 3. What value of t does 2009 correspond to? Substitute that value into your regression equation for school A to get the value of y. 4. Solve $1000 = a (b^t)$ for t for school B. Note that $\log_{10} 1000 = t \log_{10} (b) + \log_{10} a$. Now convert t into its corresponding year. 5. Some of the years are obviously 2001, 2002, 2003, 2004 (since they both have 500 - 1000 students in those years and hence compete in the second division). Now you need to find the values of t for school A for which y > 500 and the values of t for school B for which y < 1000 (note your answer to Q4 ....). The values of t common to both these solutions will give the values of t (and hence the years) for which both schools compete in the second division ..... 6. Draw the graphs, perhaps using technology (although you should have been taught how to draw graphs like these by hand). The t-coordinate of the point of intersection corresponds to the time at which both schools have the same numbers of students. To find the t-coordinate of the point of intersection, equate the regression equations for school A and school B and solve the resulting equation for t (you'll probably need to use technology to do this).	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9255145788192749, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/99643/why-does-bosonic-string-theory-require-26-spacetime-dimensions/99650	## Why does bosonic string theory require 26 spacetime dimensions? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I do not think it is possible really believe or experimentally check (now), but all modern physical doctrines suggest that out world is NOT 4-dimensional, but higher. The least sophisticated candidate - bosonic string theory says that out world is 26 dimensional (it is not realistic due to presence of tachion, and so there are super strings with 10 dimensions, M-theory with 11, F-theory with 12). Let us do not care about physical realities and ask: what mathematics stands behind the fact that 26 is the only dimension where bosonic string theory can live ? Definitely there is some mathematics e.g. 26 in that MO question is surely related. Let me recall the bosonic string theory background. Our real world is some Riemannian manifold M which is called TS (target space). We consider the space of all maps from the circle to M, actually we need to consider how the circle is moving inside M, so we get maps from $S^1\times [0~ T]$ to M ( here $S^1\times [0~ T]$ is called WS - world sheet); we identify the maps which differs by parametrization (that is how Virasoro comes into game and hence relation with Leonid's question). That was pretty mathematical, but now ill-defined physics begin - we need integrate over this infinite-dimensional space of maps/parametrizations with measure corresponding to exp( i/h volume_{2d}(image(WS))). This measure is known NOT to exist mathematically, but somehow this does not stop physists they do what they call regularization or renormalization or something like that and 26 appears... - 24 T h e $\$ s p a c i n g $\$ i s $\$ a n n o y i n g . . . – Asaf Karagila Jun 14 at 20:32 4 @Asaf MO does not allow title less than 15 chars. – Alexander Chervov Jun 14 at 21:18 14 Alexander- Some would have taken that as a sign that a title with one word and one number wasn't actually a good choice (which is expanded on in the "how to ask"). – Ben Webster♦ Jun 14 at 21:27 15 C'mon guys. It's a good question, with an ok title. Stop beeing arses! – André Henriques Jun 14 at 21:53 15 Isn't the title also exceedingly uninformative? It should be obvious to mathematicians that a title "Why 26?" must refer to the number of sporadic finite simple groups. I suppose chemists might start wondering about Iron's properties or something. – Jeff Burdges Jun 14 at 21:55 show 7 more comments ## 5 Answers In addition to Chris Gerig's operator-language approach, let me also show how this magical number appears in the path integral approach. Let $\Sigma$ be a compact surface (worldsheet) and $M$ a Riemannian manifold (spacetime). The string partition function looks like $$Z_{string}=\int_{g\in Met(\Sigma)}dg\int_{\sigma\in Map(\Sigma,M)}d\sigma\exp(iS(g,\sigma)).$$ Here $Met(\Sigma)$ is the space of Riemannian metrics on $\Sigma$ and $S(g,\sigma)$ is the standard $\sigma$-model action $S(g,\sigma)=\int_{\Sigma} dvol_\Sigma \langle d\sigma,d\sigma\rangle$. In particular, $S$ is quadratic in $\sigma$, so the second integral $Z_{matter}$ does not pose any difficulty and one can write it in terms of the determinant of the Laplace operator on $\Sigma$. Note that the determinant of the Laplace operator is a section of the determinant line bundle $L_{det}\rightarrow Met(\Sigma)$. The measure $dg$ is a 'section' of the bundle of top forms $L_g\rightarrow Met(\Sigma)$. Both line bundles carry natural connections. However, the space $Met(\Sigma)$ is enormous: for example, it has a free action by the group of rescalings $Weyl(\Sigma)$ ($g\mapsto \phi g$ for $\phi\in Weyl(\Sigma)$ a positive function). It also carries an action of the diffeomorphism group. The quotient $\mathcal{M}$ of $Met(\Sigma)$ by the action of both groups is finite-dimensional, it is the moduli space of conformal (or complex) structures, so you would like to rewrite $Z_{string}$ as an integral over $\mathcal{M}$. Everything in sight is diffeomorphism-invariant, so the only question is how does the integrand change under $Weyl(\Sigma)$. To descend the integral from $Met(\Sigma)$ to $Met(\Sigma)/Weyl(\Sigma)$ you need to trivialize the bundle $L_{det}\otimes L_g$ along the orbits of $Weyl(\Sigma)$. This is where the critical dimension comes in: the curvature of the natural connection on $L_{det}\otimes L_g$ (local anomaly) vanishes precisely when $d=26$. After that one also needs to check that the connection is actually flat along the orbits, so that you can indeed trivialize it. Two references for this approach are D'Hoker's lectures on string theory in "Quantum Fields and Strings" and Freed's "Determinants, Torsion, and Strings". - Pavel, thank you very much for yours wonderful answer. Let me ask something. Is it possible to clarify the general setup and general question which lead us to 26 ? I mean something like that 1) we have manifold M (e.g. all metrics) 2) we have measure on it (right ? or it is not measure?) 3) we want to push down the "measure" on the factor M/G , (in this example G is diffeomorphisms+Weyl). Now we see the phenomena - this "push down" cannot be defined in general only for 26. Is this correct setup ? What we cannot ? What are general conditions on M, g which will allow to do it ? – Alexander Chervov Jun 20 at 6:15 Before pushing forward $dg$ and $Z_{matter}$ from $M$ to $M/G$, you would like to pushforward the line bundles with connections that they are sections of. Separately $L_{det}$ and $L_g$ can not be pushed forward (anomaly), but $L_{det}\otimes L_g$ can (cancellation of anomalies). If $c\neq 26$, the curvature along the orbits is nonzero and so the line bundle cannot possibly come from $M/G$ (any such pullback carries a flat connection along the orbits). Essentially, you defer the definition of $dg$ until you work with $M/G$, which is finite-dimensional. – Pavel Safronov Jun 21 at 6:18 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. I think this is standard in some String Theory textbooks: The quantum operators form the Virasoro algebra, where the generators obey $[L_m,L_n]=(m-n)L_{m+n}+\frac{c}{12}m(m^2-1)\delta_{m+n,0}$. Here "c" is the central charge, which is the space-time dimension we are working over. We need this algebra to interact appropriately with the physical states of the system (i.e. $L_m\|\phi\rangle$ information), and only when $c=26$ do we guarantee that there are no negative-norm states in the complete physical system. [Addendum] In the method I described, $c=26$ arises correctly as the critical dimension so that no absurdities occur. What I believe David Roberts is thinking about (in his comment below) is another way to get the same answer: You consider light-cone coordinates and write down the mass-shell condition (summing over the worldsheet dimension $D−2$), and you end up with the requirement $\frac{D-2}{24}=1$. In other words, $c=D=26$. - I thought it was $c=24$, and the fact string worldsheets are codimension 2 bumped it up to 26... – David Roberts Jun 14 at 23:52 Maybe my addendum is what you were considering? – Chris Gerig Jun 15 at 1:02 That may have been it. That 13/6 turns up in the definition of the bracket of the generators seems odd to me. I know there is a continuation of the zeta function to -1 that turns up in one description, and so the famous $\sum n = -1/12$ interacts with the 24=26-2 in a nice way, but I can't remember exactly how that goes... – David Roberts Jun 15 at 1:43 ...and of course it was in the other answer, which I hadn't bothered reading. – David Roberts Jun 15 at 1:44 1 Yea, what I was referring to in the addendum involves a double sum $\sum^{D-2}_i\sum^\infty_n\alpha^i_{-n}\alpha^i_n$, and the zeta function appears. – Chris Gerig Jun 15 at 1:54 show 10 more comments The value of 26 ultimately comes from the need to rid the theory of negative-norm states, as previously noted. This involves regularizing the sum $\sum_{n=0}^\infty n,$ which of course diverges. One obtains a finite part equaling $-\frac{1}{12},$ which leads to the 24, and from there to 26 for the number of dimensions (25 space, 1 time). (Analytic continuation of the zeta function gives $\zeta(-1) = -\frac{1}{12}$). Rather than me writing the details in here, I suggest this nice introductory reference, where the number of dimensions required for consistency of the bosonic string is derived in Ch. 2: http://www.damtp.cam.ac.uk/user/tong/string/string.pdf - Thank you ! it is nice reference. – Alexander Chervov Jun 20 at 10:37 Just to correct a small misconception: Physicists aren't claiming that an integral over the space of smooth maps from $WS$ to $X$ exists. The path integral measure for the bosonic string is defined on something more like the "space of distributions on $WS$ valued in $X$". It's not a problem that no integral exists on the space of smooth maps, because this space shows up only as a convenient shorthand for discussing the renormalization procedure which is used to define the path integral measure. The quantum mechanics of the simple harmonic oscillator is subject to a similar abuse of language. You discuss the theory in terms of functions on the timeline $[0,t]$, but if you're careful, the path integral measure (aka, the Ornstein-Uhlenbeck measure) is actually defined on the space of distributions on $[0,t]$. (What makes life easy in this case is that the Wiener measure is supported on distributions which are almost everywhere continuous functions.) The situation is somewhat more complicated for the 2d nonlinear sigma model, because there isn't really anything you'd want to call the distributions valued in $X$. Instead you try to define the measure as a linear functional on observables which are well approximated by functions of the form $\phi \mapsto ev_{\sigma} \phi^*f$, where $f$ is a function on $X$ and $ev_\sigma$ evaluation at a point $\sigma \in WS$. The correlation function of observables $\hat{\mathcal{O}}_1$, $\hat{\mathcal{O}}_2$ should be approximated by integrals of the form $\int_{Map(L,X)} \mathcal{O}_1(\phi) \mathcal{O}_1(\phi) e^{i S_L(\phi)} d\phi$ for some finite set of points $L \subset WS$, and some approximation $S_L$ of the classical action $S$ defined using only finite differences among the points in $L$. When you refine the set of points $L$ to fill in $WS$, you get an expectational value functional on the set of observables. This expectation value functional should have the same properties (like OPE) that you see in QFTs where the classical fields take values in linear spaces. - @AJ Tolland Thank you for yours answer! It seems I indeed had some missunderstanding, let me ask you later. I asked Pavel some question above (after his answer) to clarify the general setup - may be you can comment on it also ? – Alexander Chervov Jun 20 at 8:14 I'm not an expert on this so bear with me, but I don't think you must require $\dim(M) = 26$, you must only require that the worldsheet is conformally invariant - i.e., the Weyl anomaly vanishes. You can do this by adding 26 bosons (which represent the coordinates of $M$) - which is called critical string theory - or you can turn on the dilaton expectation value - which is then called non-critical string theory. There's a lot of interesting research involving these non-critical string theories, for e.g. check out $c=1$ matrix models and type 0 string theories. - Thank you for the answer. Indeed i heard something about noncritical strings. This was active at early nineties. Some matrix model description werefound. I am not expert neither but it seems to me this is even more beyond math understanding than critical 26 dims. If you. Can comment more this would be great. – Alexander Chervov Sep 2 at 16:47	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 82, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9310423135757446, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/20128/where-does-the-energy-come-from-when-a-current-heats-a-wire-resistor?answertab=oldest	# Where does the energy come from when a current heats a wire (resistor)? I'm trying to figure out an example from a textbook (Demtröder -- Experimentalphysik 2, pg. 198) where the energy transport caused by a current is depicted: Assume you have a wire (with some resistance $R$) and a current $I$ flowing through the wire. The wire will emit energy in form of heat (with $\dot W = I^2 \cdot R$). It states that since the electric field $\mathbf E$ is parallel to to the wire (i.e. the current), and the magnetic field $\mathbf B$ is tangential to the wire, the Poynting vector $\mathbf S$ must point radially (and orthogonally) into the wire. The book goes on to argue, that therefore, the energy to replenish the "lost" heat-energy flows radially from the outside into the wire. My question is now, where does that energy come from? I would have thought that it comes from the source of the current (e.g. a battery) and travels through the wire to the point where the heat is emitted. However, this is explicitly stated to be wrong. Could you please shed some light on the issue? - ## 1 Answer The Poynting vector shows that the energy isn't transmitted through the wires; it's transmitted through the surrounding space, as in this picture: To simplify the picture, most of the heat in this circuit is produced by the blue light bulb on the right side. The energy flows from the battery "through the air" to the light bulb. The energy flows are given by the white arrows in red ellipses. The electric field isn't just parallel to the wire; because the current (charges) must ultimately get back somewhere in space, the electric field goes "mostly" from a piece of wire to another piece of wire which is oriented oppositely. On the picture above, the electric field is given by the thin red arrows (from the top to the bottom). Similarly, the thin green lines are magnetic field lines (around the wires, as you probably expected). http://www.furryelephant.com/content/electricity/visualizing-electric-current/surface-charges-poynting-vector/ It's important to realize that many of the flows indicated by the Poynting vector are partially fictitious. In fact, you can have just a static electric field induced by an electric charge at point $X_{\rm EL}$ on top of a static magnetic field from an end point of a long bar magnet located at $X_{\rm MG}$. And the Poynting vector will tell you that energy is running in loops around the axis connecting $X_{\rm EL}$ with $X_{\rm MG}$, in the surrounding vacuum. This is not a problem. One may also define the Poynting vector (and the whole stress-energy tensor) differently so that the flows will be different. However, it's equally important to realize that the energy is locally conserved with the Poynting vector being the flux. In particular, in the vacuum, the energy just "flows" through the vacuum and whenever the Poynting lines gets denser, they must become longer, and vice versa. There are no sinks or sources of energy in the vacuum. This conservation law $$\frac{\partial \rho_{\rm energy}}{\partial t} + {\rm div}\,\,\vec S =0$$ can be proved by multiplying Maxwell's equations by fields and combining the equations appropriately. In the wire, there is an extra term from the heat creation in the conductor etc. Again, in these situations, one shouldn't overreact. One shouldn't imagine energy as "some kind of liquid" that pushes all things. Energy is just an abstract quantity that is conserved. When it just runs in some loops in the vacuum, it is not a problem. Whether such an energy flow does some work on charges etc. depends on the detailed equations, the Lorentz force, Maxwell's equations, and so on: one shouldn't try to guess such influences from the flow of energy only. - Thanks for this explanation. It really helped me. Just one more thing: If you can reduce energy to your abstract conserved quantity, where you can take out as much as you like, as long as you put in the same amount somewhere else, how does that play with the law that an effect (e.g. heating the wire) cannot travel faster than the speed of light (measured from its source -- e.g. the battery)? (I totally understand if this is asking a bit too much in an addendum.) – bitmask Jan 28 '12 at 17:23 You're welcome, bitmask. Maxwell's equations may be explicitly proven to respect relativity, including the condition that signals never propagate faster than light. I didn't really say that you may subtract energy from any point and return it anywhere. On the contrary, I quoted the differential equation with ${\rm div}\,\,\vec S$ which is manifestly local: if energy disappears from some place, it is because it is flowing to a nearby point as shown by the $\vec S$ vector. The conservation is local. I just wanted to say that energy doesn't have to have a "material carrier". – Luboš Motl Jan 28 '12 at 17:37 In general relativity, by the way, the energy conservation law becomes problematic and when we want to include energy carried by gravitational waves etc., it cannot really be written locally in any meaningful way. But in non-gravitational physics, we can do it. When I said that energy was an abstract quantity, I meant that even in the vacuum, $\vec E, \vec B$ may be nonzero and generic, and therefore $\vec S$ is probably nonzero, too. Energy is flowing even though there are no "particles" over there. Nothing wrong about it. My focus was the word "abstract", like "numerical". – Luboš Motl Jan 28 '12 at 17:41 Oh, one more addition. In your comment (3 above me), you may have referred to my remark about the various ambiguous definitions of the Poynting vector. I meant that one may redefine the Poynting vector in such a way that the conservation law above will still hold (and the total energy will be unchanged) but the flows (circulation of energy) will be different. But this freedom to redefine $\vec S$ really means that the energy flux density vector, when loosely defined, is ambiguous and unphysical. That makes it obvious that such subtle changes of $\vec S$ can't be used to send any signals. – Luboš Motl Jan 28 '12 at 17:46 – Luboš Motl Jan 28 '12 at 17:50 show 2 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 16, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9622071981430054, "perplexity_flag": "head"}

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