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http://math.stackexchange.com/questions/194617/13-dof-kalman-filter	# 13 DOF Kalman filter I'm trying to develop a system with the following characteristics: Inputs: • 3-axis accelerometer [3 DOF] • 3-axis gyroscope [3 DOF] • GPS with three parameters (lat, lon, altitude) [3 DOF] • Barometric pressure [1 DOF] -estimates altitude • 3-axis magnetometer [3 DOF] Outputs: • lat, lon • altitude • velocity (x,y,z) • attitude • rotation speed From the very basic research I've done, I think I need a Kalman Filter to fuse the sensor data together. The lat/lon/altitude from the GPS is augmented by the data from the sensors; giving overall better GPS accuracy. Does anyone know how best to approach this problem and/or if there is any source code available? Many thanks in advance, - ## 3 Answers Keep in mind that a Kalman Filter is a method for estimating the state of a linear dynamic system. In practice that means given a state vector $x$, you have to design a Matrix $H$ such that the product $y = Hx$ describes your measurements. That is, $y$ would look like $$y = \left(\begin{array}{c} latitude \\ longitude \\ \vdots \\ rotation speed \end{array}\right)$$ in your case. After you have found a vector $x$ describing your system state and an appropriate $H$, you can then use the Kalman Filter equations to estimate the state $x$ . While have some freedom in choosing $x,y$ and $H$, this still requires some careful thinking. An alternative would be to use a non-linear estimator. This makes the problem easier in that you only have to find any, arbitrary function $h$ such that $y = h(x)$ is the measurement vector. (This is the same $y$ as above) The most popular non-linear estimator these days is probably the Unscented Kalman Filter, which has better performance than the EKF under almost every scenario. In any case, I doubt you will find published literature on this exact problem. I had a similar problem 2 years ago, and the most involved systems I could find had about three sensors or less. However, there is Sebastian Thruns free course "How to build a robotic car" on Udacity, which covers a lot of estimation theory, so I would recommend that if you want to learn a bit more. - 1 Unscented Kalman filtering doesn't work well. I suggest to use the 'particle filter' or the 'moving horizon filter'. If you have a lot of time to implement a very good nonlinear filter you can use the 'log homotopy filter' discovered by Daum. – Riccardo.Alestra Sep 12 '12 at 14:56 @Benno: I've been reading up on "Probabilistic Robotics" by Sebastian Thrun, Wolfram Burgard and Dieter Fox. I now understand why I should use an unscented kalman filter. I'm going to try and code something simple that I can extend to the 13DOF model. – Eamorr Sep 12 '12 at 16:07 @Riccardo: While a particle filter can better handle multimodal and other complex or highly non-linear functions, it is also much more expensive to compute. For many "reasonable" measurement functions, an UKF is a good tradeoff between accuracy and speed. (It is basically a particle filter with very few, very carefully placed particles) – Benno Sep 13 '12 at 1:14 Maybe you can use an $EKF$ (extended Kalman filter). You can find papers on 'Transfer alignment'. You can also find hints on the book: 'Introduction to random signals and applied Kalman filtering' by R.G. Brown and P.Y.C. Hwang and also in 'Kalman filtering' by M.S.Grewal and A.P.Andrews. In the 'Simulink' tool (MATLAB) you can find something which can help you. - Hi, many thanks for the reply. It's very useful to get these pointers. This is not as straight-forward as I thought. I'm going to consult with my physicist friend about the EKF. (I'm an engineer) – Eamorr Sep 12 '12 at 10:20 As a somewhat tangential word of advice: ditch GPS altitude. GPS altitude is based on the geoidic model used by the receiver. Not all geoids are the same. GPS altitude estimates coming off receivers are therefore typically very inaccurate -- way too inaccurate for state estimation of an aircraft. In fact, it's so inaccurate that it will corrupt your state estimator quite badly in some cases. I have analyzed GPS data for ground vehicles, and found that vehicles have made 300 foot drops over traversing 150 feet of distance, according to GPS altitude (obviously, this is false). As for a reference, I highly recommend Adaptive Control Design and Analysis by Gang Tao. He covers state estimation, and control of state-estimated systems, quite well in this book. - This is why I love stackexchange. Thanks so much for the tips. – Eamorr Sep 12 '12 at 14:27	{"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": 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": 0.940185010433197, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/44910/rigid-body-dynamics-of-tossing-of-a-coin	# Rigid body dynamics of tossing of a coin While tossing a coin, it is commonly experienced that you get a head, if you toss it up with the head side up, and a tails if you toss with the tails side up. Is there a mathematical proof of this using classical mechanics? I would like to see a simple model of the coin as a symmetric top, and consider the precision of the body axis of symmetry about the angular momentum. - 5 I have to say that your claim "you get a head, if you toss it up with the head side up, and a tails if you toss with the tails side up" does not match my experience. have you tried this for yourself? If you toss a coin from heads 100 times how many heads and tails do you get? – John Rennie Nov 23 '12 at 11:28 ## 2 Answers I will give it a shot. Spoiler: I did this in the body frame so that the moment of inertia is time independent, before you get excited... Starting with Euler's equations: $$I_i\dot{\Omega}_i+(I_j - I_k)\Omega_j \Omega_k = 0$$ and taking cyclic permutations of $i,j,k$ to get the three of them; and in the absence of torques (I ignore air friction). It's a symmetric top so $I=I_1=I_2 \neq I_3$ so write $$\dot{\Omega}_1 = -\frac{(I_3-I)}{I}\Omega_2 \Omega_3$$ $$\dot{\Omega}_2 = -\frac{(I - I_3)}{I}\Omega_1\Omega_3$$ $$\dot{\Omega}_3=0 \implies \Omega_3=k_1$$ Now for this problem the coin is spinning about one of the first two symmetric axies. I chose 1. Then consider small variations on the other two angular velocities from zero: $\Omega_2 = \delta\Omega_2$, $\Omega_3 = \delta\Omega_3$, and $\Omega_1 \rightarrow \Omega_1$. So we make small changes in how the coin is rotating about a line through its center perpendicular to the coin, and about the other symmetric axis. In other words, it was spinning ideally like a coin would, then we changed the ideal to a little weird spinning. Making the changes, and ignoring second order in perturbations: $$\dot{\Omega}_1=0 \implies \Omega_1 = k_1$$ $$\frac{d}{dt}(\delta\Omega_2)=-\frac{(I-I_3)}{I}\Omega_1 (\delta\Omega_3)$$ $$\frac{d}{dt}(\delta\Omega_3)=0 \implies \delta\Omega_3 = k_2$$ Then we can write $$\frac{d}{dt}(\delta\Omega_2)=-\frac{(I-I_3)}{I}k_1 k_2$$ Everything on the r.h.s is a number so $$\delta\Omega_2 = -\left( \frac{(I-I_3)}{I}k_1 k_2 \right) t$$ so depending on how big $I$ is compared to $I_3$ will determine how $\delta\Omega_2$ changes during the flip. If one uses a radius of $r=0.014$ m and $h=0.0015$ m for the hight of the coin, one gets a moment of inertia tensor like the following: $$I=M(0.0000491875) \quad I_3 = M(0.000098)$$ which tells me that the variations are unstable... which I don't really believe since I have seen a coin in real life. So look this over. But I can't find anything wrong so I'm going with it, and thinking that I can't really see a coin in real life up close while it's spinning... Hope this helps. - If the product $k_1 k_2<0$ is negative then the spin will be stable. Maybe your axis conventions need more careful consideration. – ja72 Dec 15 '12 at 14:43 Maybe the torque due to aerodynamic drag stabilizes things. – ja72 Dec 15 '12 at 14:51 This paper (http://www-stat.stanford.edu/~cgates/PERSI/papers/dyn_coin_07.pdf) shows that the probability distribution of getting a head, if I toss with the head side up is given by: $p(ψ, φ) =\frac{1}{2}+\frac{1}{\pi} \sin^{-1} (\cot(φ) \cot(ψ))$ if $(\cot φ)(\cot ψ) ≤ 1$, =1 if $\cot(φ) \cot(ψ) ≥ 1$ where $\phi, \psi$ are the Euler angles. -	{"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": 10, "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": 0.9276013970375061, "perplexity_flag": "head"}
http://mathhelpforum.com/pre-calculus/55369-arithmetic-sequence.html	# Thread: 1. ## Arithmetic Sequence An arithmetic sequence is given for each question below whose initial term a and common difference d are also given. What is the firth term? (1) a = 6; d = -2 (2) a = sqrt{2}; d = sqrt{2} 2. Originally Posted by magentarita An arithmetic sequence is given for each question below whose initial term a and common difference d are also given. What is the firth term? (1) a = 6; d = -2 (2) a = sqrt{2}; d = sqrt{2} To find the nth term, $a_n$, in an arithmetic sequence, use the following formula: $a_n=a_1+(n-1)d$, where the first term is $a_1$ and the common difference is $d$ (1) $a_5=6+(5-1)(-2)=??$ (2) $a_5=\sqrt{2}+(5-1)(\sqrt{2})=??$ 3. ## I get it... Originally Posted by masters To find the nth term, $a_n$, in an arithmetic sequence, use the following formula: $a_n=a_1+(n-1)d$, where the first term is $a_1$ and the common difference is $d$ (1) $a_5=6+(5-1)(-2)=??$ (2) $a_5=\sqrt{2}+(5-1)(\sqrt{2})=??$ Your easy steps and guidance has increased my understanding of precalculus. I hope you will help me when I enter calculus 1 not too long from 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": 12, "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": 0.9271393418312073, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/12931/approximating-an-integral-representation-of-the-number-partition-problem	# Approximating an integral representation of the Number Partition Problem One can write out an integral whose solution gives the number of solutions to the NP-Complete Number Partition Problem and I'm wondering if anyone has an suggestions or ideas on who to solve or approximate this integral analytically or numerically. Given $a_0, a_1, \dots, a_{n-1} \in \mathbb{Z}$, the Number Partition Problem asks for a partition of the $a_k$'s such that $\sum_{k=0}^{n-1} \sigma_k a_k = 0$, where $\sigma_k \in \{ -1,1 \}$. Consider the following Integral: $$I(a_0, a_1, \dots, a_{n-1}) = \frac{1}{2 \pi} \int_{-\pi}^{\pi} \prod_{k=0}^{n-1} (e^{ i a_k \theta } + e^{ -i a_k \theta } ) d\theta$$ $I(\dots)$ will count the number of solutions for a given instance of $(a_0, a_1, \dots, a_{n-1})$. Solving this integral is worse than NP-Hard (it's #P) so asking for a general solution is out of the question. But can one do any sort of approximation, either analytically or numerically? If you choose the $a_k$'s with some distribution, say uniform on some interval, can you exploit that randomness to help you approximate this integral? Any ideas would be appreciated. note: This has been studied by Borgs et all for the NPP Phase Transition and that's where I first saw this integral representation of the Number Partition Problem, but their analysis relies on approximating the family of instances given a uniform distribution on the $a_k$'s rather than trying to solve a particular instance, as I'm trying to do above. - 1 I do not see the point of crossposting a question like this. Answers improve one another if everyone can see everyone else's answers; insights from different answers can be combined. Splitting up the answers like this seems counterproductive to me. – Qiaochu Yuan Dec 3 '10 at 17:12 There was one answer from the mathoverflow site, otherwise it is essentially unanswered. Should I add a comment on what has been suggested already? – user4143 Dec 3 '10 at 17:17 1 Depending on how big the $a_k$ are and how many terms that product has, the integrand can be violently oscillatory, and quite a lot of the usual numerical methods will have trouble with this. Turning this into a contour integral might make it slightly more tractable, though. – J. M. Dec 4 '10 at 13:17 – user4143 Dec 4 '10 at 16:11 Presumably if $\theta$ goes from $-\pi$ to $\pi$, you don't need the additional $2\pi$ in the exponents. – mjqxxxx Jan 20 '11 at 16:06 show 3 more comments ## 1 Answer Using a discrete Fourier transformation seems more appropriate if the $a_i$ are integers. The number of solutions is the value of the convolution product $$\left(\delta_{a_0} + \delta_{-a_0}\right) * \left(\delta_{a_1} + \delta_{-a_1}\right) * ... * \left(\delta_{a_{n-1}} + \delta_{-a_{n-1}}\right)$$ evaluated at 0. This has the obvious dynamic programming solution, which takes something like $O(n A)$ time, where $A = \sum_i \|a_i\|$; i.e., it is a quasi-polynomial-time algorithm. If the $a_i$ are chosen at random from (bounded) distributions of integers instead, then the dynamic programming solution takes $O(n A^2)$ to calculate the expected number of solutions. The discrete Fourier transform translates that problem to a direct product of $n$ functions over $O(A)$ points, each of which can be computed in $O(A \log A)$ using the FFT, reducing the total time to $O(nA \log A)$. Moreover, the number of "approximate" partitions, i.e., choices of the $\sigma_i$ such that $\|\sum_{i=0}^{n-1}\sigma_i a_i\| \le k$ for some small $k$, falls out of this calculation essentially for free. - – user4143 Jan 21 '11 at 0:24	{"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": 25, "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": 0.3, "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": 0.9232707023620605, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/32665/why-position-is-not-quantized-in-quantum-mechanics/32666	# Why position is not quantized in quantum mechanics? Usually in all the standard examples in quantum mechanics textbooks the spectrum of the position operator is continuous. Are there (nontrivial) examples where position is quantized? or position quantization is forbidden for some fundamental reason in quantum mechanics (what is that reason?)? Update: By position quantization I mean, if position (of a particle say) is measured we get only a discrete spectrum (say 2.5 cm and 2.7 cm, but nothing in between, just in the same way that energy levels can be discrete). In that sense interference patter of photons on a photographic plate cannot be considered as position quantization because the probability density varies "continuously" from maximum to zero (or am I wrong?) - 3 position quantization is equivalent to periodic momentum, and this happens in a crystal. Position is quantized in a crystal, on a grid. – Ron Maimon Jul 23 '12 at 19:29 ## 5 Answers Position quantization in vacuum is forbidden by rotational, translational, and boost invariance. There is no rotationally invariant grid. On the other hand, if you have electrons in a periodic potential, the result in any one band is mathematically the theory of an electron on a discrete lattice. In this case, the position is quantized, so that the momentum is periodic with period p. ### Fourier duality The quasimomentum p in a crystal is defined as i times the log of the eigenvalue of the crystal translation operator acting eigenvector. Here I give the definition of crystal position operators and momentum operators, which are relevant in the tight-binding bands, and to describe the analog of the canonical commutation relations which these operators obey. These are the discrete space canonical commutation relations. In 1d, consider a periodic potential of period 1, the translation by one unit on an energy eigenstate commutes with H, so it gives a phase, which you write as: $$e^{ip}$$ and for p in a Brillouin zone $-\pi<0<\pi$ this gives a unique phase. The p direction has become periodic with period $2\pi$. This means that any superposition of p waves is a periodic function in p-space. The Fourier transform is a duality, and a periodic spatial coordinate leads to discrete p. In this case, the duality takes a periodic p to a discrete x. Define the dual position operator using eigenstates of position. The position eigenstate is defined as follows on an infinite lattice: $$\|x=0\rangle = \int_0^{2\pi} \|p\rangle$$ Where the sum is over the Brillouin zone, and the sum is over one band only. This state comes with a whole family of others, which are translated by the lattice symmetry: $$\|x=n\rangle = e^{iPn} \|x=0\rangle = \int_0^{2\pi} e^{inp} \|p\rangle$$ These are the only superpositions which are periodic on P space. This allows one to define the X operator as; $$X = \sum_n n \|x=n\rangle\langle x=n\|$$ The X operator has discrete eigenvalues, it tells you which atom you are bound to. It only takes you inside one band, it doesn't have matrix elements from band to band. The commutation relations for the quasiposition X and quasimomentum P is derived from the fact that integer translation of X is accomplished by P: $$X+ n = e^{-inP} X e^{inP}$$ This is the lattice analog of the canonical commutation relation. It isn't infinitesimal. If you make the translation increment infinitesimal, the lattice goes away and it becomes Heisenberg's relation. If you start with a free particle, any free $\|p\rangle$ state is also a quasimomentum p state, but for any given quasimomentum p, all the states $$\|p + 2\pi k\rangle$$ have the same quasimomentum for any integer k. If you add a small periodic potential and do perturbation theory, these different k-states at a fixed quasimomentum mix with each other to produce the bands, and the energy eigenstates $\|p,n\rangle$ are labelled by the quasimomentum and the band number n: you define the discrete position states as above for each band $$\|x,n\rangle = \int_0^{2\pi} e^{inp} \|p,n\rangle$$ These give you the discrete position operator and the discrete band number operator. $$N \|x,n\rangle = n \|x,n\rangle$$ if you further make the crystal of finite size, by imposing periodic boundaries in x, the discrete X become periodic and p becomes the Fourier dual lattice, so that the number of lattice points in x and in p are equal, but the increments are reciprocal. This is what finite volume discrete space QM looks like, and it does not allow canonical commutators, since these only emerge at small lattice spacing. - Do you mean the "momentum" with period p or the position? – anna v Jul 23 '12 at 15:26 @annav: I mean the momentum periodic with period p which is the same as the position quantized is a lattice of size 2pi/p. – Ron Maimon Jul 23 '12 at 18:39 – John Jul 24 '12 at 14:15 Actually, if posistion in space was quantized, in order for translational and rotational symmetry to hold, and therefore, for the conservation laws to hold, wouldn't momentum need to be quantized? If mass, time, and energy were all quantized, wouldn't that be possible? Or no? – John Jul 24 '12 at 14:23 2 @John: I don't know if you're serious, but yes, you lose conservation laws. It's not completely trivial in field theory, because you can have emergent rotational invariance at long distances, like an atomic lattice can have rotationally invariant sound waves. This doesn't work because the corrections are going to be too big from natural lattice scale at the Planck scale. It also doesn't work in string theory, where the space is holographicaly reconstructed from the symmetries of the boundary, SUSY and spacetime, and you don't have a lattice, and the symmetry is the fundamental ingredient – Ron Maimon Jul 24 '12 at 14:34 The answer is essentially what Kostya has pointed out: Position is quantized but has a continuous spectrum of (generalized) eigenvalues because the canonical commutation relations on position and momentum forbid that both of them be bounded operators (and act on finite-dimensional state spaces) by Stone-von Neumann theorem. This means that, given general constraints, at least one of them must be unbounded and thus must have a nonemtpy continuum spectrum, implying that the resolution of the identity must be in terms of an integral over non physical states. This is why in general a specific point-like position is not a physical observable eigenstate of a system, and must be spread over a dense interval (which is related to the instrumental resolution of the measurement): e.g. particles can have as much localized wave packets as our resolution allows, but they have no defined positions even in principle (at least in standard quantum mechanics). Thus the canonical commutation relations forbid you have discrete position if momentum is discrete by boundary conditions (e.g. particle in a box). This may seem trivial because position and momentum are the generators of translations of each other, but the point of the noncommutativity and the theorem is that one of them can indeed have discrete spectrum (usually boundary conditions discretize momentum, so position generalized eigenvalues are continuous). Concering semantics of "quantized", some observable property is quantized if it is an operator in the quantum physical state of the system, regardless of the discreteness or continuity of its spectrum of eigenvalues because, even in the continuum case, it is subject to the formalism of quantum mechanics: noncommutativity, uncertainty, probabilistic expected values ... Quantum jumps in possible values of discrete spectra are just the most remarkable property of the quantum world, without analogue in the classical world, hence the name quantum mechanics, but that is not a necessary condition. Since there are purely quantum degrees of freedom (e.g. spin), quantization is not fundamental, but they are nevertheless quantum observables and not classical in both senses: spin observables belong to an operator algebra formalism, they are not a function on classical phase space, and have discrete spectrum, thus being quantized (or more properly, "quantum-mechanical") in any meaning of the word. UPDATE on crystal lattices: what Ron is calling realization of discrete space might be very misleading. Any nonrelativistic space lattice model (i.e., in non quantum gravity theories) is an effective discrete model for constrained main expectation values of real position observables. The quasi-position/momentum of crystals, and thus crystal lattices of condensed matter, are an emergent property out of the symmetries of the approximate arrangement of equilibrium positions of many atoms. Any position measurement on any atom is however not point-like defined, but just highly localized around the lattice nodes. From a very rigorous point of view, one should distinguish the fundamental degrees of freedom of the system from effective quantities emergent from the system's symmetries. In the case of solid state matter, the system is composed of a huge collection of atoms which whose collective interactions constrain their localization around well-defined points of equilibrium. Therefore at the structural level we can talk about quasi-classical atoms at fixed positions (the maxima of their position probability distributions), thus creating an effective lattice of discrete quasi-position where the rest of the constraints and properties of the system (periodic potentials, momentum...) give an emergent model for quasi-particles which may seem like discretized space and momentum simultaneously. I defend that thinking of this as purely quantum mechanical position is misleading, because the lattice as a whole is classical, although discrete, as its structure is not subject to superpositions in a Hilbert space, to the noncommutativity and uncertainty principle, and its (constant) structure is not a static solution of a collective system's Schrödinger equation. The fundamental degrees of freedom are the quantum position and momentum operators of each atom, always subject to canonical commutation relations and thus continuum spectrum of one of the two, along with uncertainty in position-momentum. A quasi-position and quasi-momentum of a quasi-particle on a crystal lattice is NOT realized by the eigenstates of any actual quantum particle. In this sense, crystal lattices do not serve as example of "quantized position" (meaning discrete spectrum), as long as one retains the noun "position" to mean ordinary position. Real atoms in any crystal have nonzero bounded temperature, forbidding them to have definite constant positions: upon measuring position of any atoms in a crystal, the atom will appear localized with a distribution probability around the lattice node, but the real observable position of the atom need not be the node itself (in fact is always a region, since point-like positions are not eigenstates, only wave-packets). The quantum model Ron talks about in his answer is useful and nice for the effective lattice I talked about above. I just do not call position to something which is not what we observe when measuring position on quantum particles, that is why I talk about quasi-position in that case. In absence of evidence for a discrete quantum spacetime, quantized phase space is what is traditionally called in theoretical physics as far as I know. Therefore, any real quantum position and momentum, in the sense of the quantum operators whose expectation values "obey" (in the sense of Ehrenfest Theorem) classical Hamilton equations, and whose Schrödinger evolution "obey" (in the sense of eikonal approximations) classical Hamilton-Jacobi equations, are not simultaneously discrete. The only theories where quantum position gets discretized spectrum is in quantum gravity theories like loop quantum gravity, where the fundamental degrees of freedom of space (and time) itself get quantized (do not confuse quantum position as before, which is newtonian position related to reference bodies, with position as understood in general relativity, where spacetime is the gravitational field). There you get a granular graph of nodes forming space itself, and our traditional continuum positions are just approximations (even at the atomic level) of the relational position given by a flat gravitational field. - I stated clearly what I meant by the word quantization in the question. I did not mean the quantization "procedure" by which one promote the classical momentum for example into an operator which equals $-i \hbar \nabla$. In that sense Spin is not quantized because it has no classical analog. Anyway, again, I was talking about the quantization of spectrum. – Revo Jul 23 '12 at 17:31 I have added more to my answer at the beginning, hoping to answer more concretely the discrete spectrum issue for position due to the canonical commutation relations. – Javier Álvarez Jul 23 '12 at 18:32 1 -1: This answer is basically saying [x,p]=i means no quantized x. This is mathematically true, but both obvious and ridiculous--- the very idea of an infinitesimal generator of x translations means x is continuous from the beginning. The deduction is false for the example of a quasiparticles on a crystal lattice of finite volume where the quasiparticle momentum and position are both discrete. It's still QM, but canonical commutation fails, you get the discrete version with exponentials of P instead, and this is the right mathematical statement of discrete x. The rest is wordy and irrelevant. – Ron Maimon Jul 23 '12 at 19:33 1 On the other hand, disqualifying other people's answer as wordy and irrelevant is subjective and unnecessary. Indeed it was a wordy answer (now trimmed to the obvious and ridiculous) but not irrelevant at all if the person who wants to know about quantization, spectra and position likes a deeper interrelation of ideas. For example a careful distinction between quantized observable and discrete spectrum and its relation to the different mathematical setting of classical and quantum mechanics. However, to satisfy users like Ron Maimon, some of us prefer to remain brief, obvious and ridiculous. – Javier Álvarez Jul 23 '12 at 20:10 1 @JavierÁlvarez: Ok, I read the extended answer--- I agree with the additions for the most part, but there is one point which I disagree very strongly--- even if you quantize the atomic lattice, so that you consider quanutm nuclei interacting with quantum electrons, you still get the discrete lattice emerging in a band. This holds even at finite temperature, up to the melting temperature, it's an example of spontaneous symmetry breaking of translation symmetry, and it doesn't depend on having an external potential. The only thing it requires is a very large system, so that you approximate SSB. – Ron Maimon Jul 24 '12 at 12:27 show 9 more comments It depends on what one defines as "position". In crystals, for example, there exists a three dimensional grid on which the atoms are allowed , stacked in unit cells, so there is quantization in space to be observed, and quantum mechanical solutions are involved . More numerous are the interference solutions of qm waves which also display a quantization of space, where some positions are more probable than othes. So it is not true that position is not quantizable. All the solutions where the energy is quantized involve matter and potentials also. A free particle does not display a quantization of energy as it does not display quantization of position. If the question is addressing whether intrinsically space is quantized, as it would be also if one asked if energy is intrinsically quantized, i.e. comes in a minimum packet, that is another question. - So, suppose you have an eigenstate of $\hat{x}$: $$\hat{x}\|\psi\rangle = x\|\psi\rangle$$ Now let us act with $\hat{x}$ on $e^{i\hat{p}\delta}\|\psi\rangle$, and use this formula (I have $\hbar=1$): $$\hat{x}e^{i\hat{p}\delta}\|\psi\rangle=e^{i\hat{p}\delta}\left(\hat{x}+i\delta[\hat{p},\hat{x}]+\frac{i\delta}{2!}[\hat{p},[\hat{p},\hat{x}]]+\frac{i\delta}{3!}[\hat{p},[\hat{p},[\hat{p},\hat{x}]]]+...\right)\|\psi\rangle=$$ $$=e^{i\hat{p}\delta}(\hat{x}+\delta)\|\psi\rangle=(x+\delta)e^{i\hat{p}\delta}\|\psi\rangle$$ So you have another eigenstate, shifted at arbitrary distance $\delta$. It seems that this result holds for any pair of observables with constant commutator. - From this derivation it seems it works for $[P,X]=f(X)$, where $f(X)$ might be more complicated than $1$. Then you don't get the $+\delta$, but a shift by a function of $\delta$ and the eigenvalue $x$. (I don't know about convergence and reality of the series though.) – Nick Kidman Jul 23 '12 at 20:46 This also assumes constant commutator. The commutation relation has to fail for discrete space time--- it fails for position and momentum operators for quasiparticles in the lattice, hence you get Brillouin zones. Reducing the dynamics of each band, you get the right lattice QM formulation. It's only physical problem is lack of symmetry, no rotations and no boosts. – Ron Maimon Jul 24 '12 at 1:50 2 @RonMaimon I'm really puzzled about this crystal thing everyone is talking about. As far as I remember the commutation relations for crystal modes are still $[x_k,q_l]=i\delta_{kl}$, aren't they? – Kostya Jul 24 '12 at 9:05 @RonMaimon: I have the same problem as Kostya and have asked for references on details of the whole definition of the so called x/p operators in the case of crystals. I want to see please their deduction and interpretation. However discrete lattice anyone talks about, that cannot be fundamental, as atoms vibrate around equilibrium positions coming from main expectation values of quantized phase space position, thus subject to ordinary canonical commutation relations and the uncertainty principle. Misunderstanding a discrete model for collective latticed equilibrium averages with real position. – Javier Álvarez Jul 24 '12 at 9:20 1 @JavierÁlvarez: I dislike it not because of "conceptualization of scientific models of theories", but because the assumption of canonical commutation is exactly equivalent to the assumption of continuous space, and it is obvious that one implies the other. If someone is asking why X is discrete, one cannot use canonical commutators to argue it, because it is an equivalent assumption. You need to understand what a discrete X QM looks like, and argue that it doesn't correspond to nature, not that it is forbidden by a mathematical identity which is derived from continuity assumption. – Ron Maimon Jul 24 '12 at 11:38 show 6 more comments One must take into consideration the fact that the notion of 'particle position' in quantum mechanics is meaningless. One cannot talk about the position of a particle just like one cannot talk about a specific path taken by the particle. The position of a particle in quantum mechanics is not a dynamic variable like it is in Newtonian mechanics, it does not exist as such. The commutation relation [p,x]=ih(bar) used by some tells us, that it is not possible to measure both x and p with arbitrarily high accuracy, hence Heisenberg's uncertainty principle results from it. A good clue whether position is quantised or not can be found by observing the TDSE (time dependent Schrodinger equation.) The mathematical structure of the equation, being a differential equation in x,y,z and t, requires that both position and time must be continuous variables. This requirement is a necessity for the definition of the wave function, but nobody knows where the particle described by the wave function actually is, let alone whether its position is quantised or not! One should not confuse the quantisation of the electronic orbitals in atoms with position quantisation. - In non-relativistic quantum mechanics position is a dynamical variable, according to the usual definition: an operator that evolves according to the equation of motion $\dot{x} = i [H,x]$ in the Heisenberg picture. It is only in QFT that position is demoted to a continuous index, like time in non-relativistic QM. Also, it is not "meaningless" to talk about the position of a particle, it is simply that you cannot know the position to arbitrary precision, due to the commutation relations as you mentioned. – Mark Mitchison Jan 9 at 22:26 1 Thanks for you response. The question is both mathematical and physical. Ehrenfest's theorem can give the evolution of the mean position, <x>, of the particle accurately. If as you say "..., it is simply that you cannot know the position to arbitrary precision, ..." then, is it still meaningful asking where the particle is? The impossibility of knowing exactly where the particle is, is not subject to improvents by higher precision measurements, but it requires a compromise in the precision of the momentum. Only the wave function can tell us where the particle can be, but probabilistically. – JKL Jan 13 at 12:02 I would argue that position is a meaningful concept, it just has to be revised from the naive classical concept. It is surely meaningful to ask where an electron is. For example, Young's double slit experiment relies on distinguishing where electrons land on a screen (obviously with some intrinsic uncertainty); this is how one builds up the famous interference pattern. Etcetera etcetera. But anyhow, this is more a matter of terminology/philosophy than physics, I just wanted to bring to your attention the possibility that your phrasing could be a bit misleading if taken literally. – Mark Mitchison Jan 13 at 13:42 Thanks for that. Bohr and Einstein argued, about the meaning of QM and whether objective reality of things makes any sense, untill the end of their lives! The thing that is important is what is demonstrated by the experiment. To say that the particle has gone through one slit or the other in Young's DSE is one thing. When you do the experiment to test this out, you soon find that the interference pattern is gone and for a good reason. The spot of the electron position on the screen is only a result of its interaction, w-f collapse as they call it. So far Bohr seems to have won the argument! – JKL Jan 13 at 20:32	{"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": 18, "mathjax_display_tex": 11, "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": 0.9277620315551758, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/47782/moser-interation-for-nonlinear-elliptic-equation	# Moser interation for nonlinear elliptic equation Let $\Omega$ be and bounded open set in $\mathbb{R}^n$ and $u\in H^1(\Omega)\cap L^{\infty}(\Omega)$ be the weak subsolution of the following nonlinear and heterogeneous elliptic equation: $$-\nabla\cdot(\|\nabla u\|^{p-2}\nabla u)+c(x)\|u\|^{p-2}u=f$$ where $0\le c(x)\le M$ and $f\in L^\infty(\Omega)$. That is, for every $\phi\in H^1(\Omega)\cap L^{\infty}(\Omega)$, $$\int_\Omega\|\nabla u\|^{p-2}\nabla u\cdot\nabla\phi+c(x)\|u\|^{p-2}u\phi\,dx\le\int_\Omega f\phi\,dx$$ Show that there exists a constant $R_0=R_0(M)>0$, such that for every $0<R\le R_0$ and $\forall x_0\in\Omega$, $$\sup_{B_{R/2}}u\le C\left(\frac{1}{R^n}\int_{B_R}u^p\,dx\right)^{1/p}+C\\|f\\|_\infty$$ , where $B_R=B_R(x_0)\subset\Omega$ and $C=C(n,R_0,M)$. I think this can be done by a generalized version of Moser iteration, extended from linear case to nonlinear one. Can anyone give some clues or references? Thank you~ - 1 – Siminore Jul 14 '12 at 8:01	{"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": 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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9217755794525146, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/60397/list	Return to Answer 2 edited body Note that $\min(w,L) = \frac{1}{2} (w+L) + - \frac{1}{2} \|w-L\|$, so the main issue is to establish that the map $w \mapsto \|w\|$ is bounded on $H^{1,2}$. But this follows from the diamagnetic inequality $\|\nabla \|w\|\| \leq \|\nabla w\|$ (in the sense of distributions), which is obvious formally, but can be established rigorously by approximating the absolute value $\|x\|$ by the smoothed variant $(\|x\|^2+\varepsilon^2)^{1/2}$ and then letting $\varepsilon$ go to zero. (The diamagnetic inequality can be found for instance in the text of Lieb and Loss; it is more commonly applied in the context of covariant differentiation, but already has usefully non-trivial content for ordinary differentiation.) More generally, composition with Lipschitz functions will preserve all $W^{s,p}(\Omega)$ spaces for $0 \leq s \leq 1$ and $1 < p < \infty$ by the chain rule (if $s=1$) or fractional chain rule (if $s<1$), using regularisation arguments as necessary to make the argument rigorous. (See for instance Taylor's book "Tools for PDE" for this sort of thing.) 1 Note that $\min(w,L) = \frac{1}{2} (w+L) + \frac{1}{2} \|w-L\|$, so the main issue is to establish that the map $w \mapsto \|w\|$ is bounded on $H^{1,2}$. But this follows from the diamagnetic inequality $\|\nabla \|w\|\| \leq \|\nabla w\|$ (in the sense of distributions), which is obvious formally, but can be established rigorously by approximating the absolute value $\|x\|$ by the smoothed variant $(\|x\|^2+\varepsilon^2)^{1/2}$ and then letting $\varepsilon$ go to zero. (The diamagnetic inequality can be found for instance in the text of Lieb and Loss; it is more commonly applied in the context of covariant differentiation, but already has usefully non-trivial content for ordinary differentiation.) More generally, composition with Lipschitz functions will preserve all $W^{s,p}(\Omega)$ spaces for $0 \leq s \leq 1$ and $1 < p < \infty$ by the chain rule (if $s=1$) or fractional chain rule (if $s<1$), using regularisation arguments as necessary to make the argument rigorous. (See for instance Taylor's book "Tools for PDE" for this sort of thing.)	{"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": 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": 0.9244510531425476, "perplexity_flag": "head"}
http://mathoverflow.net/questions/37610?sort=newest	Demonstrating that rigour is important Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Any pure mathematician will from time to time discuss, or think about, the question of why we care about proofs, or to put the question in a more precise form, why we seem to be so much happier with statements that have proofs than we are with statements that lack proofs but for which the evidence is so overwhelming that it is not reasonable to doubt them. That is not the question I am asking here, though it is definitely relevant. What I am looking for is good examples where the difference between being pretty well certain that a result is true and actually having a proof turned out to be very important, and why. I am looking for reasons that go beyond replacing 99% certainty with 100% certainty. The reason I'm asking the question is that it occurred to me that I don't have a good stock of examples myself. The best outcome I can think of for this question, though whether it will actually happen is another matter, is that in a few months' time if somebody suggests that proofs aren't all that important one can refer them to this page for lots of convincing examples that show that they are. Added after 13 answers: Interestingly, the focus so far has been almost entirely on the "You can't be sure if you don't have a proof" justification of proofs. But what if a physicist were to say, "OK I can't be 100% sure, and, yes, we sometimes get it wrong. But by and large our arguments get the right answer and that's good enough for me." To counter that, we would want to use one of the other reasons, such as the "Having a proof gives more insight into the problem" justification. It would be great to see some good examples of that. (There are one or two below, but it would be good to see more.) Further addition: It occurs to me that my question as phrased is open to misinterpretation, so I would like to have another go at asking it. I think almost all people here would agree that proofs are important: they provide a level of certainty that we value, they often (but not always) tell us not just that a theorem is true but why it is true, they often lead us towards generalizations and related results that we would not have otherwise discovered, and so on and so forth. Now imagine a situation in which somebody says, "I can't understand why you pure mathematicians are so hung up on rigour. Surely if a statement is obviously true, that's good enough." One way of countering such an argument would be to give justifications such as the ones that I've just briefly sketched. But those are a bit abstract and will not be convincing if you can't back them up with some examples. So I'm looking for some good examples. What I hadn't spotted was that an example of a statement that was widely believed to be true but turned out to be false is, indirectly, an example of the importance of proof, and so a legitimate answer to the question as I phrased it. But I was, and am, more interested in good examples of cases where a proof of a statement that was widely believed to be true and was true gave us much more than just a certificate of truth. There are a few below. The more the merrier. - 10 There's a clear advantage to knowing a 'good' proof of a statement (or even better, several good proofs), as it is an intuitively comprehensible explanation of why the statement is true, and the resulting insight probably improves our hunches about related problems (or even about which problems are closely related, even if they appear superficially unrelated). But if we are handed an 'ugly' proof whose validity we can verify (with the aid of a computer, say), but where we can't discern any overall strategy, what do we gain? – Colin Reid Sep 3 2010 at 13:53 8 What kind of person do you have in mind who would suggest proofs are not important? I can't imagine it would be a mathematician, so exactly what kind of mathematical background do you want these replies to assume? – KConrad Sep 3 2010 at 15:33 7 Colin Reid- I think one can differentiate between a person understanding and a technique understanding. The latter applies even if we cannot understand the proof. We know that the tools themselves "see enough" and "understand enough", and that in itself is a significant advance in our understanding. But we still want a "better proof", because a hard proof makes us feel that our techniques aren't really getting to the heart of the problem- we want techniques which understand the problem more clearly. – Daniel Moskovich Sep 3 2010 at 16:26 12 Concerning the Zeilberger link that Jonas posted, sorry but I think that essay is absurd. If Z. thinks that the fact that only a small number of mathematicians can understand something makes it uninteresting then he should reflect on the fact that most of the planet won't understand a lot of Z's own work since most people don't remember any math beyond high school. Therefore is Z's work dull and pointless? He has written other essays that take extreme viewpoints (like R should be replaced with Z/p for some unknown large prime p). – KConrad Sep 5 2010 at 1:39 14 Every proof has it's own "believability index". A number of years ago I was giving a lecture about a certain algorithm related to Galois Theory. I mentioned that there were two proofs that the algorithm was polynomial time. The first depended on the classification of finite simple groups, and the second on the Riemann Hypothesis for a certain class of L-functions. Peter Sarnak remarked that he'd rather believe the second. – Victor Miller Sep 6 2010 at 15:56 show 20 more comments 37 Answers I think, the Cosmonut mentioning of Stokes' theorem (as a response to Gowers' specification of the answer) must be generalized to a wider statement which can't be bypassed here: in fact, exactly the impossibility to make Calculus rigorous can be considered as a cause of why Analysis appeared. In modern language the problem is the following. One could expect that the operations over what is called "elementary functions" in Calculus can be axiomatized independently from the axioms of real numbers, so that one gets a closed purely algebraic system, where the equalities of elementary functions are derived from the list of "axiomatic identities" between $x^y$, $\sin x$, etc., and the operations like derivative and integral are conceived in purely algebraic way - the formulation of the problem can be found at page 197 in my textbook in arxiv: http://arxiv.org/abs/1010.0824 (unfortunately, in Russian). But as it turns out, this is impossible, at least for the complete list of elementary functions: even the equality of elementary functions can't be defined axiomatically. And, what is amazing, this is not a classical result, it is quite new. However, I can't give a reference, what I write here is what Sergei Soloviev from Toulouse told me not long ago. - 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. Allow me to quote part of the introduction of chapter 9 of Lovász: Combinatorial Problems and Exercises. The chromatic number is the most famous graphical invariant; its fame being mainly due to the Four Color Conjecture, which asserts that all planar graphs are 4-colorable. This has been the most challenging problem of combinatorics for over a century and has contributed more to the development of the field than any other single problem. A computer-assisted proof of this conjecture was finally found by Appel and Haken in 1977. Although today chromatic number attracts attention for several other reasons too, many of which arise from applied mathematical fields such as operations research, attempts to find a simpler proof of the Four Color Theorem is still an important motivation of its investigation. So here it's not so much the proof but the search for a proof that has given something extra over just believing the theorem. Does that still count as an answer to this question? - I have found that a strong indicator of research ability is a student wanting to know why something is true. There is also an interesting distinction between an explanation and a proof. (I gave up using the word "proof" for first year students of analysis, and changed it to "explanation", a word they could understand. This was after a student complained I gave too many proofs!) A proof takes place in a certain conceptual landscape, and the clearer and better formed this is the easier it is to be sure the proof is right, rather than a complicated manipulation. So part of the work of a mathematician is to develop these landscapes: Grothendieck was a master of this! Of course the more professional a person is in an area the nearer an explanation comes to a rigorous proof. But in fact we do not write down all the details. It is more like giving directions to the station and not listing the cracks in the pavement, though warning of dangerous holes. The search for an explanation is also related to the search for a beautiful proof. So we should not neglect the aesthetic aspect. - Michael Atiyah's discussion of the "proof" and it role seems to relevant to be posted here. Taken from: "Advice to a Young Mathematician in the Princeton Companion to Mathematics." http://press.princeton.edu/chapters/gowers/gowers_VIII_6.pdf This link was provided by "mathphysicist" in answer on another MO question: http://mathoverflow.net/questions/2144/a-single-paper-everyone-should-read-closed Quotation from M. Atiyah: "In fact, I believe the search for an explanation, for understanding, is what we should really be aiming for. Proof is simply part of that process, and sometimes its consequence." "... it is a mistake to identify research in mathematics with the process of producing proofs. In fact, one could say that all the really creative aspects of mathematical research precede the proof stage. To take the metaphor of the “stage” further, you have to start with the idea, develop the plot, write the dialogue, and provide the theatrical instructions. The actual production can be viewed as the “proof”: the implementation of an idea. In mathematics, ideas and concepts come ﬁrst, then come questions and problems. At this stage the search for solutions begins, one looks for a method or strategy. Once you have convinced yourself that the problem has been well-posed, and that you have the right tools for the job, you then begin to think hard about the technicalities of the proof." - Many examples that have been given are of statements that one could at least formulate, and conjecture, without rigorous proof. However, one of the most important benefits of rigorous proof is that it allows us to step surefootedly along long, intricate chains of reasoning into regions that we previously never suspected existed. For example, it is hard to imagine discovering the Monster group without the ability to reason rigorously. In any other field besides mathematics, as soon as a line of abstract argument exceeds a certain (low) threshold of complexity, it becomes doubtful, and unless there is some way to corroborate the conclusion empirically, the argument lapses into controversy. If you are trying to search a very large space of possibilities, then it is indispensable to be able to close off certain avenues definitively so that the search can be focused effectively on the remaining possibilities. Only in mathematics are definitive impossibility proofs so routine that we can rely on them as a fundamental tool for discovering new phenomena. The classification of finite simple groups is a particularly spectacular example, but I would argue that almost any unexpected mathematical object—the BBP formula for $\pi$, the Lie group $E_8$, the eversion of the sphere, etc.—is the product of a sustained search involving the systematic and rigorous elimination of dead end after dead end. Of course, once an object is discovered, you might try to argue that mathematical rigor was not really necessary and that someone could have stumbled across it with a combination of luck, persistence, and insight. However, I find such an argument disingenuous. Mathematical rigor allows us to distribute the workload across the entire community; each reasoner can contribute his or her piece without worrying that it will be torn to shreds by controversy. Searches can therefore be conducted on a massively greater scale than would be possible otherwise, and the productivity is correspondingly magnified. - In my experience, the two greatest difficulties in mathematics are: 1. The obvious is not always true. 2. The truth is not always obvious. Rigour is the essence of mathematics. A rigorous proof provides an explanation of why a particular mathematical statement is true, and, at the same time, takes care of all the "Yes, but what if"s. Rigour and proof provide the guarantee of correctness and reliability. Rigour and proof refine our mathematical insights and instincts so that the superficially "obvious" misleads us less frequently. When I pose the problem "1, 2, 3, x Find x." the initial response is usually a derisory laugh, of disbelief that I am serious, because "the answer is obviously 4". It is easy to demonstrate using practical examples that this statement is, as it stands, nonsense. A rigorous analysis is required. - This is not strictly an answer to the question since it is a hypothetical rather than an example, but perhaps relevant nonetheless: Suppose that you have some computer program-say for keeping an airliner stable under gusts of wind-and it relies on a numerical algorithm, proved to converge under reasonable assumptions about air pressure and wind velocity, so on. A faster and more stable algorithm is developed for which no proof of convergence is known, though all the researchers in the field assure you that it always converges and that they are certain that it will always converge given the plausible scenarios your code is likely to be used for. I think it is clear that you should not trust their judgment, but rather retain the old code, despite the clear desirability of having a faster numerical algorithm. So from the perspective of the researchers, a proof might not be all that important; it may have only told them what they already knew and generated no new insights in the process. But from the perspective of the consumers of mathematics, knowledge that a proof exists may lead to incremental improvements in technology that would otherwise not happen. Of course, at this point we've come full circle and it becomes important to the researchers to supply a proof. My second point again distinguishes between consumers of mathematics and researchers: It is sometimes much easier to become 100% certain of something than 99% certain. 99% certainty that a given statement is true requires thinking about many concrete examples and developing intuition, whereas 100% certainty requires logically assenting to the statements contained in a proof. By this standard, I would say that I am 100% certain about the bulk of my mathematical knowledge, and not 99% certain. Perhaps this is a lamentable state of affairs, but time is finite. We cannot hope to develop intuition about all the statements we wish to use while working on problems we do wish to develop intuition about. In that sense, proofs encode in a few kb's the vast amount of information stored as the intuitions of all the researchers working on a particular problem. Again under this model, the purpose of proofs is for the convenience of non-researchers. - A rich source of examples may be found in the study of finite element methods for PDEs in mixed form. Proving that a given mixed finite element method provided a stable and consistent approximation strategy was usually done 'a posteriori': one had a method in mind, and then sought to establish well-posedness of the discretization. This meant a proliferation of methods and strategies tailored for very specific examples. In the bid for a more comprehensive treatment and unifying proofs, the finite element exterior calculus was developed and refined (eg., the 2006 Acta Numerica paper by Arnold, Falk and Winther). The proofs revealed the importance of using discrete subspaces which form a subcomplex of the Hilbert complex, as well as bounded co-chain projections (we now call this the 'commuting diagram property). These ideas, in turn, provided an elegant design strategy for stable finite element discretizations. - Mathematics wasn't that rigorous before N. Bourbaki: in the Italian school of Algebraic Geometry of the beginning of the XXth century the standard procedure was Theorem, Proof, Counterexample. Also at the time of Cauchy some theorems in analysis began like "If the reader doesn't choose a specially bad function we have..." The use of rigour in analysis, which Cauchy began, avoided that by being able to explain what was a "good function" in each case: analytic, $C^{\infty}$, being able to do term-by-term derivation in its expansions series... - 1 @Gabriel : This seems like a delicate historical claim. I would actually be interested to see some sources on the evolution of standards of rigor within the mathematical community. – Andres Caicedo Dec 6 2010 at 23:11 1 @Andres www-history.mcs.st-and.ac.uk/Biographies/… "Nicolas Bourbaki, a project they began in the 1930s, in which they attempted to give a unified description of mathematics. The purpose was to reverse a trend which they disliked, namely that of a lack of rigour in mathematics. The influence of Bourbaki has been great over many years but it is now less important since it has basically succeeded in its aim of promoting rigour and abstraction." – Gabriel Furstenheim Dec 7 2010 at 8:26 2 The unified description of mathematics was not the initial intent of the Bourbaki project. By all accounts, it was to write an up to date analysis textbook (losing a whole generation to World War 1 had left a gap). Of course, the whole thing got out of hand pretty quickly... – Thierry Zell Mar 19 2011 at 5:15 show 3 more comments In response to the request for an example of a statement that was widely but erroneously believed to be true: does Gauss's conjecture that $\pi(n) < \operatorname{li}(n)$ for every integer $n \geq 2$, disproved by Littlewood in 1914, qualify? - 2 It's a good example of the need for rigor. But I've always been skeptical of the story that it was widely believed to be true, since any competent mathematician familiar with Riemann's 1857 explicit formula for π(n) would have realized that there are almost certainly going to be occasional exceptions. (Littlewood removed the word "almost" by a more careful analysis.) – Richard Borcherds Sep 8 2010 at 19:58 One can rigorously prove that pyramid schemes cannot run forever, and that no betting system with finite monetary reserves can guarantee a profit from a martingale or submartingale. But there are countless examples of people who have suffered monetary loss due to their lack of awareness of the rigorous nature of these non-existence proofs. Here is a case in which having a non-rigorous 99% plausibility argument is not enough, because one can always rationalise that "this time is different", or that one has some special secret strategy that nobody else thought of before. In a similar spirit: a rigorous demonstration of a barrier (e.g. one of the three known barriers to proving P != NP) can prevent a lot of time being wasted on pursuing a fruitless approach to a difficult problem. (In contrast, a non-rigorous plausibility argument that an approach is "probably futile" is significantly less effective at preventing an intrepid mathematician or amateur from trying his or her luck, especially if they have excessive confidence in their own abilities.) [Admittedly, P!=NP is not a great example to use here as motivation, because this is itself a problem whose goal is to obtain a rigorous proof...] - 8 I doubt if the main problem is that people are not aware of the rigorous nature of non-existence proofs. First, for most people the meaning of a regorous mathematical proof makes no sense and has no importance. The empirical fact that pyramid schemes always ended in a collapse in the past should be more convincing. But even people who realize that the pyramid scheme cannot run forever (from past experience or from mathematics) may still think that they can make money by entering early enough. (The concept "run forever" is an abstraction whose relevance should also be explained.) – Gil Kalai Sep 8 2010 at 7:01 2 @Gil: this is where a proof can give more than what you set out to prove. For the pyramid scheme, not only can we prove it cannot run forever, but we can also extract quantitative evidence to show that the odds of you getting in early enough are close to zero. Of course, this will not convince the numerically illiterate, but I'm convinced there is a non-negligible portion of the population that you could reach in this way. – Thierry Zell Mar 19 2011 at 5:20 show 1 more comment Tim Gowers wrote: But I was, and am, more interested in good examples of cases where a proof of a statement that was widely believed to be true and was true gave us much more than just a certificate of truth. How about Stokes' Theorem ? The two-dimensional version involving line and surface integrals is "proved" in most physics textbooks using a neat little picture dividing up the surface into little rectangles and shrinking them to zero. Similarly, the Divergence Theorem related volume and surface integrals is demonstrated with very intuitive ideas about liquid flowing out of tiny cubes. But to prove these rigorously requires developing the theory of differential forms whose consequences go way beyond the original theorems - 1 In fact, the original motivation for the Bourbaki project was the scandalous state of affairs that no rigorous proof of the Stokes theorem could be located in the (French?) literature. The rest is history... – Victor Protsak Sep 8 2010 at 8:44 2 I don't see how this example demonstrates the importance of rigour. Most engineers or physicists (e. g. doing electrodynamics) can get by perfectly well with "tiny-cubes" proof of Stokes' theorem and nothing ever gets wrong from using this intuitive approach. – Michal Kotowski Sep 8 2010 at 22:35 3 @MichalKotowski: that is not inconsistent with the notion that mathematics has benefited hugely from a rigorous theory of differential forms. – gowers Sep 10 2010 at 17:19 show 2 more comments I tend to think that mathematics or---better---the activity we mathematicians do, is not so much defined by (let me use what's probably nowadays rather old fashioned language) its material object of study (whatever it may be: it is surely very difficult to try to pin down exactly what it is that we are talking about...) but by its formal object, by the way we know what we know. And, of course, proofs are the way we know what we know. Now: rigour is important in that it allows us to tell apart what we can prove from what we cannot, what we know in the way that we want to know it. (By the way, I don't think that it is fair to say that, for example, the Italians were not rigorous: they were simply wrong) - 3 I once heard a less accurate but more punch version of this in some idiosyncratic history of mathematics lectures: "in mathematics, we don't know what it is we are doing, but we know how to do it". – Yemon Choi Sep 8 2010 at 5:08 3 Yemon, there is a famous definition of mathematics, perhaps, from "What is mathematics?" by Courant and Robbins: mathematics is what mathematicians do. – Victor Protsak Sep 8 2010 at 8:53 show 1 more comment Claim The trefoil knot is knotted. Discussion One could scarcely find a reasonable person who would doubt the veracity of the above claim. None of the 19th century knot tabulators (Tait, Little, and Kirkman) could rigourously prove it, nor could anybody before them. It's not clear that anyone was bothered by this. Yet mathematics requires proof, and proof was to come. In 1908 Tietze proved the beknottedness of the trefoil using Poincaré's new concept of a fundamental group. Generators and relations for fundamental groups of knot complements could be found using a procedure of Wirtinger, and the fundamental group of the trefoil complement could be shown to be non-commutative by representing it in $SL_2(\mathbb{Z})$, while the fundamental group of the unknot complement is $\mathbb{Z}$. In general, to distinguish even fairly simple knots, whose difference was blatantly obvious to everybody, it was necessary to distinguish non-abelian fundamental groups given in terms of Wirtinger presentations, via generators and relations. This is difficult, and the Reidemeister-Schreier method was developed to tackle this difficulty. Out of these investigations grew combinatorial group theory, not to mention classical knot theory. All because beknottedness of a trefoil requires proof. Claim Kishino's virtual knot is knotted. Discussion We are now in the 21st century, and virtual knot theory is all the rage. One could scarecely find a reasonable person who would argue that Kishino's knot is trivial. But the trefoil's lesson has been learnt well, and it was clear to everyone right away that proving beknottedness of Kishino's knot was to be a most fruitful endeavour. Indeed, that is how things have turned out, and proofs that Kishino's knot is knotted have led to substantial progress in understanding quandles and generalized bracket polynomials. Summary Above we have claims which were obvious to everybody, and were indeed correct, but whose proofs directly led to greater understanding and to palpable mathematical progress. - 10 I very much like your answer although I'd put what I consider to be a different spin on it. To me the key point of interest in showing the trefoil is non-trivial is that it shows that one can talk in a quantitative, analytical way about a concept that at first glance seems to have nothing to do with standard conceptions of what mathematics is about. A trefoil has no obvious quantitative, numerical thing associated to it. In contrast, the statement $\pi > 3$ is very much steeped in traditional mathematical language so it's rather unsurprising that mathematics can say things about it. – Ryan Budney Sep 6 2010 at 20:18 7 Let me make my point more hyperbolically. That one can rigorously show that a trefoil can't be untangled, this is one of the most effective mechanisms one can use to communicate to people that modern mathematics deals with sophisticated and substantial ideas. Mathematics as a subject wasn't solved with the development of the Excel spreadsheet. :) – Ryan Budney Sep 7 2010 at 0:28 5 =HOMFLYPOLYNOMIAL(A3:B5) – Mariano Suárez-Alvarez Jan 2 2012 at 3:54 show 3 more comments I have the tendency to think that the need for absolute certainty is related to the arborescent structure of mathematic. The mathematics of today rest upon layers of more ancient theories and after piling up 50 layers of concepts, if you are only sure of the previous layers with a confidence of 99%, a disaster is bound to happen and a beautiful branch of the tree to disappear with all the mathematicians living on it. This is rather unique in natural sciences with the exception of extremely sophisticated computer programms but, in mathematics, you will have to fix by yourself an equivalent of 2K bug. Of course, there are people who are willing to take the risk to see what they have achieved collapse in front of their eyes by working under the assumption that an unproven, but highly plausible, result is true (like Riemann hypothesis or Leopoldt conjecture). In some cases this is actually a good way to be on top of the competition (think of the work of Skinner and Urban on the main conjecture for elliptic curves which rests upon the existence of Galois representations that were not proven to exist before the completion of the proof of the Fundamental Lemma). - I think the recent work on compressed sensing is a good example. As I understand from listening to a talk by Emmanuel Candes - please correct me if I get anything wrong - the recent advances in compressed sensing began with an empirical observation that a certain image reconstruction algorithm seemed to perfectly reconstruct some classes of corrupted images. Candes, Romberg, and Tao collaborated to prove this as a mathematical theorem. Their proof captured the basic insight that explained the good performance of the algorithm: $l_1$ minimization finds a sparse solution to a system of equations for many classes of matrices. It was then realized this insight is portable to other problems and analogous tools could work in many other settings where sparsity is an issue (e.g., computational genetics). If Candes, Romberg, and Tao had not published their proof, and if only the empirical observation that a certain image reconstruction works well was published, it is possible (likely?) that this insight would never have penetrated outside the image processing community. - The fundamental lemma is an example that most believed and on whose truth several results depend. According to Wikipedia, Professor Langlands has said ... it is not the fundamental lemma as such that is critical for the analytic theory of automorphic forms and for the arithmetic of Shimura varieties; it is the stabilized (or stable) trace formula, the reduction of the trace formula itself to the stable trace formula for a group and its endoscopic groups, and the stabilization of the Grothendieck–Lefschetz formula. None of these are possible without the fundamental lemma and its absence rendered progress almost impossible for more than twenty years. and Michael Harris has also commented that it was a "bottleneck limiting progress on a host of arithmetic questions." - 1 Fundamental lemma is not even 50% obvious by any stretch of imagination. Like most of the Langlands program, it is not a specific result that admits a precise, uniform statement; rather, it is a guiding principle that needs to be fine-tuned in order to be compatible with other things that we, following Langlands, would like to believe in $-$ then, and only then, does it becomes meaningful to talk about proving it. – Victor Protsak Sep 6 2010 at 21:35 1 Thank you, Victor. I'm not proposing that the Fundamental Lemma is obvious, but it seems that is was accepted as likely to be true, because others based new work on it. PC below gives the example of Skinner and Urban, and Peter Sarnak says here time.com/time/specials/packages/article/…, that "It's as if people were working on the far side of the river waiting for someone to throw this bridge across," ... "And now all of sudden everyone's work on the other side of the river has been proven." – Anthony Pulido Sep 6 2010 at 22:45 2 I was under the impression that the fundamental lemma, at least, is a set of statements clear enough to be amenable to proof attempts. I think we have on page 3 here arxiv.org/abs/math/0404454 and Theorems 1 and 2 here arxiv.org/abs/0801.0446 precise statements for the various fundamental lemmas... It's possible I'm misunderstanding you. – Anthony Pulido Sep 6 2010 at 22:55 4 The fundamental lemma had a precise formulation, due to Langlands and Shelstad, in the 1980s (following earlier special cases). It is a collection of infinitely many equations, each side involving an arbitrarily large number of terms (i.e. by choosing an appropriate case of the FL, you can make the number of terms as large as you like). It was universally believed to be true because otherwise the theory of the trace formula (some of which was proved, but some of which remained conjectural), as developed by Langlands and others, would be internally inconsistent, something which no-one ... – Emerton Sep 7 2010 at 4:05 3 ... believed could be true. This a typical phenomenon in the Langlands program, I would say: certain very general principles, which one cannot really doubt (at least at this point, when the evidence for Langlands functoriality and reciprocity seems overwhelming), upon further examination, lead to exceedingly precise technical statements which in isolation can seem very hard to understand, and for which there is no obvious underlying mechanism explaining why they are true. But one could note that class field theory (which one knows to be true!) already has this quality. – Emerton Sep 7 2010 at 4:13 show 5 more comments Circle division by chords, http://mathworld.wolfram.com/CircleDivisionbyChords.html, leads to a sequence whose first terms are 1, 2, 4, 8, 16, 31. It's simple and effective to draw the first five cases on a blackboard, count the regions, and ask the students what's the next number in the sequence. - 7 I like that example and have used it myself. However, the conjecture that the number of regions doubles each time has nothing beyond a very small amount of numerical evidence to support it, and the best way of showing that it is false is, in my view, not to count 31 regions but to point out that the number of crossings of lines is at most n^4, from which it follows easily that the number of regions grows at most quartically. – gowers Sep 5 2010 at 15:23 13 More mischievously, the sequence is 1, 2, 4, 8, 16, ... , 256, ... – Richard Borcherds Sep 6 2010 at 4:28 3 That I never knew. What a great fact! – gowers Sep 6 2010 at 19:25 I would like to preface this long answer by a few philosophical remarks. As noted in the original posting, proofs play multiple roles in mathematics: for example, they assure that certain results are correct and give insight into the problem. A related aspect is that in the course of proving an intuitively obvious statement, it is often necessary to create theoretical framework, i.e. definitions that formalize the situation and new tools that address the question, which may lead to vast generalizations in the course of the proof itself or in the subsequent development of the subject; often it is the proof, not the statement itself, that generalizes, hence it becomes valuable to know multiple proofs of the same theorem that are based on different ideas. The greatest insight is gained by the proofs that subtly modify the original statement that turned out to be wrong or incomplete. Sometimes, the whole subject may spring forth from a proof of a key result, which is especially true for proofs of impossibility statements. Most examples below, chosen among different fields and featuring general interest results, illustrate this thesis. 1. Differential geometry a. It had been known since the ancient times that it was impossible to create a perfect (i.e. undistorted) map of the Earth. The first proof was given by Gauss and relies on the notion of intrinsic curvature introduced by Gauss especially for this purpose. Although Gauss's proof of Theorema Egregium was complicated, the tools he used became standard in the differential geometry of surfaces. b. Isoperimetric property of the circle has been known in some form for over two millenia. Part of the motivation for Euler's and Lagrange's work on variational calculus came from the isoperimetric problem. Jakob Steiner devised several different synthetic proofs that contributed technical tools (Steiner symmetrization, the role of convexity), even though they didn't settle the question because they relied on the existence of the absolutely minimizing shape. Steiner's assumption led Weierstrass to consider the general question of existence of solutions to variational problems (later taken up by Hilbert, as mentioned below) and to give the first rigorous proof. Further proofs gained new insight into the isoperimetric problem and its generalizations: for example, Hurwitz's two proofs using Fourier series exploited abelian symmetries of closed curves; the proof by Santaló using integral geometry established more general Bonnesen inequality; E.Schmidt's 1939 proof works in $n$ dimensions. Full solution of related lattice packing problems led to such important techniques as Dirichlet domains and Voronoi cells and the geometry of numbers. 2. Algebra a. For more than two and a half centuries since Cardano's Ars Magna, no one was able to devise a formula expressing the roots of a general quintic equation in radicals. The Abel–Ruffini theorem and Galois theory not only proved the impossibility of such a formula and provided an explanation for the success and failure of earlier methods (cf Lagrange resolvents and casus irreducibilis), but, more significantly, put the notion of group on the mathematical map. b. Systems of linear equations were considered already by Leibniz. Cramer's rule gave the formula for a solution in the $n\times n$ case and Gauss developed a method for obtaining the solutions, which yields the least square solution in the underdetermined case. But none of this work yielded a criterion for the existence of a solution. Euler, Laplace, Cauchy, and Jacobi all considered the problem of diagonalization of quadratic forms (the principal axis theorem). However, the work prior to 1850 was incomplete because it required genericity assumptions (in particular, the arguments of Jacobi et al didn't handle singular matrices or forms. Proofs that encompass all linear systems, matrices and bilinear/quadratic forms were devised by Sylvester, Kronecker, Frobenius, Weierstrass, Jordan, and Capelli as part of the program of classifying matrices and bilinear forms up to equivalence. Thus we got the notion of rank of a matrix, minimal polynomial, Jordan normal form, and the theory of elementary divisors that all became cornerstones of linear algebra. 3. Topology a. Attempts to rigorously prove the Euler formula $V-E+F=2$ led to the discovery of non-orientable surfaces by Möbius and Listing. b. Brouwer's proof of the Jordan curve theorem and of its generalization to higher dimensions was a major development in algebraic topology. Although the theorem is intuitively obvious, it is also very delicate, because various plausible sounding related statements are actually wrong, as demonstrated by the Lakes of Wada and the Alexander horned sphere. 4. Analysis The work on existense, uniqueness, and stability of solutions of ordinary differential equations and well-posedness of initial and boundary value problems for partial differential equations gave rise to tremendous insights into theoretical, numerical, and applied aspects. Instead of imagining a single transition from 99% ("obvious") to 100% ("rigorous") confidence level, it would be more helpful to think of a series of progressive sharpenings of statements that become natural or plausible after the last round of work. a. Picard's proof of the existence and uniqueness theorem for a first order ODE with Lipschitz right hand side, Peano's proof of the existence for continuous right hand side (uniqueness may fail), and Lyapunov's proof of stability introduced key methods and technical assumptions (contractible mapping principle, compactness in function spaces, Lipschitz condition, Lyapunov functions and characteristic exponents). b. Hilbert's proof of the Dirichlet principle for elliptic boundary value problems and his work on the eigenvalue problems and integral equations form the foundation for linear functional analysis. c. The Cauchy problem for hyperbolic linear partial differential equations was investigated by a whole constellation of mathematicians, including Cauchy, Kowalevski, Hadamard, Petrovsky, L.Schwartz, Leray, Malgrange, Sobolev, Hörmander. The "easy" case of analytic coefficients is addressed by the Cauchy–Kowalevski theorem. The concepts and methods developed in the course of the proof in more general cases, such as the characteristic variety, well-posed problem, weak solution, Petrovsky lacuna, Sobolev space, hypoelliptic operator, pseudodifferential operator, span a large part of the theory of partial differential equations. 5. Dynamical systems Universality for one-parameter families of unimodal continuous self-maps of an interval was experimentally discovered by Feigenbaum and, independently, by Coullet and Tresser in the late 1970s. It states that the ratio between the lengths of intervals in the parameter space between successive period-doubling bifurcations tends to a limiting value $\delta\approx 4.669201$ that is independent of the family. This could be explained by the existence of a nonlinear renormalization operator $\mathcal{R}$ in the space of all maps with a unique fixed point $g$ and the property that all but one eigenvalues of its linearization at $g$ belong to the open unit disk and the exceptional eigenvalue is $\delta$ and corresponds to the period-doubling transformation. Later, computer-assisted proofs of this assertion were given, so while Feigebaum universality had initially appeared mysterious, by the late 1980s it moved into the "99% true" category. The full proof of universality for quadratic-like maps by Lyubich (MR) followed this strategy, but it also required very elaborate ideas and techniques from complex dynamics due to a number of people (Douady–Hubbard, Sullivan, McMullen) and yielded hitherto unknown information about the combinatorics of non-chaotic quadratic maps of the interval and the local structure of the Mandelbrot set. 6. Number theory Agrawal, Kayal, and Saxena proved that PRIMES is in P, i.e. primality testing can be done deterministically in polynomial time. While the result had been widely expected, their work was striking in at least two respects: it used very elementary tools, such as variations of Fermat's little theorem, and it was carried out by a computer science professor and two undergraduate students. The sociological effect of the proof may have been even greater than its numerous consequences for computational number theory. - 3 I meant the inspirational effect due to (a) elementary tools used; and (b) the youth of 2/3 of the authors. – Victor Protsak Sep 7 2010 at 8:09 28 It is indeed great and inspiring to see very young people cracking down famous problems. Recently, I find it no less inspiring to see old people cracking down famous problems. – Gil Kalai Sep 7 2010 at 8:57 1 As far as Euler's formula goes, the book Proofs and Refutations by Imre Lakatos en.wikipedia.org/wiki/Proofs_and_Refutations shows how many interesting questions and new considerations can be derived from a seemingly-obvious formula. – Thierry Zell Mar 19 2011 at 5:30 show 1 more comment Here's an example: In the Mathscinet review of "Y-systems and generalized associahedra", by Sergey Fomin and Andrei Zelevinsky, you find: Let $I$ be an $n$-element set and $A=(a_{ij})$, $i,j\in I$, an indecomposable Cartan matrix of finite type. Let $\Phi$ be the corresponding root system (of rank $n$), and $h$ the Coxeter number. Consider a family $(Y_i(t))_{i\in I,\,t\in\Bbb{Z}}$ of commuting variables satisfying the recurrence relations $$Y_i(t+1)Y_i(t-1)=\prod_{j\ne i}(Y_j(t)+1)^{-a_{ij}}.$$ Zamolodchikov's conjecture states that the family is periodic with period $2(h+2)$, i.e., $Y_i(t+2(h+2))=Y_i(t)$ for all $i$ and $t$. That conjecture claims that an explicitly described algebraic map is periodic. The conjecture can be checked numerically by plugging in real numbers with 30 digits, and iterating the map the appropriate number of times. If you see that time after time, the numbers you get back agree with the initial values with a 29 digit accuracy, then you start to be pretty confident that the conjecture is true. For the $E_8$ case, the proof presented in the above paper involves a massive amount of symbolic computations done by computer. Is it really much better than the numerical evidence? Conclusion: I think that we only like proofs when we learn something from them. It's not the property of "being a proof" that is attractive to mathematicians. - 4 Gian-Carlo Rota would agree, for he said (in "The Phenomenology of Mathematical Beauty") that we most value a proof that enlightens us. – Joseph O'Rourke Sep 5 2010 at 3:00 2 That's a very interesting example, even if it is of the opposite of what I asked. My instinct is to think it's good that there's a proof, but I'm not sure how to justify that. And obviously I'd prefer a conceptual argument. – gowers Sep 5 2010 at 15:26 13 And now, of course, we finally understand exactly what Gian-Carlo meant: A proof enlightens us if: 1) it is the first proof, 2) it is accepted, and 3) it has at least 10 up votes! – Gil Kalai Sep 6 2010 at 21:05 "Sufficient unto the day is the rigor thereof."-E.H.Moore. There's a lot of discussion over not only the role of rigor in mathematics,but whether or not this is a function of time and point in history.Clearly,what was a rigorous argument to Euler would not pass muster today in a number of cases. Passing from generalities to specific cases,I think the prototype of statements which were almost universally accepted as true without proof was the early-19th century notion that globally continuous real valued functions had to have at most a finite number of nondifferentiable points.Intuitively,it's easy to see why in a world ruled by handwaving and graph drawing,this would be seen as true. Which is why the counterexamples by Bolzano and Weirstrauss were so conceptually devastating to this way of approaching mathematics. Edit: I see Jack Lee already mentioned this example "below" in his excellent list of such cases. But to be honest,I don't think his first example is really about rigor so much as a related but more profound change in our understanding how mathematical systems are created. The main reason no one thought non-Euclidean geometries made any sense was because most scientists believed Euclidean geometry was an empirical statement about the fundamental geometry of the universe.Studies of mechanics supported this until the early 20th century;as long as one stays in the "common sense" realm of the world our 5 senses perceive and we do not approach relativistic velocities or distances,this is more or less true. Eddington's eclipse experiments finally vindicated not only Einstein's conceptions,but indirectly,non-Euclidean geometry-which until that point,was greeted with skepticism outside of pure mathematics. - 1 This seems more like a discussion on Jack Lee's post than an answer. – Jonas Meyer Sep 4 2010 at 23:27 show 3 more comments Based on the recent update to the question, Fermat's Last Theorem seems like the top example of a proof being far more valuable than the truth of the statement. Personally it's a rare occurrence for me to use the nonexistence of a rational point on a Fermat curve but for instance it is quite common for me to use class numbers. - Here is an example: 19 century geometers extended Euler's formula V-E+F=2 to higher dimensions: the alternating sum of the number of i-faces of a d-dimensional polytope is 2 in odd dimensions and 0 in even dimensions. The 19th centuries proofs were incomplete and the first rigorous proof came with Poincare and used homology groups. Here, what enabled a rigorous proof was arguably even more important than the theorem itself. - 1 This is not so much an example of increased care in an arguement as the development of critical technology needed to prove a result.The need for such technology is not always clear to mathematicains when they begin to formulate such arguments.The important question here is whether or not the general form of Euler's formula could have been proven WITHOUT it.In dimensions less then or equal to 3,there are many alternate proofs using purely combinatorial arguements.I'm not sure it can be proven without homology in higher-dimensional spaces. – Andrew L Sep 4 2010 at 23:39 1 Andrew: Yes, it can be proven without homology for higher dimensional polytopes: see ics.uci.edu/~eppstein/junkyard/euler/shell.html – David Eppstein Sep 5 2010 at 5:02 2 It does happen that techniques developed in order to give a proof are more very important. In this case, as it turned out a few decades after Poincare, the high dimensional Euler's theorem can be proved without algebtaic topology, and in the 70s even the specific gaps in the 19th century proofs was fixed, but the new technology allows for extensions of the theorem that cannot be proved by elementary method and it also shed light on the original Euler theorem: That the Euler characteristic is a topological invariant. – Gil Kalai Sep 5 2010 at 5:10 The way current computer algebra systems (that I know of) are designed is a compromise between ease of use and mathematical rigor. Although in practice, most of the answers given by CASes are correct, the lack of rigor is still a problem because the user cannot fully trust the results (even under the assumption that the software is bug-free). Now, it might sound like just another case of "99% certainty is enough," but in practice it means having to verify the results independently afterwards, which could be considered unnecessary extra work. The root of the problem seems to be that a CAS manipulates expressions when it should output theorems instead. In many cases, the expressions simply don't have any rigorous interpretation. For example, variables are usually not introduced explicitly and thus not properly quantified; in the result of an indefinite integral they might even appear out of nowhere. Dealing with undefinedness is another problem. All of this is inherent in the architecture of computer algebra systems, so it cannot be fixed properly without switching to a different design. The extra 1% of certainty may indeed not justify such a change. But if rigor had been emphasized more from the start, maybe we would have trustworthy CASes now. I think this line of thought can be generalized. (As a non-mathematician) I can't help but wonder how mathematics would have progressed without the widespread introduction of rigor in the 19th century. I can't really imagine what things would be like if we still didn't have a proper definition of what a function is. So maybe rigor is indeed not strictly necessary in particular cases, but it has shaped mathematical practice in general. - 2 I tend to not use CAS packages that aren't open-source. Knowing the precise details of the implementation of the algorithm allows you to understand its limitations, when it will and will not function. There is still uncertainty in this of course -- did I really understand the algorithm? Is the computer hardware faulty? Does the compiler not compile the code correctly? And so on. Open-source also has the advantage that the algorithms are re-usable and not "canned". – Ryan Budney Sep 4 2010 at 20:40 I think that the question itself is entirely misleading. It tacitly assumes as if mathematics could be separated into two parts: mathematical results and their proofs. Mathematics is nothing other than the proofs of mathematical results. Mathematical statements lacks any value, they are neither good nor bad. From the mathematical point of view, it is entirely immaterial whether the answer to a mathematical question like `Is there an even integer greater than two that is not the sum of two primes?' is yes or no. Mathematicians simply do not interested in the right answer. What they would like to do is to solve the problem. That is the main difference between natural sciences or engineering on the one hand, and mathematics on the other. A physicist would like to know the right answer to his question and he does not interested in the way it is obtained. An engineer needs a tool that he can use in the course of his work. He does not interested in the way a useful device works. Mathematics is nothing other than a specific set consisting of different solutions to different problems and, of course, some unsolved problems waiting to be solved. Proofs are not important for mathematics, they constitute the body of knowledge we call mathematics. - 7 This is a very Bourbakist view. Much of interesting mathematics does not conform to it, because ideas and open problems are just as important in mathematics as rigorous proofs (even leaving aside the distinction between theory and proofs that is not appreciated by non-mathematicians). – Victor Protsak Sep 5 2010 at 4:27 3 Victor, Gyorgy's point of view does not conflict with the importance of ideas and open problems. Still for a large part of mathematics proofs is the essential part. The relation between a mathematical result and its proof can often be compared to the relation between the title of a picture or a poem or a musical piece and the content. – Gil Kalai Sep 5 2010 at 5:20 4 Gil, gowers' question addressed the distinction between an intuitive proof and a rigorous proof and my comment was written with that in mind. Using your artistic analogies, let me say that a piece of music cannot be reduced to the set of notes, nor a poem to the set of words, that comprise it (the analogy obviously breaks down for "modern art", such as atonal music and dadaist poetry). – Victor Protsak Sep 5 2010 at 7:28 i once got a letter from someone who had overwhelming numerical evidence that the sum of the reciprocals of primes is slightly bigger than 3 (he may have conjectured the limit was π). The sum is in fact infinite, but diverges so slowly (like log log n) that one gets no hint of this by computation. - 42 This reminds me of a classical mathematical foklore "get rich" scheme (or scam). The ad in a newspaper says: send in 10 dollars and get back unlimited amount of money in monthly installments. The dupe follows the instructions and receives 1 dollar the first month, 1/2 dollar the second month, 1/3 dollar the third month, ... – Victor Protsak Sep 6 2010 at 2:37 20 I remember the first time I learned that the harmonic series is divergent. I was in high school, in my first calculus class; it was in 2001. I was really surprised and couldn't really believe that it could be divergent, so I programmed my TI-83 to compute partial sums of the harmonic series. I let it run for the entire day, checking in on the progress periodically. If I recall correctly, by the end of the day the partial sum remained only in the 20s. Needless to say, I was not convinced of the divergence of the series that day. – Kevin Lin Sep 6 2010 at 7:09 9 If one wants to carry this to the extreme, any divergent series with the property that the n-th term goes to zero will converge on a calculator as the terms will eventually fall below the underflow value for the calculator, and hence be considered to be zero. – Chris Leary Jan 1 2012 at 23:54 The evidence for both quantum mechanics and for general relativity is overwhelming. However, one can prove that without serious modifications, these two theories are incompatible. Hence the (still incomplete) quest for quantum gravity. - 28 Devils Advocate: Couldn't one argue that this demonstrates the opposite? Our theories are mathematically incompatible, yet they can compute the outcome of any fundamental physical experiment to a half dozen digits. Clearly, this shows that mathematical consistency is overrated! – David Speyer Sep 6 2010 at 12:06 16 @David, you may be right: quantum mechanics and general relativity are incompatible, and the first time that they come into mathematical conflict the universe will end. This should be around the time when the first black hole evaporates, around $10^{66}$ years from now. – Peter Shor Sep 6 2010 at 14:01 3 Indeed nonrigirous mathematical computations and heuristic arguments in physics are spectacularly successful for good experimental predictions and even for mathematics. Yet the accuracy David talked about is only in small fragments of standard physics. Of course, just like asking what is the purpose of rigor we can also ask what is the purpose gaining the 7th accurate digit in experimental predictions. The answer that allowing better predictions like rigorous proofs enlighten us is a good partial answer. – Gil Kalai Sep 7 2010 at 5:17 1 There were intensice discussions regarding the role of rigor over physics weblogs. Here is an example from Distler's blog "musing": golem.ph.utexas.edu/~distler/blog/archives/… – Gil Kalai Sep 7 2010 at 8:07 show 1 more comment [Edited to correct the Galileo story] An old example of a plausible result that was overthrown by rigor is the 17th-century example of the hanging chain. Galileo once said (though he later said otherwise), and Girard claimed to have proved, that the shape was a parabola. But this was disproved by Huygens (then aged 17) by a more rigorous analysis. Some decades later, the exact equation for the catenary was found by the Bernoullis, Leibniz, and Huygens. In the 20th century, some people thought it plausible that the shape of the cable of a suspension bridge is also a catenary. Indeed, I once saw this claim in a very popular engineering mathematics text. But a rigorous argument shows (with the sensible simplifying assumption that the weight of the cable is negligible compared with the weight of the road) that the shape is in fact a parabola. - 17 The story about Galileo and the hanging chain is a myth: he was well aware that it is approximately but not exactly a parabola, and even commented that the approximation gets better for shorter chains. If you take a very long chain with the ends close together, which Galileo was perfectly capable of doing, it is obviously not a parabola – Richard Borcherds Sep 3 2010 at 21:43 3 It may be a mistranslation: my guess is he meant that a catenary RESEMBLES a parabola. Later on in the same book he makes it clear that he knows they are different. – Richard Borcherds Sep 3 2010 at 22:44 2 Years ago, I saw in "Scientific American" a 2-page ad for some calculator. One of the two pages was a photograph of a suspension bridge. Across the top was the equation of a catenary. – Andreas Blass Sep 4 2010 at 1:27 9 I was very fascinated by Richard Borcherds's comments and looked at two different translations of the Galileo's book (I also found a quote from the original text, but my Italian is not good enough). The hanging line is definitely described to take the shape of a parabola, but this statement is given in a section describing quick ways of sketching parabolae. Indeed, later in the book, Galileo talks about the shape of a hanging chain being parabolic only approximately: "the Fitting shall be so much more exact, by how much the describ'd Parabola is less curv'd, i.e. more distended". – Aleksey Pichugin Sep 4 2010 at 11:15 4 @J.M.:Jungius may have been the first to publish a proof that the catenary is not a parabola (1669), but the proof of Huygens in his letter to Mersenne was earlier (1646). – John Stillwell Sep 8 2010 at 15:10 show 4 more comments This question is begging for someone to state the obvious, so here goes. Take for example the existence and uniqueness of solutions to differential equations. Without these theorems, the mathematical models used in many branches of the physical sciences are incapable of making actual predictions. If potentially the DE has no solutions, or the model provides infinitely many solutions, your model has no predictive power. So the model isn't really science. In that regard, the point of proof in mathematics is to create a foundation that allows for quantitative physical sciences to exist to have a firm philosophical foundation. Moreover, the proofs of existence and uniqueness shed light on the behaviour of solutions, allowing one to make precise predictions about how good various approximations are to actual solutions -- giving a sense for how computationally expensive it is to make reliable predictions. - 6 I regret to say that most physicists seem to neither know nor care about rigorous proofs of existence and uniqueness theorems. But they seem to have no trouble doing good physics without them. – Richard Borcherds Sep 3 2010 at 21:55 4 In physics, having a formula or approximation scheme depending on N parameters (= dim. of phase space) shows existence and uniqueness locally, with global questions of singularities, attractors, topology etc understood by calculation and simulation. For classical ODE this is almost always enough and there are very few cases where careful analysis and error estimates overturned accepted physics ideas. There are more cases where physics heuristics drove the mathematics and some where they changed intuitions that prevailed in the math community. – T. Sep 4 2010 at 7:58 4 @T it sounds like you're assuming existence and uniqueness. What are you approximating? Approximations don't matter if you don't know what you're approximating. Moreover, if you're approximating one of infinitely-many solutions, this gives you no sense for how all the solutions (with certain initial conditions) behave, and limits your ability to predict anything. – Ryan Budney Sep 4 2010 at 18:26 4 May I point out that models which do not satisfy existence or uniqueness can be very useful. For an example, consider the Navier-Stokes equations (you get the Clay Prize for proving existence). It is quite possible that there are initial conditions where the solution does not exist. This could happen because the Navier-Stokes equations assume that the fluid is incompressible, and all real fluids are (at least to some very small degree) compressible. Even if existence were to fail to be satisfied, these equations would still have enormous predictive power, and be real science. – Peter Shor Sep 5 2010 at 17:04 5 From a physical point of view, uniqueness is a claim about causality. Given our belief in causality, any proposed law of nature that does not obey uniqueness may well be considered physically defective. But perhaps more important than uniqueness is the stronger property of continuous dependence on the data. (The former asserts that the same data will lead to the same solution; the latter asserts that nearby data leads to nearby solutions.) Once one has this, one has some confidence that one's model can be numerically simulated with reasonable accuracy, and also be resistant to noise. – Terry Tao Sep 8 2010 at 5:46 show 8 more comments When I teach our "Introduction to Mathematical Reasoning" course for undergraduates, I start out by describing a collection of mathematical "facts" that everybody "knew" to be true, but which, with increasing standards of rigor, were eventually proved false. Here they are: 1. Non-Euclidean geometry: The geometry described by Euclid is the only possible "true" geometry of the real world. 2. Zeno's paradox: It is impossible to add together infinitely many positive numbers and get a finite answer. 3. Cardinality vs. dimension: There are more points in the unit square than there are in the unit interval. 4. Space-filling curves: A continuous parametrized curve in the unit square must miss "most" points. 5. Nowhere-differentiable functions: A continuous real-valued function on the unit interval must be differentiable at "most" points. 6. The Cantor Function: A function that is continuous and satisfies f'(x)=0 almost everywhere must be constant. 7. The Banach-Tarski paradox: If a bounded solid in R^3 is decomposed into finitely many disjoint pieces, and those pieces are rearranged by rigid motions to form a new solid, then the new solid will have the same volume as the original one. - 3 Regarding 5: cf the comment here: mathoverflow.net/questions/22189/… gives a strictly increasing function whose derivative is zero almost everywhere. Intuitively such a thing shouldn't exist, but applying the definitions rigourously shows it is true. – David Roberts Sep 4 2010 at 4:08 19 Historical examples tend to retroactively attribute stupid errors that were not the original, and still subtle, issue. In 3 and 7 the equivalences are not geometric (see Feynman's deconstruction of Banach-Tarski as "So-and-So's Theorem of Immeasurable Measure"). For #1, Riemannian geometry doesn't address historical/conceptual issue of non-Euclidean geometry, which was about logical status of the Parallel Axiom, categoricity of the axioms, and lack of 20th-century mathematical logic framework. Zeno's contention that motion is mysterious remains true today, despite theory of infinite sums. – T. Sep 4 2010 at 8:15 3 It seems to me that items 3-7 are regarded by most people as "monsters" and as such not really worthy of serious consideration. As for items 1 and 2, I think that not only have most people not heard of them, when they do hear of them, they regard them either as jokes or don't really get the point at all. So it doesn't seem to me that these are convincing arguments for most people. (They are, to be sure, convincing arguments for me.) To some extent, I'm sure this is something that can only be appreciated by some experience. I think for instance that the Pythagorean theorem is [out of space – Carl Offner Apr 28 2011 at 2:58	{"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": 34, "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": false, "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": 0.953877866268158, "perplexity_flag": "middle"}
http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/Maxima_and_minima	# All Science Fair Projects ## Science Fair Project Encyclopedia for Schools! Search Browse Forum Coach Links Editor Help Tell-a-Friend Encyclopedia Dictionary # Science Fair Project Encyclopedia For information on any area of science that interests you, enter a keyword (eg. scientific method, molecule, cloud, carbohydrate etc.). Or else, you can start by choosing any of the categories below. # Maxima and minima In mathematics, a point x* is a local maximum of a function f if there exists some ε > 0 such that f(x) ≥ f(x) for all x with \|x-x\| < ε. Stated less formally, a local maximum is a point where the function takes on its largest value among all points in the immediate vicinity. On a graph of a function, its local maxima will look like the tops of hills. A local minimum is a point x* for which f(x) ≤ f(x) for all x with \|x-x\| < ε. On a graph of a function, its local minima will look like the bottoms of valleys. A global maximum is a point x* for which f(x) ≥ f(x) for all x. Similarly, a global minimum is a point x for which f(x) ≤ f(x) for all x. Any global maximum (minimum) is also a local maximum (minimum); however, a local maximum or minimum need not also be a global maximum or minimum. The concepts of maxima and minima are not restricted to functions whose domain is the real numbers. One can talk about global maxima and global minima for real-valued functions whose domain is any set. In order to be able to define local maxima and local minima, the function needs to take real values, and the concept of neighborhood must be defined on the domain of the function. A neighborhood then plays the role of the set of x such that \|x - x\| < ε. One refers to a local maximum/minimum as to a local extremum (or local optimum), and to a global maximum/minimum as to a global extremum (or global optimum). Contents ## Finding maxima and minima Finding global maxima and minima is the goal of optimization. For twice-differentiable functions in one variable, a simple technique for finding local maxima and minima is to look for stationary points, which are points where the first derivative is zero. If the second derivative at a stationary point is positive, the point is a local minimum; if it is negative, the point is a local maximum; if it is zero, further investigation is required. If the function is defined over a bounded segment, one also need to check the end points of the segment. ## Examples • The function x2 has a unique global minimum at x = 0. • The function x3 has no global or local minima or maxima. Although the first derivative (3x2) is 0 at x = 0, the second derivative (6x) is also 0. • The function x3/ 3 - x has first derivative x2 - 1 and second derivative 2x. Setting the first derivative to 0 and solving for x gives stationary points at -1 and +1. From the sign of the second derivative we can see that -1 is a local maximum and +1 is a local minimum. Note that this function has no global maxima or minima. • The function \|x\| has a global minimum at x = 0 that cannot be found by taking derivatives, because the derivative does not exist at x = 0. • The function cos(x) has infinitely many global maxima at 0, ±2π, ±π, ..., and infinitely many global minima at ±π, ±3π, ... . • The function 2cos(x) - x has infinitely many local maxima and minima, but no global maxima or minima. • The function x3 + 3x2 - 2x + 1 defined over the closed interval (segment) [-4,2] (see graph) has two extrema: one local maximum in $x = (-1-\sqrt{15})/3$, one local minimum in $x = (-1+\sqrt{15})/3$, a global maximum on x=2 and a global minimum on x=-4. ## Functions of more variables For functions of more variables similar concepts apply, but there is also the saddle point. ## See also 03-10-2013 05:06:04	{"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": 2, "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": 0.8693269491195679, "perplexity_flag": "middle"}
http://nrich.maths.org/4769	### Balances and Springs Balancing interactivity with springs and weights. ### Inside Outside Balance the bar with the three weight on the inside. ### Balance Point Attach weights of 1, 2, 4, and 8 units to the four attachment points on the bar. Move the bar from side to side until you find a balance point. Is it possible to predict that position? # Swings and Roundabouts ##### Stage: 4 Challenge Level: This text is usually replaced by the Flash movie. $O$ is the centre of the balance, and $X$ is the position of the initially empty hook. If you leave the hook at $X$ empty, what angle does $OX$ make to the vertical once the balance settles? If you were to attach a weight of $1$ unit at $X$ , what do you think the angle might be? Were you right? If you were to attach a weight of $2$ units at $X$ what do you think the angle might be? Right that time? You only need the $1$ and $2$ unit weights for the problem set but the others are there to let you explore, conjecture and test - have fun. The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.	{"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": 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": 0.9410585761070251, "perplexity_flag": "middle"}
http://mathematica.stackexchange.com/questions/tagged/algebraic-manipulation?page=2&sort=active&pagesize=30	# Tagged Questions The art of manipulating an algebraic expression into the desired form. 3answers 211 views ### Is it possible to use Composition for polynomial composition? I want to do this: $P = (x^3+x)$ $Q = (x^2+1)$ $P \circ Q = P \circ (x^2+1) = (x^2+1)^3+(x^2+1) = x^6+3x^4+4x^2+2$ I used Composition for testing if that could ... 3answers 238 views ### Limiting form of a polynomial expression When simplifying an expression by hand, a trick that is often used is to remove terms that are lower powers of the independent variable, for instance, as $x \rightarrow \infty$, $x^2 + x$ becomes ... 1answer 162 views ### Finding mappings between expressions Suppose we have an expression of the form: $j=\frac{A\left(t\right)}{B\left(t\right)}=\frac{C\left(s\right)}{D\left(s\right)}$ That is, $j$ can be expressed either as a function of $t$, or as a ... 3answers 460 views ### What function can I use to evaluate $(x+y)^2$ to $x^2 + 2xy + y^2$? What function can I use to evaluate $(x+y)^2$ to $x^2 + 2xy + y^2$? I want to evaluate It and I've tried to use the most obvious way: simply typing and evaluating $(x+y)^2$, But it gives me only ... 1answer 187 views ### RootSum result manipulation/simplification Consider the sum sum1 = Sum[ k/( k^7 - 2 k + 3), {k, Infinity}] ... 4answers 299 views ### “Evaluating” polynomials of functions (Symbols) I want to implement the following type evaluation symbolically $$(f^2g + fg + g)(x) \to f(x)^2 g(x) + f(x) g(x) + g(x)$$ In general, on left hand side there is a polynomial in an arbitrary number of ... 2answers 1k views ### How to convert a system of parametric equations to a normal equation? For example, I have a system of parametric equations (R is a constant number) : ... 2answers 160 views ### What does MinimalPolynomial do? suppose $x^5-x-1=0$, $y^7-y-1=0$ and $z=x+y$. I want to find a minimal polynomial expression of $z$, such that $p(z)=0$, and the code can be written like this: ... 2answers 278 views ### expanding a polynomial and collecting coefficients I'm trying to expand the following polynomial ... 4answers 854 views ### Checking if two trigonometric expressions are equal Say I have two trigonometric expressions which are a bit complicated. Is there a quick way to check if they reduce to the same thing (that they are equal) using Mathematica? I was solving this: \$y'' ... 6answers 900 views ### Replacing composite variables by a single variable To replace a single variable by another variable, one can simply use the the replace all (/.) operator (e.g., ... 4answers 407 views ### Is there a way to Collect[] for more than one symbol? Oftentimes you find yourself looking for polynomials in multiple variables. Consider the following expression: a(x - y)^3 + b(x - y) + c(x - y) + d as you can ... 3answers 253 views ### Distances between points in periodic cube How can one implement more efficiently/elegantly/memory savvily the following function which returns a matrix of all Euclidian distances between points in 3D within a cube of width ... 2answers 279 views ### How do I make Mathematica understand roots of unity? Try the following: z = Exp[2 \[Pi] I / 5] Z = { { z, 1 }, { 0, 1/z } } Simplify[MatrixPower[Z, 5]] //MatrixForm Of course, this should return a 2 x 2 identity ... 4answers 171 views ### I need to move all the variables(x,y,z) to one side of an equality for the equation of a plane I have 5(x-1)-(y+1)-(z+1)==0. If I use FullSimplify on it it returns 5x==7+x+y. This is not ... 2answers 293 views ### Manipulating an equation into standard quadratic form? Say I have an equation of the form $$u s + \frac{1}{v} + \frac{1}{p s + q} = 0$$ (or any form that can be written as a standard quadratic, really, the above form is just an example; they'll all be ... 2answers 261 views ### Shaping/simplifying equations in a certain way A problem I am occasionally facing is to simplify an equation not to it's shortest form but to a form that is simple by other means. Often, this is grouping the term according to certain functions, ... 2answers 294 views ### How to define a non-standard algebra in Mathematica? I want to define an algebra, where there are three elements: 0, 1 and $\infty$ and two operations, addition and multiplication defined, both commutative: \begin{align*} 0+0&=0\\ 0+1&=1\\ ... 6answers 761 views ### How do I replace a variable in a polynomial? How do I substitue z^2->x in the following polynomial z^4+z^2+4? z^4+z^2+4 /. z^2->x ... 2answers 474 views ### Expand modulus squared Is it possible to make a function in Mathematica that expands expressions of the form $$\|z + w\|^2 = \|z\|^2 + 2\text{Re} \overline{z}w + \|w\|^2?$$ Preferably it should also be able to handle things ...	{"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": 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": 0.8967083096504211, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2007/06/27/categorification/?like=1&source=post_flair&_wpnonce=966197e62c	# The Unapologetic Mathematician ## Categorification Categorification is a big, scary word that refers to a much simpler basic idea than you’d expect, and it’s also huge business right now. There’s a lot of this going on at The n-Category Café, for instance. Both knot Floer homology and Khovanov homology are examples of categorifications that seem to be popular with the crew at the Secret Blogging Seminar. John Baez has a bunch of papers circling around “higher-dimensional algebra”, which is again about categorification. So what is categorification? Well, as I said the core idea’s not really that hard to grasp, but it’s even easier to understand by first considering the reverse process: decategorification. And decategorification is best first approached through an example. Since this post will be going to this fortnight’s Carnival, I’ll be saying a fair amount that I’ve said before and linking back to definitions most regular readers should know offhand. It might seem slower, but I’ll be back up to the regular rhetorical pace tomorrow. Or you could take it as a bit of review, just to make sure you’re on top of things. Because of this, the post is going to get long, so I’ll put it behind a jump. Let’s consider the category $\mathbf{FinSet}$ of finite sets and functions between them. We’ve identified various kinds of functions that are of special interest to us: one-to-one functions (“injections” or “monomorphisms”), onto functions (“surjections” or “epimorphisms”), and functions which are both (“bijections” or “isomorphisms”). Right now we’re particularly interested in isomorphisms in this category — bijections of finite sets. More to the point, we’re interested in getting rid of them. “Isomorphic” means “essentially the same”, so why bother with a bunch of different copies of what are essentially the same set over and over again? We can set up a bijection from the set $\{a,b,c\}$ to the set $\{1,2,3\}$, so let’s just say they’re the same thing. “What? Madness!”, you say, and it’s easy to say that when you’re holding the category in your hands. The dirty little secret is that we do this all the time without saying so explicitly, and we teach all our children to do the same thing. The term we tell them is “counting”, and the more technical term is cardinality. We can set up a bijection between two finite sets exactly when they have the same number of elements. Identifying such sets amounts to forgetting everything about them except how many elements they have. The tautology “3=3″ expands to, “the number of elements in $\{a,b,c\}$ is the same as the number of elements in $\{1,2,3\}$,” which really means, “there is a bijection between $\{a,b,c\}$ and $\{1,2,3\}$.” This is an example of decategorification: we take some category — here it’s $\mathbf{FinSet}$ — and break it down to the set of isomorphism classes of its objects — here it’s the set $\mathbb{N}$ of natural numbers. This loses a lot of information. In particular, we only say that two objects are isomorphic instead of how they are isomorphic. Categorification is the (somewhat loosely-defined) reverse process of trying to replace equations by explicit isomorphisms. There’s a story (fable?) I remember from when I was first learning about the abstract notion of “number”. A shepherd lets his flock out from the pen to graze during the day, but he wants to make sure that they all come back at night. To handle this, as they leave the pen in the morning he picks up a little stone as each sheep leaves the pen and puts it in a bag. In the evening, each time a sheep enters the pen he takes a stone out of the bag. If the bag is empty when the sheep are all in, he’s got the same number in the pen as he did in the morning. What’s “really” going on here? There are three sets we’re looking at: $M=\{\mathrm{sheep\ in\ the\ morning}\}$, $S=\{\mathrm{stones\ in\ the\ bag}\}$, and $E=\{\mathrm{sheep\ in\ the\ evening}\}$. When the shepherd puts a stone in the bag for each sheep leaving in the morning he’s setting up an explicit bijection from $M$ to $S$. Then when he takes a stone out for each returning sheep he’s setting up an explicit bijection from $S$ to $E$. We can compose these two bijections to get a bijection from $M$ to $E$. That is, the sets $M$ and $E$ are now known to be in the same isomorphism class in $\mathbf{FinSet}$, and isomorphism classes are determined by cardinalities, so $M$ and $E$ have the same cardinality — the same number of elements. Now, this would be just a curiosity if that’s all there was to it, but it goes deeper still. Remember that the natural numbers aren’t just a set. There’s a lot of extra structure there: $\mathbb{N}$ is a commutative ordered rig. We have an addition and a multiplication of natural numbers, which satisfy a number of algebraic properties. We have a concept of which numbers are bigger than other numbers, which plays nicely with the addition and the multiplication. All of this structure lifts up to $\mathbf{FinSet}$. We can see that $\|A\|\leq\|B\|$ if and only if there is a monomorphism (one-to-one function) from $A$ to $B$. The category of finite sets has finite products and coproducts. The product $A\times B$ is the regular Cartesian product of $A$ and $B$ — the set of pairs $(a,b)$ with $a\in A$ and $b\in B$. The coproduct $A\coprod B$ is the disjoint union of $A$ and $B$ — the set containing all elements in both sets, but counting a copy of an element in $A$ as different from a copy of “the same” element in $B$. If we write $\|A\|$ for the cardinality of $A$, we can verify that $\|A\times B\|=\|A\|\cdot\|B\|$, and that $\|A\coprod B\|=\|A\|+\|B\|$. That is, coproducts categorify the addition of cardinal numbers, while products categorify their multiplication. The initial and terminal objects of $\mathbf{FinSet}$ are the set $\varnothing$ with no elements and a set $$ with a single element, respectively. Note that even though the terminal object is only specified up to isomorphism, we’re just going to identify any sets that are isomorphic when we decategorify so they all become the cardinal $1$. The empty set, of course, becomes the cardinal ${}0$. We know that $n\cdot1=n$ for all numbers $n$. In $\mathbf{FinSet}$ this says that $A\times\cong A$ for all finite sets $A$. We can verify this identity using the universal property of the product: $A$ has the identity function to $A$ and the (unique) function to the terminal object $$. Given any other set $X$ with a function $f:X\rightarrow A$ and the unique function $X\rightarrow$, we can just use $f$ as the unique function required by the product. Thus $A$ satisfies the universal property of $A\times*$, and so must be in that isomorphism class. We can dualize this argument to show that $A\coprod\varnothing\cong A$. This shows that we can categorify the additive and multiplicative identities. The same sorts of arguments can be used to verify all the other basic properties of the natural numbers: • Addition and multiplication are associative: $(a+b)+c=a+(b+c)$ and $(a\cdot b)\cdot c=a\cdot(b\cdot c)$. • Addition and multiplication are commutative: $a+b=b+a$ and $a\cdot b=b\cdot a$. • Multiplication distributes over addition: $a\cdot(b+c)=a\cdot b+a\cdot c$. • Addition and multiplication preserve order: if $a\leq b$ then $a+c\leq b+c$ and $a\cdot c\leq b\cdot c$. Try to at least write down the categorified statements of each of these properties. If you’re up for it, prove them using similar techniques to those used to prove the identity properties above. If you’re really up for a challenge, show that all the isomorphisms you’re writing down are natural. In studying algebraic gadgets as sets with extra structure — and especially in using them to attack other problems — we write down a lot of equations when “really” we just have equivalences. Categorification seeks to replace these equations by isomorphisms. Sets become categories. Functions between sets become functors between categories. Equations between functions become natural isomorphisms between functors. Extra structures on sets become extra structures on categories. Of course it immediately gets more complicated than that. Down in set-land equations are equations are equations, but in category-land we can run into trouble. We have to pick the natural isomorphisms to categorify our equations “coherently” so that as we string them together we don’t end up in any contradictions. And then immediately we’ll want to take whatever we get and categorify that, which will lead to even more convoluted constructions in general… However, there is hope, and that hope is why we categorify in the first place. There is usually some way of looking at a structure that makes it perfectly clear how to categorify it, and there’s in a sense in which this perspective is the way the structure “should” be viewed. As Urs Schreiber said, “One knows one is getting to the heart of the matter when the definitions in terms of which one conceives the objects under consideration categorify effortlessly.” I rephrase this as, “when you really understand a mathematical concept, its categorification should be readily apparent”, which then becomes the directive: If you want to understand something, try to categorify it! ### Like this: Posted by John Armstrong \| Category theory ## 13 Comments » 1. Meaning, I really need to get about categorifying supersymmetric quantum mechanics. Comment by \| June 28, 2007 \| Reply 2. Meaning, I really need to get about categorifying supersymmetric quantum mechanics. It might sound funny now, but isn’t Gukov attempting to do some toy models of this? Comment by \| June 28, 2007 \| Reply 3. [...] But this all then raises a natural question: we’re looking at microstates (statistical mechanics) and declare examples of them to be “the same” if their macrostates are the same. But we don’t like “the same” — we like “isomorphic. So, is the passage from statistical mechanics to thermodymanics a decategorification? [...] Pingback by \| July 11, 2007 \| Reply 4. Over at the Carnival they’re claiming you did a post on “categorization”. Grr. Comment by \| July 18, 2007 \| Reply 5. Yes, despite my telling them the right term, it being in the post title, in the URL, and so on… Honestly, the Carnival has been getting pretty disappointing of late. In the last two weeks I think there’s been one other serious mathematics post other than mine. Then there’s a bunch of little toys and games and math-ed things. They’re useful and all, but I think the mathematicians have generally abandoned it, especially as Alon has less and less to do with the blogosphere. Comment by \| July 18, 2007 \| Reply 6. Categories and Surreals: Disordered Thoughts I’ve been going through André Joyal’s category-theoretic construction of the surreal numbers, futzing around to see how restricting the construction to yield only the real numbers affects things. (See my earlier remarks on this and/or Mar… Trackback by \| August 1, 2007 \| Reply 7. [...] Armstrong wrote a nice little introduction to category theory a few weeks ago, showing how numbers, addition, and subtraction are really just the [...] Pingback by \| August 15, 2007 \| Reply 8. [...] if I recall), let’s just consider which representations are isomorphic. That is, let’s decategorify this [...] Pingback by \| October 30, 2008 \| Reply 9. [...] Instead, I’ve been sucked into the vortex of math blogs and in particular this useful post on categorification, from which I learned quite a bit. I haven’t got much to say about the gist of the post, [...] Pingback by \| December 10, 2008 \| Reply 10. [...] and Philosophy An aspiring philosopher of mathematics got hold of my post on categorification, and it’s leading to (what I find to be) an interesting discussion of how Hume’s [...] Pingback by \| December 10, 2008 \| Reply 11. “Equations between functions become natural isomorphisms between categories.” Presumably you mean “natural isomorphisms between functors”? Comment by Tom \| December 16, 2008 \| Reply 12. You know, you’re right. I’m surprised nobody caught that earlier. Thanks. Comment by \| December 16, 2008 \| Reply 13. [...] me as being worth mentioning, though. John Armstrong at The Unapologetic Mathematician writes about categorification: the process of recasting a mathematical abstraction into the language of category theory, as a [...] Pingback by \| April 30, 2010 \| 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": 71, "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": 0.9365189671516418, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?t=11392	Physics Forums ## Notation for recursion - better typesetting I know this is nothing new to most people here, but my question is really more about the notation than the mathematics. And since this question is really about notation, and it’s my first use of the LaTex equation formatter, I get to experiment with something new. That’s always fun. But first, a little background. I was playing around with ways to generated ordered list of subsets of a given set, and collected the subsets by size. For a set of size n, there is one null set, and there are n subsets of size one, and of subsets of two elements there are $\frac{n!}{2!(n-2)!}$ or $\binom{n}{2}$ or “n choose 2” subsets, and so on, and if all these are added up there is a total of $2^n$ subsets for a set of n elements. There are always $2^n$ subsets of a set on n elements. It’s well known that $\binom{n}{m}$ are also the binomial coefficients, since $$(a+b)^n = \sum_{m=0}^n a^{(n-m)}\binom{n}{m} b^m$$ and if a = b = 1, and since 1 to any power is still just 1, then we have $$2^n = \sum_{m=0}^n \binom{n}{m}$$ So, what would $3^n$ look like? We use the formula immediately above with a=1 and b=2 to get $$\sum_{m=0}^n 1^{(n-m)} \binom{n}{m} 2^m$$ But $2^m$ was given as the original summation above, so we can write $$3^n = \sum_{m=0}^n \left( \binom{n}{m} \cdot \sum_{p=0}^m \binom{m}{p} \right)$$ I liked that, because we can write an equivalent form for $3^n$ in which the value “3” never appears. We can do the same thing for $4^n$ and so on, and in general (and I do hope I can use LaTex to make this come out right) $$X^n = \underbrace{\sum_{m=0}^n \left( \binom{n}{m} \cdot \left( \sum_{p=0}^m \binom{m}{p} \cdot \left( \sum_{q=0}^p \binom{p}{q} \cdot \left( \sum_{r=0}^q \binom{q}{r} \cdot \cdot \cdot \cdot \binom{v}{w} \right) \right) \right) \right) \cdot \cdot \cdot}$$ I think that’s cool, although the notation is awkward. {And even LaTex, though it does have an \underbrace statement, doesn’t let me specify the number of times to repeat the operation, and lining it up in plain text was impossible) So there must be a better way, and the whole process looks like it’s a perfect candidate for recursion. All we need to use are subscripted variables to control the summations instead of the awkward n, m, p, q, r … form. Let’s see if LaTex can handle this rewrite, as $$X^n = \underbrace{\sum_{a_1=0}^n \binom{n}{a_1} \cdot \left( \sum_{a_2=0}^{a_1} \binom{a_1}{a_2} \cdot \left( \sum_{a_3=0}^{a_2} \binom{a_2}{a_3} \cdot \left( \sum_{a_4=0}^{a_3} \binom{a_3}{a_4} \cdot \cdot \cdot \cdot \binom{a_{X-2}}{a_{X-1}} \right) \right) \right) \cdot \cdot \cdot}$$ Drats, on my screen the typesetting is truncated. The last part is $\binom{a_{X-2}}{a_{X-1}}$ That’s better, but it still isn’t tidy. Is there a better way? And that’s really my question. Is there a convenient notation to indicate such recursive processes? If anyone knows of it and can inform me, I’d really appreciate it. If not, I’m suggesting such a notation, and I‘d like your input on it. Now I’m really hoping LaTex can display it as I intend to—but I won’t know until I post it to see if I got it right. So I’ll have to take a look and post a follow up (or edit this post to correct it until I get it right, something else I haven’t done yet as a newbie). Okay, here it is: $$X^n = ^{{^}X [a_0=n]} \sum_{a_{(+1)}=0}^{a_} \left( \binom{a_}{a_{(+1)}} \cdot \right)$$ There, nice and concise, neat. Does that make sense? The prefacing superscript * designates the recursion structure, and the following value gives the recursion count, or it could be a criteria like n=1, whatever might apply, such as the initial value for $a_0$ as used. So what do you think of my suggested notation? Does it seem understandable, simple and usable? Flexible and adaptable? Does it not interfere with any other established use for such notation? Can it be improved? And for those who enjoy recursion as much as I do, keep in mind that $$\binom{n}{m} = \binom{n-1}{m-1} + \binom{n-1}{m}$$ So that the entire exponentiation process can be reduced recursively to a series of additions and simple multiplications and nothing more. No factorials, no divisions at all. Granted, it’s going to be a whole lot of additions, but it’s the principle of the thing that matters, right? Thread Tools \| \| \| \| \|------------------------------------------------------------------\|-----------------------------------\|---------\| \| Similar Threads for: Notation for recursion - better typesetting \| \| \| \| Thread \| Forum \| Replies \| \| \| Math & Science Learning Materials \| 776 \| \| \| General Math \| 2 \| \| \| General Math \| 10 \| \| \| General Math \| 3 \| \| \| Introductory Physics Homework \| 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": 11, "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": 0.9390967488288879, "perplexity_flag": "middle"}
http://physics.stackexchange.com/users/2190/david-bar-moshe?tab=activity&sort=comments&page=5	David Bar Moshe reputation 1924 bio website location age member for 2 years, 2 months seen 8 hours ago profile views 892 \| \| \| bio \| visits \| \| \| \|-------------\|------------------\|---------\|----------\|-------------\|-------------------\| \| \| 7,647 reputation \| website \| \| member for \| 2 years, 2 months \| \| 1924 badges \| location \| \| seen \| 8 hours ago \| \| 128 Comments \| \| \| \| \|-------\|---------\|---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------\| \| Nov14 \| comment \| Gauge symmetry is not a symmetry?Cont. Thus this current is conserved when the gauge field satisfies its equation of motion, the matter field needs not satisfy its equation of motion for the conservation. Thus, one may say that the current conservation requires only the gauge fields to be on-shell. But this is not the whole story; the time component of the gauge field equations of motion is the Bianchi identity (or the Gauss law). \| \| Nov14 \| comment \| Gauge symmetry is not a symmetry? \| \| Nov13 \| comment \| Majorana-like representation for mixed symmetric states?@ Piotr I have only a partial answer for your question. I am adding an update to the previous answer trying to explain this point. \| \| Nov13 \| comment \| Geometric quantization of identical particles@Squark Thank you for your answer \| \| Nov9 \| comment \| Geometric quantization of identical particles@Dan Thanks, your remark is very helpful. I guess that the orbifold fundamental group should be the relevant object to the orbifold quantization. \| \| Nov9 \| comment \| Geometric quantization of identical particlesThank you for your answer and reference \| \| Oct25 \| comment \| Edge theory of FQHE - Unable to produce Green's function from anticommutation relations and equation of motion? \| \| Oct25 \| comment \| Edge theory of FQHE - Unable to produce Green's function from anticommutation relations and equation of motion? \| \| Oct21 \| comment \| Which exact solutions of the classical Yang-Mills equations are known?If you allow the function $f$ to be time dependent $f(r,t)$, it is known that this Ansatz admits regular solutions. I couldn't find references for these solutions, but I think that it is no to hard to reproduce them. \| \| Oct10 \| comment \| Covariant derivatives \| \| Oct10 \| comment \| Orbits of maximally entangled mixed statesThanks for the remark. The criterion for maximal entanglement that I have in mind (and didn't spell out explicitely in the question) is the one given by Bengtsson (in the second reference) following equation (22), namely,a state within a given orbit (of the biparticle density matrix) is to be called maximally entangled if its partial trace with respect to system 2 is maximally mixed (maximal Von Neumann entropy) relative to all states within the same orbit. I'll write an update to the question for clarification. \| \| Sep21 \| comment \| Galilean invariance of classical lagrangian \| \| Sep21 \| comment \| Galilean invariance of classical lagrangianJohn, The relativistic free particl Lagrangian L = -mc^2\sqrt{1-\frac{v^2}{c^2}} is not Lorentz invariant, only the full action $S = \int L dt$, where the noninvariance of $dt$ cancels that of the Lagrangian. In the nonrelativistic limit we take the small speed limit but in addition we assume the time is invariant, this is technically the reason why the invariance is lost. \| \| Sep19 \| comment \| Trouble with constrained quantization (Dirac bracket)@Greg,if you change the relative sign between the derivative terms, you would have got an interesting problem, where one can exercise the treatment of Dirac's first class constraints, or equivalently symplectic reduction or the projection onto the lowest landau level. \| \| Sep13 \| comment \| Equivalence of canonical quantization and path integral quantization@Greg, Thank you, Please be free to accept other answers, I don't mind. I don't know exactly what is the new question that you have in mind. If I have information beyond what I wrote here, it will be my pleasure to try to answer your new question. \| \| Sep11 \| comment \| Equivalence of canonical quantization and path integral quantization@Greg I have added an update trying to answer the questions in your comment. \| \| Aug23 \| comment \| How does the quantum path integral relate to the quantization of energy?@Piotr The form in which the "sum over states" is written is just illustrative, of course, if continuous spectrum is present, we should write an integral over the spectral projectors. What I wanted to emphasize is that we can detect the presence of discrete spectrum by observing the presence of Dirac "peaks" in the Fourier transform of the evolution kernel computed by means of a path integral. \| \| Aug23 \| comment \| The quantized energy level E depends on which power of n?@ David Zaslavsky - I deleted the answer using the delete option at the bottom of the page, I hope that's the right way to do this \| \| Aug22 \| comment \| The quantized energy level E depends on which power of n?Thanks @Nic, corrected \| \| Aug13 \| comment \| In quantum mechanics, given certain energy spectrum can one generate the corresponding potential? \|	{"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": 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": 0.902760922908783, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/275209/confusion-related-to-a-convex-optimization-problem	# Confusion related to a convex optimization problem I was going through Stephen Boyd's lecture on convex optimization. However, I am a bit confused about a problem Given Minimize $f(x) = x_1^2+x_2^2$ subject to $f_1(x) = \frac{x_1}{1+x_2^2} \leq 0$ How come $f_1(x)$ is not convex? I was going through Stephen Boyd's book related to convex optimization http://www.stanford.edu/~boyd/cvxbook/bv_cvxbook.pdf Here is the exact screenshot of the page - 1 I cant understand you question, what do you want to know? Do you want to know if $f_1$ is convex? – Tomás Jan 10 at 16:01 – whuber Jan 10 at 16:19 The constraint only says $x_1 \le 0$. So looks like min is $0$ when $(x_1,x_2)=(0,0)$. – coffeemath Jan 10 at 17:23 Compare $f(1,0)$ with $\frac12\big(f_1(1,1)+f_1(1,-1)\big)$. Or fix $x_1=1$ and check the sign of $\frac{\mathrm d^2}{\mathrm dx_2^2}f(1,x_2)$. – Rahul Narain Jan 10 at 19:38 @Tomás. Yeah I want to know why $f_1(x)$ is not convex? – user34790 Jan 10 at 19:43 ## 1 Answer Look at $f_1$ along any "vertical" line $x_1=c$ where $c\neq0$. For positive $c$, you get a failure of convexity near the $x_1$-axis, and for negative $c$ you get a failure of convexity far from the $x_1$-axis. -	{"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": 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": 0.9099162817001343, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/124044/show-int-infty-infty-frac-sinh-ax-sinh-pi-z-cos-bx-dx-frac/124156	# Show $\int_{-\infty}^{\infty} \frac{\sinh ax}{\sinh \pi z} \cos bx \ dx = \frac{\sin a}{\cos a + \cosh b} \ -\pi < a < \pi$ I want to show that $\displaystyle \int_{-\infty}^{\infty} \frac{\sinh ax}{\sinh \pi z} \cos bx \ dx = \frac{\sin a}{\cos a + \cosh b} \ -\pi < a < \pi$ So I let $f \displaystyle (z) = e^{ibx} \frac{\sinh az}{\sinh \pi z}$ and integrated under the rectangular contour with vertices at $R, R+ i, R + i$, and $-R$, and and an indentation at $z=i$ to avoid the simple pole. I end up with $\displaystyle \text{PV} \left( \int_{-\infty}^{\infty} f(x) \ dx + e^{-b} \cos a \int_{-\infty}^{\infty} f(t) \ dt + i e^{-b} \sin a \int_{-\infty}^{\infty} e^{ibt} \frac{\cosh at}{\sinh \pi t} \ dt \right) = e^{-b} \sin a$ And if I equate real parts, $\displaystyle (1+ e^{-b} \cos a) \int_{-\infty}^{\infty} \frac{\sinh ax}{\sinh \pi x} \cos bx \ dx - e^{-b}\sin a \int_{-\infty}^{\infty} \frac{\cosh at}{\sinh \pi t} \sin bt \ dt = e^{-b} \sin a$ Did I use wrong contour or the wrong function? Or is there a relationship between $\displaystyle \int_{-\infty}^{\infty} \frac{\sinh ax}{\sinh \pi x} \cos bx \ dx$ and $\displaystyle \int_{-\infty}^{\infty} \frac{\cosh at}{\sinh \pi t} \sin bt \ dt$? - – Sam Mar 24 '12 at 20:58 Thanks. It's essentially the same problem. But how did he evaluate that infinite series? – Random Variable Mar 24 '12 at 21:15 $(-1)^n e^{-n\pi\omega}\sin(\pi \kappa n)$ is the imaginary part of $\exp(-n\pi \omega + i \pi \kappa n + i\pi n) = e^{-n\pi (\omega +i (\kappa + 1))}$, which is a geometric series. – Sam Mar 24 '12 at 22:26 I should have realized that. – Random Variable Mar 24 '12 at 23:06 But I think the contour should be a rectangle with vertices at $N + \frac{1}{2}, (N+ \frac{1}{2})+i(N+\frac{1}{2}),-(N+\frac{1}{2}) + i(N+ \frac{1}{2})$, and $-N-\frac{1}{2}$ where $N$ is an integer – Random Variable Mar 24 '12 at 23:13 ## 1 Answer I found another approach which is much longer but easier to justify. Let $\displaystyle f(z) = \frac{e^{(a+ib)z}}{\sinh \pi z}$ Notice that $\displaystyle f(z) = \frac{(\cosh az + \sinh az)e^{ibz}}{\sinh \pi z}$ Now integrate along the rectangle with vertices at $-R, R, R+i,$ and $-R+i$, and indentations at $z=0$ and $z=i$. After a lot of work, you will end up not only showing that $\displaystyle \int_{-\infty}^{\infty} \frac{\sinh ax}{\sinh \pi x} \cos bx = \frac{\sin a}{\cos a + \cosh b}$ , but also that $\displaystyle \int_{-\infty}^{\infty} \frac{\cosh ax}{\sinh \pi x} \sin bx = \frac{\sinh b}{\cos a + \cosh b}$. -	{"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": 23, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9365078806877136, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/49189/a-problem-in-probability-theory	# a problem in probability theory Let $X_{i}$ be independent and identically distributed random variables such that $S_n/n\to 0$ almost surely where $S_n=X_1+\dots+X_n$. How to prove that $E\|X_1\|<\infty$ and therefore $EX_1=0.$ - ## 2 Answers For every positive $n$ call $U_n=S_n/n$. There are two simple facts at the basis of the characterization of the behaviour of the sequence $(U_n)$ when the increments $(X_n)$ are i.i.d. but not integrable. (1) Stochastic fact: if $(X_n)$ is i.i.d. and $X_1$ is not integrable then, for every finite $x$, infinitely many events $A_n=[\|X_n\|\ge xn]$ are realized, almost surely. (2) Deterministic fact: consider a deterministic sequence $(x_n)$ and for every positive $n$ define $u_n=(x_1+\cdots+x_n)/n$. If there exists a positive $x$ such that $\|x_n\|\ge xn$ for infinitely many indices $n$, then the sequence $(u_n)$ is divergent. Putting facts (1) and (2) together yields that $(U_n)$ is almost surely divergent, in particular the event $[U_n\to0]$ has probability zero. To prove fact (1), note that $(A_n)$ is an independent sequence of events and that the series $\sum P(A_n)$ is divergent hence Borel-Cantelli lemma yields the result. To prove fact (2), assume that $(u_n)$ is convergent and that $u_n\to u$. Then, $$x_n/n=u_n-u_{n-1}(n-1)/n\to u-u=0,$$ hence $\|x_n\|\le\frac12xn$ for every $n$ large enough, which is a contradiction. Added later on Thought I might as well recall the proof that, in (1), the series $\sum P(A_n)$ in divergent. The idea is that $P(A_n)=P(\|X_1\|\ge xn)$ and that $$\sum_{n\ge0}P(\|X_1\|\ge xn)=\sum_{n\ge0}(n+1)P(x(n+1)>\|X_1\|\ge xn)\ge x^{-1}E(\|X_1\|).$$ And finally, this upper bound loses almost nothing, since one also has $$\sum_{n\ge1}P(\|X_1\|\ge xn)=\sum_{n\ge1}nP(x(n+1)>\|X_1\|\ge xn)\le x^{-1}E(\|X_1\|).$$ - Suppose for a contradiction that ${\rm E}\|X_1\| = \infty$. If ${\rm E}(X_1^ + )=\infty$ (respectively, ${\rm E}(X_1^ + ) < \infty$) and ${\rm E}(X_1^ - ) < \infty$ (respectively, ${\rm E}(X_1^ - ) = \infty$), then, a.s., $S_n / n \to \infty$ (respectively, $S_n / n \to -\infty$) as $n \to \infty$. If ${\rm E}(X_1^ + )=\infty$ and ${\rm E}(X_1^ - )=\infty$, then one of the following holds: 1) $S_n / n \to \infty$ a.s.; 2) $S_n / n \to -\infty$ a.s.; 3) $\lim \sup _{n \to \infty } n^{ - 1} S_n = \infty$ and $\lim \inf _{n \to \infty } n^{ - 1} S_n = -\infty$ a.s. Hence, $S_n/n \to 0$ a.s. implies ${\rm E}\|X_1\| < \infty$. EDIT: For further details, including a reference, see the first three paragraphs of my answer to this question. - Obviously, this is not the intended solution (as it relies on a "heavy" theorem, which you are not allowed to use), but it was relevant to post it here. (Note that it gives much more than what you actually asked for.) – Shai Covo Jul 3 '11 at 12:30	{"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": 46, "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": 0.3, "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": 0.9471621513366699, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/tagged/mad?sort=active&pagesize=15	# Tagged Questions Median Absolute Deviation (MAD) is a measure of variability in a sample of data, and is often used as an alternative to measures like standard deviation since it is more resistant to outliers. 2answers 61 views ### Median + MAD for skewed data I am trying to figure out what happens if you apply hampel's outlier detection technique based on the median and the MAD to data that is skewed. Apparently, the advantage of hampel's method over ... 3answers 292 views ### Mean$\pm$SD or Median$\pm$MAD to summarise a highly skewed variable? I'm working on highly skewed data, so I'm using the median instead of the mean to summarise the central tendency. I'd like to have a measure of dispersion While I often see people reporting mean ... 1answer 63 views ### Propagation of errors with median absolute deviation from the median? Is there a theoretically-sound way to perform propagation of errors with robust statistics? I am trying to characterize the errors inherent in a measurement and propagate the uncertainty through ... 4answers 463 views ### How to calculate overall standard deviation from standard deviation of sub-periods? If I have the mean, s.d., median and count for sample A, and the same for sample B, can I throw away the samples, and calculate exactly mean, median and s.d. for the combination? Well the mean is ... 2answers 366 views ### Outlier detection for heavy-tailed data Applying modified z-score for outlier elimination on some data (Iglewicz and Hoaglin, 1993), I discovered that a big proportion of the data (~10%) was outside the range ... 5answers 776 views ### Use of robust spread measures such median average deviation and median filters for time series I have a time series where I need to detect gross anomalies due to coding errors, not small shifts in the structure of the series. I am interested in the most recent data points, not historical data ... 1answer 479 views ### Median of medians as robust mean of means? The location and scale of a normally distributed data can be estimated by sampling the data then taking the mean of the sample means and standard deviations, respectively. For non-normal ... 2answers 283 views ### MAD equivalent for standard error As far as I know, one can calculate the relative standard error from the standard deviation of a data sample. I am looking for the Median Absolute Deviation equivalent for standard error. Does one ... 1answer 137 views ### Using MAD as a way of defining a threshold for significance testing If I have a set of terms each term having a particular frequency associated with it (the number of the times the term has appeared in fixed corpus of papers), then is the following method of ...	{"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": 0.18, "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": 0.9288244843482971, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/180351/induction-commutes-with-duals-and-tensor-products	# Induction commutes with duals and tensor products I need references for the following two results. Let $G$ be a finite group and let $H$ be a subgroup of $G$. Let $V$ be a finite-dimensional representation of $H$. 1. $(\text{Ind}_H^G V)^{\ast} \cong \text{Ind}_H^G V^{\ast}$. 2. Given a finite-dimensional representation $W$ of $G$, we have $(\text{Ind}_H^G V) \otimes W \cong \text{Ind}_H^G (V \otimes \text{Res}_H^G W)$. I know references for these that prove them with character theory, but I need them for general fields (not just fields of characteristic $0$). - Is $G$ supposed to be finite? In any case, I would imagine that these kinds of results follow from Frobenius reciprocity. – Qiaochu Yuan Aug 8 '12 at 16:46 The group $G$ is finite. However, the version of Frobenius reciprocity I know (eg the one in Serre's book) is a theorem about characters, so I don't think it applies here. – Juan Aug 8 '12 at 16:51 – Qiaochu Yuan Aug 8 '12 at 17:23 @QiaochuYuan : Thanks. I just spent 15-20 minutes trying to prove these with that version of Frobenius reciprocity. I was unsuccessful, but that doesn't mean it can't be done. – Juan Aug 8 '12 at 18:38	{"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": 0.9590017795562744, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?t=199160&page=2	Physics Forums Pitch vs Noise? Hello everyone. I'm not a Physics major: I'm a Music major. And I have a question. I've read various books on acoustics, but I still haven't found the answer. Ok. The way I see it, there are 2 basic types of sounds (I didn't get this out of a book, it's my own conclusion from everyday observation). One is pitch, the other is (for lack of a better word) noise. Now every sound is a noise, right? True. But by pitch I mean a sound whose pitch we can determine. If I play a note on the piano, you can tell me "oh, that's a C", or you can give me some mathematical frequency for what we call "C". By noise, I mean a sound whose pitch(es) cannot be determined. - And maybe I'm wrong about this, because I know next to nothing about Physics. But for example, say an airplane takes off and you have the unfortunate experience of standing behind it and you hear this EEEEEEEEEEEAAHHHHHHHHHHHH!!! or whatever. Now let's say you go to the piano and try to reproduce the "pitch(es)" of the engine going off. YOU CAN'T! Because the sound has no pitch we can determine. It's just noise. Now here's my question. 1) Is everything I said above true? and 2) If it is, how do we describe, in scientific terms, what I have termed "noise?" If it has no defined pitch, what is it? What is the technical term for it, and how can we analyze it? PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus What you call noise, rather than having a single pitch, has a spread of them. That is, the sound waves are a superposition of many waves with well defined pitches. Relating back to music, have you ever heard two piccolo's (or even flutes, though not as bad) that are only slighly out of tune. They produce an awful wobbling sound (a beat). Imagine that with lots of different frequencies and you have your jet engine. Quote by NeoDevin What you call noise, rather than having a single pitch, has a spread of them. That is, the sound waves are a superposition of many waves with well defined pitches. Relating back to music, have you ever heard two piccolo's (or even flutes, though not as bad) that are only slighly out of tune. They produce an awful wobbling sound (a beat). Imagine that with lots of different frequencies and you have your jet engine. 1)So is there a technical term for this spread of pitches? 2)How can we determine what pitches are in the spread? Would a spectrogram analyze it? Pitch vs Noise? NQ: You're pretty much spot on. (ND: I wonder how much of music/noise is due to phase effects? ) There's a nice Wikipedia article you should look at: http://en.wikipedia.org/wiki/Colors_of_noise Also- look at http://hypertextbook.com/physics/waves/music/ Quote by christianjb NQ: You're pretty much spot on. (ND: I wonder how much of music/noise is due to phase effects? ) There's a nice Wikipedia article you should look at: http://en.wikipedia.org/wiki/Colors_of_noise Also- look at http://hypertextbook.com/physics/waves/music/ Thank you. I had actually read those articles before, and found the 2nd one especially helpful. The Wikipedia article about white, red, and pink noise makes no sense to me, because it never really puts it into layman's terms by providing an example. In the case of the jet engine, does white/pink/red/etc noise occur? In what way? Are these "noise colors" in anyway related to the colors shown in some spectrograms? I came across this spectrogram software called Raven (developed at Cornell), and I was wondering why they are in shades of color? Check it out: http://www.birds.cornell.edu/brp/raven/Raven.html 1) No, a fair bit was untrue. 2) Sound is a type of energy, measured in decibels. Loosely, in music, the pitch is equal to the peak frequency, which is to say, it is the component of the sound with the most energy. In reality, an A played on a piano will have a frequency spectrum with its peak frequency at 440 Hz (or 220 Hz, or 880 Hz, etc... depending on octave) - the sound will not be composed of pure 440 Hz energy. Why do you think an A sounds different on a piano than it does on a guitar, or a violin? Because the frequency spectrum of these instruments is different. In scientific terms, pitch means the same thing as frequency. Noise is a type of sound, although its definition is subjective, it is simply undesired signal. For example, if you made a recording and you noticed some background hiss, you might consider that to be noise. Noise will still have a frequency spectrum, sometimes, if the noise is of characteristic frequency, it is possible to apply some kind of "bandpass" filter to your recording so as to remove the noise. Quote by billiards 1) No, a fair bit was untrue. 2) Sound is a type of energy, measured in decibels. Loosely, in music, the pitch is equal to the peak frequency, which is to say, it is the component of the sound with the most energy. In reality, an A played on a piano will have a frequency spectrum with its peak frequency at 440 Hz (or 220 Hz, or 880 Hz, etc... depending on octave) - the sound will not be composed of pure 440 Hz energy. Why do you think an A sounds different on a piano than it does on a guitar, or a violin? Because the frequency spectrum of these instruments is different. In scientific terms, pitch means the same thing as frequency. Noise is a type of sound, although its definition is subjective, it is simply undesired signal. For example, if you made a recording and you noticed some background hiss, you might consider that to be noise. Noise will still have a frequency spectrum, sometimes, if the noise is of characteristic frequency, it is possible to apply some kind of "bandpass" filter to your recording so as to remove the noise. Thank you! Your description of pitch/frequency was quite clear. As for your definition of "noise", I am familiar with that textbook definition of noise as "unwanted sound" and that is not at all what I'm refering to. I used the term "noise" because I can't think of a better term for it, but I want to diferentiate it somehow from specific pitch. My main question is how can we analyze the frequencies of something like a jet engine roar?. In the case of a Mozart symphony, I can easily tell you note-by-note what the frequencies are. I can even write them down for you in musical notation. But how can we "describe" a get engine roar, if we cannot pin-point a specific pitch at any specific time? Quote by billiards 1) No, a fair bit was untrue. a few things need also to be fixed in your understanding, bill. or at least how you expressed it. 2) Sound is a type of energy, measured in decibels. sound can be measured lotsa different ways to get various parameters. dB is one expression of a parameter called "loudness". dB SPL (Sound Pressure Level) is a different (but similar) expression that is the logarithm of physical amplitude of the sound intensity (power per unit area). the difference between the two is shown in the Fletcher-Munson relationship: http://en.wikipedia.org/wiki/Fletcher-Munson_curves (looks like ISO updated the data a little bit). Loosely, in music, the pitch is equal to the peak frequency, no, that is incorrect. on many levels. just as dB SPL is a physical quantity while dB loudness is a perceptual quantity, pitch is a perceptual paramter (something that we measure with our ears and brains), while frequency is physical parameter and something that we (can) measure with electronic instruments. in the case of a quasi-periodic tone (quasi-periodic tones are also harmonic, all frequency components are very close to integer multiples of a common fundamental frequency), the pitch we perceive (relative to some standard reference pitch) is closely related to the logarithm of the fundamental frequency. the might be zero energy at the fundamental (so it certainly will not be the peak frequency), but if the other odd-numbered harmonics do have sufficient energy, that fundamental still is deterministic. if $f_0$ is the fundamental of a tone that might be expressed as: $$x(t) = \sum_{n=1}^{+\infty} r_n \cos(2 \pi n f_0 t + \phi_n)$$ then the pitch, measured in octaves and relative to A440, would be: $$\log_2 \left( \frac{f_0}{\mathrm{440 Hz}} \right)$$ but even that expression becomes less accurate as the pitch gets very high in the musical scale (and slightly sharpened). i.e. the A that is 3 octaves above A440 is actually a teeny-weeny bit higher in fundemental frequency than 23 x 440 Hz. that is purely because of perceptual reasons and not because of physics. it's how a piano tuner tunes it. which is to say, it is the component of the sound with the most energy. NO! this is not true. not true at all. In reality, an A played on a piano will have a frequency spectrum with its peak frequency at 440 Hz (or 220 Hz, or 880 Hz, etc... depending on octave) - the sound will not be composed of pure 440 Hz energy. Why do you think an A sounds different on a piano than it does on a guitar, or a violin? Because the frequency spectrum of these instruments is different. In scientific terms, pitch means the same thing as frequency. No. it is related, but it is not the same thing. it does not mean the same thing. Noise is a type of sound, although its definition is subjective, it is simply undesired signal. For example, if you made a recording and you noticed some background hiss, you might consider that to be noise. Noise will still have a frequency spectrum, sometimes, if the noise is of characteristic frequency, it is possible to apply some kind of "bandpass" filter to your recording so as to remove the noise. "noise" means a lot of things, a lot of different things to different people in different contexts. if "music" (particularly electronically synthesized music) or "pitch" is not the context, if you're just talking with some Joe on the street about other stuff, usually "noise" is synonymous with "racket" or "din" and is usually undesired. sometimes, in Communication Systems Theory (something that EE study in school), noise is this additive error signal that is also undesired (and we are trying to recover the signal or parameters about it before the noise was added by the transmission channel). in acoustics, audio, and music synthesis, then often "noise" is "white noise" or "pink noise" or some other flavor and is a random process with certain statistical properties (as is the noise in Communication Theory), but in music, it might well be a component that is deliberately synthesized and made part of a signal. sometimes these noisy signals have a perceived pitch. and, remember: pitch is a perceptual parameter, in its definition, not a physical one. if you generated some broadbanded noise (which would be so wide and broadbanded, it would be hard to say it has any "pitch") and run that through a bandpass filter or a comb filter with a decent amount of resonance, the output of that filter would be "noisy" but still have a sense of pitch for people. that noise has pitch and it is not exactly a periodic function (but would have a strong component in it that has some statistical elements of periodicity). BTW, guys, this is what i work in. i am not a physicist and i tread more lightly in the physics issues (particularly at the Relativity Forum), but in this area, i might be the resident guru here (unless someone else i know in the area is lurking). Quote by rbj a few things need also to be fixed in your understanding, bill. or at least how you expressed it. sound can be measured lotsa different ways to get various parameters. dB is one expression of a parameter called "loudness". dB SPL (Sound Pressure Level) is a different (but similar) expression that is the logarithm of physical amplitude of the sound intensity (power per unit area). the difference between the two is shown in the Fletcher-Munson relationship: http://en.wikipedia.org/wiki/Fletcher-Munson_curves (looks like ISO updated the data a little bit). etc. That's excellent, rbj. Thank you so much for taking the time to explain clarify these matters (and thank you again billiards for your reply!) If you are indeed the guru on this matter (I do not doubt it!), then I would very much appreciate it if you would answer my main question: how can we analyze the sound of something like a jet engine? Another member here said that the roar of a jet engine is actually a "spread" of different pitches, resulting in no definite pitch. If that is so, is there some way we can discover (via some kind of software?) the exact pitches that make up this "spread" and graph how they change in time (as the jet becomes more distant)? Quote by NoiseQuestion I would very much appreciate it if you would answer my main question: how can we analyze the sound of something like a jet engine? depends on what you mean by "analyze". one thing we can do is window off a segment (of finite length) of the jet engine sound and run it through a Fourier Transform of some manner. sometimes this is called the Short-Time Fourier Transform: http://en.wikipedia.org/wiki/Short-t...rier_transform . Another member here said that the roar of a jet engine is actually a "spread" of different pitches, that is what Fourier analysis is all about. it is about taking some arbiatrary waveform and breaking it up into component frequencies. more precisely: expressing the arbitrary waveform as some (possibly infinite) sum of sinusoidal waveforms, each having their own frequency parameter. resulting in no definite pitch. that may or may not be true. i tend to hear a "whine" from the engines in a commecial jet that could be assigned a pitch. definitely i can hear the pitch of a two-cycle motorcycle engine ("vrin, din, din, din"). the roar of fighter jets as they blast by me (Vermont Air National Guard) seem to be so broadbanded, i can't tell a specific pitch although whatever amorphous, indeterminite pitch they have does tend to lower (because of doppler) as they fly by. so do they have a pitch or not? that's like the old Clarol commercial (with a touch of sexual inuendo) those of us that are in our 50s remember: "Does she or doesn't she?" (hope so! ) If that is so, is there some way we can discover (via some kind of software?) the exact pitches that make up this "spread" and graph how they change in time (as the jet becomes more distant)? yes, it's called a "spectrum analyzer" and its output is called a "spectogram". and it based, at least in some sense, on the STFT (this is true, even if your spectrum analyzer is a "filter bank", a collection of bandpass filters that are all tuned to variously spaced frequencies). Mentor Quote by rbj i tend to hear a "whine" from the engines in a commecial jet that could be assigned a pitch. I agree, jet engines "whine". I spent a very uncomfortable flight right next to an engine that was running at a frequency that was about 1-2 Hz away from the other engine. The resulting "beat" was very steady and irritating. I guess I "whine" too Recognitions: Science Advisor Jet engine noise is a very similar situation to a regular sound from any other source. It is the same in that it has components that will be picked up, i.e. fan tip noise, burner rumble and so on. However there is also turbulence induced noises that you simply will see no dominant source. For example, on an FFT, you will see a nice solid peak at, what is referred to as blade pass frequency of the fan. It is one of the major noise contributors. However, the turbulent noise created at the rear of the engine due to exhaust is extremely random. If you looked at an FFT and a trace in the time domain, you won't be able to tell the difference. They are both a mess. So in that respect, I would say that noise would have to have some reference to randomness in the signal. But like has already been mentioned, "noise" is a pretty subjective term to my knowledge. Quote by rbj depends on what you mean by "analyze". one thing we can do is window off a segment (of finite length) of the jet engine sound and run it through a Fourier Transform of some manner. sometimes this is called the Short-Time Fourier Transform: http://en.wikipedia.org/wiki/Short-t...rier_transform . that is what Fourier analysis is all about. it is about taking some arbiatrary waveform and breaking it up into component frequencies. more precisely: expressing the arbitrary waveform as some (possibly infinite) sum of sinusoidal waveforms, each having their own frequency parameter. that may or may not be true. i tend to hear a "whine" from the engines in a commecial jet that could be assigned a pitch. definitely i can hear the pitch of a two-cycle motorcycle engine ("vrin, din, din, din"). the roar of fighter jets as they blast by me (Vermont Air National Guard) seem to be so broadbanded, i can't tell a specific pitch although whatever amorphous, indeterminite pitch they have does tend to lower (because of doppler) as they fly by. so do they have a pitch or not? that's like the old Clarol commercial (with a touch of sexual inuendo) those of us that are in our 50s remember: "Does she or doesn't she?" (hope so! ) yes, it's called a "spectrum analyzer" and its output is called a "spectogram". and it based, at least in some sense, on the STFT (this is true, even if your spectrum analyzer is a "filter bank", a collection of bandpass filters that are all tuned to variously spaced frequencies). Thanks once again. By "analyze", I mean somehow capturing the unique characteristics of a given sound (jet engine, motorcylce, etc) in time in some kind of notation (like a graph). It is easy to notate music, because the pitches are very easy to distinguish. But in the case of an engine, it is impossible, at least in music notation. Can you recommend any spectogram/STFT software that you use? I have a Mac, and I need something user friendly (i.e. not MatLab!!! remember I'm not a Physics major, just a musician.) Recognitions: Science Advisor There are a few shareware frequency analyzers out there that let you take in a sound through your sound card and then process it to give you the frequency spectrum of your signal. http://www.visualizationsoftware.com/gram.html http://www.sharewareconnection.com/a...cillometer.htm Quote by FredGarvin Jet engine noise is a very similar situation to a regular sound from any other source. It is the same in that it has components that will be picked up, i.e. fan tip noise, burner rumble and so on. However there is also turbulence induced noises that you simply will see no dominant source. For example, on an FFT, you will see a nice solid peak at, what is referred to as blade pass frequency of the fan. It is one of the major noise contributors. However, the turbulent noise created at the rear of the engine due to exhaust is extremely random. If you looked at an FFT and a trace in the time domain, you won't be able to tell the difference. They are both a mess. So in that respect, I would say that noise would have to have some reference to randomness in the signal. But like has already been mentioned, "noise" is a pretty subjective term to my knowledge. Hmm. Very interesting. So you mean that there are some sounds (like turbulence) that have no dominant (outstanding) pitches? But if the signals of the noise are "random", doesn't that still imply that they exist? And if so, isn't there some kind of analysis software that could capture the sound and follow the exact pitch signals, however random they may be? Quote by FredGarvin There are a few shareware frequency analyzers out there that let you take in a sound through your sound card and then process it to give you the frequency spectrum of your signal. http://www.visualizationsoftware.com/gram.html http://www.sharewareconnection.com/a...cillometer.htm Thank you! I'm not sure these are Mac compatable, though. I know the second one isn't. Have you ever tried Raven? http://www.birds.cornell.edu/brp/raven/Raven.html Recognitions: Science Advisor I just grabbed a couple of quick links. It appears that the Raven program will do exactly what you need. Thread Tools \| \| \| \| \|--------------------------------------\|--------------------------------------------\|---------\| \| Similar Threads for: Pitch vs Noise? \| \| \| \| Thread \| Forum \| Replies \| \| \| General Physics \| 3 \| \| \| Advanced Physics Homework \| 4 \| \| \| General Physics \| 3 \| \| \| Introductory Physics Homework \| 4 \| \| \| Set Theory, Logic, Probability, Statistics \| 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": 1, "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": 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": 0.96009361743927, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/70920/showing-that-1-cos-t-cos2-t-dots-cos6-t-is-a-linearly-independ/70925	Showing that $\{ 1, \cos t, \cos^2 t, \dots, \cos^6 t \}$ is a linearly independent set The problem is: Show that $\{ 1, \cos t, \cos^2 t, \dots, \cos^6 t \}$ is a linearly independent set of functions defined on $\mathbb{R}$. The problem expects the student to use a computer program such as Matlab. To solve the problem I created a matrix in Matlab with $t$ ranging from $1$ to $7$, then row reduced it and got the identity matrix as result. I.e. the columns of the matrix is linearly independent. Is it enough to just show for $t=1\to 7$; could it not possibly break down at some other number? And is there a more elegant way of solving this without the use of "brute force"? Here's the Matlab code if necessary: ````for t = 1:7 A(t,:) = [1 cos(t) (cos(t))^2 (cos(t))^3 (cos(t))^4 (cos(t))^5 (cos(t))^6]; end ```` - 2 It will break down at some other numbers (extreme example $t=\pi, 2\pi, \dots, 7\pi$, but there are more subtle ones). But if you can find some collection of seven $t$'s, that settles it. There is a much easier (for me) way that involves no Matlab. – André Nicolas Oct 8 '11 at 20:29 2 Simply composing each of the functions with $\cos^{-1}$ on the right gives 7 polynomials on [-1,1], which must be independent because they have different degree. Since function composition preserves linear combinations, the original functions must also be independent. – Henning Makholm Oct 8 '11 at 20:48 2 Those polynomials Henning is referring to are termed Chebyshev polynomials. – J. M. Oct 9 '11 at 2:08 4 Answers The computational approach that you used works well. For our particular collection of functions, or more generally for the powers of a single function $f(t)$, there is a straightforward non-computational approach. As a bonus, the argument below works just as well for $666$ as it does for $6$. Suppose to the contrary that our collection $\{1,\cos t, \cos^2 t, \dots, \cos^6 t\}$ of functions is not linearly independent. Then there are constants $a_0, a_1, \dots, a_6$, not all $0$, such that $$a_0+a_1 \cos t +a_2 \cos^2 t +\cdots +a_6 \cos^6 t=0 \quad\text{ for all $t$}.$$ Let $P(x)=a_0+a_1x+a_2x^2+\cdots +a_6x^6$. Not all the coefficients $a_i$ are $0$, so the equation $P(x)=0$ has at most $6$ solutions. (A non-zero polynomial of degree $\le n$ has at most $n$ roots.) Thus if we can show that the function $\cos t$ can take on more than $6$ different values, we obtain the desired contradiction. But $\cos t$ takes on infinitely many different values as $t$ ranges over the reals, or indeed over any interval of non-zero length. This completes the proof. - Thank you for your answer, André! Forgive me for asking, but it is not obvious to me why the fact that $\cos t$ can take on more than 6 different values makes a contradiction? – Jodles Oct 10 '11 at 23:47 1 We are supposing that there are $a_i$ not all $0$ such that $P(\cos t)=0$, where $P(x)$ is the polynomial of my post. Since $\cos t$ can take on more than $6$ values, that would mean that $P(x)$ would be $0$ at more than $6$ places. But a non-zero polynomial of degree $\le 6$ can be $0$ at at most $6$ places. The only way out is to conclude that $a_i=0$ for all $i$, meaning that your $7$ functions are linearly independent. For the same reason, $1$, $\tan t$, $\tan^2 t$, and so on up to $\tan^{101} t$ is a linearly independent collection. – André Nicolas Oct 11 '11 at 1:03 Two comments: 1. Suppose that you have functions $f_1(t)$, $f_2(t),\ldots,f_k(t)$, and that they are linearly dependent in the vector space of all real-valued functions. That means that you can find real numbers $\alpha_1,\ldots,\alpha_k$ such that $$\alpha_1 f_1(t)+\cdots + \alpha_k f_k(t) = 0.$$ That means that for any value $a$ of $t$, you will have $$\alpha_1f_1(a) + \cdots + \alpha_k f_k(a) = 0.$$ In particular, if you pick $k$ different values for $t$, $a_1,\ldots,a_k$, then you have: $$\begin{align} \alpha_1f_1(a_1) + \cdots + \alpha_k f_k(a_1) &= 0\\ \alpha_1f_1(a_2) + \cdots + \alpha_k f_k(a_2) &= 0\\ &\cdots\\ \alpha_1f_1(a_k) + \cdots + \alpha_k f_k(a_k) &= 0 \end{align}$$ which in turn means that: $$\alpha_1 \left(\begin{array}{c}f_1(a_1)\\f_1(a_2)\\ \vdots \\ f_1(a_k)\end{array}\right) +\alpha_2\left(\begin{array}{c}f_2(a_1)\\f_2(a_2)\\\vdots\\f_2(a_k)\end{array}\right) + \cdots + \alpha_k\left(\begin{array}{c}f_k(a_1)\\f_k(a_2)\\\vdots\\f_k(a_k)\end{array}\right) = \left(\begin{array}{c}0\\0\\\vdots\\0\end{array}\right),$$ so the columns of the corresponding matrix are linearly dependent, so the corresponding matrix is singular. By contrapositive, if the matrix you get by evaluating the functions at $k$ different points as you did is nonsingular, then the functions have to be linearly independent, as you conclude. However, the converse is not true: it may be that you get "unlucky" and evaluate at points where the matrix is singular, even if the functions are not. If instead of using $t=1,2,\ldots,7$ you had used $t=\pi,2\pi,\ldots,7\pi$, your matrix would have been singular, even though the functions are, as you concluded, linearly independent. So the matrix being nonsingular is sufficient, but it is not necessary for the functions to be linearly independent. 2. As to other methods of doing this... (comment two): the standard way when your functions can be differentiated enough times (as they can be here), is to use the Wronskian. - Of course, compute the Wronskian. – GEdgar Oct 8 '11 at 21:14 Thank you for your answer, Arturo! I think my problem might stem from a misunderstanding of what linear independence requires. I thought for functions to be linearly independent, they need to be so for all $\mathbb{R}$; however, it seems like it's the other way around: i.e. if a set of functions are not linearly dependent for all $\mathbb{R}$, then they are linearly independent. So if we can show one example of linear independence, then we're done. Is that the gist of what you were saying? – Jodles Oct 10 '11 at 23:51 @Jodles: I honestly don't understand your comment. By definition, there are only two possibilities for any set of vectors: either they are linearly independent, or they are linearly dependent. It is always the case that if a set of vectors is not linearly dependent, then it is linearly independent; and that if it is not linearly independent, then it is linearly dependent. For functions, equality means equality as functions, so when we write that $a_1f_1(t)+\cdots+a_nf_n(t)=0$, we mean that this holds for all values of $t$. – Arturo Magidin Oct 11 '11 at 4:12 The equality $$\cos x=\frac{e^{ix}+e^{-ix}}{2}$$ shows that $(\cos x)^n$, being a polynomial of degree $n$ in $e^{ix}$ and $e^{-ix}$, is not a linear combination of lower powers of $\cos x$. The fact (implicitly used) that the functions $e^{inx}$ are linearly independent follows from the equality $$\frac{d}{dx}\ e^{inx}=in\ e^{inx}.$$ - The system of functions $f_i,i=1,2,...,n$ is linearly independent on some set $A$ iff the condition $\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \displaystyle \sum\limits_{i}\alpha_i f_i(x) = 0$ for all $x\in A$ is satisfied only for $\alpha_i = 0$ for all $i=1,...,n$. So, you should solve as many equations, as there are elements in $A$. On the other hand, it's sufficient to take just some elements from $A$ to show that $$\sum\limits_{i}\alpha_i f_i(x_j) = 0$$ for all $j=1,...,m$ implies that $\alpha_i = 0$ for all $i$. I would say that in your case it's sufficient to consider $8$ points $x_j$ to obtain the over-determined system of linear equations on $\alpha_i$ since $n=7$ in your case and to obtain the over-determined system usually it is sufficient to have $n+1$ equation. -	{"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": 84, "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": 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": 0.9446815252304077, "perplexity_flag": "head"}
http://mathoverflow.net/questions/102463?sort=votes	## Ideals of etale structure sheaves ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Is it known whether or not every sheaf of ideals of the etale structure sheaf of a Noetherian scheme is generated by finitely many of its sections? Of course it is trivially true for some widely used special cases. But is it known one way or the other, in this generality? - 2 what is the etale structure sheaf? – Yosemite Sam Jul 17 at 16:00 1 The sheaf $U \mapsto \mathcal{O}_X(U)$ defined on étale opens $U \to X$. – Martin Brandenburg Jul 17 at 18:01 1 Yes, though I would write $U\mapsto \mathscr{O}_U(U)$ to define it. – Colin McLarty Jul 17 at 19:17 1 OK guys, I'm being really really thick right now. What's the difference with the Zariski structure sheaf? Aren't the two categories (etale/zariski quasi-coherent modules) equivalent for schemes (via the forget morphism)? – Yosemite Sam Jul 17 at 22:44 But I need not only quasi-coherent (sheaves of) ideals. I want to know this for all (sheaves of) ideals. – Colin McLarty Jul 17 at 23:11 show 5 more comments ## 1 Answer I must apologize for posting a false answer. in writing up a proof i discovered a gap which grew to a counterexample. In fact not every sheaf of ideals of an etale structure sheaf is finitely generated. I have added a counterexample to the end of my ArXiv paper on cohomology in second order arithmetic arXiv:1207.0276v2. Intuitively, an etale ideal can hold information about arbitrarily high degree extensions which is not reducible to information about any fixed degree so the ideal is not finitely generated. The counterexample shows that given any non-zero element $x$ of an algebraically closed field $k$ a single etale sheaf of ideals on the punctuated line $k\0$ can pick one $2^n$-th root of $x$ for every $n$. The example has $x=1$ but rescaling $k\0$ takes 1 to any other $a\neq 0$. -	{"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": 0.89659583568573, "perplexity_flag": "head"}
http://nrich.maths.org/1968/solution?nomenu=1	## 'Voting Paradox' printed from http://nrich.maths.org/ ### Show menu Many of the solutions sent in this month applied values to a candidate's position in a voters' list. For example, if someone voted BCA, B would get $2$ points, C $1$ point and A $0$ points. With one of these models applied to the example all candidates would score a total of 3 and therefore come out equal. However, the paradox does not depend on assigning values to positions rather who does better overall and the paradox is that the answer is "no one" not because they all score the same but that everyone does better than everyone else! Amongst the solutions we have recieved, two were from Justin of Skyview High School and Ling Xiang from Tao Nan School. I have used these as the basis of the following explanation. As each voter can choose any one of the six orders ABC, ACB, BAC, BCA, CAB and CBA, there are altogether (666 = 216) different combinations of votes that could be cast. We say the results are intransitive if, for example, A beats B and B beats C but A loses to C so that it is impossible to decide on a winner. So what combinations of votes would the voters need cast to make up a set of three that are intransitive? We have to find the total number of intransitive combinations of votes over the total number of combinations of votes (216) to get the probability of the paradox arising. • If the first voter votes ABC, the second voter has to vote either BCA or CAB for the results to be intransitive. • If the second voter then votes BCA, then CAB is the only order for the third voter which makes the results intransitive. • If the second voter votes CAB, then BCA is the only order for the third voter which makes the results intransitive. So, if the first voter votes ABC, there would be $2$ possible combinations of votes making the results intransitive. As the first voter can vote 6 possible votes (ABC, ACB, BAC, BCA, CAB, CBA), the total number of intransitive combinations of votes possible is ($6\times2 = 12$). So, the probability of this paradox of collective choice arising is $$\frac{12}{216}=\frac{1}{18}$$ However, if we are going to be really water-tight we should really prove that the results are intransitive if and only if no two voters agree on their first choice, nor on their second, nor on their third. The proof might look something like this: \| \| \| \| \|-------------\|--------------\|----------------------------------\| \| First voter \| Second Voter \| Third voter \| \| ABC \| ACB \| ABC or ACB or BAC, BCA, CAB, CBA \| For all the six possibilities forwhat the third voter might vote, all the results would be transitive. So, the results are transitive if two voters agree on their first choice. If two or more voters agree on their second choice, then either they also agree on the first, which we have just shown to be transitive or if not (for example ABC and CBA) then the 3 candidates are all exactly equal as a result of these two votes. The outcome will be decided by the order chosen by the third voter and the result is transitive. If two or more voters agree on the third choice, suppose B is placed third by two voters, then A beats B and C beats B by at least 2 choices to 1. The result is transitive whether A beats C (when A beats C, C beats B and A beats B) or alternatively C beats A (when C beats A, A beats B and C beats B). This means that the results are intransitive if and only if no two voters agree on their first choice, nor on their second, nor on their third.	{"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": 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": 0.949815034866333, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/138377/n-distinct-objects-in-k-boxes-and-methods-for-solving-combinatorics-problem	# $n$ distinct objects in $k$ boxes, and methods for solving combinatorics problems I am a beginner in the subject of combinatorics and would like to know a few things. A) Someone please explain me how we get the K to the nth number of ways of distributing n distinct objects in k boxes. B) When do we use the following methods: a+b+c+..+n = x ; and (1 + x + x^2 + x^3+...) to the power something - 1 I don't understand part (B). The first is rather specific, in the second are you asking about virtually the entire theory of generating functions or...? – anon Apr 29 '12 at 10:03 Maybe a book like Path to Combinatorics for Undergraduates will help? – Eisen Apr 29 '12 at 12:28 ## 1 Answer About 1: You get $k^n$ since for each of the $n$ elements you have $k$ choices (you choose the box), so you have $k\cdot k\cdots k$ exactly $n$ times (this is "the product principle" of combinatorics in action). In 2 I think you're confusing the notation. The basic idea here is this: suppose you have the equation $a+b=n$ where $a,b$ can get every nonnegative integer value. One way to find the number of solutions is to compute the coefficient of $x^n$ in $(x^0+x^1+x^2+x^3+\dots)(x^0+x^1+x^2+x^3+\dots)$. Here the first parenthesis represents the values that can $a$ can get, in the exponent of $x$; the second parenthesis represents values $b$ can get. Now, to generalize this, all you need to do is add more parenthesis - one for each variable in the original equation - and maybe remove some elements from specific parenthesis. For example, if we have the equation $a+b+c=n$ and we demand that $a$ is odd and $b$ is even (and $c$ whatever), this is described by $$(x^1+x^3+x^5+\dots)(x^0+x^2+x^4+\dots)(x^0+x^1+x^2+x^3+\dots)$$ (Note the exponents). - ## protected by Zev Chonoles♦Apr 20 at 0:47 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": 18, "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": 0.9187430739402771, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/201335/bessel-and-cosine-function-identity-formula	# Bessel and cosine function identity formula by expanding into series ( sorry i have tried but get no answer) how could i prove that $$\sqrt \pi\frac{d^{1/2}}{dx^{1/2}}J_{0} (a\sqrt x) = \frac{\cos(a\sqrt x)}{\sqrt x}$$ - ## 2 Answers A related problem. The power series of $J_{0} (a\sqrt x)$ is $$J_{0} (a\sqrt x)= \sum _{{\it k}=0}^{\infty }{\frac { \left( -1 \right) ^{{\it k}}a^{2k} x^{{\it k}}}{ 2^{2k}\Gamma \left( 1+ {\it k} \right) ^{2}}}$$ Applying the formula for fractional derivative of a monomial $$\frac{d^q}{dx^q} x^m = \frac{\Gamma(m+1)}{\Gamma(m-q+1 )} x^{m-q}\,,$$ to the above series yields $$\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}}J_{0} (a\sqrt x) =\sum _{{\it k}=0}^{\infty }{\frac { \left( -1 \right) ^{{\it k}}a^{2k} \Gamma(k+1)\,x^{{\it k-\frac{1}{2}}}}{ 2^{2k}\Gamma(k+\frac{1}{2})\Gamma\left( 1+{\it k}\right) ^{2}}} \,.$$ Simplifying the above series and using the identity $\Gamma(2k+1)=\frac{1}{\sqrt{\pi}}2^{2k}\Gamma(k+1)\Gamma(k+\frac{1}{2})$, we get $$\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}}J_{0} (a\sqrt x) =\frac{1}{\sqrt{\pi}\sqrt{x}}\,\sum _{{\it k}=0}^{\infty }{\frac { \left( -1 \right)^{{\it k}} \,(a\sqrt{x})^{2k}}{ \Gamma(2k+1)}} \,.$$ Multiplying both sides of the above equation by $\sqrt{\pi}$, we reach the desired result follow $$\sqrt{\pi} \frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}}J_{0} (a\sqrt x) =\frac{1}{\sqrt{x}}\,\sum _{{\it k}=0}^{\infty }{\frac { \left( -1 \right)^{{\it k}} \,(a\sqrt{x})^{2k}}{ \Gamma(2k+1)}}=\frac{\cos(a\sqrt x)}{\sqrt x} \,.$$ - For simplicity let $a=1$. Using the definition of the fractional derivative we find that if we can show that $$\int_0^x \frac{J_0(\sqrt{t})}{(x-t)^{1/2}} dt = 2\sin\sqrt{x},$$ then we will have the quoted result. (The derivative of this result is the fractional derivative of interest.) A straightforward approach is to expand $J_0$ in small $t$ and integrate term by term. The resulting integrals are related to the beta function, and after some manipulation we arrive at the series expansion for $2\sin\sqrt{x}$. Finally, we have $$\begin{eqnarray} \sqrt \pi\frac{d^{1/2}}{dx^{1/2}}J_{0} (\sqrt x) &=& \frac{d}{dx} 2\sin\sqrt{x} \\ &=& \frac{\cos{\sqrt{x}}}{\sqrt{x}}. \end{eqnarray}$$ To recover $a$, let $x\to a x$. - OK thaks to oen and Mhenni i messed myself up :D with the power series :D and all the stuff.. – Jose Garcia Sep 24 '12 at 8:04 @JoseGarcia: Glad to help. – oen Sep 24 '12 at 10:54	{"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": 9, "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": 0.9068204164505005, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/9608/determining-distribution-of-maximum-of-dependent-normal-variables?answertab=active	# Determining distribution of maximum of dependent normal variables I have a stochastic variable x with this property: if it's measured at t1 and again at t2, then x(t2)-x(t1) has a normal distribution with mean 0 and standard deviation Sqrt[t2-t1]. I want to find the distribution of the maximum value this variable reaches between t1 and t2, or confirm my intuition that this is not well-defined. My approach: break [t1,t2] into multiple intervals and integrate the following (PDF[NormalDistribution[x,y]] = normal probability distribution with mu of x and standard deviation of y): PDF[NormalDistribution[0,Sqrt[1/4]]][x0]* PDF[NormalDistribution[x0,Sqrt[1/4]]][x1]* PDF[NormalDistribution[x1,Sqrt[1/4]]][x2]* PDF[NormalDistribution[x2,Sqrt[1/4]]][x3] where each xi is integrated from -Infinity to m. This specific example computes the probability that the maximum on [0,1] is less than m by breaking [0,1] into 4 parts. Breaking [0,1] into more parts should yield more accurate results, although I slightly suspect that the limit diverges. Mathematica slows to a crawl even breaking [0,1] into 5 or more parts. I've tried replacing the normal distribution with others (uniform, DeltaDirac, C/(1+x^2), etc), with no better luck. Googling yields many results (this appears to be a "Wiener Process"), but I can't find the actual distribution of the maximum anywhere (nor does it say anywhere that such a maximum doesn't exist). Ultimate goal is to price box options: http://money.stackexchange.com/questions/4312/calculating-fair-value-of-an-oanda-com-box-option - ## 2 Answers You have a stochastic process $\lbrace{X(t):t \geq 0\rbrace}$ with the property that $X(t_2)-X(t_1) \sim N(0,t_2 - t_1)$, which is quite obviously supposed to be Brownian motion (BM). Suppose first that you want to find the distribution function of the running maximum $M(t)=\mathop {\max }\limits_{0 \le s \le t} X(s)$ (the maximum exists, since BM has continuous sample paths). There is a very simple formula for that, namely: $${\rm P}(M(t) \le x) = \sqrt {\frac{2}{{\pi t}}} \int_0^x {e^{ - u^2 /(2t)} {\rm d}u}, \;\; x \geq 0.$$ The situation is a little more complicated if you want to find the distribution function of $M(t_1 ,t_2 ) = \mathop {\max }\limits_{t_1 \le s \le t_2 } X(s)$ (i.e., the maximum of $X$ over the time interval $[t_1,t_2]$). For this purpose, we need to condition on the initial value $X(t_1)$. Since $X(t_1) \sim N(0,t_1)$, it has density function $f(u;t_1) = \frac{1}{{\sqrt {2\pi t_1 } }}e^{ - u^2 /(2t_1 )}$, and by the law of total probability we have $${\rm P}(M(t_1 ,t_2 ) \le x) = \int_{ - \infty }^\infty {{\rm P}(M(t_1 ,t_2 ) \le x\|X_{t_1 } = u)f(u;t_1) {\rm d}u}.$$ Now, as follows from basic properties of BM, conditioned on $X_{t_1}=u$, $M(t_1 ,t_2 )$ can be replaced by $u + M(0,t_2 - t_1)$, i.e. by $u + M(t_2 - t_1)$ (more precisely, by $u$ plus an independent copy of $M(t_2 - t_1)$, which is independent of $X_{t_1}$). This leads to $${\rm P}(M(t_1 ,t_2 ) \le x) = \int_{ - \infty }^\infty {{\rm P}(M(t_2 - t_1 ) \le x - u)f(u;t_1 ){\rm d}u}.$$ Finally, since $M(t_2 - t_1 )$ cannot be negative, we have to integrate only from $-\infty$ to $x$. That is, $${\rm P}(M(t_1 ,t_2 ) \le x) = \int_{ - \infty }^x {{\rm P}(M(t_2 - t_1 ) \le x - u)f(u;t_1 ) {\rm d}u}, \;\; x \in {\bf R}.$$ So, we have a double integral with an elementary integrand. Maybe one can simplify it. Also, maybe one can find the result in the literature. - Mathematica simplifies your first equation to Erf[x/Sqrt[2t]], which seems simpler. Is this a valid simplification? – barrycarter Nov 10 '10 at 3:50 Erf[x/Sqrt[2t]] differs from my first equation only by a simple change of variable, hence it is not a simplification. However, the representation in terms of (the error function) Erf might be more useful, since Erf is a "common function". – Shai Covo Nov 10 '10 at 13:58 Wasn't being critical, just hoping to find a closed form for the second, more general, integral. – barrycarter Nov 10 '10 at 16:39 As I noted at the end of my edited answer, maybe one can simplify the second (double) integral. I will think about it a little. – Shai Covo Nov 10 '10 at 17:19 1 @ronaf: It is quite obvious that the process is supposed to be a Brownian motion. In fact, it is quite difficult to construct a counterexample (assuming that $X$ is continuous and $X_0 = 0$). – Shai Covo Nov 10 '10 at 17:34 show 1 more comment OK, I think I've found a closed form (someone double check my math here). If the CDF of x (for x>0) is: $\text{erf}\left(\frac{x}{\sqrt{2} \sqrt{t}}\right)$ the PDF (for x>0) is clearly: $\frac{\sqrt{\frac{2}{\pi }} e^{-\frac{x^2}{2 t}}}{\sqrt{t}}$ To compute the probability that the maximum between t1 and t2 is exactly x, we compute the probability that x[t1] == y (normal distribution w SD of Sqrt[t1]): $\frac{e^{-\frac{y^2}{2 \text{t1}}}}{\sqrt{2 \pi } \sqrt{\text{t1}}}$ and then multiply the probability that the max will be x-y between t1 and t2 (only valid for y<=x): $\frac{\sqrt{\frac{2}{\pi }} e^{-\frac{(x-y)^2}{2 (\text{t2}-\text{t1})}}}{\sqrt{\text{t2}-\text{t1}}}$ So we integrate this wrt y from -Infinity to x: $\frac{e^{-\frac{(x-y)^2}{2 (\text{t2}-\text{t1})}-\frac{y^2}{2\text{t1}}}}{\pi \sqrt{\text{t1}} \sqrt{\text{t2}-\text{t1}}}$ and finally get: $\frac{e^{-\frac{x^2}{2 \text{t2}}} \left(\text{erf}\left(\frac{x\sqrt{\frac{1}{\text{t1}}-\frac{1}{\text{t2}}}}{\sqrt{2}}\right)+1\right)}{\sqrt{2 \pi } \sqrt{\text{t2}}}$ I realize all of this was in the previous post, but I wanted to play around with LaTex. - What you did above seems to correspond to the density function (rather than distribution function) of the maximum between $t_1$ and $t_2$. This is why your final answer is in terms of one integral only (erf); however, for the distribution function the answer is still in terms of a double integral. – Shai Covo Nov 10 '10 at 21:13	{"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": 29, "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": 0.926418662071228, "perplexity_flag": "head"}
http://quant.stackexchange.com/questions/2360/what-exactly-is-meant-by-microstructure-noise/3277	# What exactly is meant by “microstructure noise”? I see that term tossed around a lot, in articles relating to HFT, and ultra high frequency data. It says at higher frequencies, smaller intervals, microstructure noise is very dominant. What is this microstructure noise that they refer to? - ## 5 Answers The term has a different meaning to different people. 1. to econometricians, microstructure noise is a disturbance that makes high frequency estimates of some parameters (e.g. realized volatility) very unstable. Generally this strand of the literature professes agnosticism as to the its origin; 2. to market microstructure researchers, microstructure noise is a deviation from fundamental value that is induced by the characteristics of the market under consideration, e.g. bid-ask bounce, the discreteness of price change, latency, and asymmetric information of traders. The last example is frequently cited but I don't think it is accurate. Asymmetric information does not have to be a microstructure phenomena, although it can be partly driven by architectural properties of the market. - I presume the paper I saw clobbered with the term was referring to the first usage. What is the cause of this disturbance which affects HF estimates and not less frequent data? – Palace Chan Nov 11 '11 at 21:07 1 Group 1 being agnostic means that they can't be bothered, as long as they have a good stochastic model for it, and as long as they can filter it out. If you care about the origin, and want to develop some intuition for it, then read some Group 2 papers. It's a fun read, but it won't help you much with 1. The most typical culprit for HF noise is the so called bid-ask bounce. – Ryogi Jan 26 '12 at 19:06 Some cynical but functional definitions: • It's what you can't model if you're not using tick by tick data • It's what proper quant pricing theory doesn't know how to model yet • It's information (order book behavior) that reflects momentary fluctuations in the supply/demand of a given contract, rather than its underlying value (eg an arbitrage free price) The third is the most accurate but the first is the most useful, IMHO. The second is the most common meaning in academic papers. Some examples: • When prices hit certain levels (often round number prices eg multiples of 5), they might move suddenly as a large number of traders have punched in stops at these levels • In the minutes before a news announcement, spreads widen as some market makers pull their orders in anticipation of upcoming volatility As you might expect, it is more significant in illiquid markets where there are insufficient players to make efficient market assumptions hold. - There are rigorous econometric definitions, as has already been eluded to by others. For practical purposes, microstructure noise is a component of a price process that exhibits mean reversion on some (possibly time-varying) frequency. This reversion is particularly attractive to liquidity provisioners, who seek to profit from this noise component (along with other sources of revenue). In a strict sense, liquidity providers do not want to take directional bets on an asset (though in practice this certainly need not be the case). Since high-frequency market makers are effectively today's liquidity providers in many electronic markets (replacing the traditional NYSE specialists that had a fiduciary responsibility to provide liquidity), it's quite common to see 'HFT' intrinsically linked to 'microstructure noise'. Be sure to check out the late Fischer Black's article, 'Noise', from JoF, 1985: http://www.jstor.org/pss/2328481 Although the paper's direction differs somewhat from your question, it should be of interest. - It seems that your question refers to the microstructure noise defined in papers about intraday volatility estimates. Originally, it comes from the bid-ask bounce, i.e. the fact that even if the volatility is zero, you have buyers and sellers at this price and consequently you observe prices at Bid or Ask prices, and not at mid-price. Because of that, if you use the classical quadratic variation estimate for the squared volatility: even with an underlying volatility of zero, you will measure a lot of time $(S_{ask}-S_{bid})^2$. If you model this using an additive noise $\epsilon$, meaning that when the mid-price between $0$ and $T$ follows a discretized Arithmetic Brownian motion ($S_{(k+1)\delta}=S_{k\delta}+\sigma\sqrt{\delta}\xi_{k+1}$), the price you observe at $k\delta$ is $S_{k\delta}+\epsilon_k$ (remember that $\epsilon_k$ is something around half a bid-ask spread, explaining why the traded price is not exactly the mid). Consequently when you sample $[0,T]$ in $N$ slices (i.e. $\delta=T/N$), the expectation of the quadratic variation, that should be an estimate of $\int_0^T \sigma_t^2 dt$ is: $$U(N)=\mathbb{E}\left(\sum_{n=1}^{N} (S_{n\delta}+\epsilon_n - (S_{(n-1))\delta} +\epsilon_{n-1}))^2 \right)= \sum_{n=1}^{N} \mathbb{E}(S_{n\delta} - S_{(n-1))\delta} )^2 + \mathbb{E}(\epsilon_n -\epsilon_{n-1})^2$$ since the microstructure noise is assumed to be independent of the price, moroever it is centered and has a variance $v$ so $\mathbb{E}(\epsilon_n -\epsilon_{n-1})^2=2v$. Our estimate is now: $$U(N)\mathop{\rightarrow}_{N\longrightarrow +\infty}\int_0^T \sigma_t^2 dt+2Nv$$ It means that the estimate of the squared volatility increases linearly with the sampling rate! And technically this comes only from the microstructure noise (with variance $v$). For more information on microstructure in general (and volatility estimate specifically), you can read Market microstructure: confronting many view points (Abergel, Bouchaud, Foucault, Lehalle and Rosenbaum Ed). - Noise is a data anomaly that typically occurs in the absence of signal. In the context of HFT it may refer to order activity (microstructure) motivated by a competing methodology, such as the rebalancing of a mutual fund portfolio to match the composition of the S&P. - I wouldn't call an index rebalance "noise" since that's exactly the kind of activity that market makers are looking to provide liquidity for. To me, "noise" implies unpredictability. – chrisaycock♦ Feb 12 at 12:24	{"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": 15, "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": 0.9492446184158325, "perplexity_flag": "middle"}
http://en.wikiversity.org/wiki/Advanced_Classical_Mechanics/Dissipative_Forces	# Advanced Classical Mechanics/Dissipative Forces From Wikiversity Dissipative forces are forces of such nature that energy is lost from a system when motion takes place. Of course energy is in general conserved but it is lost from the degrees of freedom of interest into heat (the random motion of internal degrees of freedom) or radiation (the motion of new particles created by the motion -- light usually). The force can often be represented by $\left \|\vec F \right \| = a\left ( \vec r, t \right ) \left \| \vec v \right \|^n.$ Depending on the value of the index $n$ we have different types of dissipative forces. ## Types of Dissipative Forces ### Frictional Forces (Dry Friction) The force of kinetic friction is supposed to be proportional to the normal force and independent of area of contact or speed. It is simply a property of the materials in contact. Specifically, we have $\left\| \vec F \right \| = \mu \left \| F_{\rm normal} \right \|$ so $n=0$. The frictional force opposes the relative motion of two surfaces. ### Viscous Forces Here the frictional force increases as the first power of the relative speed between the surfaces and opposes the relative motion. Viscous friction is important for wet surfaces at small relative velocties. ### High-Velocity Friction At higher velocities, the force of friction increases as a higher power of the relative velocity. For example, $n=2$ gives a good approximation to the dissipative force experiences by objects travelling through fluids at high Reynolds number $Re = (vL)/\nu$ where $\nu$ is the viscousity of the fluid. ## Determining the Generalized Force The Lagrangian method requires us to find the generalized forces corresponding to the different coordinates that we have used to characterize the degrees of freedom of the motion. There are two general ways to determine the generalized forces. 1. Calculate them in Cartesian coordinates and convert to the generalized coordinates 2. Use the power function ### Partial derivatives For any force that we know in Cartesian (or any other set of coordinates for that matter), we can find the generalized force by using the definition of the generalized coordinates in terms of the coordinates in which the force is known. The definition of the generalized force is $Q_j = \sum_i \vec F_i \cdot \frac{\partial \vec r_i}{\partial q_j}.$ To do this you have to first determine the dissipative force in Cartesian coordinates and the full transformation between the Cartesian coordinates and the generalized coordinates -- it can be pretty painful for even simple problems. #### Damped Pendulum Let's look at a pendulum of length $l$. We will use the angle $\theta$ and the vertical to describe the single degree of freedom. The position of the bob is given by $x=l \sin \theta, y=-l \cos \theta\,$ and the velocities are $\dot x= l \cos \theta \dot \theta, \dot y = l \sin \theta \dot \theta \, .$ The viscous force opposes the direction of motion and it is proportional to the velocity so we have $F_x= - a l \cos \theta \dot \theta, F_y = - a l \sin \theta \dot \theta \, .$ We also need the partial derivatives for the transformation $\frac{\partial x}{\partial \theta}= l \cos \theta, \frac{\partial y}{\partial \theta} = l \sin \theta \,$ to calculate the generalized force $F_\theta = F_x \frac{\partial x}{\partial \theta} + F_y \frac{\partial y}{\partial \theta} = - a l^2 \cos^2 \theta \dot \theta - a l^2 \sin^2 \theta \dot \theta = - a l^2 \dot \theta$ ### Power function The great advance of using the potential instead of the generalized forces directly was that the definition of the partial derivative took care of all of the heavy lifting involved in going from one system of coordinates to another. It turns out that one can use an analogous quantity called the power function to do the same thing (sometimes this is called the dissipation function). Like for the potential the power function for various simple forms of forces can simply be written down and used. In analogy with the potential we define the power function such that force on particle $i$ in the $x$-direction is $F_{i,x} = \frac{\partial P}{\partial {\dot x}_i}$ and similarly for the other directions and particles. Of course, not all forces can be written in this way but many dissipative forces can. Let's write out the generalized force using the standard formula $Q_j = \sum_i \vec F_i \frac{\partial \vec r_i}{\partial q_j} = \sum_i \vec F_i \frac{\partial v_i}{\partial \dot q_j} = \sum_i \frac{\partial P}{\partial \vec v_i} \frac{\partial v_i}{\partial \dot q_j} = \frac{\partial P}{\partial \dot q_j}$ where we used the equality $\partial \vec r_i/\partial q_j = \partial \vec v_i/\partial \dot q_j$ in the first step, definition of the power function in the second step, and the definition of the partial derivative in the final step. #### Damped Pendulum Let's look at the pendulum again. The viscous force is $\vec F = -a \vec v$ so the power function is $P = -\frac{1}{2} a v^2 = -\frac{1}{2} a \left ( l \dot \theta \right )^2$ and $F_\theta = \frac{\partial P}{\partial \dot \theta} = -a l^2 \dot \theta.$ Let's combine the results for the power function with the Lagrangian for the pendulum, we have $L = T - V = \frac{1}{2} m l^2 \dot \theta^2 + g m l \cos \theta$ and Lagrange's equations including the power function are $\frac{d}{dt}\| \frac{\partial L}{\partial \dot \theta} - \frac{\partial L}{\partial \theta} = Q_{{\rm NC},\theta} = \frac{\partial P}{\partial \dot \theta}$ which yields $m l^2 \ddot \theta + g m l \sin \theta = -a l^2 \dot \theta.$ If we take the small angle limit of this equation we get $\ddot \theta = - \frac{g}{l} \theta - \frac{a}{m} \dot \theta,$ the equation for a damped linear oscillator. In this example, we have included all of the forces that we could have included in the Lagrangian, leaving only the dissipative force to be included in the power function. Of course, one could have included non-dissipative forces in the power function as well, but one must be careful not to include the same force twice, so it is generally a good idea to include as much as possible in the Lagrangian, leaving only the forces that cannot be included in the Lagrangian for the power function. This has the added advantage that one can still look at the Lagrangian for first integrals if the power function does not depend on a particular coordinate. #### More Generally If the dissipative force is given by $\left \| \vec F \right \| = a \left \| \vec v \right \|^n$ or more precisely $\vec F = -a \left \| \vec v \right \|^{n-1} \vec v$ then the power function is given by $P = -\frac{1}{n+1} a \left \| \vec v \right \|^{n+1}.$ More generally if the dissipative force points in the direction of the relative velocity it can also be written as a power function. If $\vec F_i = \phi_i \left ( \vec r_i, \left \| \vec v_i \right \|, t \right ) \frac{\vec v_i}{\left \| \vec v_i \right \|}$ then $P = \sum_i \int \phi_i d \left \| \vec v_i \right \|$ ## A Useful Example with Dry Friction It turns out that the simplest looking type of friction ('dry friction') actually can yield some useful surprises. Let's calculate the frictional force between a package and a moving conveyor belt. File:Convevor.png Conveyor Belt Let's assume that the belt is moving along at a velocity $v_y$ in the $y-$direction and calculate the frictional forces that resist moving the package across the belt in the $x$ and $y$ directions. Let's write the power function for the dry friction between the belt and the package, $P = -\mu m g v$ where $\mu$ is the coefficient of kinetic friction between the belt and the package, $m$ is the mass of the package and the relative speed between the package and the belt is given by $v = \sqrt{ \dot x^2 + \left ( \dot y - v_y \right )^2 } .$ We can calculate the frictional force resisting motion in the two directions $F_x = \frac{\partial P}{\partial \dot x} = -\mu m g \frac{\dot x}{\sqrt{ \dot x^2 + \left ( \dot y - v_y \right )^2 }}, F_y = \frac{\partial P}{\partial \dot y} = -\mu m g \frac{\dot y - v_y}{\sqrt{ \dot x^2 + \left ( \dot y - v_y \right )^2 }}.$ Let's imagine that $v_y$ is much larger $\dot x$ and $\dot y$ then we have $F_x \approx -\mu m g \frac{\dot x}{v_y}, F_y \approx \mu m g.$ The $y-$ component indicates a force that tries to drag the package along the belt. It is approximately equal to the kinetic friction that you are used to. The $x-$ component is much smaller by a factor of $\dot x/v_y$ which could be large. You can try this at the supermarket by blocking a big box of corn flakes on its side as the belt is moving. Even though the friction between the belt and the box is large (you can feel this force with the hand that impedes the motion of the box), the force to move the box across the belt is quite small. This effect is used for more important purposes when removing a cork from a bottle. It is much easier to removing the cork while twisting it round than to pull it out directly.	{"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": 48, "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": 0.9187282919883728, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/66047/markov-processes-driven-by-the-noise	# Markov processes driven by the noise Let $\xi_n\in \Xi$ be a sequence of iid random variables with $n \in\mathbb N\cup\{0\}$, which we call a noise process. Construct a process $$Z_{n+1} = f(Z_n,\xi_n)\quad(\star)$$ with $Z_0\in E$ and $f:E\times\Xi\to E$ is a measurable function of two variables. Clearly, $Z$ is a Markov process. On the other hand, consider a Markov process $X$ on $E$ given by its transition kernel $P(x,A)$. Is it always possible to find a noise process (on some set $\Xi$) and a measurable function $f:E\times \Xi\to E$ such that given any initial condition (maybe random) $X_0\in E$ $$\operatorname{Law}(X_1) = \operatorname{Law}(f(X_0,\xi_0))$$ Briefly speaking, if any Markov process can be presented in the form $(\star)$? - ## 1 Answer Yes, this is the representation of a discrete time Markov process as a randomized dynamical system as found in Proposition 8.6 (page 145) from Foundations of Modern Probability (2nd edition) by Olav Kallenberg: Let $X$ be a process on $\mathbb{Z}_+$ with values in a Borel space $S$. Then $X$ is Markov iff there exist some measurable functions $f_1,f_2,\dots:S\times[0,1]\to S$ and iid $U(0,1)$ random variables $\xi_n$ independent of $X_0$ such that $X_n=f_n(X_{n-1},\xi_n)$ almost surely for all $n\in\mathbb{N}$. Here we may choose $f_1=f_2=\cdots =f$ iff $X$ is time homogeneous. - – Tim Nov 18 '12 at 18:35	{"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": 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": 0.3, "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": 0.9222081899642944, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/37730/conservation-of-energy-in-a-magnet/37732	# Conservation of Energy in a magnet When a permanent magnet attracts some object, lets say a steel ball, energy is converted into for instance kinetic energy and heat when attraction happens, and they eventually collide. Does this imply that energy is drawn from the magnetic field and the magnet is depleted, making it weaker and weaker for each magnetic attraction performed? (if the answer requires Quantum Mechanical explanations, please elaborate :)) - ## 2 Answers No, there is no need for the permanent magnets to lose any internal energy or strength when they are used to do work. Sometimes they may weaken but they don't have to. The energy needed to do the work is extracted from the energy stored in the magnetic field (mostly outside the magnets), $\int B^2/2$, and if the magnets are brought to their original locations, the energy is returned to the magnetic field again. The process may be completely reversible and in most cases, it is. Imagine two (thin) puck-shaped magnets with North at the upper side and South at the lower side. If you place them on top of each other, the magnetic field in the vicinity of the pucks is almost the same as the magnetic field from one puck – the same strength, the same total energy. However, these two pucks attract because if you want to separate them in the vertical direction, you are increasing the energy. In particular, if you separate them by a distance much greater than the radius of the puck's base, the total magnetic field around the magnets will look like two copies of a singlet magnet's field and the energy doubles. If the magnets are close, the magnetic energy is $E$; if they're very far in the vertical direction, it's $2E$. You may consider this position-dependent energy to be a form of potential energy (although there are some issues with this interpretation in the magnetic case when you consider more general configurations: in particular, a potential-energy description becomes impossible if you also include electric charges), potential energy that is analogous to the gravitational one. Gravitational potential energy may be used to do work but it may be restored if you do work on it (think about a water dam where water can be pumped up or down). Nothing intrinsic has to change about the objects (water) and the same is true for the magnets. Let me mention that for a small magnet with magnetic moment $\vec m$ in a larger external magnetic field, the potential energy is simply $$U = -\vec m\cdot \vec B$$ Independently of the magnetic field at other points, the potential energy is given simply by $\cos\theta$ from the relative orientation of the magnetic moment and the external magnetic field (times the product of absolute values of both of these vectors). - That's what I have first thought about, but it does not stand when we compare this solution to the electromagnet's solution. Griffith shows the generator provides the extra work. Magnetic force does not work. – Shaktyai Sep 19 '12 at 9:52 @Luboš Motl any insight on that comment? – Yngve B. Nilsen Sep 19 '12 at 10:22 It's been discussed in detail in a previous Physics SE question I can't find right now. It's a misconception that magnets never do work. Of course that they do. They only don't do any work when acting on a charged particle without a magnetic moment: the magnetic Lorentz force is orthogonal to the velocity so the speed doesn't change. But magnets do work on magnets etc. It's very transparent e.g. exactly in the case of a particle with a magnetic moment for which the potential energy was written above. – Luboš Motl Sep 19 '12 at 11:25 – Luboš Motl Sep 19 '12 at 11:28 1 Thanks! Great explanation, and good discussion in the referenced thread :) – Yngve B. Nilsen Sep 19 '12 at 12:49 show 4 more comments I had the same problem a couple of month ago. If instead of a permanent magnet one uses an electromagnet, Griffith shows that the generator provides the extra energy. So for a permanent magnet it is reasonable to assume the energy is not drawn from the field (if it was the case,the magnetic field would work). For a permanent magnet, the most plausible source of energy is magnetic domain reconfiguration. In other word, everytime you use a permanent magnet it looses some 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": 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": 0.9316471815109253, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/51551/block-slides-on-smooth-triangular-wedge-kept-on-smooth-floor-find-velocity-of-we/52214	# block slides on smooth triangular wedge kept on smooth floor.Find velocity of wedge when block reaches bottom Find the velocity of the triangular block when the small block reaches the bottom: Here is what I did: The final velocity(at the bottom)of the small block of mass m is $\sqrt{2gh}$ along the plane of the incline with respect to the triangle (due to uniform acceleration $g\sin a$ covering distance $\frac{h}{\sin a}$). Let the velocity of the triangular wedge be $V$. Since there is a net external force in the vertical direction, linear momentum is conserved only in the horizontal direction. Then,the velocity of the small block with respect to ground is $$\Bigl(\sqrt{2gh} \cos a\ + V \Bigr) ,$$ we are not considering the direction of $V$, which intuitively should be leftward, but we take rightward. Afterward we should get a negative sign indicating the left direction. Applying conservation of linear momentum in the horizontal direction we get $$MV + m \Bigl(\sqrt(2gh) \cos a\ + V \Bigr) = 0 .$$ Thus we find that $$V = \frac{-m \Bigl(\sqrt(2gh) \cos a\ \Bigr)}{m+M}.$$ However, my book mentions that the answer is something different. I wouldn't like to mention it here because I do not want reverse-engineering from the answer. Please help and explain where I may be wrong. - even using conservation of energy gives the same answer – Surya Prakash Jan 18 at 14:38 – Joe Jan 26 at 8:01 ## 2 Answers The trouble is because you assumed that the final velocity of the small block is $\sqrt{2gh}$. This is true only if the wedge was stationary (in a frame of reference that is inertial), then what happens is that that the normal force from the wedge on the mass completely balances $mg\cos\alpha$, leaving the component $mg\sin\alpha$ down the wedge as you said. But the situation is a little more complicated now, because the wedge is moving simultaneously as the small block slides down. So the forces don't balance out as described in the previous paragraph. One can look at it in terms of energy to gain a better idea. The earth-wedge-mass system is isolated, so its total energy is conserved. The wedge doesn't gain or lose any potential energy, so the only change in potential energy comes from the mass. The change is $- mgh$. This must be distributed to the kinetic energies of BOTH the wedge and the mass. That is, \begin{align} &\Delta K + \Delta U = \Delta E = 0 \nonumber \\ \implies & \Delta K_{wedge} + \Delta K_{block} - mgh = 0. \end{align} if the wedge wasn't moving at all, we would then have $\Delta K_{wedge} = 0$, so \begin{align} \frac{1}{2}mv^2 = mgh \implies v = \sqrt{2gh} \end{align} like you said. But we see that if the wedge was moving, it 'eats' up some of the potential energy that would otherwise have gone to the mass. In other words, the small mass' speed will NOT be $v = \sqrt{2gh}$ at the bottom. Having identified the flaw in your argument, how do we solve the question? There are a few ways. You can draw your force diagrams, carefully balancing out the forces and finding the geometric relation how the position of the mass relates to the position of the wedge. This analysis is perhaps easier in the wedge's frame of reference, but then you would have to add a fictitious force as it is not an inertial frame. But the easiest analysis would be in terms of energy conservation, like the equation I gave you. We have \begin{align} \frac{1}{2}MV^2 + \frac{1}{2}mv^2 - mgh = 0. \end{align} Now all you have to do is find how $V$ is related to $v$. This is simple from conservation of momentum and some trigonometry, try it. (Edit) I noticed after posting that you specifically highlighted the fact that $v = \sqrt{2gh}$ is with respect to the wedge. Lest you start pointing that out, this is not true, because the force the mass feels down the wedge is not $mg\sin\alpha$, because in this frame (wedge's frame, which is not inertial), there is the fictitious force. - Well,thanks a lot.I actually gave a thought about this at the first attempt,but I was hasty about momentum conservation of the system as the internal forces dont matter,and there isn't any external force horizontally.So,I set out to do as above.Now I need to add a 'pseudo' force due to acceleration of the frame,which is in turn again due to the normal force,and after getting the velocity with respect to the frame,I need to get the velocity with respect to ground.Or,I will try the energy conservation way. – Surya Prakash Jan 19 at 3:22 @nerxxx im not getting the correct answer.My first method of solving was what you said,Used conservation of energy and conservation of momentum properly.Answer seemed somewhat close to the correct answer.Then i tried the 2nd method,where i did my previous solution posted above,this time adding the fictitious force.Again,the answer was close but wrong.The correct answer must be $$\sqrt { \frac {2 m^2 gh \cos^2a}{(M+m)(M+ m\ sin^2 a)}}$$ – Surya Prakash Jan 19 at 18:31 in both the methods that i tried,i got the numerator correctly,also i got the (M+m) in the denominator,but the remaining denominator part was different for the two different method.Please try the solution according to the above methods – Surya Prakash Jan 19 at 18:38 what were the answers you got for the two methods you used? What were your energy and momentum equations? I have the same answer as the one you wrote. – nervxxx Jan 20 at 23:12 the answers were : using energy conservation and consevation of momentum horizontally-- same as above,but instead $${\sin^2 a}$$ in the denominator, i got $${ (\cos a -1)^2 }$$ the other answer i did by the first method corrected with pseudo force is $${ \sqrt {\frac {4m^2gh \cos^2 a }{(m+M)^2 (1+ \sin^2 a)} } }$$ – Surya Prakash Jan 21 at 8:57 show 2 more comments $A =$ acceleration of $M$ to the left $a =$ acceleration on m down the incline $a'$ = acceleration of $m$ relative to $M$ along the incline N= normal reaction between block and plane $PLANE -N sin(x) = -MA$ $N = MA/sin(x)$ BLOCK: $F$ along incline: $mg sin(x)=m(a'-A cos(x))$ $a' = g sin (x) + A cos (x)$ $F$ perpendicular to incline: $$N-mg cos (x) = m(-A sin (x))\tag 1$$ Substituting $N=MA/sin(x)$ in eq.(1), we get $A= \frac{mg cos(x)sin(x)}{M+m sin^2 (x)}$ $t$ = time taken by block to slide down the incline $\frac{h}{sin(x)}= \frac{a't^2}2 = \frac{(g sin(x)+A cos(x))t^2}2$ Solving for $$t= \sqrt{\frac{2h}{sin(x) (g sin(x) + A cos(x)}}\tag 2$$ Substitute the expression for $A$ in eq. (2) $V=$ velocity of plane $= At$ Simplify to get $\sqrt{\frac{2gh}{(m+M)(M+m sin^2(x))}} \times m cos(x)$ (Sorry, I wanted to show the FBD's but unable to upload images) - Hi user. Welcome to Physics.SE. Here, we use an unique TeX markup called MathJax, same as Math.SE. The markup is very much helpful in understanding equations, etc. Please have a look here for an introductory, or atleast have a look at our FAQ for an overview. For now, I've revised your post. You can revise it again, if you think I've missed something. BTW, Don't worry that you can't post images. New users don't have the privilege. Just give me the link of the image. I'll add it to your answer :-) – Ϛѓăʑɏ βµԂԃϔ Jan 26 at 4:53 Thanks a lot,thats how i got the answer,and also through similar ways,refer my comments on the above answer.the question is solved – Surya Prakash Jan 26 at 8:36	{"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": 35, "mathjax_display_tex": 9, "mathjax_asciimath": 1, "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": 0.9375462532043457, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/36428/in-what-way-do-cooper-pairs-of-electrons-bond-and-stay-bonded-in-superconductors?answertab=oldest	# In what way do Cooper pairs of electrons bond and stay bonded in superconductors? I understand how electrons initially move into another's vicinity, but nowhere can I find a fathomable answer to this. Also, does the pairs forming 'a condensate' mean a Bose-Einstein condensate? - ## 3 Answers They don't really bind, the interaction can be too weak to bind, but the pairs condense anyway. The theory is interesting precisely because the condensation is happening from an interaction too weak to form actual pairs. When the interaction is strong, the pairs can be considered as effective particles, and the theory of superconductivity is different and simpler--- it's a BEC of the pairs. The standard description is BCS theory, and I'll explain it Bogoliubov's way, which is found in countless places. I'll consider the electron field interacting with a nonrelativistic instantaneous potential. This is not accurate in traditional superconductors already, because the interaction is phononic and retarded, but whatever, the phenomenon doesn't care about this very much. you have an electron field $\psi(k)$, and the interaction Hamiltonian is $$V(k_\mathrm{in}-k_\mathrm{out}) sum_{s} \bar{\psi}^{s_1}_{k_1}\bar{\psi}^{s_2}_{k_2} \psi^{s_3}_{k_3} \psi^{s_4}_{k_4} \delta(\sum k)$$ Where the $\delta$ is multiplied by 3 $2\pi$ factors as always, I assume V is spherical symmetric, and there are two spin fluids which are degenerate in energy. The main assumption, which is justified self-consistently is that the field $\psi^0_{k}\psi^1_{-k}$ has an expectation value. You can see this is self consistent from the fact that if V is negative at $2\|k_f\|$, this expected value lowers the energy classically. The exected value of $\psi\psi$ is a type of condensation, but it isn't local condensation because there is no real local bilinear field that it is the exectation value of. In the limit that the coupling is strong, you can make a local field which creates a bound Cooper pair, but this is only linked adiabatically to the BCS picture. - You truly have overestimated my aptitude, yet thanks, however, for the bits that I could understand. – Alyosha Sep 14 '12 at 19:35 1 @Alyosha: Sorry, it's not so hard--- the condensate is of pairs, but they aren't close to each other, they are nonlocally paired, the condensate is in k-space. You can think of it as follows: there is a fermi sea of electrons, and which electron is making the bound state with each other electron is not well defined, as they swap in and out with each other, but the end result is that there is a BEC anyway. – Ron Maimon Sep 14 '12 at 19:53 ... also, I tried to make a much longer answer, but this site is having incredible lag problems regarding typing TeX heavy answers. It used to be that turning off the preview fixed this, but no more. – Ron Maimon Sep 14 '12 at 19:56 I have, bar the mathematics, understood that far. But what stops the electrons from doing this at relatively high temperatures (i.e. forgetting high temperature superconductors)? It can't just be the disruptive thermal motions, as not all materials superconduct. Or are the electrons which demonstrate this effect in the conduction band beforehand in superconducting materials, even when hot, and just need to have the thermal motion removed to begin condensing? – Alyosha Sep 15 '12 at 20:34 And, remembering the forgotten, why do some materials' electrons condense at higher temperatures (or: what mechanism DO they use, then, if not a BEC, to superconduct?)? – Alyosha Sep 15 '12 at 20:38 show 4 more comments Yes, the cooper pairs form a BEC. See Kardar's Statistical Physics of Particles for a detailed derivation of this. I think Kittel's Thermal Physics might also have some explanation (on an easier level). For a BEC to form, the bosons should be non-interacting. To good approximation, this is the case for cooper pairs. - How do they actually 'stick together', though? Is it essentially the way that all BEC stick together, and I need to read into it more? – Alyosha Sep 14 '12 at 18:44 This is not really true--- the field can be considered a BEC of Cooper pairs in a certain sense, it is the expected value of a fermion bilinear, but the pairs wouldn't bind in isolation from the Fermi sea, and the description is entirely weak coupling, you never have to make a binding potential. This is not the correct explanation, although it is what happens at strong coupling and the two are continuously linked by varying the coupling. – Ron Maimon Sep 14 '12 at 18:50 It is also not true that a BEC requires the bosons to be noninteracting. Also interaction (repulsive) bosons form a BEC. In order for the BEC to be superfluid there is even interaction needed! – Fabian Sep 14 '12 at 21:49 Of course there are more interesting BECs. The simplest however is just a bose gas, which is more-or-less a good intuitive model of cooper pairs. – user404153 Sep 14 '12 at 22:03 1 @user404153: It's a good intuitive model, but it's not the real deal. The OP asked what's going on, not the simplest model. I agree that this is the simplest model, it's essentially Landau's. The BCS model though is more interesting because of the nonlocal nature of the bilinear, which allows an arbitrarily weak interaction to cause condensation, when an arbitrarily weak interaction does not cause binding. – Ron Maimon Sep 15 '12 at 5:09 show 1 more comment The attraction is caused by screening, which means that when two electrons come together, they tend to push away the nearby electrons, which leaves a net positive charge in the area that keeps those two electrons together. - 1 The attraction is usually by phonons, which is the deformation of the heavy nuclei, not the other electrons. You can tell it's this by the isotope effect--- the critical temperature depends on the nuclear mass. This means the nuclear inertia is important, that the nuclei are moving during the binding interaction. – Ron Maimon Sep 15 '12 at 0:47 – jcohen79 Sep 16 '12 at 22:21 Phonons play absolutely no role in HighTc materials, this is certain, because the phonon pairing is no stronger in highTc materials, and phonons always gives S-wave pairing since it is approximately isotropic in the scales where it is stronger than Coulomb repulsion. The pairing in HighTc is due to a mechanism that I think I know, but nobody else agrees with me on. I can tell you if you ask a question on it. The phonons are the source of the force in regular superconductivity for sure, since the critical temperature isotope dependence is correctly predicted from this hypothesis. – Ron Maimon Sep 16 '12 at 23:45	{"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": 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": 0.9386520981788635, "perplexity_flag": "middle"}
http://terrytao.wordpress.com/tag/i-method/	What’s new Updates on my research and expository papers, discussion of open problems, and other maths-related topics. By Terence Tao # Tag Archive You are currently browsing the tag archive for the ‘I-method’ tag. ## Resonant decompositions and the I-method for the cubic nonlinear Schrodinger equation on R^2 24 April, 2007 in math.AP, paper \| Tags: dispersive equations, I-method, NLS, Schrodinger equation \| by Terence Tao \| 2 comments My paper “Resonant decompositions and the I-method for the cubic nonlinear Schrodinger equation on ${\Bbb R}^2$“, with Jim Colliander, Mark Keel, Gigliola Staffilani, and Hideo Takaoka (aka the “I-team“), has just been uploaded to the arXiv, and submitted to DCDS-A. In this (long-delayed!) paper, we improve our previous result on the global well-posedness of the cubic non-linear defocusing Schrödinger equation $i u_t+ \Delta u = \|u\|^2 u$ in two spatial dimensions, thus $u: {\Bbb R} \times {\Bbb R}^2 \to {\Bbb C}$. In that paper we used the “first generation I-method” (centred around an almost conservation law for a mollified energy $E(Iu)$) to obtain global well-posedness in $H^s({\Bbb R}^2)$ for $s > 4/7$ (improving on an earlier result of $s > 2/3$ by Bourgain). Here we use the “second generation I-method”, in which the mollified energy $E(Iu)$ is adjusted by a correction term to damp out “non-resonant interactions” and thus lead to an improved almost conservation law, and ultimately to an improvement of the well-posedness range to $s > 1/2$. (The conjectured region is $s \geq 0$; beyond that, the solution becomes unstable and even local well-posedness is not known.) A similar result (but using Morawetz estimates instead of correction terms) has recently been established by Colliander-Grillakis-Tzirakis; this attains the superior range of $s > 2/5$, but in the focusing case it does not give global existence all the way up to the ground state due to a slight inefficiency in the Morawetz estimate approach. Our method is in fact rather robust and indicates that the “first-generation” I-method can be pushed further for a large class of dispersive PDE. Read the rest of this entry »	{"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": 11, "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": 0.884933352470398, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?t=24177	Physics Forums Thread Closed Page 1 of 3 1 2 3 > ## Meaning of operators for observables. I understand that observables in quantum mechanics are represented by hermitian operators, which are converted into a matrix when expressed in a particular basis. I also understand that when the basis used is an eigenbasis of the operator, the matrix becomes diagonal, having the eigenvalues as diagonal elements, and those eigenvalues are the possible results of the measurement. But I still think I am missing something, that I fail to visualize the connection between the math and the physical process. I know of three things you can do to a state: (1) You can unitarily change it, like under a force or just letting time pass (2) You can make a measurement, which would change the state by eliminating some of its components. (3) You can make a change of basis, which would leave the state alone and just change your description of it by rotating the frame of reference (in Hilbert space). But I dont understand how the observable operators relate to the three points above. I can see how the operator can, in its eigenbasis and when braketed with the state vector, give a weighted sum of all the possible measurement resuts, which would correspond to the expectation value. But that's the most I can make out of it. And, again, I think I am missing somethhing. If a state (pure) is represented by a state vector, and operators operate on state vectors to yield other state vectors; what is that operators corresponding to measurables do to the state vector? It can't be a change of basis because this is accomplished by a unitary operator. It can't be a real measurement because the operator is not projecting the state vector on a basis, unless it is already in its eigenbasis. And the meaning of these operators become even less clear when they appear multiplied times each other like in the conmutators. I have started to explore this problem with spin operators, but an elementary treatment has not answered my questions. I realize that after reading a lot about angular momentum I might get to the point I understand. But I thought there should be some explanation of this independent of the particular states under consideration. I am really confused. All I have read doesn't seem to clarify this puzzle. --Alex-- PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus The description of an observable as a Hermitian operator is not strictly necessary from the point of view of measurement theory alone. One could simply describe a measurement by a collection of projectors onto the relevant eigenstates instead. However, the formalism is useful because of the dual role that observables play in quantum mechanics. They can be things that are measured, but also they can be things that are conserved by the unitary part of the dynamics. The latter means that if a system starts in a particular eigenstate of an operator representing a conserved quantity, then it will remain in that eigenstate. It is easy to check that if H is a Hermitian operator, then e^(iH) is a unitary operator that conserves H. Hence, the Hermitian operators are useful because they can be used to describe both the unitary part of quantum dynamics and the measurement part. Thank you sylboy. You mention a dual role of operators. Probably I can explore both roles separate, as otherwise it may become overwhelming. With respect to the role they play in the time evolution, I have read (but never quite understood) that hermitian operators conserve probability. I think I need to study this more, but I think I would rather start with the other role you mention. (measurement) After reading your response, I was thinking the following: To give you the different eigenvalues of a particular possible measurement, the operator's matrix needs to be diagonalized. When you diagonalize the operator, you are changing the basis, the same unitary operator used to diagonalize the observables operator is used on the state, to express it in the new basis. If the operator't matrix in a certain basis is not diagonal, the matrix itself contains all the information needed to construct the unitary matrix needed to diagonalize it. So, could we say that the function of the observable's operator is to provide all the information needed for the basis change, even if the operator used to accomplish that change of basis is not the measurable's operator itself but the unitary operator that is built based on information provided by the observable's matrix? Maybe I am going in the wrong direction, but I am trying to build a conceptual scafold that can make sense to me, and which I can use as a frame of reference to later interpret and fill-in all the details. I can also understand that a task like this is somewhat personal, with each person having different preferences, depending on their personalities. I understand most people may prefer to just work with the operators, without worrying too much about their meaning, and later their meaning becomes apparent through their use. I can understand how that may work for them. I have never considered myself very good and efficient at heavy symbol manipulation, and for this reason I rely more on deeper conceptual understanding and sometimes on paradigms. Sylboy, thanks again, and I hope you'll give me a few more insights. --Alex-- ## Meaning of operators for observables. From the perspective of measurement theory, a Hermitian operator is just a convenient summary of everything you could want to know about the outcome statistics of that measurement on any state. Another convenient summary would be the set of projection operators associated with the eigenspaces, or the set of eigenstates if the operator is non-degenerate. The latter is often more convenient for discussing the foundational issues of quantum mechanics, but in practice physicists find the former more convenient because of the close relationship to conserved physical quantities. It is true that the operator contains all the information about the appropriate change of basis you need to make to figure out the outcome probabilities and post-measurement state. I don't know why you are trying to attach such a great significance to this piece of linear algebra, but you may choose to understand it in this way if you wish. It is not true that Hermitian operators conserve probability, but unitary ones certainly do. This just means that the inner product of the state with itself remains constant. The interpretation of "conserve probability" in this statement is just that the probability that the system is in some state is always 1. To see why this is true you just have to note that the question "Is the system in some state?" corresponds to the identity operator, so it always has the value 1 under unitary evolution. Recognitions: Homework Help Quote by alexepascual I know of three things you can do to a state: ... (2) You can make a measurement, which would change the state by eliminating some of its components. The measurement "rotates" the state. Perhaps it is bad to think of the measurement as "eliminating components." Quote by alexepascual ... what is that operators corresponding to measurables do to the state vector? ... It can't be a real measurement because the operator is not projecting the state vector on a basis, unless it is already in its eigenbasis. You lost me on this one. The basis is a mathematical abstraction. I don't see how that can get in the way of a physical measurement. Quote by alexepascual ... I thought there should be some explanation of this independent of the particular states under consideration. The theory/formalism is much more clear when demonstrated in the abstract. The three things you mentioned above do not depend on a particular state. Are you looking to add as many more things to this list as possible? Quote by alexepascual To give you the different eigenvalues of a particular possible measurement, the operator's matrix needs to be diagonalized. ... I think you may be blurring the critical distinction between an operator and a matrix. There is much overlap in the sense that you can express (and even think of) the operator in terms of a matrix and you can use a matrix to find the various properties of an operator. But the operator itself is not strictly a matrix. As I understand it, the fundamental distinction between an operator and a scalar dynamical variable is twofold: 1) the commutation issue and 2) the fact that an operator is awaiting a state on which to operate whereas a scalar dynamical variable stands alone. A matrix also stands alone (can be treated without multiplying it on a state), so it may not be providing you with a comprehensive picture. In otherwords, when you are considering an observable Ω as a matrix, then you have already, whether explicitly or not, performed the operations <φi\|Ω\|φj>. Sylboy: I guess every person has a different way of learning, and I can understand how we can get impatient when other people try to grasp the problem from a different angle than the one we have chosen. I appreciate your comments though, and I have printed a copy to go over them as I keep thinking about this issue. Turin: With respect to your first comment: Quote by Turin The measurement "rotates" the state. Perhaps it is bad to think of the measurement as "eliminating components." Rotating the state can give you the state expressed in the eigenbasis of the measurement operator, but if the state corresponds to a superposition of different eigenstates, the rotation doesn't choose one of the eigenstates as the result of your measurement. I cound understand how rotating the states (in Hilbert space) can be part of the measurement, but I would bet it can't be the whole story. You would need a projector or something to complete the measurement. A rotation of the basis would not change the state, but a measurement does, as it chooses one eigenstate and sets the others to zero. Even if we ignore the selecting of a particular outcome and consider the rotation part, a unitary operator allows you to rotate the state, so why would you need the hermitian operator at all? I am not really questioning . I know that there must be a reason, it is just that I think I am missing something. Quote by Turin You lost me on this one. The basis is a mathematical abstraction. I don't see how that can get in the way of a physical measurement. I just meant that the operator operating on a state does not give you the eigenvectors or the eigenvalues unless the state is already expressed in the eigenbasis of the operator. Do I make any nore sense? In other words: when you find the eigenbasis of the operator, and diagonalize it using the needed unitary operators, only then you get to see the eigenvalues in the diagonal, and only when the state vector is expressed in this eigenbasis do the coefficients relate to the probability of getting the eigenvalues as results of the measurement. Quote by Turin The theory/formalism is much more clear when demonstrated in the abstract. The three things you mentioned above do not depend on a particular state. Are you looking to add as many more things to this list as possible? I agree with you. I would like to understand this issue without having to refer to a particular system. But as I have failed, I am trying to use a two state system as an example. If I can understand these concepts without getting into angular momentum, etc. I would prefer to do that. I am trying to keep the "list" as short as possible. With respect to your comment about my blurring the distiction about an operator and a matrix, I understand everything you are saying and totally agree. Maybe I didn't express myself clearly, but I understand the distinction. I understand that the operator can stand on its own and not become a matrix untill you express it in some basis by bracketing with the basis vectors. That's what I meant by "the matrix of an operator". Like the Pauli matrices would be the matrices of the spin operators expressed in the z (+/-) basis. Recognitions: Homework Help Quote by alexepascual ... if the state corresponds to a superposition of different eigenstates, the rotation doesn't choose one of the eigenstates as the result of your measurement. This statement makes me wonder if we might be talking about two different things. I am picturing an arbitrary state and an arbitrary measurement. Lets say we have an operatoer, Ω, that has an eigenbasis {\|ωi>}, and let it operate on a state, \|ψ>. Mathematically, we would get: Ω\|ψ> = Σωi\|ωi>. Physically, however, we would get: Ω\|ψ> ---> ωi\|ωi> with a corresponding probability: √(<ψ\|ωi><ωi\|ψ>) which is generally nonunity. Since it is nonunity, the rotation is nondeterministic. Is that what you mean - how can the selection of eigenstate be nondeterministic? Quote by alexepascual A rotation of the basis would not change the state, but a measurement does, as it chooses one eigenstate and sets the others to zero. Even if we ignore the selecting of a particular outcome and consider the rotation part, a unitary operator allows you to rotate the state, so why would you need the hermitian operator at all? I didn't mean a rotation of the basis, I meant a rotation of the state. You need the hermiticity as opposed to the unitarity because 1) the formalism requires real eigenvalues for observables, and 2) the value of the resultant eigenvalue (usually) contains all of the experimental information, which would be trivial if it were always unity. Therefore, you cannot get a meaningful formalism based on eigenvalued results by only allowing hermitian unitary transformations. I hope I'm not mixing stuff up; it's been over a year since my last QM instruction. Quote by alexepascual I just meant that the operator operating on a state does not give you the eigenvectors or the eigenvalues unless the state is already expressed in the eigenbasis of the operator. ... when you find the eigenbasis of the operator, and diagonalize it using the needed unitary operators, only then you get to see the eigenvalues in the diagonal, and only when the state vector is expressed in this eigenbasis do the coefficients relate to the probability of getting the eigenvalues as results of the measurement. My point was that the operation inherently returns an eigenvalue of the operator, according to the formalism. Whether or not it looks that way when you write it down on paper is a different issue. The physical interpretation of the QM state is that its "length" is arbitrary and its "direction" alone completely determines the physical state. When the operator acts on the state (when you "look" at the state), it necessarily returns an eigenvalue and points the state in the corresponding direction. Quote by alexepascual ... I am trying to use a two state system as an example. I believe you mean that you are trying to use a 2 dimensional Hilbert space. The system can be in a superposition of two states, the superposition being its own distinct state, and therefore allowing an infinite number of distinct states. (I'm assuming you're talking about spin 1/2 particles.) Turin, I was as a matter of fact missinterpreting you with respect to the rotation. I was talking about a rotation of the basis and you were talking about a rotation of the state vector. I think this is crucial and deserves some discussion. A rotation of the state vector with respect to a particular basis would imply a chance in the coefficients (components) of the vector. This would represent a change in the physical object described by the vector. What I mean by a rotation of the basis may a sloppy way of expressing it. (although I have seen it in some books). What I actually mean is this: Every time you describe the vector you use some basis. So you could say that you are using a reference frame that coincides with (is aligned with) that basis. A change to another basis could be seen as a rotation of your reference frame so that it now aligns with the new basis. The position of the state vector here remains fixed with respect to any of the possible bases (with respect to the Hilbert space). It is just your point of view that changes. So the physical situation does not change, only your description of it changes. With respect to the equations you list, the first one is: Quote by Turin Ω\|ψ> = Σωi\|ωi>. Are you sure this is correct? Wouldn't it be: Ω\|ψ> = Σi ωi \|ψi> or... Ω\|ψ> = Σi ωi \|ωi><ωi\|ψ> or... Ω\|ψ> = Σi ωi <ωi\|ψ>\|ωi> or... Ω\|ψ> = Σi ωi ci \|ωi> Some other questions (if you don't mind): (1) to write these equations I had to cut and paste from yours. How do you make greek letters and subscripts on this forum? (2) I noticed that your notation is very similar to Shankar. Is that the book you used?. (3) I would assume you studied Quantum Mechanics in school (graduate studies?) What are you doing these days? anything related to physics? I am myself presently unemployed. I used to work as an engineer but have a hard time finding a job now. I got a bachelor's degree in physics which doesn't help. I was planning to get my master's in physics because that is what I like, but now I am reconsidering. Even if I don't continue my formal studies in physics, I'll keep studying QM on my own. Turin: thanks for your replies, they are helping me a lot in thinking about these issues in QM. Things are not clear for me yet but I feel I am making some progress. Thanks again, --Alex-- It should be Ω\|ψ> = Σi ωi \|ωi><ωi\|ψ> or... Ω\|ψ> = Σi ωi <ωi\|ψ>\|ωi> presuming that ωi are the eigenvalues of Ω and \|ωi> are the corresponding eigenstates. I dont know how to wite the greek symbols (I also used cut and paste, but I do know how to do this: $$\Omega \| \psi \rangle = \sum_i \omega_i \langle \omega_i \| \psi \rangle \| \omega_i \rangle$$ which is much cooler anyway (click on the equation to find out how). I think Turin means that on measurement a state is rotated onto the eigenstate corresponding to the outcome that was obtained. The unitary operator that does this depends on both the initial state \|ψ> and the outcome \|ωi>. It is not unique, since there are many unitary operators that do this. I think it is better to regard the state update rule as a projection rather than a rotation becasuse the projection operators are unique and do not depend on \|ψ>. This is a better way of thinking about it if you ever study the more advanced aspects of measurement theory, where state vectors are generalized into things called density operators and the notion of a projective measurement is generalized into something called a POVM (Positive Operator Valued Measure). Not all physicists have to study this, but if you are interested in the foundations of quantum theory or quantum computing and information theory then you will eventually come across it. I think quantum statistical mechanics uses this formalism as well. Recognitions: Homework Help Quote by alexepascual A rotation of the state vector with respect to a particular basis would imply a chance in the coefficients (components) of the vector. This would represent a change in the physical object described by the vector. Yes. Quote by alexepascual Every time you describe the vector you use some basis. Basically. I imagine that one could probably make an argument for an exception, but I don't imagine that it would be altogether enlightening for our purposes. Quote by alexepascual A change to another basis could be seen as a rotation of your reference frame so that it now aligns with the new basis. I'm not sure I agree with this. In some ways, it does seem like a rotation, but I imagine that it could also involve reflections and discontinuities. Quote by alexepascual The position of the state vector here remains fixed with respect to any of the possible bases (with respect to the Hilbert space). It is just your point of view that changes. So the physical situation does not change, only your description of it changes. Do you mean as a consequence of a measurement. If I'm not mistaken, this is not consistent with experimental result. For instance, if the measurement were only to change the basis but leave the physical state alone, then I don't think there would be any non-trivial commutation of observables. Quote by alexepascual Are you sure this is correct? Wouldn't it be: ... You are quite right. How sloppy (incorrect) of me. I wouldn't use your first suggestion, though. Quote by alexepascual (1) to write these equations I had to cut and paste from yours. How do you make greek letters and subscripts on this forum? Concatenate these strings: "&" "alpha" ";" Quote by alexepascual (2) I noticed that your notation is very similar to Shankar. Is that the book you used?. Yes. Everything about QM that I have learned in a formal setting has been from Shankar. Quote by alexepascual (3) I would assume you studied Quantum Mechanics in school (graduate studies?) What are you doing these days? anything related to physics? Yes. I graduate tomorrow as a matter of fact. I have no job prospects, so I am stressing more than anything else. We seem to have striking similarities in background and contemporary situation. Recognitions: Homework Help Quote by slyboy The unitary operator that does this depends on both the initial state \|ψ> and the outcome \|ωi>. It is not unique, since there are many unitary operators that do this. Can you give an example of U1\|ψ> = \|φ> and U2\|ψ> = \|φ> such that U1 /= U2. For some reason, I just can't think of how to show that. Quote by slyboy I think it is better to regard the state update rule as a projection rather than a rotation becasuse the projection operators are unique and do not depend on \|ψ>. How would this then demonstrate the statistical dependence on \|ψ>? Can you give an example of U1\|ψ> = \|φ> and U2\|ψ> = \|φ> such that U1 /= U2. For some reason, I just can't think of how to show that. To specify a d-dimensional unitary operator, you have to give its action on at least d-1 linearly independent vectors (the last one will be fixed by orthogonality relations). There are no examples in 2-dimensions, which may be why you had difficulty finding an example, but here is one in a real 3-d space: Suppose \|1> and \|2> and \|3> are orthonormal basis vectors and we want to find a unitary, U, such that $$U \|1 \rangle = \frac{1}{\sqrt{2}} ( \|1 \rangle + \|2 \rangle)$$ We could choose $$U \|2 \rangle = \frac{1}{\sqrt{2}} ( \|1 \rangle - \|2 \rangle)$$ $$U \|3 \rangle = \|3 \rangle$$ but $$U \|2 \rangle = \|3 \rangle$$ $$U \|3 \rangle = \frac{1}{\sqrt{2}} ( \|1 \rangle - \|2 \rangle)$$ is an equally good choice. If you are concerned that this is just a relabeling of the basis vectors then note that $$U \|2 \rangle = \frac{1}{\sqrt{3}} ( \|1 \rangle - \|2 \rangle + \|3\rangle)$$ $$U \|3 \rangle = \frac{1}{\sqrt{6}} (\|1 \rangle - \|2 \rangle - 2 \|3\rangle)$$ is also a good choice. I'm sure you can now come up with a continuous infitinty of other possibilities. How would this then demonstrate the statistical dependence on \|ψ>? The state update rule is $$\|\psi \rangle \rightarrow \frac{\Pi \| \psi \rangle}{\|\| \Pi \| \psi \rangle \|\|}$$ where $$\Pi$$ is the projector onto the eigenspace corresponding to the eigenvalue obtained in the measurement. This clearly shows the dependence on the state, but the operator involved is not dependent on the state and is uniquely defined (modulo some mathematical technicalities that arise for continuous variables). Well guys, you gave me a lot to read. I'll have to take a break now and start working on something that perhaps can make me some money. I printed your posts and I'll read them tonight on the couch while my wife watches TV. I think I'll answer tonight or tomorrow morning. --Alex-- Recognitions: Homework Help Quote by slyboy To specify a ... unitary operator, ... Of course; that makes sense. I don't know why I couldn't think of that. Quote by slyboy The state update rule is $$\|\psi \rangle \rightarrow \frac{\Pi \| \psi \rangle}{\|\| \Pi \| \psi \rangle \|\|}$$ where $$\Pi$$ is the projector onto the eigenspace corresponding to the eigenvalue obtained in the measurement. This clearly shows the dependence on the state, but the operator involved is not dependent on the state and is uniquely defined (modulo some mathematical technicalities that arise for continuous variables). I'm still missing something. I don't see how this helps one understand the collapse. In fact, it seems more complicated than the observable operator formalism. If $$\Pi$$ does not depend on the state, then what would prevent one from specifying $$\Pi$$ to project $$\|\psi\rangle$$ onto a state $$\|\phi\rangle$$ for which $$\langle\phi\|\psi\rangle = 0$$? Slyboy: sorry for misspelling your name. It must be dyslexia. Other things came up and I was not able to take the time to think about your argument about the different unitary operators. Turin: States in Hilbert space could be visualized as points on a sphere. to one state, there corresponds only one point on the sphere. When we move the state vector, we are going prom point a to point b on the sphere. This motion can always be accomplished by a one of two rotations in the plane determined by both points and the origin. If you accomplish the change by a continuous motion or a sudden jump is not made explicit by the unitary operator. In 2-D real space, a rotation matrix takes you form point a to point b on the unit circle, but although we call the matrix a "rotation" matrix, that matrix does not tell you that you are smoothly moving the thing through all the intermediate angles. The rotation matrix could very well represent the point dissapearing form position "a" and appearing at position "b". In this case, or in the Hilbert space case, I wonder if the matrix for a rotation of 180 degrees and that for a reflection would look different. On the other hand, if what we are doing is changing basis, where each eigenvector is represented by a point on the sphere, as we are talking about many points, reflection would look different than rotation, but still the rotation matrix would not necessarily mean a smooth motion. Now, when I suggested viewing a basis change as a rotation, I was not trying to be precise. I was just trying to find some way to visualize the thing, even if later I have to replace that concept by a more precise one. I understand orthogonal matrices are not only rotation matrices but also reflection and inversion matrices, and the same must be true for unitary matrices I guess. Quote by Turin Do you mean as a consequence of a measurement. If I'm not mistaken, this is not consistent with experimental result. For instance, if the measurement were only to change the basis but leave the physical state alone, then I don't think there would be any non-trivial commutation of observables. I agree with you and I think you missinterpreted me. It was exactly my point that a change of basis alone can not represent a measurement. On some equations you posted previously you said: "mathematically it would be like this.." and "physically it would be like this..". Without getting into the details and more on a philosofical note, I think that what we are trying to do with the math is represent as close as possible what is happening physically. I do understand that some times we may use terms whose "physical existence" is questionable and in other cases there may be no good math to represent the physical process. But I would guess in most of quantum mechanics there is a good correspondence between the "physical" and the "mathematical", this as long as we consider things like a state as "physical". I am not trying here to get into deep philosofical discusion, all I mean is that I was somewhat suspicious about your assertion about something being one way "mathematically" and another way "physically". Slyboy and Turin, To summarize a little our discussion, I think we kind of agree on the following: (1) A measurement "could" be represented by a rotation of the state vector (which does not need to be continuous) to a position where it coincides with one of the eigenvectors of the measurement operator. (2) The previous representation may not be the best. A projection may be a better way to visualize it or to carry out the calculation. (3) If we have a measurement operator and a generic state vector, in order to carry out the multiplication you express both the operator and the vector in some basis. If the basis you chose is not an eigenbasis of the operator, when you multiply, you are not going to get each component of the state vector multiplied times the eigenvalue. In order to get that you have to make a change of basis to the eigenbasis of the operator. You achieve this change of basis using a unitary operator, which you apply to both the matrix and the vector (unless ine of them is already expressed in the eigenbasis). This change of basis does nothing to the physical state itself, it is only the description that changes. Once you have everything in the eigenbasis of the operator, (the matrix is now diagonal) then multiplying the measurement matrix times the state vector gives you a vector where each original component is now multiplied times the eigenvalue. Do we agree up to this point? Well even if this is correct, I fail to see the significance in this product. It appears that the change in basis was more crucial (it gave us the eigenvalues and the eigenvectors). But multiplying the matrix times the vector doesn't seem to accomplish much other than murking things up by mixing amplitudes with eigenvalues by multiplying them together. The only use I can see in this is to put a bra on the left and then get the expectation value. Someone may say that that's exactly what the operators are for, that thy give us a bridge between that classical quantities and the quantum world. I can understand that. On the other hand, looking at the commutation relations, it appears that the operators have a greater significance than this. Maybe what I am trying to visualize can not be visualized, but I think there should be some meaning to the operators even if we stay in the quantum world and not try to make comparisons with the classical. I understand the correspondence principle has been important, but probably that was just a crutch. So in summary, I take a state, I multiply it times a measurement matrix (yes I say matrix, not operator) and the matrix is not in the eigenbasis of the operator. What have I done to the state? If I multiply the state vector times two measurement matrices (like in a commutator), what have I really done? Recognitions: Homework Help alexepascual and slyboy, You have converted me. I don't know what I was thinking. I now believe that the exact arguments that I made in favor of the rotation picture actually serve to better promote the projection picture. (I'm still awaiting slyboy's reply with interest.) alexepascual, Perhaps I was unclear when I distinquished the "mathematical" quality from the "physical" quality. What I meant was: - when you "write it down" or "do the calculation" the writing contains a distribution and there is no way for the calculation to make a selection of a particular eigenvalue. The calculation can show a favored eigenvalue, but there is no way to be sure unless the state is prepared as an eigenstate. - when you "do the experiment" the "display on the experimental apparatus" will "show only one of the eigenvalues" and will not give a distribution. Of course, if you "do the experiment" repeatedly, the apparatus will eventually develop the distrubution and so "agree with the math." I didn't intend anything philosophical. I agree with (1) and (2). I still don't quite follow the point of (3). Maybe you could give an example. I think what is confusing me about your (3) is that I see no need to first express the state in some obscure basis and then subsequently transform to the eigenbasis of the operator. Furthermore, if the state is an eigenstate (of the operator), then it does not matter whether or not it is expressed in the eigenbasis (of the operator); the operator will still return the original state multiplied by the eigenvalue. If the state is not an eigenstate, it can still be decomposed into a superposition of eigenstates (without applying any unitary operators?) (See your post, post #8). Turin, When I said I was not intending a deep philosophical discussion, I was talking about my own argument. I didn't mean that your arguments appeared to be philosofical. Actually I do like to think and discuss the philosofical fouindations and implications of quantum mechanics, but here I was trying to not to fall into that temptation as I wanted to focus on the math and its relation to the physical entities it describes. Some of what you said in your last post is giving me food for though. I'll have to take some time to digest it. I also want to go over the Slyboy's discussion about the unitary operators ( I haven't had time yet). Thread Closed Page 1 of 3 1 2 3 > Thread Tools Similar Threads for: Meaning of operators for observables. Thread Forum Replies Quantum Physics 5 Advanced Physics Homework 7 General Physics 13 General Physics 13 General Physics 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": 17, "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": 0.9568358063697815, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/169475/integrability-of-the-poisson-kernel	# Integrability of the Poisson kernel Let $$P_t(x)=\frac{2t}{\omega_{n+1} (\|x\|^2+t^2)^\frac{n+1}{2}}$$ denote to the Poisson kernel on the upper half space $\mathbb{R}^n\times \mathbb{R}_+$. How do I see $P_t \in L^1 \cap L^\infty$ and why can I conclude that $P_t \in L^p\ \forall\ 1\leq p \leq \infty$? - In spherical coordinates the problem reduces to a one-dimensional integral. – Andrew Jul 11 '12 at 16:40 ## 2 Answers Being $P_t$ continuous, in order to prove $P_t\in L^1\cap L^{\infty}$ it is sufficient the relation $$P_t(x)=\mathcal{O}\left(\frac{1}{\|x\|^{n+1}}\right),\text{ as }\|x\|\to\infty.$$ About your latest question, it is a special instance of the interpolation theorem. Let be $E$ a measurable subset of $\mathbb{R}^N$ and $p,q,r\in[1,\infty]$ with $p\leq r\leq q.$ If $f\in L^p(E)\cap > L^q(E)$ then $f\in L^r(E),$ with $$\|\|f\|\|_r\leq\|\|f\|\|_p^{\alpha}\|\|f\|\|_q^{1-\alpha}$$ where $\alpha\in > [0,1]$ is determined through $\frac{\alpha}{p}+\frac{1-\alpha}{q}=\frac{1}{r}.$ As you can figure out it is an easy application of Hoelder inequality. - As for boundedness, can you see which value of x yields the maximum of the kernel? x is in the denominator... And once a function f is bounded, say, by M, you get $\|f\|^p\le M^{...}\|f\|$. (Trying to leave something for you to work out. ) -	{"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": 14, "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": 0.9081682562828064, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/18514-linear-approximation-application.html	# Thread: 1. ## Linear Approximation Application I'm having some trouble applying linear approximation and changes, so I'm stuck on this question. The diameter of a tree = 10 in. Circumference increases by 2 in About how much did the tree's diameter grow? How much did the tree's cross section area grow? I get the concept of linearization, but I have no idea how to apply it in questions like this. Thanks again for your time. 2. Originally Posted by Wyau I'm having some trouble applying linear approximation and changes, so I'm stuck on this question. The diameter of a tree = 10 in. Circumference increases by 2 in About how much did the tree's diameter grow? How much did the tree's cross section area grow? I get the concept of linearization, but I have no idea how to apply it in questions like this. Thanks again for your time. The linear approximation to a change in the argument of a function f(x) is $f(x + \Delta x) \approx f(x) + f^{\prime}(x) \cdot \Delta x$ So we know that the initial diameter of the tree is d = 10 in. And we know that the circumference of the tree grows by 2 in. The formula for the circumference is $C = \pi d$ so $d(C) = \frac{C}{\pi}$ I'm assuming you have to use the linear approximation equation here (though the answer turns out to be exact, not an approximation) so $C = \pi d = \pi \cdot 10~in = 31.415~in$ and $d^{\prime}(C) = \frac{1}{\pi}$ Thus $d(C + \Delta C) = \frac{C}{\pi} + \frac{1}{\pi} \cdot \Delta C$ $d(31.415~in + 2~in) = \frac{31.415~in}{\pi} + \frac{1}{\pi} \cdot 2~in = 10.637~in$ For the second problem, we know that the cross-section is a circle and the area of a circle is: $A = \pi r^2 = \pi \left ( \frac{d}{2} \right )^2 = \frac{\pi}{4}d^2$ So $A^{\prime}(d) = \frac{\pi}{2}d$ The change in d is: $\Delta d = 10.637~in - 10~in = 0.637~in$ So $A(d + \Delta d) \approx A(d) + A^{\prime}(d) \cdot \Delta d$ $\approx \frac{\pi}{4}(10~in)^2 + \frac{\pi}{2}(10~in) \cdot (0.637~in)$ $\approx 88.540 ~ in^2$ (The actual area is $A = 88.8581~in^2$, so this is a good approximation.) -Dan	{"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": 14, "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": 0.932580292224884, "perplexity_flag": "head"}
http://www.acs.psu.edu/drussell/Demos/reflect/reflect.html	# Acoustics and Vibration Animations Daniel A. Russell, Graduate Program in Acoustics, The Pennsylvania State University All text and images on this page are ©1997-2013 by Daniel A. Russell and may not used in other web pages or reports without permission. The content of this page was originally posted on March 21, 1997. The HTML code was modified to be HTML5 compliant on March 18, 2013. # Reflection of Waves from Boundaries These animations were inspired in part by the figures in chapter 6 of Introduction to Wave Phenomena by A. Hirose and K. Lonngren, (J. Wiley & Sons, 1985, reprinted by Kreiger Publishing Co., 1991) When an object, like a ball, is thrown against a rigid wall it bounces back. This "reflection" of the object can be analyzed in terms of momentum and energy conservation. If the collision between ball and wall is perfectly elastic, then all the incident energy and momentum is reflected, and the ball bounces back with the same speed. If the collision is inelastic, then the wall (or ball) absorbs some of the incident energy and momentum and the ball does not bounce back with the same speed. Waves also carry energy and momentum, and whenever a wave encounters an obstacle, they are reflected by the obstacle. This reflection of waves is responsible for echoes, radar detectors, and for allowing standing waves which are so important to sound production in musical instruments. ## Wave pulse traveling on a string The animation at left shows a wave pulse travelling on a string. The speed, c, with which the wave pulse travels along the string depends on the elastic restoring force (tension, T) and inertia (mass per unit length, μ) according to $c=\sqrt{\frac{T}{\mu }}$ ## Reflection from a HARD boundary The animation at left shows a wave pulse on a string moving from left to right towards the end which is rigidly clamped. As the wave pulse approaches the fixed end, the internal restoring forces which allow the wave to propagate exert an upward force on the end of the string. But, since the end is clamped, it cannot move. According to Newton's third law, the wall must be exerting an equal downward force on the end of the string. This new force creates a wave pulse that propagates from right to left, with the same speed and amplitude as the incident wave, but with opposite polarity (upside down). • at a fixed (hard) boundary, the displacement remains zero and the reflected wave changes its polarity (undergoes a 180o phase change) ## Reflection from a SOFT boundary The animation at left shows a wave pulse on a string moving from left to right towards the end which is free to move vertically (imagine the string tied to a massless ring which slides frictionlessly up and down a vertical pole). The net vertical force at the free end must be zero. This boundary condition is mathematically equivalent to requiring that the slope of the string displacement be zero at the free end (look closely at the movie to verify that this is true). The reflected wave pulse propagates from right to left, with the same speed and amplitude as the incident wave, and with the same polarity (right-side up). • at a free (soft) boundary, the restoring force is zero and the reflected wave has the same polarity (no phase change) as the incident wave ## Reflection from an impedance discontinuity When a wave encounters a boundary which is neither rigid (hard) nor free (soft) but instead somewhere in between, part of the wave is reflected from the boundary and part of the wave is transmitted across the boundary. The exact behavior of reflection and transmission depends on the material properties on both sides of the boundary. One important property is the characteristic impedance of the material. The characteristic impedance of a material is the product of mass density and wave speed, Z= ρ c. If a wave with amplitude ξ1 in medium 1 encounters a boundary with medium 2, the amplitudes of the reflected wave is given by ${\xi }_{r}=\frac{{Z}_{1}/{Z}_{2}-1}{{Z}_{1}/{Z}_{2}+1}{\xi }_{1}$ and the amplitude of the wave transmitted into medium 2 is given by ${\xi }_{2}=\frac{2}{{Z}_{1}/{Z}_{2}+1}{\xi }_{1}$ In the animations below, two strings of different densities are connected so that they have the same tension. The density μ of the thick string is 4 times that of the thin string. • If the speed of waves on a string is related to density and tension by $c=\sqrt{\frac{T}{\mu }}$ how do the wave speeds compare for the two strings? ### From high speed to low speed (low density to high density) In this animation the incident wave is travelling from a low density (high wave speed) region towards a high density (low wave speed) region. • How do the amplitudes of the reflected and transmitted waves compare to the amplitude of the incident wave? • How do the polarities of the reflected and transmitted waves compare to the polarity of the incident wave? • How do the widths of the reflected and transmitted waves compare to the width of the incident wave? ### From low speed to high speed (high density to low density) In this animation the incident wave is travelling from a high density (low wave speed) region towards a low density (high wave speed) region. • How do the amplitudes of the reflected and transmitted waves compare to the amplitude of the incident wave? • How do the polarities of the reflected and transmitted waves compare to the polarity of the incident wave? • How do the widths of the reflected and transmitted waves compare to the width of the incident wave? Back to the Acoustics and Vibration Animations Page	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 4, "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": 0.9047350883483887, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/35562/is-a-1d-vector-also-a-scalar/35600	# Is a 1D vector also a scalar? A vector in one dimension has only one component. Can we consider it as a scalar at the same time? Why time is not a vector, although it can be negative and positive (when solving for time the kinematics equation for example)? - ## 4 Answers A scalar is defined to be invariant under transformations of the coordinate system. Thus, a vector in one dimension is not a scalar. Time is a "parameter", or a component of a 4-vector in special relativity. In classical mechanics, it is essentially a one-dimensional vector. - [I realize this in an old question, but this comment is meant for future readers.] Your first line is a bit vague. You're talking about orthogonal transformations, living in $O(1) = \{ \pm 1 \}$. Restricting as usual to $SO(1)$ leaves you with nothing but the identity transformation, so any 1D vector would indeed be a scalar. – Vibert Feb 27 at 22:28 A scalar with a unit is a 1-dimensional (axial) vector; changing the basis corresponds to changing the unit. A number (without a unit) is not a 1-dimensional vector in the terminology used by physicists. However, it is a 1-dimensional vector in the terminology used in linear algebra. - A vector in a $1D$ space is not a scalar. But if we choose a basis (which in this case consists only in one vector, say $E$), any other vector is of the form $vE$, with $v$ a scalar. So we can identify the $1D$ vector space with $\mathbb R$, but the identification depends on the choice of $E$. In the case of the time, things are similar. For the Minkowski spacetime, consider an orthonormal basis, consisting in three spacelike vectors and one timelike vector. This basis allows us to express any event in terms of three space coordinates, and one time coordinate. The time coordinate is a scalar, and can indeed be positive or negative. For curved spacetime, the things get more difficult. The spacetime is no longer a vector space. We use coordinates, which are no longer obtained from vector frames. Sometimes they can't even be global, so we have to take them local (limited to a subset of the spacetime). - The perspectives of the other answers I do not think are the correct ones. Here is one from the perspective of representation theory. Whether a quantity is a "scalar" or a "vector" (or something more exotic) is a question of what representation of the group of isometries it resides in. For n-dimensional Euclidean space, this is the group O(n). For n=1, O(n) has just the elements 1 and -1. A vector acts nontrivially under -1, while a scalar is unchanged. Speaking more broadly, we can consider antisymmetric tensor fields (sections of the exterior powers of the tangent bundle). The top exterior power, the so-called tangent frames (or if you prefer their duals, the volume forms), are in bijection with the group of scalars if our space is orientable. That is, fixing an orientation (which is a global section of this bundle) O, every other top rank tensor is of the form f(x)O for some scalar function f. If we're in Euclidean space, only the parity transformation -1 can act nontrivially on one of these. It acts trivially iff the dimension is even, so scalars are top tensor fields in even dimensions and psuedoscalars are top tensor fields in odd dimensions. - Your view is in conflict with the standard convention that allows one to talk about polar and axial vectors, which have different behavior under eflection. A scalar is an axial vector in 1D. – Arnold Neumaier Sep 5 '12 at 8:53 In my answer, polar and axial vectors correspond to 1-frames (ordinary vectors) and n-1-frames. – user404153 Sep 5 '12 at 19:30	{"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": 9, "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": 0.9064446091651917, "perplexity_flag": "head"}
http://physics.aps.org/articles/print/v2/32	# Viewpoint: Weak measurements just got stronger , H. H. Wills Physics Laboratory, University of Bristol, Bristol BS8 1TL, UK Published April 27, 2009 \| Physics 2, 32 (2009) \| DOI: 10.1103/Physics.2.32 In the weird world of quantum mechanics, looking at time flowing backwards allows us to look forward to precision measurements. In 1964 when Yakir Aharonov, Peter Bergman, and Joel Lebowitz started to think seriously about the issue of the arrow of time in quantum mechanics [1]—whether time only flows from the past to the future or also from the future to the past—none of them could have possibly imagined that their esoteric quest would one day lead to one of the most powerful amplification methods in physics. But in the weird, unpredictable, yet wonderful way in which physics works, one is a direct, logical, consequence of the other. As reported in Physical Review Letters by P. Ben Dixon, David J. Starling, Andrew N. Jordan, and John C. Howell at the University of Rochester this amplification method makes it possible to measure angles of a few hundred femtoradians and displacements of 20 femtometers, about the size of an atomic nucleus [2]. In classical mechanics if we know the initial conditions of a particle—its initial position and velocity—and the forces that act on it at all times, then we know everything about the particle. No measurement performed in the future can tell us anything new—we can compute its results from what we already know. Similarly, we don’t need to be told anything about the past—we can compute that as well. Quantum mechanically, however, the situation is dramatically different. Suppose we know the initial conditions, the wave function $\|Ψ〉$ at time $t0$, and know the dynamics (the Hamiltonian) at all times. Still, in general, we cannot predict the result of a measurement performed at a later time $t1$. We can calculate the probabilities for the different possible outcomes, but not which of them will actually occur. So the measurement at $t1$ yields new information, information that was not available at any other earlier times. All the above was well known from the early days of quantum mechanics. What Aharonov, Bergman, and Lebowitz realized, however, is that the outcome of the later measurement not only can be used for calculating the behavior of the particle for even later times $t>t1$, as it was universally considered, but also has implications about the past too. Indeed, suppose we start with an ensemble of particles, each prepared at time $t0$ in the same initial state $\|Ψ〉$. At some intermediate time $t$, where $t0<t<t1$, we subject each particle to some measurement and later, at $t1$, we perform a final measurement. We can now split the original ensemble into sub-ensembles according to the result of the final measurement. Each of these sub-ensembles is usually called a “pre- and post-selected ensemble.” The statistics of the results of the measurements at the intermediate time $t$ are, in general, different in each pre- and post-selected ensemble, and different from the statistics over the initial, pre-selected-only ensemble. So the results at time $t$ depend not only on what happened at the earlier time $t0$, but also on what happens at the later time $t1$. By pre- and post-selection one can prepare strange sub-ensembles. We could, for example, start with spin 1/2 particles polarized “up” along the $z$ direction and at $t1$ perform a measurement of the spin in the $x$ direction and select only the cases when the spin turned out to be up along $x$. Then, at any intermediate time $t$, the spin components in both the $z$ and the $x$ directions (two noncommuting observables) are completely determined—something of a quantum impossibility. But how do we know that both spin components are completely defined? Well, suppose the Hamiltonian is zero (no magnetic field) so that the spin doesn’t precess. If at time $t$ we measure the spin in the $z$ direction we must find it up—it was prepared so at $t0$. On the other hand, if instead we measure the spin along $x$, we must also find it up—otherwise the measurement at $t1$ couldn’t find it up. One may legitimately wonder, however, if this isn’t completely trivial. Indeed, we can play a similar game in classical physics as well, with any system—coins, dice, and so on, where the results are probabilistic. Selecting according to the results of a later experiment can affect the statistics of an earlier experiment, and there is no mystery here—no future affecting the past. The difference, however, is that if we would know everything at the initial time, say, exactly how the coin is thrown, the friction with the air, etc., we could predict everything at the initial moment. In quantum mechanics there is no way to do this—we need to wait until the later time. Still, could there be a more compelling argument that at intermediate times the spin is indeed up in both the $x$ and $z$ direction? About two decades after the paper by Aharonov and his colleagues, he made a breakthrough. He asked, “What if at time $t$, I measure the spin along a direction $φ$ in the $x$-$z$ plane?” Quantum mechanically, $Sφ=Sxcosφ+Szsinφ$. If it is indeed true that both $Sx=+1/2$ and $Sz=+1/2$, it must be that $Sφ=(1/2)cosφ+(1/2)sinφ$. There was a little problem though: this could yield a value larger that $1/2$, for example, $Sπ/4=2/2$, an impossible result. Indeed, the spin in any direction can only take the values of $±1/2$; Aharonov was predicting a value different from all possible eigenvalues and in fact larger that the largest of them. The idea of the future affecting the past in quantum mechanics seemed doomed. But Aharonov, together with David Albert and Lev Vaidman [3] understood where the problem was. They realized that measuring $Sπ/4$ was tantamount to simultaneously measuring the spin along the $x$ and $z$ directions, an impossible task since they would disturb each other. But one could in fact limit the mutual disturbance if one is willing to pay the penalty of allowing the measurements be less precise. They called such measurements “weak” because of the reduced disturbance and these outcomes are called “weak values.” When such measurements are performed, $Sπ/4$ does indeed take the impossible value of $2/2$. Viewed from one angle, this story is all about fundamental philosophical ideas. Does the spin indeed have a value larger than $1/2$ or is the result simply an error in the imprecise measuring device used? Does the spin indeed have both the $x$ spin component and the $z$ one well defined? And, above all, does time indeed flow in two directions in quantum mechanics? To be sure, the strange outcome of the measurement of $Sπ/4$ in this pre- and post-selected ensemble could indeed be obtained as an error in the measurement, an error in which the pointer of the measuring apparatus moved more than it should have. The explanation can be fully given by standard quantum mechanics, involving regular past-to-future-only flow of time. But the explanation is cumbersome and involves very intricate interference effects in the measuring device. Assuming that time flows in two directions tremendously simplifies the problem. As far as I can tell, Aharonov, Albert, and Vaidman hold the view that one should indeed accept this strange flow of time. I fully agree. Not everybody agrees though, and this is one of the most profound controversies in quantum mechanics. Viewed from another angle, however, what you see in the above example is an amplification effect: for all the particles in the pre-and post selected ensemble, the pointer of the measuring device moves more than usual. Forget all the philosophy; the amplification still remains. And it doesn’t even need to be quantum—the quantum interference responsible for the movement of the pointer can be mimicked by classical waves. And amplify it does. In a recent paper, Hosten and Kwiat et al. [4] use this technique to amplify the displacement of a laser beam by a factor of 10000, which allowed them to measure displacements of $1Å$ and to confirm the existence of the Hall effect for light. Dixon et al. [2] used a Sagnac interferometer (Fig. 1) with a moving mirror that can be rotated by a piezoelectric crystal. The pre-selection is done by simply sending photons into the interferometer. Once inside, a photon can move clockwise or counterclockwise; this is the quantum variable that is measured. When the mirror is tilted from 45 degrees, the clockwise and counterclockwise moving photons are deflected up or down, respectively, from the central trajectory. The deflection is the “pointer,” which registers which way (clockwise or counterclockwise) the photon went. The post-selection is made by considering only photons emerging towards the detector; this is a post-selection because photons could also exit the interferometer through the other face of the beam splitter. As in the case of the spin $1/2$ particle described above, the post-selected beam is deflected more than it would be in the absence of the post-selection. Examined in detail, this extra deflection results from an interference effect between two overlapping beams, one due to the photons that moved clockwise and one due to the counterclockwise ones. Each of them is moved only a little—up or down—with the expected “normal” shift, but they interfere destructively in most of the regions and only a lower intensity beam survives, moved off center by a much larger amount. The output beam falls on a quadrant detector that registers its displacement. The experiment looks deceptively simple. It is nothing but. The amplification obtained is only of the order of 100, much smaller than in the experiment of Hosten and Kwiat [4]. But the amplification is added on top of an arrangement that was already very sensitive due to a number of clever tricks (though not all details are given). Conceptually they have also accomplished quite a feat: the mapping of the experiment onto the theory is highly nontrivial. In particular, they used an arrangement in which the “weak value” is a purely imaginary number—yes, that is possible as well. So while the quest for understanding the flow of time continues, high-precision measurements just got better. ### References 1. Y. Aharonov, P. G. Bergmann, and J. Lebowitz, Phys. Rev 134, B1410 (1964). 2. P. B. Dixon, D. J. Starling, A. N. Jordan, and J. C. Howell, Phys. Rev. Lett. 102, 173601 (2009). 3. Y. Aharonov, D. Z. Albert, and L. Vaidman, Phys. Rev. Lett. 60, 1351 (1988). 4. O. Hosten and P. Kwiat, Science 319, 787 (2008). ### Highlighted article #### Ultrasensitive Beam Deflection Measurement via Interferometric Weak Value Amplification P. Ben Dixon, David J. Starling, Andrew N. Jordan, and John C. Howell Published April 27, 2009 \| PDF (free) ### Figures 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": 50, "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": 0.9408493041992188, "perplexity_flag": "head"}
http://crypto.stackexchange.com/questions/3229/is-the-encryption-of-a-hash-a-good-mac	# Is the encryption of a hash a good MAC? At university we were told that it is a bad idea to implement a MAC by simply concatenating a key with the data to sign and to run it through a hash function (e.g. $s = \mathrm{hash}(k\|\|\mathrm{data})$ or $s = \mathrm{hash}(\mathrm{data}\|\|k)$). The next ideas that were presented then were HMAC and CBC-MAC which are a lot more complex (but standardized). Now I'm wondering what the security of the following "idea" would be (I'm sure that there a good reasons why it is not used as it is more simple than HMAC or CBC-MAC): 1. Compute the hash value $h = \mathrm{hash}(\mathrm{data})$. 2. Compute the ciphertext $c = \mathrm{enc}_k(h) = \mathrm{enc}_k(\mathrm{hash}(\mathrm{data}))$. 3. Set signature $s = c$. Where $\mathrm{hash}$ is an arbitrary hash function and $\mathrm{enc}$ an arbitrary, symmetric block cipher using a key $k$. I was not able to find any statements about such schemes, maybe I searched in the wrong place. However, it would be nice if you could help me with that question. - 2 It certainly has worse properties compared to HMAC, thanks to low collision resistance of this scheme. In particular 128 bit MACs which are fine with HMAC, are too weak with your scheme. – CodesInChaos Jul 13 '12 at 13:33 You're also relying on the strength of the cipher for both encryption and authentication, so given a non-key-disclosing attack on the block cipher you could also inject packets. With HMAC you'd only be able to read the stream unless you could recover the key. – Polynomial Jul 13 '12 at 15:46 1 Note that hash(k\|\|data) is secure for the next generation of hash functions, including all SHA-3 finalists. – CodesInChaos Jul 13 '12 at 19:10 2 I would also question whether HMAC is considerably more complex that the present idea. Is $Hash( K_1 \| Hash (K_2 \| Data))$ really more complex than $Encrypt( Hash( Data ))$? – poncho Jul 15 '12 at 21:11 ## 2 Answers No, in general, this is not secure, unless you make additional assumptions on the encryption method beyond the standard assumption of privacy. To simplify things a bit, the assumption of privacy means that given a ciphertext $C$, the attacker has no information about what the plaintext might be. However, in your case, we don't really care if the attacker can figure out what the plaintext of the encryption function; we also give him the data, and he can compute $hash(data)$ himself, should he care to. What we are concerned with is (again, to simplify a bit) that an attacker, given a message M and a valid tag for that message, cannot come up with another message, and a valid tag for that message. Translating that into your proposal, if the attacker was given $M$, and $E(Hash(M))$, can he pick another message $M'$, and come up with $E(Hash(M'))$? Well, for a lot of encryption methods, he can. For example, if we consider a block cipher in counter mode, well, if you flip a bit in the ciphertext, the corresponding bit in the plaintext also flips. What that means that if the attacker computes $E(Hash(M)) \oplus Hash(M) \oplus Hash(M')$, well, that turns out to be precisely $E(Hash(M'))$, and so the attacker has won. The additional property that we need to assume for the encryption method is nonmalleability; that is, given $M$ and the corresponding encryption $E(M)$, the attacker cannot modify the encryption so that it decrypts to any other specific message. Of the standard encryption modes, well, ECB actually is nonmalleable, if (and this is a big if) the hash fits entirely within a single block output. Given that 128 bit hashes are vulnerable to collisions (and a hash collision would be another way of producing a forgery), this means using a nonstandard block cipher (for example, Rijndael with a 256 bit block size). Authenticated encryption modes are also nonmalleable. However, this may be considered cheating; authenticated encryption modes work by effectively using a MAC internally; if the point of the exercise is to create a crypto primitive from other crypto primitives, well, this didn't do it. - It turns out that this is actually secure, up to the length of the block cipher, if $\mathrm{enc}(\cdot)$ is a secure block cipher (a pseudorandom permutation) and if $\mathrm{hash}(\cdot)$ is collision-resistant. However, there is a catch. (You just knew there had to be one, didn't you?) The catch is that typical block ciphers have too narrow of a block width for this to be adequately secure. In other words, the catch arises when you try to work out quantitative security level afforded by this construction. For instance, let's say you use AES as your $\mathrm{enc}(\cdot)$ and SHA1 truncated to 128 bits (to match AES's 128-bit block width) as your $\mathrm{hash}(\cdot)$. Well, then you only get 64-bit security. This is vulnerable to collision-finding attacks of complexity approximately $2^{64}$. After examining about $2^{64}$ messages, you expect to find a pair of messages $m,m'$ such that $\mathrm{hash}(m) = \mathrm{hash}(m')$, through a simple birthday argument. To be secure against such attacks, you'd need a hash function and block cipher whose block width is at least 160 bits. But few modern block ciphers support such a block width -- and this was especially the case when HMAC was defined. Therefore, HMAC is a better fit for the common primitives typically available today. There is a second reason why HMAC was defined. HMAC was designed to be robust: to minimize the assumptions it makes about the hash function. In particular, the HMAC construction was designed so that HMAC would have a chance of remaining a secure MAC construction, even if someone happens to discover collision attacks on the hash function. This turned out to be a prescient design strategy. For instance, MD5-HMAC was widely used -- and then folks discovered feasible collision attacks on MD5. Fortunately, despite the fact that the collision-resistance of MD5 is totally broken, MD5-HMAC still appears to be secure: no one knows a way to break it. So, the designers of HMAC were pretty successful in making the HMAC construction resilient to certain kinds of failures of the hash function. In contrast, your construction does not have that kind of resilience. This is a second reason why one might prefer HMAC over your construction. -	{"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": 28, "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": 0.955670952796936, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/46320/terminology-for-opposite-null-lines/46349	# Terminology for opposite null lines Is there a name for two null lines that lie on the opposite sides of the null cone? Each line can be obtained from the other by reflection in the axis of the null cone (the time-axis). In terms of world-lines, this corresponds to two photons moving in the opposite directions. If there is not a standard name, what would you choose to call them as a pair? - Would "bilateral null lines" be appropriate? – Andrey Sokolov Dec 9 '12 at 0:06 – Sklivvz♦ Dec 9 '12 at 11:32 @Skliwz: I don't think geodesics is appropriate. My questions concerns the Minkowski space, which is flat. – Andrey Sokolov Dec 10 '12 at 2:55 ## 2 Answers There isn't an official standard name for opposite null lines. Note that opposite null lines are not a coordinate-independent geometric (invariant) notion, and hence it is not a very useful concept. If two null lines happen to lie on opposite sides of the light-cone in one reference frame, then they may not lie on opposite sides of the light-cone wrt. a boosted reference frame. Conversely, two different non-opposite null lines wrt. one reference frame may be boosted in such a way that they become opposite null lines wrt. the new reference frame. - Is "opposite null lines" a standard term? – Andrey Sokolov Dec 9 '12 at 1:01 1 I doubt that there is an official standard terminology for this non-geometric notion. However, for a fixed reference frame, the term opposite null lines seems to be rather natural. – Qmechanic♦ Dec 9 '12 at 1:29 If your answer is "there is no name", then you should say so clearly - if you don't know then this should be a comment, really :-) – Sklivvz♦ Dec 9 '12 at 14:22 I updated the answer. – Qmechanic♦ Dec 9 '12 at 14:48 To expand on Qmechanic's point, we can demonstrate how null directions get moved around even in flat space by Lorentz transformations: Given a Lorentz vector $X^a$ you can construct a 2x2 Hermitian complex matrix $$X^{AA'} = \frac{1}{\sqrt{2}}\left(\begin{array}{cc}X^0+X^3 & X^1+iX^2 \\ X^1-iX^2 & X^0-X^3\end{array}\right)$$ The Lorentz norm of $X^a$ is then just the determinant of this matrix. If $X^A$ is null, then $X^{AA'}$ factorizes into the product of a pair of length 2 complex vectors (spinors) $$X^{AA'}=s^A\bar{s}^{A'}$$ so the null direction is determined (up to a complex scaling factor) by the pair of complex numbers making up the spinor $s^A$. Now the space of pairs of complex numbers ($\mathbb{C}^2$) modulo a scale is just the Riemann sphere ($\mathbb{C}P_1$). Lorentz transformations preserving the norm of $X^a$ are represented by 2x2 complex matrices with unit determinant acting on $X^{AA'}$:$$X^{AA'} \rightarrow \Lambda^{A}_{\;B}X^{BB'}\bar{\Lambda}^{A'}_{\:B'}$$ These matrices, elements of $SL(2;\mathbb{C})$ act on individual spinors $$s^A \rightarrow \Lambda^{A}_{\;B} s^B$$ and since these spinors represent null directions, this is how the null directions get scrambled around by Lorentz transformations. The "opposite null directions" you would like would be antipodal points on this Riemann sphere, but these are not preserved by the general $SL(2;\mathbb{C})$ (i.e. Lorentz) transformation (although an individual specific transformation will have fixed points). The easiest way to see where points on the Riemann sphere get sent to is to think of the mapping as a Mobius transformation. Since it would have no invariant meaning to do so, there is no standard terminology for null lines corresponding to antipodal points on the Riemann sphere (which is the nearest concept I can envisage which corresponds to your "opposite null lines). - You are labouring in vain. I haven't asked whether these lines are invariant. Qmechanic already explained why there is no standard terminology, in flat spacetime, and suggested a term for these lines. You do not offer any alternatives. – Andrey Sokolov Dec 10 '12 at 3:06 – Qmechanic♦ Dec 20 '12 at 11:00	{"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": 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": 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": 0.8928354382514954, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/8580/where-can-i-find-the-measured-magnetic-moments-of-the-quarks?answertab=votes	# Where can I find the measured magnetic moments of the quarks? I read that the magnetic moment of the proton can be expressed as the sum of the magnetic moments of the three quarks like $\mu_p = \frac{4}{3}\mu_u - \frac{1}{3}\mu_d$. But I couldn't find a table listing the measured magnetic moments of the quarks. Can you point me to a reference? - The magnetic moments of the nucleons are largely not the sum of the magnetic moments of the valence quarks in any microscopic model (the "constituent quark model" that Johannes talks about is a crude tool long since abandoned for most purposes). – dmckee♦ Apr 13 '11 at 18:30 ## 2 Answers Since a quark $q_i$ is a spin $\frac{1}{2}$ Dirac particle with charge $z_i e$ and constituent mass $m_i$ with $i=u,d,s...$ theory predicts its magnetic moment as $\mu_i = \frac{z_i e \hbar}{2 m_i}$. Note that the formula in your question is only valid in the constituent quark model where a baryon consists only of three constituent quarks. Therefore you have to use the constituent quark mass in the formula for the magnetic moment too. A list of constituent quark masses is eg. found here. These masses are obtained from hadron spectroscopy. I am not aware of any direct measurements of quark magnetic moments. - Is $z_i = \pm 1/3,\pm 2/3$ ? What about the $g$-factor for the different quarks? – asmaier Apr 13 '11 at 20:55 1 Yes it is. The $g$-factor in this model is 2 for all quarks, because they are considered point particles with spin 1/2 obeying the Dirac equation (no QED). – Johannes Apr 13 '11 at 21:47 Can you point me to a reference describing the constituent quark model, where the equations for the magnetic moment according to this model are explained? – asmaier Apr 14 '11 at 12:17 You can read about this in just about every introductory particle physics book. For example the books of Griffiths (chapter 5.10) and Povh (chapter 15.4) cover it. – Johannes Apr 14 '11 at 14:37 Quarks are confined. Any "measurements" come from the measurements of seen particles, utilizing the Dirac equation, Quantum Field Theory and the symmetries of the standard model to constrain the quantities of the quarks, be it masses or moments or charges. The standard model was extremely well verified by the measurements at LEP up to the energies of LEP. -	{"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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.895664632320404, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/102595/list	## Return to Answer 3 meeting objections Here are two inequivalent binary codes with the same distance enumerator. Code A consists of the codewords $a=00000000000$, $b=11110000000$, $c=11111111100$, $d=11000110011$; code R consists of the codewords $r=0000000000$, $s=1111000000$, $t=0111111111$, $u=1000111100$. Writing $xy$ for the Hamming distance between words $x$ and $y$, we have $ab=rs=4$, $bc=ru=5$, $cd=tu=6$, $ac=rt=9$, and $ad=bd=st=su=7$, so the distance enumerators are identical. But code A has the 7-7-4 triangle $adb$, while code R has the 7-7-6 triangle $tsu$, so there is no distance-preserving bijection between the two codes. I'm sure there are shorter examples. EDIT: a much shorter example: Code A consists of the codewords $a=000$, $b=100$, $c=011$, $d=111$; code R consists of the codewords $r=0000$, $s=1000$, $t=0100$, $u=0011$. Both codes have two pairs of words at each of the distances 1, 2, and 3; R has a 1-1-2 triangle, A doesn't. MORE EDIT: In response to Jyrki's objections, here's one where both codes are of the same length, and no component is constant over all code words. $a=0000$, $b=1000$, $c=1110$, $d=1111$; $r=0000$, $s=0111$, $t=1110$, $u=1111$. Each code has two pairs at distance 1 and two at distance 3, and one each at distances 2 and 4; first code has a 1-3-4 triangle, $abd$, second code doesn't. 2 added 285 characters in body Here are two inequivalent binary codes with the same distance enumerator. Code A consists of the codewords $a=00000000000$, $b=11110000000$, $c=11111111100$, $d=11000110011$; code R consists of the codewords $r=0000000000$, $s=1111000000$, $t=0111111111$, $u=1000111100$. Writing $xy$ for the Hamming distance between words $x$ and $y$, we have $ab=rs=4$, $bc=ru=5$, $cd=tu=6$, $ac=rt=9$, and $ad=bd=st=su=7$, so the distance enumerators are identical. But code A has the 7-7-4 triangle $adb$, while code R has the 7-7-6 triangle $tsu$, so there is no distance-preserving bijection between the two codes. I'm sure there are shorter examples. EDIT: a much shorter example: Code A consists of the codewords $a=000$, $b=100$, $c=011$, $d=111$; code R consists of the codewords $r=0000$, $s=1000$, $t=0100$, $u=0011$. Both codes have two pairs of words at each of the distances 1, 2, and 3; R has a 1-1-2 triangle, A doesn't. 1 Here are two inequivalent binary codes with the same distance enumerator. Code A consists of the codewords $a=00000000000$, $b=11110000000$, $c=11111111100$, $d=11000110011$; code R consists of the codewords $r=0000000000$, $s=1111000000$, $t=0111111111$, $u=1000111100$. Writing $xy$ for the Hamming distance between words $x$ and $y$, we have $ab=rs=4$, $bc=ru=5$, $cd=tu=6$, $ac=rt=9$, and $ad=bd=st=su=7$, so the distance enumerators are identical. But code A has the 7-7-4 triangle $adb$, while code R has the 7-7-6 triangle $tsu$, so there is no distance-preserving bijection between the two codes. I'm sure there are shorter examples.	{"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": 79, "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": 0.9224819540977478, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/45899/what-are-the-coefficients-and-in-this-equation-for-decompression-induced?answertab=active	# What are the coefficients α, β, and κ in this equation for decompression-induced gas bubble growth? I am reading a text about gas nuclei I encountered the following formula: $$r = \alpha + \beta \left( \frac{T}{P} \right)^\frac{1}{3} + \kappa \left( \frac{T}{P} \right)^\frac{2}{3}$$ $r$ is the critical radius of the nucleus, while $T$ and $P$ are the current temperature and pressure respectively. I have no idea what $\alpha$, $\beta$ or $\kappa$ are. Also, I don't even known the name of such formula and the principle behind it. Can anybody explain what $\alpha$, $\beta$ or $\kappa$ are and what the underlying principle is? The text is REDUCED GRADIENT BUBBLE MODEL, page 12 equation 61. UPDATE: The problem is that I'm not able to find the values of the three constants. If I use the values from the first three rows of the table below the equation: • $T = 293 K, P = 13 fsw, r = 2.10$ • $T = 293 K, P = 33 fsw, r = 1.36$ • $T = 293 K, P = 53 fsw, r = 1.34$ I get (approximately) $\alpha = 4.395, \beta = -3.260, \kappa = 0.866$. However such values do not work for e.g. $P = 273 fsw$. If I use my constants, I get $r = 1.966$, while the table reports $r = 0.80$. Also, the table shows that $r$ decreases for $P \in [13, 273]$, my function increases approximately at $P = 73$ (where my $r$ is 1.403 against 1.32). I think $\alpha$, $\beta$ and $\kappa$ must be functions of pressure. The Wikipedia page about Equation of State shows $\alpha$ and $\kappa$ as functions of pressure and temperature, but there are no traces of $\beta$. - 3 It would help it you could tell us what the title of this text is. My first thought is that alpha, beta, and kappa are empirical coefficients, with no particular meaning otherwise. – AlanSE Dec 4 '12 at 15:03 Comment to the question(v6): It would be great if OP (or someone else) could create a more substantive title. – Qmechanic♦ Dec 5 '12 at 14:59 @Qmechanic: As shown in my answer below, it is an ordinary virial equation of state truncated at quadratic order and applied to bubbles (spherical symmetry) with reparametrization to critical radius. I do not have edit privileges. – juanrga Dec 6 '12 at 12:15 @juanrga: My comment refers to a previous version(v6) of the title. The title is okay now(v8). – Qmechanic♦ Dec 6 '12 at 12:20 @Qmechanic: Yes, it is ok now, but when I submitted my comment I saw still the old title "What's the meaning of this equation?". – juanrga Dec 6 '12 at 13:03 ## 3 Answers Setting $Y \equiv (T/P)^\frac{1}{3}$ the expression looks as a virial equation of state in a power series $$r = \alpha + \beta Y + \kappa Y^2$$ with $\alpha$, $\beta$, and $\kappa$ the coefficients on the expansion. The meaning of the coefficients is the usual: e.g., $\alpha$ is the radius when $Y=0$... EDIT: I have found a paper by the same authors confirming that the above is a virial equation of state truncated to quadratic order. They start from the ordinary virial equation of state at arbitrary order and substitute volume for systems with spherical symmetry $V=(4\pi/3) r^3$. Then take the cubic root and obtain (I am using their own notation in this paper) $$r = \sum_{i=0}\beta_i \left( \frac{nRT}{P} \right)^{i/3}$$ Next, they truncate the expansion to quadratic order $$r = \beta_0 + \beta_1 \left( \frac{nRT}{P} \right)^{1/3} + \beta_2 \left( \frac{nRT}{P} \right)^{2/3}$$ re-parametrize from total bubble radius to excitation radius and rewrite the expression by introducing the $nR$ factors into new $\alpha$, $\beta$, and $\kappa$, which they again name "virial coeficients". - -1: The linked text says it is valid for all $T$, that is generally not true for power series. Do you have any proof for your claim? – Bernhard Dec 5 '12 at 11:39 @Bernhard: Thank you for the negative point, but you are completely wrong. It is trivial to see that the above is a virial equation truncated up to quadratic order, but in any case I have found a paper by the same authors confirming what I said in my answer. I am editing it. – juanrga Dec 6 '12 at 11:40 Thanks for the additional information. Makes things clear! – Bernhard Dec 6 '12 at 12:14 @Bernhard: Thank you very much by undo the negative point! – juanrga Dec 6 '12 at 13:11 That's beautiful! I haven't tried finding the betas (to see if everything fits) however it makes much more sense now. Many thanks also for providing the link to the paper! – user16538 Dec 7 '12 at 18:41 show 1 more comment Without reading the entire paper, it looks like the authors are exploring, among other things, the growth of gas bubbles resulting from decompression in the context of underwater diving. More specifically, an underwater diver experiences an additional atmosphere of pressure for each 34 feet she descends below the surface of the water. When its time to swim back to the surface, the diver must be very careful not to ascend too quickly or else she might experience Decompression Sickness. Essentially, dissolved gases in the body can start to bubble out as the pressure decreases, causing potentially serious health issues. The process of forming bubbles is called "nucleation". The equation you asked about gives the minimum nuclei radius such that the bubble will grow in size as the pressure is decreased from $P$ at temperature $T$. According to the paper you provided, the greek letters are "Equation of State skin constants" which probably means (as AlanSE said) they are experimentally determined coefficients...nothing profound. The equation is useful because one must have information about bubble growth in order to figure out a safe rate of ascent to the surface. The paper contains additional modeling that further develops the analysis of safe ascent. It does not seem that the authors have named the formula. EDIT Sorry, I think I missed the intent of your question. You probably already know what I wrote! Anyway, since you're trying to find the skin constants I put the values from Table 5 into MS Excel and did a second order polynomial fit at $T=293K$ and got ${\alpha}=0.458$, ${\beta}=0.465$, ${\kappa}=0.034$. The R-Squared value is $0.90$. I'm not sure how accurate you need this to be...if you know the temperature/depth profile of the water you are modeling then we could improve the fit. I know you'd prefer to have the constants as functions of temperature and pressure but unfortunately I couldn't find that info. Maybe someone else will have more success doing the research. Hopefully this rough estimate helps somewhat. - As stated in your text, $\alpha, \beta$, and $\kappa$ are equation-of-state constants, or in otherwords are constants based on the pressure, temperature, volume, etc of the substance being dealt with. Wikipedia has some good background information and examples on how some similar values are calculated.	{"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": 47, "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": 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": 0.9423153400421143, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/tagged/diagnostic+logistic	# Tagged Questions 1answer 63 views ### How do I compute a cutoff based on sensitivity/specificity when the characteristics of my sample is different from the population? I have a dataset containing the performance of a novel instrument to screen for disease A. The novel instrument uses a scoring system to score the subject to determine if they have disease A. I then ... 2answers 533 views ### Residuals for Logistic Regression and Cooks Distance 1) Are there any particular assumptions regarding the errors for logistic regression such as the constant variance of the error terms and the normality of the residuals? 2) Also typically when you ... 1answer 184 views ### Logistic regression and complementary log log model is there like a diagnostic checking(like those in the time series) done for logistic regression and complementary loglog model? 1answer 814 views ### Explanation of R diagnostic plot for logistic regression I'm hoping someone can explain this bit of R code for me related to glm(). I don't understand the diagnostic plot that has been suggested. It seems a more ... 0answers 138 views ### “Brute force” expected deviance for logistic regression? A commonly used goodness of fit statistic for logistic regression is the deviance. This is also known as the likelihood ratio chi-square statistic. It is defined as: $$D=\sum_{i=1}^{N}d_i^2$$ ... 2answers 7k views ### Likelihood ratio test in R I'm not sure if I'm asking something stupid or off topic here, but I can't think where can I ask this question. suppose I am going to do a univariate logistic regression on several independent ...	{"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": 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": 0.9403715133666992, "perplexity_flag": "middle"}
http://gafferongames.com/virtualgo/the-shape-of-the-go-stone/	# The Shape Of The Go Stone February 9, 2013 Hello, I’m Glenn Fiedler and welcome to Virtual Go, my project to simulate a Go board and stones. Our first goal is to develop a physically accurate simulation of the motion of a go stone. The first step towards this goal is to study and understand the shape of a go stone in detail. If you’ve ever played go, you know that a go stone has a particularly interesting wobble when placed on the board. This wobble is a direct consequence of its unique shape. It’s very important to understand this shape so we can simulate it properly. So lets get started! ## Slate And Shell In Japan, Go stones are traditionally made out of slate and clam shell. Clam shell stones come in several grades of quality the highest being yuki or “snow” grade with fine regularly spaced lines. Go stones also come in various sizes, the choice of which is mostly personal preference. The thicker the stone the more expensive it is as only a small portion of the clam shell is suitable. At first the go stone seems like an ellipse, but side-on you can see that it is not. Instead it is a very interesting shape called a biconvex solid. This shape is interesting because it is the intersection of two spheres. We can study this shape in 2D by looking at the intersection of two circles: By varying the radius of the circle and how far apart the center of the circles are, we can generate go stones of different sizes. This is interesting, but when I’m creating a go stone I don’t really want it to be parameterized this way. Instead I’d like to say, “Hey, I want a stone this width and this height” and have a function that calculates what the radius of the circles are and how far apart they should be. In order to write this function we need to do some math: First notice that the point Q lies on the generating circle, so the line CQ has length r: $d + h/2 = r$ $d = r - h/2$ The point P is also on the generating circle so the green line CP has length r as well. Using Pythagoras theorem and substituting for d: $r^2 = d^2 + (w/2)^2$ $r^2 = ( r - h/2 )^2 + (w/2)^2$ $r^2 = ( h^2/4 - hr + r^2 ) + w^2/4$ $r^2 = h^2/4 - hr + r^2 + w^2/4$ $0 = h^2/4 - hr + 0 + w^2/4$ $hr = h^2/4 + w^2/4$ $r = ( h^2 + w^2 ) / 4h$ Which gives us everything we need to write the function: void calculateBiconvex( float w, float h, float & r, float & d ) { r = ( ww + hh ) / ( 4h ); d = r - h/2; } Now that we can mathematically define a go stone parameterized by its width and height, there is just one problem: the edge is very sharp! In order for our our stone to be aesthetically pleasing we need to round off the edge with a bevel. Lets parameterize the bevel by its height b: In three dimensions the bevel is actually a torus (donut) around the edge of the go stone. We need to calculate the major and minor radii r1 and r2 of the torus as a function of b and the dimensions of the go stone: The key to solving this is to realize that if the go stone and the bevel are to match perfectly then the tangent of the two circles must be equal at the point P. If the tangent is equal then the normal must be equal as well. This means that the center of the bevel circle lies at the intersection of the line CP and the x axis. We already know C so if we can find the point P then we can find this intersection point. Once we know the intersection point we can find r1 and r2. Since P is at the start of the bevel: $P_y = b/2$ Because P lies on the biconvex circle with center C and radius r we can use the equation of the circle to find x as a function of y: $x^2 + y^2 = r^2$ $x = \sqrt{ y^2 - r^2 }$ We need y relative to the circle center C, not in go stone coordinates, so we add d and substitute y’ for y: $y' = b/2 + d$ $P_x = \sqrt{ ( b/2 + d )^2 + r^2 }$ We can now find r1 by similar triangles: $r_1/P_x = d / ( d + b/2 )$ $r_1 = P_x d / ( d + b/2 )$ and q by Pythagoras theorem: $q^2 = d^2 + r_1^2$ $q = \sqrt{ d^2 + r_1^2 }$ Because line CP has length r and substituting for q: $q + r_2 = r$ $r_2 = r - q$ $r_2 = r - \sqrt{ d^2 + r_1^2 }$ Now we have everything we need to write the function: void calculateBevel( float r, float d, float b, float & r1, float & r2 ) { const float y = b/2 + d; const float px = sqrt( yy + rr ); r1 = px d / ( d + b/2 ); r2 = r - sqrt( dd + r1r1 ); } An important note: we’re only going to apply this bevel to the go stone that we render. All physics calculations will use the biconvex solid with sharp edges. Why? Well, the bevel is very small and it has very little effect on how the stone moves. Plus, seeing as how the go stone typically spends most of it’s time wobbling about on the sphere surface cutting a small corner here (or not) seems a perfectly good approximation. Using approximate collision detection like this is actually quite common in video games. For example, a highly detailed triangle mesh is typically approximated with one or more convex shapes because doing exact triangle-triangle collision detection is very slow. In our case, it’s a wise trade-off because it simplifies the mathematics. # Next: Tessellating The Go Stone Donations offset hosting costs and encourage me to write more articles! { 4 comments… read them below or add one } Tajger I have a question – I want to make a go board in Little Big Planet but I have problems with LOGIC of surrounding go stones. If you could describe this I would be happy Glenn Fiedler Sorry I don’t have a copy of that game anymore	{"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": 21, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9261761903762817, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/179989/liar-functions-is-this-a-known-concept	# “Liar functions”: is this a known concept? While reading this comment and thinking about how you could change the functions without convergence (because the Lebesgue integral doesn't care about changes at countably many places), I just arrived at a concept which could be called "liar functions", for reasons which should be obvious after reading this. The concept is based on the known fact that the set of all real numbers you can name (no matter in which way) is countable, because the number of names is countable. Therefore you can change a function in all "named" positions, without changing its integral (or any other property which is only dependent on values "almost everywhere). Let $N\subset\mathbb R$ be the set of named values (that is, the set of values you can uniquely specify). Then I call a "liar function" a function $f:\mathbb R\to \mathbb R$ which has the following properties: • The restriction of $f$ to $N$ has an obvious extension $f_1$ to all of $\mathbb R$. • The restriction of $f$ to $\mathbb R\setminus N$ also has an obvious extension $f_2$ to $\mathbb R$. • $f_1$ and $f_2$ are significantly different. As a simple example, consider the function $$f(x)=\cases{1 & for $x$ in $N$\\ 0 & otherwise}$$ This function simply lies about its value: Whenever you evaluate it at a position you can specify, it gives $1$. However, it is actually $0$ almost everywhere, except where you can look. Another function, which lies more subtly, would be the function $$g(x)=\cases{1/q & for $x=p/q\in \mathbb{Q}$, $\operatorname{lcd}(p,q)=1$\\ 0 & for $x\in N\setminus \mathbb Q$\\\frac{1}{1+x^2} & otherwise}$$ This on named values looks like the well known function which is continuous on all irrational numbers but discontinuous at all rational numbers. However actually it equals the Lorentz function almost everywhere (which e.g. means its integral over $\mathbb R$ exists, but is not $0$ as one might infer if one onle knows it on named values), and it is actually nowhere continuous (but almost everywhere equal to a continuous function). Now my question: Is this concept of "liar functions" (probably under another name) already known? Also, does there exist a liar function which is discontinuous in any $x\in N$, but continuous for any $x\notin N$? - I also know little on this subject, but your 'known number' seems to mean the concept of computable number. According to Wikipedia, there has been some attempts to develop analysis based on the field of computable numbers, so it may be helpful to search in these literature. Finally, I guess that the function $f(x)$ that assigns the reciprocal of the minimal Gödel number representing $x$ for computable $x$, and $0$ otherwise seems work for you last question. – sos440 Aug 7 '12 at 16:38 This isn't quite the same as the computable numbers. For instance, Chaitin's constant $\Omega$ is nameable but not computable. I think you'd need to fix a language in which you're doing your naming before starting to ask questions like the last one. Once you've done that, as long as your language has at most countably many symbols, you might try defining a function that's $1/n$ at numbers with $n$-symbol names and 0 at unnameables. But the continuity argument seems tricky. – Kevin Carlson Aug 7 '12 at 16:48 – celtschk Aug 7 '12 at 16:50 ... define your number as "0." + the concatenation of all programs that always terminate, in lexicographic order. That's a well-defined number, but you cannot compute it (because that would, by design, involve solving the halting problem). – celtschk Aug 7 '12 at 16:52 The concept of "nameable numbers" is problematic on its face. Let $x$ be the smallest positive integer not nameable in $100$ characters or less. (Uh-oh.) – mjqxxxx Aug 7 '12 at 16:53 show 11 more comments ## 1 Answer For any sequence of reals $r_1,r_2,\dots$, there is a function that is discontinuous precisely at elements of this sequence. For example, define $f(x)$ to be the sum of $2^{-n}$ over all $n$ such that $r_n<x$. - Nice construction. That definitely answers my second question, thank you. – celtschk Aug 7 '12 at 17:05	{"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": 32, "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": 0.9462786316871643, "perplexity_flag": "head"}
http://www.sagemath.org/doc/prep/Advanced-2DPlotting.html	# Tutorial for Advanced 2d Plotting¶ This Sage document is one of the tutorials developed for the MAA PREP Workshop “Sage: Using Open-Source Mathematics Software with Undergraduates” (funding provided by NSF DUE 0817071). It is licensed under the Creative Commons Attribution-ShareAlike 3.0 license (CC BY-SA). Thanks to Sage’s integration of projects like matplotlib, Sage has comprehensive two-dimensional plotting capabilities. This worksheet consists of the following sections: This tutorial assumes that one is familiar with the basics of Sage, such as evaluating a cell by clicking the “evaluate” link, or by pressing Shift-Enter (hold down Shift while pressing the Enter key). ## Graphing Functions and Plotting Curves¶ ### Cartesian Plots¶ A simple quadratic is easy. ```sage: plot(x^2, (x,-2,2)) ``` You can combine “plot objects” by adding them. ```sage: regular = plot(x^2, (x,-2,2), color= 'purple') sage: skinny = plot(4x^2, (x,-2,2), color = 'green') sage: regular + skinny ``` Problem : Plot a green $$y=\sin(x)$$ together with a red $$y=2\,\cos(x)$$. (Hint: you can use pi as part of your range.) Boundaries of a plot can be specified, in addition to the overall size. ```sage: plot(1+e^(-x^2), xmin=-2, xmax=2, ymin=0, ymax=2.5, figsize=10) ``` Problem : Plot $$y=5+3\,\sin(4x)$$ with suitable boundaries. You can add lots of extra information. ```sage: exponential = plot(1+e^(-x^2), xmin=-2, xmax=2, ymin=0, ymax=2.5) sage: max_line = plot(2, xmin=-2, xmax=2, linestyle='-.', color = 'red') sage: min_line = plot(1, xmin=-2, xmax=2, linestyle=':', color = 'red') sage: exponential + max_line + min_line ``` You can fill regions with transparent color, and thicken the curve. This example uses several options to fine-tune our graphic. ```sage: exponential = plot(1+e^(-x^2), xmin=-2, xmax=2, ymin=0, ymax=2.5, fill=0.5, fillcolor='grey', fillalpha=0.3) sage: min_line = plot(1, xmin=-2, xmax=2, linestyle='-', thickness= 6, color = 'red') sage: exponential + min_line ``` ```sage: sum([plot(x^n,(x,0,1),color=rainbow(5)[n]) for n in [0..4]]) ``` Problem : Create a plot showing the cross-section area for the following solid of revolution problem: Consider the area bounded by $$y=x^2-3x+6$$ and the line $$y=4$$. Find the volume created by rotating this area around the line $$y=1$$. ### Parametric Plots¶ A parametric plot needs a list of two functions of the parameter; in Sage, we use square brackets to delimit the list. Notice also that we must declare t as a variable first. Because the graphic is slightly wider than it is tall, we use the aspect_ratio option (such options are called keywords ) to ensure the axes are correct for how we want to view this object. ```sage: t = var('t') sage: parametric_plot([cos(t) + 3 cos(t/9), sin(t) - 3 * sin(t/9)], (t, 0, 18pi), fill = True, aspect_ratio=1) ``` Problem : These parametric equations will create a hypocycloid. $x(t)=17\cos(t)+3\cos(17t/3)$ $y(t)=17\sin(t)-3\sin(17t/3)$ Create this as a parametric plot. Sage automatically plots a 2d or 3d plot, and a curve or a surface, depending on how many variables and coordinates you specify. ```sage: t = var('t') sage: parametric_plot((t^2,sin(t)), (t,0,pi)) ``` ```sage: parametric_plot((t^2,sin(t),cos(t)), (t,0,pi)) ``` ```sage: r = var('r') sage: parametric_plot((t^2,sin(rt),cos(rt)), (t,0,pi),(r,-1,1)) ``` ### Polar Plots¶ Sage can also do polar plots. ```sage: polar_plot(2 + 2cos(x), (x, 0, 2pi), color=hue(0.5), thickness=4) ``` Although they aren’t essential, many of these examples try to demonstrate things like coloring, fills, and shading to give you a sense of the possibilities. More than one polar curve can be specified in a list (square brackets). Notice the automatic graded shading of the fill color. ```sage: t = var('t') sage: polar_plot([cos(4t) + 1.5, 0.5 * cos(4t) + 2.5], (t, 0, 2pi),\ ... color='black', thickness=2, fill=True, fillcolor='orange') ``` Problem: Create a plot for the following problem. Find the area that is inside the circle $$r=2$$, but outside the cardiod $$2+2\cos(\theta)$$. ### Interactive Demonstration¶ It may be of interest to see all these things put together in a very nice pedagogical graphic. Even though this is fairly advanced, and so you may want to skip the code, it is not as difficult as you might think to put together. ```sage: html('<h2>Sine and unit circle (by Jurgis Pralgauskis)</h2> inspired by <a href="http://www.youtube.com/watch?v=Ohp6Okk_tww&feature=related">this video</a>' ) sage: # http://www.sagemath.org/doc/reference/sage/plot/plot.html sage: radius = 100 # scale for radius of "unit" circle sage: graph_params = dict(xmin = -2radius, xmax = 360, ... ymin = -(radius+30), ymax = radius+30, ... aspect_ratio=1, ... axes = False ... ) sage: def sine_and_unit_circle( angle=30, instant_show = True, show_pi=True ): ... ccenter_x, ccenter_y = -radius, 0 # center of cirlce on real coords ... ... sine_x = angle # the big magic to sync both graphs :) ... current_y = circle_y = sine_y = radius sin(anglepi/180) ... circle_x = ccenter_x + radius cos(anglepi/180) ... graph = Graphics() ... # we'll put unit circle and sine function on the same graph ... # so there will be some coordinate mangling ;) ... # CIRCLE ... unit_circle = circle((ccenter_x, ccenter_y), radius, color="#ccc") ... # SINE ... x = var('x') ... sine = plot( radius sin(xpi/180) , (x, 0, 360), color="#ccc" ) ... graph += unit_circle + sine ... # CIRCLE axis ... # x axis ... graph += arrow( [-2radius, 0], [0, 0], color = "#666" ) ... graph += text("$(1, 0)$", [-16, 16], color = "#666") ... # circle y axis ... graph += arrow( [ccenter_x,-radius], [ccenter_x, radius], color = "#666" ) ... graph += text("$(0, 1)$", [ccenter_x, radius+15], color = "#666") ... # circle center ... graph += text("$(0, 0)$", [ccenter_x, 0], color = "#666") ... # SINE x axis ... graph += arrow( [0,0], [360, 0], color = "#000" ) ... ... # let's set tics ... # or http://aghitza.org/posts/tweak_labels_and_ticks_in_2d_plots_using_matplotlib/ ... # or wayt for http://trac.sagemath.org/sage_trac/ticket/1431 ... # ['$-\pi/3$', '$2\pi/3$', '$5\pi/3$'] ... for x in range(0, 361, 30): ... graph += point( [x, 0] ) ... angle_label = ". $%3d^{\circ}$ " % x ... if show_pi: angle_label += " $(%s\pi) $"% x/180 ... graph += text(angle_label, [x, 0], rotation=-90, ... vertical_alignment='top', fontsize=8, color="#000" ) ... # CURRENT VALUES ... ... # SINE -- y ... graph += arrow( [sine_x,0], [sine_x, sine_y], width=1, arrowsize=3) ... graph += arrow( [circle_x,0], [circle_x, circle_y], width=1, arrowsize=3) ... graph += line(([circle_x, current_y], [sine_x, current_y]), rgbcolor="#0F0", linestyle = "--", alpha=0.5) ... ... # LABEL on sine ... graph += text("$(%d^{\circ}, %3.2f)$"%(sine_x, float(current_y)/radius), [sine_x+30, current_y], color = "#0A0") ... # ANGLE -- x ... # on sine ... graph += arrow( [0,0], [sine_x, 0], width=1, arrowsize=1, color='red') ... # on circle ... graph += disk( (ccenter_x, ccenter_y), float(radius)/4, (0, anglepi/180), color='red', fill=False, thickness=1) ... graph += arrow( [ccenter_x, ccenter_y], [circle_x, circle_y], ... rgbcolor="#cccccc", width=1, arrowsize=1) ... ... ... if instant_show: ... show (graph, graph_params) ... return graph sage: ##################### sage: # make Interaction sage: ###################### sage: @interact sage: def _( angle = slider([0..360], default=30, step_size=5, ... label="Pasirinkite kampą: ", display_value=True) ): ... ... sine_and_unit_circle(angle, show_pi = False) ``` ## Plotting Data¶ ### Plotting Data Points¶ Sometimes one wishes to simply plot data. Here, we demonstrate several ways of plotting points and data via the simple approximation to the Fibonacci numbers given by $F_n=\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^n\; ,$ which is quite good after about $$n=5$$. First, we notice that the Fibonacci numbers are built in. ```sage: fibonacci_sequence(6) <generator object fibonacci_sequence at ...> ``` ```sage: list(fibonacci_sequence(6)) [0, 1, 1, 2, 3, 5] ``` The enumerate command is useful for taking a list and coordinating it with the counting numbers. ```sage: list(enumerate(fibonacci_sequence(6))) [(0, 0), (1, 1), (2, 1), (3, 2), (4, 3), (5, 5)] ``` So we just define the numbers and coordinate pairs we are about to plot. ```sage: fibonacci = list(enumerate(fibonacci_sequence(6))) sage: f(n)=(1/sqrt(5))((1+sqrt(5))/2)^n sage: asymptotic = [(i, f(i)) for i in range(6)] sage: fibonacci [(0, 0), (1, 1), (2, 1), (3, 2), (4, 3), (5, 5)] sage: asymptotic [(0, 1/5sqrt(5)), (1, 1/10(sqrt(5) + 1)sqrt(5)), (2, 1/20(sqrt(5) + 1)^2sqrt(5)), (3, 1/40(sqrt(5) + 1)^3sqrt(5)), (4, 1/80(sqrt(5) + 1)^4sqrt(5)), (5, 1/160(sqrt(5) + 1)^5sqrt(5))] ``` Now we can plot not just the two sets of points, but also use several of the documented options for plotting points. Those coming from other systems may prefer list_plot. ```sage: fib_plot=list_plot(fibonacci, color='red', pointsize=30) sage: asy_plot = list_plot(asymptotic, marker='D',color='black',thickness=2,plotjoined=True) sage: show(fib_plot+asy_plot, aspect_ratio=1) ``` Other options include line, points, and scatter_plot. Having the choice of markers for different data is particularly helpful for generating publishable graphics. ```sage: fib_plot=scatter_plot(fibonacci, facecolor='red', marker='o',markersize=40) sage: asy_plot = line(asymptotic, marker='D',color='black',thickness=2) sage: show(fib_plot+asy_plot, aspect_ratio=1) ``` ## Contour-type Plots¶ ### Contour Plots¶ Contour plotting can be very useful when trying to get a handle on multivariable functions, as well as modeling. The basic syntax is essentially the same as for 3D plotting - simply an extension of the 2D plotting syntax. ```sage: f(x,y)=y^2+1-x^3-x sage: contour_plot(f, (x,-pi,pi), (y,-pi,pi)) ``` We can change colors, specify contours, label curves, and many other things. When there are many levels, the colorbar keyword becomes quite useful for keeping track of them. Notice that, as opposed to many other options, it can only be True or False (corresponding to whether it appears or does not appear). ```sage: contour_plot(f, (x,-pi,pi), (y,-pi,pi),colorbar=True,labels=True) ``` This example is fairly self-explanatory, but demonstrates the power of formatting, labeling, and the wide variety of built-in color gradations (colormaps or cmap). The strange-looking construction corresponding to label_fmt is a Sage/Python data type called a dictionary , and turns out to be useful for more advanced Sage use; it consists of pairs connected by a colon, all inside curly braces. ```sage: contour_plot(f, (x,-pi,pi), (y,-pi,pi), contours=[-4,0,4], fill=False,\ ... cmap='cool', labels=True, label_inline=True, label_fmt={-4:"low", 0:"medium", 4: "hi"}, label_colors='black') ``` Implicit plots are a special type of contour plot (they just plot the zero contour). ```sage: f(x,y) -x^3 + y^2 - x + 1 ``` ```sage: implicit_plot(f(x,y)==0,(x,-pi,pi),(y,-pi,pi)) ``` A density plot is like a contour plot, but without discrete levels. ```sage: density_plot(f, (x, -2, 2), (y, -2, 2)) ``` Sometimes contour plots can be a little misleading (which makes for a great classroom discussion about the problems of ignorantly relying on technology). Here we combine a density plot and contour plot to show even better what is happening with the function. ```sage: density_plot(f,(x,-2,2),(y,-2,2))+contour_plot(f,(x,-2,2),(y,-2,2),fill=False,labels=True,label_inline=True,cmap='jet') ``` It can be worth getting familiar with the various options for different plots, especially if you will be doing a lot of them in a given worksheet or pedagogical situation. Here are the options for contour plots. • They are given as an “attribute” - no parentheses - of the contour_plot object. • They are given as a dictionary (see the programming tutorial). ```sage: contour_plot.options {'labels': False, 'linestyles': None, 'region': None, 'axes': False, 'plot_points': 100, 'linewidths': None, 'colorbar': False, 'contours': None, 'aspect_ratio': 1, 'legend_label': None, 'frame': True, 'fill': True} ``` Let’s change it so that all future contour plots don’t have the fill. That’s how some of us might use them in a class. We’ll also check that the change happened. ```sage: contour_plot.options["fill"]=False sage: contour_plot.options {'labels': False, 'linestyles': None, 'region': None, 'axes': False, 'plot_points': 100, 'linewidths': None, 'colorbar': False, 'contours': None, 'aspect_ratio': 1, 'legend_label': None, 'frame': True, 'fill': False} ``` And it works! ```sage: contour_plot(f,(x,-2,2),(y,-2,2)) ``` We can always access the default options, of course, to remind us. ```sage: contour_plot.defaults() {'labels': False, 'linestyles': None, 'region': None, 'axes': False, 'plot_points': 100, 'linewidths': None, 'colorbar': False, 'contours': None, 'aspect_ratio': 1, 'legend_label': None, 'frame': True, 'fill': True} ``` ### Vector fields¶ The syntax for vector fields is very similar to other multivariate constructions. Notice that the arrows are scaled appropriately, and colored by length in the 3D case. ```sage: var('x,y') (x, y) sage: plot_vector_field((-y+x,yx),(x,-3,3),(y,-3,3)) ``` ```sage: var('x,y,z') (x, y, z) sage: plot_vector_field3d((-y,-z,x), (x,-3,3),(y,-3,3),(z,-3,3)) ``` 3d vector field plots are ideally viewed with 3d glasses (right-click on the plot and select “Style” and “Stereographic”) ### Complex Plots¶ We can plot functions of complex variables, where the magnitude is indicated by the brightness (black is zero magnitude) and the argument is indicated by the hue (red is a positive real number). ```sage: f(z) = exp(z) #z^5 + z - 1 + 1/z sage: complex_plot(f, (-5,5),(-5,5)) ``` ### Region plots¶ These plot where an expression is true, and are useful for plotting inequalities. ```sage: region_plot(cos(x^2+y^2) <= 0, (x, -3, 3), (y, -3, 3),aspect_ratio=1) ``` We can get fancier options as well. ```sage: region_plot(sin(x)sin(y) >= 1/4, (x,-10,10), (y,-10,10), incol='yellow', bordercol='black', borderstyle='dashed', plot_points=250,aspect_ratio=1) ``` Remember, what command would give full information about the syntax, options, and examples? ## Miscellaneous Plot Information¶ ### Builtin Graphics Objects¶ Sage includes a variety of built-in graphics objects. These are particularly useful for adding to one’s plot certain objects which are difficult to describe with equations, but which are basic geometric objects nonetheless. In this section we will try to demonstrate the syntax of some of the most useful of them; for most of the the contextual (remember, append ?) help will give more details. #### Points¶ To make one point, a coordinate pair suffices. ```sage: point((3,5)) ``` It doesn’t matter how multiple point are generated; they must go in as input via a list (square brackets). Here, we demonstrate the hard (but naive) and easy (but a little more sophisticated) way to do this. ```sage: f(x)=x^2 sage: points([(0,f(0)), (1,f(1)), (2,f(2)), (3,f(3)), (4,f(4))]) ``` ```sage: points([(x,f(x)) for x in range(5)]) ``` Sage tries to tell how many dimensions you are working in automatically. ```sage: f(x,y)=x^2-y^2 sage: points([(x,y,f(x,y)) for x in range(5) for y in range(5)]) ``` #### Lines¶ The syntax for lines is the same as that for points, but you get... well, you get connecting lines too! ```sage: f(x)=x^2 sage: line([(x,f(x)) for x in range(5)]) ``` #### Balls¶ Sage has disks and spheres of various types available. Generally the center and radius are all that is needed, but other options are possible. ```sage: circle((0,1),1,aspect_ratio=1) ``` ```sage: disk((0,0), 1, (pi, 3pi/2), color='yellow',aspect_ratio=1) ``` There are also ellipses and various arcs; see the full plot documentation. #### Arrows¶ ```sage: arrow((0,0), (1,1)) ``` #### Polygons¶ Polygons will try to complete themselves and fill in the interior; otherwise the syntax is fairly self-evident. ```sage: polygon([[0,0],[1,1],[1,2]]) ``` #### Text¶ In 2d, one can typeset mathematics using the text command. This can be used to fine-tune certain types of labels. Unfortunately, in 3D the text is just text. ```sage: text('$\int_0^2 x^2\, dx$', (0.5,2))+plot(x^2,(x,0,2),fill=True) ``` ### Saving Plots¶ We can save 2d plots to many different formats. Sage can determine the format based on the filename for the image. ```sage: p=plot(x^2,(x,-1,1)) sage: p ``` For testing purposes, we use the Sage standard temporary filename; however, you could use any string for a name that you wanted, like "my_plot.png". ```sage: name = tmp_filename() # this is a string sage: png_savename = name+'.png' sage: p.save(png_savename) ``` ```sage: pdf_savename = name+'.pdf' sage: p.save(pdf_savename) ``` Notably, we can export in formats ready for inclusion in web pages. ```sage: svg_savename = name+'.svg' sage: p.save(svg_savename) ``` ### Table Of Contents #### Previous topic Sage Introductory Programming Tutorial ### Quick search Enter search terms or a module, class or function name.	{"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": 9, "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": 0.8881027102470398, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/18197?sort=newest	## restriction of a representation of GL(n) to GL(n-1) ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $R$ be real numbers and consider an irreducible unitary representation (\pi,V) of $GL_n(R)$ in some Hilbert space $V$, now $GL_{n-1}(R)$ embeds in $GL_n(R)$ on the left upper diagonal block. Now I wanna ask properties of the restrictions of $V$ to $GL_{n-1}$. This representation is not irreducible, so we may ask is it possible to say something about the closed invariant subspaces? Does this restriction have a multiplicity free decomposition? Note that we can ask the same question for other irreducible Banach or Frechet representation of $GL_n$, or replacing real numbers by complex or p-adic numbers. - In the case of $GL_n$ of a $p$-adic field, this question is related to the theory of derivatives, due to Bernstein and Zelevinsky. You may want to look at their papers on the representation theory of $p$-adic $GL_n$ (which are beautiful in any case). – Emerton Mar 14 2010 at 19:23 ## 2 Answers Though Peter's answer addresses the finite-dimensional representation theory, I believe that the question asks about the unitary representations on Hilbert spaces, and more general irreps on Banach and Frechet spaces. This question has been the subject of much recent work by Avraham Aizenbud, Dmitry Gourevitch, Steve Rallis, Gerard Schiffmann, and Eitan Sayag. In particular, Aizenbud and Gourevitch prove the following in their paper "Multiplicity One Theorem for $(GL_{n+1}(R), GL_n(R))$": Let $F = R$ or $F = C$. Let $\pi$ and $\tau$ be irreducible admissible smooth Fr\'echet representations of $GL_{n+1}(F)$ and $GL_n(F)$, respectively. Then $$dim \left( Hom_{GL_n(F)}(\pi, \tau) \right) \leq 1.$$ This paper is on the ArXiv, and now published in Selecta, according to Aizenbud's webpage. Zhu and Binyong have also proved this, I believe. The result has also been proven for irreducible smooth repreesentations of $GL_{n+1}(F)$ and $GL_n(F)$, when $F$ is a $p$-adic field by Aizenbud-Gourevitch-Rallis-Schiffmann. Considering the smooth Fr\'echet case should suffice for the case of unitary representations on Hilbert spaces, I believe, by considering the subspace of smooth vectors and Garding's theorem. I'd guess it would also work for Banach space representations, but I'm not an expert on these analytic things. It's important to note that semisimplicity may be lost when one restricts smooth representations in these settings -- so their theorem says something about occurrences of quotients after restriction. It's important to be careful about the meaning of "multiplicity-free" in these situations. - ### 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. If you use complex numbers instead of real numbers, it is true that the restriction is multiplicity free. This is an important fact, and is used to construct the Gelfand-Zetlin basis for V. This is discussed is Fulton and Harris' book "representation theory", section 25.3. They also discuss generalizations to other classical algebras -	{"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": 23, "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": 0.9207607507705688, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/101994-integration-trigonometric-substitution.html	# Thread: 1. ## Integration by trigonometric substitution I'm new to this type of question and I can only get so far into the question before I don't know what to do - I haven't been taught more but I very much would like to know. Use the substitution $x$ = $3\sin\theta$ to show that $\int_{1.5}^{3} \sqrt{9 - x^2} dx$ = $\int_{a}^{b} k \cos^2\theta d\theta$ where the values of a, b and k are to be found. Hence evaluate $\int_{1.5}^{3} \sqrt{9 - x^2} dx$. So far I can say that $x$ = $3\sin\theta$ and $dx$ = $3\cos\theta d\theta$ $\sqrt{9 - (3\sin\theta)^2}$ = $\sqrt{9 - 9\sin^2\theta}$ = $\sqrt{9 (1 - \sin^2\theta)}$ = $\sqrt{9\cos^2\theta}$ = $3\cos\theta$ And this is as far as I can get. Could someone please help me finish the question and understand how it works from this point onwards? Any help is greatly appreciated 2. $\int_{1.5}^{3} \sqrt{9 - x^2} dx$ = $\int_{a}^{b} k \cos^2\theta d\theta$ where the values of a, b and k are to be found. As you have done we let: $x$ = $3\sin\theta$ and $dx$ = $3\cos\theta d\theta$ so $\int_{1.5}^{3} \sqrt{9 - x^2} dx$ = $\int_{1.5}^{3}\sqrt{9 - (3\sin\theta)^2} 3\cos\theta d\theta$ = $\int_{1.5}^{3} 3\cos \theta \cdot 3\cos\theta d\theta$ = $\int_{1.5}^{3} 9 \cos^2\theta d\theta$ so a=1.5 b=3 k=9 now to evaluate this integral you can use the identity: $\cos^2 x = \frac{1+cos(2x)}{2}$	{"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": 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": 0.9600411057472229, "perplexity_flag": "head"}
http://mathoverflow.net/questions/21663/recursions-which-define-polynomials	## Recursions which define polynomials ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) There are many examples (Somos sequences, special polynomials related to rational solutions of the Painleve equations) when a recurrence relation, which a priori produces a sequence of rational functions, in reality results in a polynomial sequence. In one of my last projects (joint work with Ole Warnaar) we "naturally" arrived at solution to the following problem which does not fit classes of sequences known to me. Problem. The sequence of rational functions $P_0(t),P_1(t),\dots$ is defined by the recurrence relation $$P_n(t)=P_{n-1}(t)\cdot\frac{4t}{1+t}+\binom{2n}n\frac{1+t^{n+1}}{1+t} \quad\text{for $n\ge1$}$$ and initial condition $P_0(t)=1$. Show that $P_n(t)$ are polynomials with positive coefficients. I know that Sloane's Encyclopedia of Integer Sequences allows one to guess the polynomials; proving then is a usual machinery. I wonder on what is actually known about nonhomogeneous recurrences $P_n(t)=a(t)P_{n-1}(t)+b_n(t)$, where $a(t)$ and $b_1(t),b_2(t),\dots$ are given rational functions and $a(t)$ is not a polynomial, whose solutions are polynomials. Have you seen other examples? For higher-order recursions? What about the positivity aspect (as in the problem above)? Edit. In order to make my question complete, I add the solution to the problem: $$P_n(t)=\sum_{k=0}^nA_{k,n-k}t^k, \qquad\text{where}\quad A_{k,m}=\frac{(2k)!(2m)!}{k!(k+m)!m!}.$$ It is an exercise in number theory to verify that all $A_{k,m}$ are integers. These numbers are in a certain sense very close to the binomial coefficients $B_{k,m}=\dfrac{(k+m)!}{k!m!}$ (so that the analogue of $P_n(t)$ is $(1+t)^n$), although no combinatorial interpretation is known for general $k,m$. I. Gessel in [J. Symbolic Computation 14 (1992) 179--194] addresses this combinatorial problem and gives several hypergeometric proofs of the integrality. - 3 If you replace "polynomials" with "rational functions" you could look up the Laurent phenomenon. arxiv.org/abs/math/0104241 – John Mangual Apr 17 2010 at 14:23 John, thanks for the hint. It's indeed an interesting piece, although related more to Somos sequences rather to my nonhomogeneous linear question. – Wadim Zudilin Apr 17 2010 at 23:13 1 Could you expand a bit on what you expect? In the case at hand, if B(x) is the generating function for the bn, P(x) is just (B(x)+P0-b0)/(1-a x), so you could interpret Pn (roughly) as a B times a sequence of a's. Are you asking when these objects, with the objects in "B" and the object "a" having rational weights, have polynomial weight? Doesn't that sounds too general? – Martin Rubey Apr 18 2010 at 7:50 I could not find any other example of a polynomial sequence generated in this way. I simply look for another such sequence of polynomials, so if you've ever seen something similar, please let me know. One more example would be a good answer to my question. – Wadim Zudilin Apr 18 2010 at 23:08 ## 3 Answers I have no answer to your question, but some related examples. Consider the q-Fibonacci polynomials defined by f(0, x, s)=0, f(1, x, s)=1 and f(n, x, s)=x f(n-1, x, s)+q^(n-2) s f(n-2, x, s). Then the subsequences f(k n, x, s) satisfy a homogeneous recursion with rational coefficients which for k>2 are not polynomials (see e.g. my paper in arXiv 0806.0805). More precisely f(k, x, q^k s) f(k n, x, s) – f(2 k, x, s) f(k (n-1), x, q^k s) +(-1)^k q^(k(3k-1)/2) s^k f(k, x, s) f(k (n-2), x, q^(2k) s)= 0 or equivalently f(k, x, q^(n-2k) s) f(k n, x, s) – f(2 k, x, q^(k (n-2)) s) f(k (n-1), x, s) +(-1)^k q^(-k (3k+1)/2) q^(k^2 n) s^k f(k, x, q^(k (n-1)) s) f(k (n-2), x, s)= 0. Analogous results are true for powers of q-Fibonacci polynomials. - @Johahn: thank you very much for this family of $q$-examples and the reference to your work on that. Even this is not an answer to my problem, the "integrality" phenomenon looks very similar. I have to think of this interesting direction as well. – Wadim Zudilin Apr 27 2010 at 12:13 1 I had forgotten to mention that in the course of Gosper's algorithm such recursions occupy a prominent role, cf. the book a=b by Petkovsek, Wilf and Zeilberger. – Johann Cigler Apr 28 2010 at 8:00 ### 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. Multiplying the recurrence relation $P_n(t)=a(t)P_{n-1}(t)+b_n(t)$ by $x^n$ and summing up over $n=1,2,\dots,\infty$, we get $${\cal P}(x,t) - P_0(t) = a(t){\cal P}(x,t)x + {\cal B}(x,t)$$ where ${\cal P}(x,t) = \sum_{n=0}^{\infty} P_n(t) x^n$ and ${\cal B}(x,t) = \sum_{n=1}^{\infty} b_n(t) x^n$ are generating functions for $P_n(t)$ and $b_n(t)$ respectively. Therefore, $${\cal P}(x,t) = \frac{{\cal B}(x,t) + P_0(t)}{1-a(t)x}$$ and $P_n(t)$ can be expressed as the coefficients of $x^n$ in the r.h.s., that is $$P_n(t) = P_0(t) a(t)^n + \sum_{k=1}^n b_k(t) a(t)^{n-k}.$$ - Max, I am sorry for being so late with your answer. Your formula indicates that it's quite unlikely to have $P_n(t)$ polynomials, isn't it? And your machinery could hardly prove the polynomiality for the above choice of $a(t)$ and $b_n(t)$. Can you give me some hints on how to generate (say, nontrivially) polynomials from your final formula, having a rational function $a(t)$? Thanks! – Wadim Zudilin Jul 2 2010 at 9:08 How about a simple relation like this? `$$P_{n}(t) = \frac{P_{n-1}(t)}{1+t} + \frac{t}{1+t} $$` with `$P_0(t) = 1$`, `$b_{n}(t) = t/(1+t)$` and `$a(t) = 1/(1+t) $`? The solution is `$P_n(t) = 1$` which is a polynomial with positive coefficient. Or another example: `$P_{n}(t) = (1+t)^n$`, a(t) = 1/(1-t), `$b_n(t) = t^2(1+t)^{n-1}/(t-1)$` In general, you can find examples quite easily, given that you have `$P_n(t)$` in mind, and function a(t) which is rational, and you solve for `$b_n(t)$`. `$$b_n(t) = P_n(t)-a(t)P_{n-1}(t)$$` As long as `$b_n(t)$` is rational as you required, you have a valid example. - 1 Ross, this question is asking for recurrences for which there is no a priori reason for the solution to be polynomial, meaning that the result must be surprising in some sense. Your answer defeats this purpose, since you are devising the recurrence starting from a polynomial sequence. – Gjergji Zaimi Aug 6 2010 at 7:17 "I could not find any other example of a polynomial sequence generated in this way. I simply look for another such sequence of polynomials, so if you've ever seen something similar, please let me know. One more example would be a good answer to my question.", from author's comment. He just asked for 1 more example. And there is no mention of surprise? – Ross Tang Aug 6 2010 at 8:01 From another comment: "Can you give me some hints on how to generate (say, nontrivially) polynomials from your final formula, having a rational function a(t) ? Thanks! " He asked for method to generate polynomial from Max Alekseyev's formula. So I think providing a method is good enough. – Ross Tang Aug 6 2010 at 8:03 1 I was merely suggesting that your answer is tautological for it is claiming that the recurrences which have polynomial solutions are the recurrences that are satisfied by a sequences of polynomials... – Gjergji Zaimi Aug 6 2010 at 8:27	{"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": 5, "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": 0.9317647218704224, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/35920/maximum-possible-information-in-the-universe	# Maximum Possible Information in the universe? I remember hearing about this in one of the programs in discovery science. The physicist claimed that the maximum possible information in the universe is (10)^(10^123) whereas the maximum possible information that can be known by man is (10)^(10^90). Can anyone explain to me how can we arrive at such a specific number, and also how can information be represented by only numbers? - The information in the Universe measured in bits or nats is just $10^{123}$ from the holographic bound; the actual entropy of the known matter except black holes - is about $10^{90+}$ and is dominated by the cosmic microwave background. However, this guess for the entropy is obsolete because most of the entropy we know in the Universe is carried by the black holes, bringing us above $10^{100}$. If you have one more exponentiation, 10 to those large numbers, it doesn't measure the information but the number of possibilities one may distinguish (8 bits = byte distinguish 256 states...). – Luboš Motl Sep 9 '12 at 8:11 ## 2 Answers The number 10^123 emerges as (roughly) the number of Planck areas contained within the boundary of the observable universe. If each Planck area can be (roughly) in two states, a total of 10^123 yes/no questions suffice to describe the boundary of the universe and - via the (still speculative) holographic principle - the whole universe. In other words, if the universe is a hologram, about 10^123 bits of information are needed to describe it. - 2 Why is the holographic principle "speculative"? It isn't 1990 anymore. – Ron Maimon Sep 8 '12 at 15:47 3 @RonMaimon It is speculative since it is not experimentally verified. And physics is an experimental science. Speculative is neither incompatible with natural nor synonymous of bizarre. I wonder why you claim it is not speculative. – drake Sep 8 '12 at 18:39 If I encode the whole universe from t=0 up to now in a digits sequence, there is more informations contained in this code than if I just encode the whole universe right now, but where in Planck aeras is stored the history of the universe ? – Shaktyai Sep 8 '12 at 20:40 @drake: Physics always involves extrapolation from data, and in this case the extrapolation is relatively safe, since there is no hint of an alternative theoretical construction which even begins to work. When there is only one known possibility, and it is sure to work (although for all we know there is something else which works too, but for the life of me I can't think of what it could be) we generally say it's "established". – Ron Maimon Sep 9 '12 at 3:00 1 @RonMaimon I agree that physics involves extrapolation from data, but holography (especially some versions or it) requires a change of paradigm as I guess you will agree. It is not just an extension of a previous idea, it is not neutrinos' masses. So, in my opinion, the position you and most of people take is very dogmatic even though holography may be very natural. And dogmatism is very dangerous for a scientific area with lack of experiments/observations. Why not be more open minded? – drake Sep 9 '12 at 20:45 Also, it is estimated that there are approximately 10^80 protons (or electrons or neutrons) and approximately 10^90 photons in the observable universe. So this is how many "objects" you have to work with. I don't know how to get from this to the maximum amount of information that can be known by man. -	{"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": 0.9382382035255432, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/2014/if-you-break-a-stick-at-two-points-chosen-uniformly-the-probability-the-three-re/75876	## If you break a stick at two points chosen uniformly, the probability the three resulting sticks form a triangle is 1/4. Is there a “nice” proof of this fact? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) There is a standard problem in elementary probability that goes as follows. Consider a stick of length 1. Pick two points uniformly at random on the stick, and break the stick at those points. What is the probability that the three segments obtained in this way form a triangle? Of course this is the probability that no one of the short sticks is longer than 1/2. This probability turns out to be 1/4. See, for example, problem 5 in these homework solutions. It feels like there should be a nice symmetry-based argument for this answer, but I can't figure it out. I remember seeing once a solution to this problem where the two endpoints of the interval were joined to form a circle, but I can't reconstruct it. Can anybody help? - This seems to be related to a recent math.stackexchange question: math.stackexchange.com/questions/72977/… – Michael Lugo Nov 2 2011 at 16:10 ## 13 Answers Here's what seems like the sort of argument you're looking for (based off of a trick Wendel used to compute the probability the convex hull of a set of random points on a sphere contains the center of the sphere, which is really the same question in disguise): Connect the endpoints of the stick into a circle. We now imagine we're cutting at three points instead of two. We can form a triangle if none of the resulting pieces is at least 1/2, i.e. if no semicircle contains all three of our cut points. Now imagine our cut as being formed in two stages. In the first stage, we choose three pairs of antipodal points on the circle. In the second, we choose one point from each pair to cut at. The sets of three points lying in a semicircle (the nontriangles) correspond exactly to the sets of three consecutive points out of our six chosen points. This means that 6 out of the possible 8 selections in the second stage lead to a non-triangle, regardless of the pairs of points chosen in the first stage. - 3 That doesn't seem like the argument I remember, but it's satisfying nonetheless. – Michael Lugo Oct 23 2009 at 2:47 For the above-mentioned result about spheres, see math.stackexchange.com/questions/1400/… – ACL Nov 23 2011 at 8:28 ### 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. Consider an equilateral triangle with altitude 1. It is not hard to show that if you choose a point randomly in this triangle, the distances to the three sides gives the same distribution of lengths that you obtain by breaking a stick at two random points. Now, the locus of points for which no distance is longer than 1/2 is the smaller equilateral triangle formed by joining the midpoints of the edges, which has area 1/4 that of the original triangle. - 3 I can't quite see this statement that isn't hard to show. Hint? – Qiaochu Yuan Jun 3 2011 at 14:14 @Qiaochu: Suppose you know the length ℓ of one of the three pieces. What is the distribution of the remaining two? In the triangle version, this corresponds to keeping the middle point the same height above the base, which yields a uniform distribution from 0 to 1−ℓ of the length of the second piece. In the original question, this corresponds to keeping the length of one of the three intervals fixed, and it is easy to see that this gives a uniform distribution on the length of the other two intervals. – Peter Shor Jun 4 2011 at 11:22 @Qiaochu (continued): In this case, the knowledge of the marginal distribution of the other two pieces, given the length of one, determines the distribution. You can see this because the stochastic process of holding the length of one edge fixed, then randomly selecting the lengths of the other two, and then repeating with the length of a different edge fixed, is essentially the same as the Metropolis algorithm for uniformly sampling points inside the triangle. – Peter Shor Jun 4 2011 at 11:22 @Qiaochu: The distances correspond to the barycentric coordinates of the point. More generally, choosing a point uniformly in a $k$-simplex corresponds to independently uniformly choosing $k$ points in the unit interval and using the resulting $k+1$ interval lengths as barycentric coordinates. – joriki Mar 9 at 1:54 Is the argument you remember along the lines of: picking three points on a circle, what is the probability they lie in the same semicircle? The problem is discussed here: http://godplaysdice.blogspot.com/2007/10/probabilities-on-circle.html - 73 This would be the part where I, somewhat embarrassedly, point out that I wrote that. – Michael Lugo Oct 23 2009 at 4:13 A triangle is possible iff no part is $>{1\over2}$. With probability ${1\over2}$ both cuts are on the same side of the midpoint $M$, in which case no triangle is possible. If the cuts $x$ and $y$, $\ x < y$, are on different sides of $M$ then with probability ${1\over 2}$ the point $x$ is further left in its half than $y$ is in the right half. In this case there is no triangle possible either. It follows that only ${1\over 4}$ of all cuts admit the forming of a triangle. - Yes, here's a nice and beautiful argument! First you should draw a picture of axes `a` and `b`. You're asked to select uniformly a point in the square `[0,1]x[0,1]`. Now because of the symmetry (sic!) it's equivalent to choosing the points `a` and `b` uniformly in the triangle cut from the square by `b > a`. So you're actually uniformly selecting a point inside triangle defined by lines `a>=0`, `b<=1`, 'b>=a'. Now let's find the conditions to be able to make a triangle of short sticks. We should have `a + (1-b) > b-a`, `b-a + (1-b) > a` and `b > 1 - b` which indeed, as you say, boils down to ````b > 1/2, a < 1/2, b-a < 1/2 ```` It remains to note that those lines create inside the big triangle a small triangle which is similar to big but with all lengths `1/2` of the big, so this small triangle has area of exactly 1/4 of original! - "First, the way you define the probability, because of the symmetry (sic!) it's equivalent to first choosing first point a uniformly and then second point b uniformly on [a, 1]." I don't think that this is the case. In the original problem, the probability that there was a piece longer than 1-epsilon decayed like C/epsilon^2 as epsilon tended to 0 (a necessary condition is that no cut lies in (epsilon, 1-epsilon). In your model, it decays like c/epsilon (a sufficient condition is to have A larger than 1-ep). – Kevin P. Costello Oct 23 2009 at 4:35 Yes, you were right to point my mistake. The geometric picture is correct, but I referred to it incorrectly. Will edit. – Ilya Nikokoshev Oct 23 2009 at 15:57 5 Just rephrasing your argument: One can partition the 2-dim simplex, defined by x>=0, y>=0, z>=, x+y+z=1 into 4 identical triangles defined by adding the conditions: 1) x>=½ 2) y>=½ 3) z>=½ 4) x<=½ and y<=½ and z<=½ of which the 4th is the event that you can form a triangle. – Ori Gurel-Gurevich Nov 4 2009 at 4:09 Great way to say it indeed. – Ilya Nikokoshev Nov 4 2009 at 20:59 One reference for a solution to this problem is Carlos d'Andrea and Emiliano Gomez, "The broken spaghetti noodle", American Mathematical Monthly 113 (2006), p. 555. More generally the probability that an interval broken at n-1 points chosen uniformly at random is broken into pieces which can be rearranged to form an $n$-gon is $1 - n/2^{n-1}$. - There's something missing in your last sentence: $1-n/2^{n-1}$ is the probability of what event? – Mark Meckes Jun 25 2010 at 20:00 Mark, thanks! That's what I get for writing MathOverflow answers on an iPhone. (I was flipping through an old issue of the Monthly and wasn't near a computer.) – Michael Lugo Jun 26 2010 at 0:48 Both Kevin Costello's argument and mine can be adapted to give the $1-n/2^{n-1}$ answer in the general case. – Peter Shor Jun 4 2011 at 15:40 It seems natural to rephrase the question in terms of barycentric coordinates in a triangle. These coordinates are numbers $x$, $y$, $z$ in the interval $[0,1]$ satisfying the equation $x+y+z=1$. We are looking for triples $(x,y,z)$ of such numbers satisfying the three triangle inequalities $x \le y+z$, $y\le x+z$, and $z\le x+y$. Replacing the relations "$\le$" by "$=$", we get line segments joining the midpoints of the edges of the triangle. These line segments cut the triangle into four congruent subtriangles. The central one of these four subtriangles is the region where all three triangle inequalities hold, and this region has area equal to one quarter of the area of the big triangle. This is essentially the same argument as in the answers by Peter Shor and Ilya Nikokoshev, particularly in the reformulation of the latter answer in Ori Gurel-Gurevich's comment - SOLUTION USING ONLY ORIGINAL LINE: Call the end points A and B and the midpoint M. Let N be the midpoint of MB. Call P the first random point and Q the second. In full generality, we can consider P as lying between M and B. On AM, let H be the point for which HP is half the length of the original stick. For the 3 sticks to make a triangle, it is necessary and sufficient that the total length of any 2 sticks be greater than the third. Therefore no piece can exceed half the length of the original stick. That tells us immediately that Q cannot lie on AH or PB, immediately eliminating half the possible points for Q. Likewise all points on MP are also eliminated. Only points on HM qualify as possible Q points. If we let AB=1, the probability that the 3 sticks can form a triangle is then the average length of HM for all Ps. Note that if P=N, the midpoint of MB, then HM=MP, so the qualifying HM points comprise 1/4 the length of the stick and p=1/4 that a triangle can be formed after randomly choosing a Q for this P. Similarly, for every P to the right of N there is a matching P an equal distance to the left of N, so that the average probability (HM length) for these two Ps is 1/4. That yields an overall average probability for all Ps of 1/4. - My answer is 1/4. Consider a stick of length L. We make a cut at distance x from one end. Second cut at distance y from the same end. No side of a triangle can be greater than the sum of the other two sides, so both x and (y-x) have to be less than L/2. Probability is the number of favorable outcomes divided by total number of outcomes, so we get ((L/2) * (L/2))/(L*L) = 1/4! - Unless I'm missing something, why do you assume the number of outcomes $= L^2$? – Michael Kissner Sep 19 2011 at 16:31 @Michael Kissner: Maybe I'm messing it up, but I kinda figured that the minimum value the product of x and (y-x) can take is 0 and the maximum value it can take is L times L, which is L^2. Actually, if we take the first cut at x and the second cut at a distance dx from x, we can integrate the product of x and dx between the limits of 0 and L/2 for the favorable outcomes, and between 0 and L for the total outcomes. This would also give us the answer as 1/4. OMG, if this logic is flawed, then I've stumbled on to this answer by accident, which is kinda cool in some weird way! - 2 It's probably better to edit, or add to, your original answer, rather than start a new one – Yemon Choi Sep 19 2011 at 22:57 Let AB be the stick. WLOG we may assume AB=1(Since the probability won't depend on the length of AB). Let the points at which the stick is broken be P and Q. AP=x,PQ=y and QB=z. Since $0\leq AP,PQ,QB \leq 1$ we need to consider all the points inside the $1\times 1\times 1$ cube. Futhermore the points lie on the x+y+z=1 plane. x+y+z=1 plane(click on the link to see the image of the plane) On applying the triangle inequalities (i.e $x+y>z,y+z>x\text{ and }x+z>y$) we find that the net area of points satisfying the condition of forming a triangle is the shaded potion. Shaded Area(Click on the link to see the shaded area) Since the points $J,K,I$ are the midpoints of the sides of the of the triangle ACE.The probability = $\dfrac{\text{Area of }\Delta JKI}{\text{Area of }\Delta ACE}=\dfrac{1}{4}$ - Your answer is similar to Allen Hatcher's: mathoverflow.net/questions/2014/… – Joel Reyes Noche Mar 22 at 23:08 Sorry,didn't see that. – Shaswata Mar 23 at 6:09 Practically, the likelihood is >25% because there would be a natural tendency to break the larger piece after the first break. This increases the likelihood of having no piece longer than 1/2 because the probability weighted range for the second break is decreased. If we are breaking sticks (typically considered small in length), it is also not practical to break into extra short sections because sticks are not infinitely thin. It is very difficult to break stick into smaller pieces than 2-3 times their thickness. This tilts the likelihood well into the >25% range. - OP made precise how the stick is broken. First, two point are so to say marked on the stick then it is broken there; so your break first and whathever is not relevant to the question at hand. If you prefer you can also replace 'break' by 'cut with a high-precision saw' to minimise you concerns. In any case, this is an idealized model, of course. Now, for the good side of your question: it is true that one needs to be careful to make precise how exactly something is chosen at random, sometimes there is more then one natural interpretation. Yet here this was made precise in qu. Moreover,... – quid Nov 15 at 22:05 1 it can also be sometimes interesting for such questions to investigate actual real worl situations and/or more refined models that are closer to reallity (eg, how fair is flipping a coin really is not clear). Yet all this is irrelevant to the question at hand. In that sense your answer is besides the point. – quid Nov 15 at 22:09 another way to get the answer is to write a computer program to generate 2 random numbers from unif(0,1), say x1,x2, whenever x1x2 swap their values), let l1=x1,l2= x2 - x1 , l3= 1 - x2 , make a function that returns 1 when l1 +l2 >l3 && l1 +l3 >l2 && l3 +l2 >l1 , 0 otherwise . now repeat this experiment n times , as ntends to infinity by the frequentists definition of probability we get a good estimate of the probable value. . . . . . it turns out to be 1/4 . - 2 Monte Carlo methods like this are great when there's no hope for an exact analysis of what's going on. However in this case we have several exact analyses. – James Cranch Nov 23 2011 at 9:11	{"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": 35, "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": 0.9306986331939697, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/8633/is-energy-exchange-quantized	# Is energy exchange quantized? In the photoelectric effect there is a threshold frequency that must be exceeded, to observe any electron emission, I have two questions about this. I) Lower than threshold: What happen with lesser frequency/energy photons? I mean there is no energy transferred? If some energy were transferred, (and it must be transfered quantized), how can be experimentally known it was quantized, if there is no electron emission? II) Intensity dependence: There is any known dependency with intensity? (I mean a dependency not about the number of electrons (I am already aware of this) but about the energy of them) I thought energy of electron emited was independent with intensity, but then I found this link dependency with intensity (is a paper to buy that I haven't read) but relates energy with intensity. - ## 2 Answers Energy exchange is quantized when moving a electron from one bound state to another bound state. This isn't because the exchange is inherently quantized, but because the states the electron may occupy are quantized. Thus the standard photo-electric effect in which a photon can not excite an atom unless it has a minimum energy. However,... There are multi-photon processes by which sub-threshold light can excite transitions. Cross-section for them go by intensity-squared (or worse) and are very small for any reasonable light intensity. To study or employ them you get powerful, short pulsed laser systems. Where short pulsed means nano-second or faster pulses and powerful means "Do not look into beam with remaining eye". Even then you don't get a lot of rate. These processes are utterly negligible for the kind of benchtop experiment we use to teach the photoelectric effect: you just can't get enough intensity. (See below for how negligible.) The conceptual model here is that the first photon bumps the electron to a short-lived, unstable state without well defined quantum numbers, and the second comes along before that state decays and finishes the job. We're currently exploring the application of such a process to calibrating light yields, opacities in a large volume of scintillating material. From New J.Phys.12:113024, 2010: For gases the one-photon absorption cross-section $\sigma_1$ is typically of the order of $10^{−17}\text{ cm}^2$, whereas the two-photon and the three-photon cross-sections are of the order of $\sigma_s = W/F_2 \approx 10^{−50}\text{ cm}^4\text{ s}$ and $\sigma_3 = W/F_3 \approx 10^{−83}\text{ cm}^6\text{ s}^2$, respectively. Where $F$ is intensity in photons/second and W is excitation rate in reciprocal seconds. - Hello @dmckee I think this statement "the first photon bumps the electron to a short-lived, unstable state without well defined quantum numbers, and the second comes along before that state decays and finishes the job" is very is bold but doubtful, if that is possible it seems to be a direct contradiction with the "photon concept", because you could gradually add energy without needing a single particle-like action. – HDE Apr 14 '11 at 18:13 @HDE: See for instance for an application of this technique. Though this paper actually uses three photons with an intermediate step at a well defined energy level on the way to ionization, and therefore goes by intensity cubed. – dmckee♦ Apr 14 '11 at 18:27 Further, this process could not be mimicked by a continuous process, as there is a energy threshold for the one process, a energy threshold for the two photon process, an so on; and the rate are determined in a quantum mechanical way. – dmckee♦ Apr 14 '11 at 18:33 Amazing, multiphoton ionization.. I didn't knew anything of this!( perhaps I have stop reading Louis de Broglie and get something newer haha..) I wonder if there is not a chance of subdividing again the threshold of every "photon process" into photons of successively small energy, if possible, where is the limit? – HDE Apr 14 '11 at 20:12 1 @HDE: The limit is the cross-section. Even for immensely intense light sources the production rate is dropping like a stone. Each additional photon brings in another factor of the fine structure constant and another short lifetime for the unstable state (and the further they are from a stable state, the shorter the lifetime). Look at the behavior quoted above. If you need more than few photons, it just won't happen often enough to measure, much less use. – dmckee♦ Apr 14 '11 at 20:18 If the frequency of the incident radiation is lower than the threshold frequency then either a photon is fully absorbed or not absorbed at all. It is absorbed only if it has an energy which is just enough to excite the electron to the higher state but not enough (less than the work function) to make the electron to leave the surface of the material. The only way we know that the energy of the electron is quantized in an atom is by analyzing its spectrum. For a particular frequency above the threshold, the number of the electrons emitted will linearly increase with the intensity of the incident radiation. - 1 This is the first order effect, and is correct for essentially every purpose. However, there are contributions from higher order graphs which allow multi-photon effects. It takes a pulsed, high power laser to get the intensity needed to study them. – dmckee♦ Apr 14 '11 at 17:21	{"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": 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": 0.9477199912071228, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2009/11/10/the-cross-product-and-pseudovectors/?like=1&source=post_flair&_wpnonce=9916d21231	# The Unapologetic Mathematician ## The Cross Product and Pseudovectors Finally we can get to something that is presented to students in multivariable calculus and physics classes as if it were a basic operation: the cross product of three-dimensional vectors. This only works out because the Hodge star defines an isomorphism from $A^2(V)$ to $V$ when $\dim(V)=3$. We define $u\times v=(u\wedge v)$ All the usual properties of the cross product are really properties of the wedge product combined with the Hodge star. Geometrically, $u\times v$ is defined as a vector perpendicular to the plane spanned by $u$ and $v$, which is exactly what the Hodge star produces. We choose which perpendicular direction by the “right-hand rule”, but this is only because we choose the basis vectors $e_1$, $e_2$, and $e_3$ (or as these classes often call them: $\hat{\imath}$, $\hat{\jmath}$, and $\hat{k}$) by the same convention, and this defines an orientation we have to stick with when we define the Hodge star. The length of the cross product is the area of the parallelogram spanned by $u$ and $v$, again as expected from the Hodge star. Algebraically, the cross product is anticommutative and linear in each variable. These are properties of the wedge product, and the Hodge star — being linear — preserves them. The biggest fib we tell students is that the value of the cross product is a vector. It certainly looks like a vector on the surface, but the problem is that it doesn’t transform like a vector. Before the advent of thinking of all these things geometrically, people thought of a vector quantity as a triple of real numbers that transform in a certain way when we change to a different orthonormal basis. This is inspired by the physical world, where there’s no magic orthonormal basis floating out somewhere to pick out coordinates. We should be able to turn our heads and translate the laws of physics to compensate exactly. These rotations form the special orthogonal group of orientation- and inner product-preserving transformations, but we can also throw in reflections to get the whole orthogonal group, of all transformations from one orthonormal basis to another. So let’s imagine what happens to a cross product when we reflect the world. In fact, stand by a mirror and hold out your right hand in the familiar way, with your index finger along one imagined vector $u$, your middle finger along another vector $v$, and your thumb pointing in the direction of the cross product $u\times v$. Now look in the mirror. The orientation has been reversed, and mirror-you is holding out its left hand! If mirror-you tried to use its version of the cross product, it would find that the cross product should go in the other direction. The cross product doesn’t behave like all the other vectors in the world, because it doesn’t reflect the same way. Physicists to this day use the old language describing a triple of real numbers that transform like a vector under rotations, but point the wrong way under reflections. They call such a quantity a “pseudovector”. And they also have a word for a single real number that somehow mysteriously flips its sign when we apply a reflection: a “pseudoscalar”. Whenever we read about scalar, vector, pseudovector, and pseudoscalar quantities, they just mean real numbers (or triples of them) and specify how they change under certain orthogonal transformations. But geometrically we can see exactly what’s going on. These are just the spaces $A^0(V)=\mathbb{R}$, $A^1(V)=v$, $A^2(V)$, and $A^3(V)$, along with their representations of the orthogonal group $\mathrm{O}(V)$. And the “pseudo” means we’ve used the Hodge star — which depends essentially on a choice of orientation — to pretend that bivectors in $A^2(V)$ and trivectors in $A^3(V)$ are just like vectors in $V$ and scalars in $\mathbb{R}$, respectively. And we can get away with it for a long time, until a mirror shows up. The only essential tool from multivariable calculus or introductory physics built from the cross product that we might have need of is the “triple scalar product”, which takes three vectors $u$, $v$, and $w$. It calculates the cross product $v\times w$ of two of them, and then the inner product $\langle u,v\times w\rangle=\langle u,(v\wedge w)\rangle$ with the third to get a scalar. But this is the coefficient of our unit cube $\omega$ in the definition of the Hodge star: $\displaystyle\langle u,(v\wedge w)\rangle\omega=u\wedge(v\wedge w)=u\wedge v\wedge w$ since $*(v\wedge w)=(-1)^{2\cdot(3-2)}v\wedge w$. That is, the triple scalar product gives the (oriented) volume of the parallelepiped spanned by $u$, $v$, and $w$, just as we remember from those classes. We really don’t need the cross product as a primitive operation at all, and in the long run it only leads to confusion as it identifies vectors and pseudovectors without the explicit use of the orientation-dependent Hodge star to keep us straight. ### Like this: Posted by John Armstrong \| rants ## 3 Comments » 1. Thank you! I’ve always wanted to see a clear explanation of this online, and I think the Wikipedia articles are a little too intimidating. Comment by \| November 10, 2009 \| Reply 2. [...] turns out that we already know of an example of a Lie algebra: the cross product of vectors in . Indeed, take three vectors , , and and try multiplying them out in all three [...] Pingback by \| May 17, 2011 \| Reply 3. [...] product on -forms, while the inner ones are the inner product on vector fields. This is exactly the cross product of vector fields on . Advertisement Eco World Content From Across The Internet. Featured on [...] Pingback by \| October 11, 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": 38, "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": 0.9291646480560303, "perplexity_flag": "head"}
http://quant.stackexchange.com/questions/4638/black-scholes-american-put-option	# Black-Scholes American Put Option Here is my question: This is a question about Black-Scholes model, but it may be applicable to more complicated models. Throughout the discussion, the strike price $K$, interest rate $r$ and volatility $\sigma$ will be assumed to be constant. The thing we are interested in is the time decay of the option. First consider a perpetual American put problem. This has an optimal exercising level which I will call $b$ and this is given by $b=\frac{K}{1+\sigma^2/2r}$ Now consider any $x>b$, this $x$ is a level, for which it would not be optimal to exercise in a perpetual option. In this case, does this necessarily mean, there exist a $T$ such that $t\in (0,T)$, the equation below holds $\mathbb{E}_x e^{-r(t\hat{}\tau_b)}(K-X_{t\hat{}\tau_b})^+ > (K-x)^+$ here $\tau_b$ denotes the first hitting time of level $b$. I think I have shown this is the case for some $t$. Here is my proof: Denote $V(x)=\mathbb{E}_x e^{-r\tau_b}(K-X_{\tau_b})^+$. It is widely known that $V(x)>(K-x)^+$. Assume the statement above is false, then $\mathbb{E}_x e^{-r(t\hat{}\tau_b)}(K-X_{t\hat{}\tau_b})^+ \leq (K-x)^+$ for all $t>0$. Then we can take the limit as $t\uparrow\infty$ on the left-hand side and apply the dominated convergence theorem, we see that $V(x)\leq (K-x)^+$ Contradicting the inequality $V(x)>(K-x)^+$ However, I have struggled to show $\mathbb{E}_x e^{-r(t\hat{}\tau_b)}(K-X_{t\hat{}\tau_b})^+ > (K-x)^+$ for small value of $t$. I believe this conjecture is true for all $t>0$ but struggled to prove it. Anyone has any ideas? - ## 1 Answer Consider a finite horizon problem with the time to maturity $t$, then there exists an optimal stopping boundary $B(t)$, such that if the stock price is at this level, we should exercise the option. Notice, this $B(t)>b$ and it is an increasing function of $t$. If we start our process, at $B(t)$ with amount of time $t$ left to run, then since we started at the optimal stopping boundary so $(K-x)^+ > \mathbb{E}_{B(t)} (K-X_\tau)^+$ for any stopping time $\tau<t$. In fact, the result I asked for is totally ridiculous, because as $t$ goes to $0$, $B(t) \uparrow K$. For small times, it is a lot better to stay where you are than let the process run on.	{"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": 34, "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": 0.9514729976654053, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2009/11/12/the-jacobian-of-a-composition/?like=1&_wpnonce=cb13afa090	# The Unapologetic Mathematician ## The Jacobian of a Composition Let’s start today by introducing some notation for the Jacobian determinant which we introduced yesterday. We’ll write the Jacobian determinant of a differentiable function $f$ at a point $x$ as $J_f(x)=\det(df(x))$. Or, in more of a Leibnizean style: $\displaystyle\frac{\partial(f^1,\dots,f^n)}{\partial(x^1,\dots,x^n)}=\det\left(\frac{\partial f^i}{\partial x^j}\right)$ We’re interested in determining the Jacobian of the composite of two differentiable functions. To which end, suppose $g:X\rightarrow\mathbb{R}^n$ and $f:Y\rightarrow{R}^n$ are differentiable functions on two open regions $X$ and $Y$ in $\mathbb{R}^n$, with $g(X)\subseteq Y$, and let $h=f\circ g:X\rightarrow\mathbb{R}^n$ be their composite. Then the chain rule tells us that $\displaystyle dh(x)=df(g(x))dg(x)$ where each differential is an $n\times n$ matrix, and the right-hand side is a matrix multiplication. But these matrices are exactly the Jacobian matrices of the functions! And since the by definition, the determinant of the product of two matrices is the product of their determinants. That is, we find the equation $\displaystyle J_h(x)=J_f(g(x))J_g(x)$ Or, we could define $y^i=g^i(x)$ and use the Leibniz notation to write $\displaystyle\frac{\partial(h^1,\dots,h^n)}{\partial(x^1,\dots,x^n)}=\frac{\partial(h^1,\dots,h^n)}{\partial(y^1,\dots,y^n)}\frac{\partial(y^1,\dots,y^n)}{\partial(x^1,\dots,x^n)}$ As a special case, let’s assume that the differentiable function $f:X\rightarrow\mathbb{R}^n$ is injective in some open neighborhood $A$ of a point $a$. That is, every $x\in A$ is sent to a distinct point by $f$, making up the whole image $f(A)$. Further, let’s suppose that the function $f^{-1}$ which sends each point $y\in f(A)$ back to the point in $A$ from which it came — $f^{-1}(y)=x$ if and only if $y=f(x)$ — is also differentiable. Then we have the composition $f^{-1}(f(x))=x$, and thus we find $\displaystyle J_{f^{-1}}(f(a))J_f(a)=1$ or $\displaystyle\frac{\partial(y^1,\dots,y^n)}{\partial(x^1,\dots,x^n)}\frac{\partial(x^1,\dots,x^n)}{\partial(y^1,\dots,y^n)}=1$ Thus, if a differentiable function $f$ has a differentiable inverse function defined in some neighborhood of a point $a$, then the Jacobian determinant of the function must be nonzero at that point. A fair bit of work will now be put to turning this statement around. That is, we seek to show that if the Jacobian determinant $J_f(a)\neq0$, then $f$ has a differentiable inverse in some neighborhood of $a$. About these ads ### Like this: Like Loading... Posted by John Armstrong \| Analysis, Calculus ## 3 Comments » 1. [...] in with a first step towards proving the inverse function theorem we talked about at the end of yesterday’s post. This is going to get [...] Pingback by \| November 13, 2009 \| Reply 2. [...] Inverse Function Theorem At last we come to the theorem that I promised. Let be continuously differentiable on an open region , and . If the Jacobian determinant at some [...] Pingback by \| November 18, 2009 \| Reply 3. [...] But now we recognize the product of the two Jacobian determinants as the Jacobian of the composition: [...] Pingback by \| January 7, 2010 \| 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. • ## RSS Feeds RSS - Posts RSS - Comments • ## Feedback Got something to say? Anonymous questions, comments, and suggestions at Formspring.me! %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": 35, "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": 0.922215461730957, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/64205/list	2 edited tags 1 # Ergodicity of Convoluted White Noise I have a question regarding ergodicity in infinite dimensional spaces. Let $\mathcal{D}$ be the space of distributions on a Schwartz space, and let $\mu$ be the white noise process which exists by the Bochner-Minlos theorem. We can define the translation $\tau_x \phi$ of a distribution $\phi$ by: $$\langle \tau_x \phi, \varphi \rangle = \langle \phi, \tau_x \varphi \rangle,$$ for all $\varphi \in C^\infty_c(\mathbb{R}^n)$, where $\tau_x \varphi$ is the usual translation of a function. The set of translations form a group acting on $\mathcal{D}$. My first question is: Is the group $\lbrace \tau_x : x\in \mathbb{R}^n \rbrace$ ergodic with respect to the white noise measure? Now let $\varphi \in C^\infty_c(\mathbb{R}^n)$, then we can define the convolution $\varphi * \zeta(\omega)$, where $\zeta$ is a white-noise distributed variable. This is a $C^\infty$ random variable. My second question is: Is $\varphi * \zeta(\omega)$ ergodic with respect to translations? My intuition is that the first answer is true, and the second answer is "true, assuming the support is sufficiently small", but I don't even know what tools to use to tackle this problem.	{"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": 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": 0.9010494947433472, "perplexity_flag": "head"}
http://mathoverflow.net/questions/98060/constructing-an-icosahedral-weight-2-eigenform/98352	## Constructing an icosahedral weight 2 eigenform? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) (This question is a spin-off from this other question, and is largely inspired by it.) Let $f \in S_1(\Gamma_1(800),\chi)$ be the weight-one "icosahedral" eigenform constructed by Buhler in his thesis [1]: $$f = q - iq^3 - ijq^7 - q^9 + jq^{13} + \cdots.$$ Here $i = \sqrt{-1}$ and $j = \frac{1+\sqrt{5}}{2}$; (and if you'd like to see the first 360 Fourier coefficients, then see page 69 of loc. cit). It's "icosahedral", because the corresponding representation $\rho_f : G_\mathbb{Q} \to GL_2(\mathbb{C})$ has projective image isomorphic to $A_5$. (If you're wondering what $\chi$ is, it is the product of the character of order 2 and conductor 4 with the character of order 5 and conductor 25 sending 2 to $\zeta_5$.) Here is my basic question: Can I use this $f$ to construct a weight 2 cuspidal eigenform $g$ whose associated mod $\lambda$ representation $\bar{\rho}_{g,\lambda}$ (for $\lambda$ some prime ideal of the coefficient field of $g$) has projective $A_5$ image? Here are some thoughts I've had: An idea I first saw in section 1 of Lecture 1 of Gelbart's article in [2] gives me some hope; pick a weight one Eisenstein series $E$ that is congruent to 1 mod $l$ (some rational prime) and consider the product $fE$, which will be a weight 2 cuspform (but not an eigenform). Applying a lifting lemma of Deligne and Serre might produce an eigenform with the desired property at some $\lambda$ lying above $l$. I say "might", because the lemma I'm looking at on page 163 in [2] is working with primes above 3, and 3 may be the only prime for which this lifting works; (and since I'm chiefly interested in characteristics 7, 19 and 61, the approach may fail). Moreover, if I want the answer as a $q$-expansion, then perhaps this lifting is not explicit enough. (I thought about being more demanding in the question and stipulating the coefficient field of $g$ and the characteristic of $\lambda$, but decided against it...) Finally, I had wanted to ask this question starting with the conductor 133 "tetrahedral" form found by Tate and some of his students, because that came first historically (see the previous question), and I'd then be asking about "tetrahedral" weight 2 forms; but I was unable to write down its $q$-expansion; MAGMA gives a Runtime error when you ask it to compute a basis of the weight one forms at level 133 (though it seems to be fine at smaller levels such as 23 and 47). And since I really like $q$-expansions, I used Buhler's form. [1]: J. Buhler: Icosahedral Galois Representations. LNM 654. [2]: G. Cornell, J. Silverman, G. Stevens (eds): Modular Forms and Fermat's Last Theorem. Springer. - 1 It will be nice if some kind soul computes the $q$-expansion of the weight-$133$ tetrahedral form and posts it below Kevin's answer (mathoverflow.net/questions/97624/…). – Chandan Singh Dalawat May 27 at 3:03 1 If Magma is really dying when you ask it to compute the level 133 form, then you should send in a bug report -- the Magma folks are very good about fixing bugs quickly. – David Loeffler May 27 at 9:39 2 I just ran a test of computing weight 1 modular forms for levels $\le 200$ with Magma, and found two other bugs in addition to this one (it fails for $N \in \{112, 124, 133, 136, 148, 168, 171, 180, 196\}$. I've sent them a full bug report. – David Loeffler May 27 at 10:35 2 It's funny that the magma code doesn't work because I thought they just pasted my code in, and my code works fine for these weights (as far as I can see -- it gives answers and doesn't crash at any rate!). Barinder: if you want to see the level 133 form then download www2.imperial.ac.uk/~buzzard/char0results.m . Take a look at it -- should be self-explanatory. – Kevin Buzzard May 27 at 14:36 2 Here is a Sage code snippet that produces the q-expansion of the level 133 non-dihedral eigenform (just a linear combination of the basis vectors for the space that Kevin gives): pastebin.com/qEcuv7EF – David Loeffler May 28 at 9:31 show 2 more comments ## 2 Answers (This answer is partly a precis of an answer I gave to Barinder in person, which I am posting here in case anyone else is interested.) The Deligne--Serre lifting lemma is a completely general statement about endomorphisms of free modules over discrete valuation rings; there is no need to assume that the residue field has characteristic 3 (it need not even be finite). Googling "Deligne Serre lifting" brings up multiple detailed accounts of this lemma and its proof. So the strategy you propose will work, as long as you can find the necessary weight 1 Eisenstein series. You say you're interested in characteristic 7; try typing EisensteinForms(Gamma1(7), 1).integral_basis()[0].qexp(50).change_ring(Zmod(7)) into Sage. So this argument will give you a weight 2 eigenform $g$ of level 931 which is congruent to $f$ mod some prime above $7$. It's still not completely clear that the mod $7$ representations of $f$ and $g$ have projective image that is tetrahedral, because the mod 7 representation attached to $f$ might have smaller image than the characteristic 0 one; but you might be able to rule this out if you think about what the smaller images might be -- the only proper quotients of $S_4$ are cyclic and dihedral groups, and these would force rather strong conditions on the reductions of $f$ and $g$ mod 7. Sadly, getting your hands on $g$ explicitly is rather hard work because the weight 2 space it lives in is rather large: you can pin down what the character of $g$ should be, but the space of weight 2 forms of level 931 with this character is 86-dimensional. But it should be possible to get a list of eigenforms in this space if you leave Sage (or Magma) running for long enough, and to check for each of them whether there is a prime of its coefficient field above $7$ with respect to which it is congruent to $f$. - ### 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. If f is an eigenform of level 800 and charachter $\chi$, then how we have $\chi(3)=0$? (since $a_{9}=-1$ and $a_{3}=-i$, then $\chi(3)=0$) -	{"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": 47, "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": 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": 0.9401476979255676, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/tagged/kinematics?sort=active&pagesize=15	# Tagged Questions The description of the movement of bodies by their position, velocity, acceleration (and possibly higher time derivatives, such as, jerk) without concern for the underlying dynamics/forces/causes. 14answers 6k views ### Why does kinetic energy increase quadratically, not linearly, with speed? As Wikipedia says: [...] the kinetic energy of a non-rotating object of mass $m$ traveling at a speed $v$ is $mv^2/2$. Why does this not increase linearly with speed? Why does it take so much ... 1answer 27 views ### Static friction force on a block in a tunnel Linked to this Comparing Static Frictions Suppose there is a cuboidal vertical tunnel , and a cubical block in it such that all surface of the block are in contact with the four walls of the ... 4answers 117 views ### Integrating radial free fall in Newtonian gravity I thought this would be a simple question, but I'm having trouble figuring it out. Not a homework assignment btw. I am a physics student and am just genuinely interested in physics problems involving ... 2answers 65 views ### Centripetal Force Acceleration Suppose you want to perform a uniform circular motion . Then a body performing uniform circular motion horizontally needs an acceleration $= \frac{v^2}{r}$ at each point on the circular path with ... 1answer 103 views ### Utilizing maximum acceleration $a$ for displacement $d$ with initial velocity $v_0$ and final velocity $v_1$ Problem My goal is to move an object from point a to b (displacement $d$) as fast as possible utilizing the maximum available acceleration $a_{max}$, taking into account the initial velocity $v_0$ ... 2answers 793 views ### Rigid body dynamics joints I can't seem to find any info on connected rigid bodies by a joint. Can someone explain the basics to me? I'm trying to do a little research to find out how feasible it would be to implement 3d ... 3answers 2k views ### Difference b/w Kinetics & Kinematics w/concrete example (I know whether I understand this or not doesn't matter much to my work & study but am just curious.) 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This came up again today when I decided to ... 1answer 31 views ### Projectile motion: Thrust in x direction = acceleration in x direction? 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I was watching a youtube video the other day where an economist said that he challenged his physics professor on this question back when he was in school. His professor said each scenario is the same, ... 2answers 79 views ### Do you use the magnitude equation to get speed from an accelerometer? A guy suggested to me that getting speed from an accelerometer required the use of this equation: $\text{speed} = \sqrt{x^2 + y^2 + z^2}$ This does not make any sense to me, all that you would get ... 2answers 90 views ### Doubt in Kinematics I know that this isn't the place for such basic questions, but I didn't find the answer to this anywhere else. It's pretty simple: some particle moves in straight line under constant acceleration from ...	{"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": 55, "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": 0.9284766912460327, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/59128/ambiguity-in-computing-straight-line-complexity-of-multivariate-polynomials?answertab=votes	# Ambiguity in computing straight-line complexity of multivariate polynomials I was wondering: I'm desiring to compute the following polynomial, $X_1^2 + 2 X_1\cdot X_2 +X_2^2$ where $X_1,X_2$ are indeterminates. I can store intermediate results and use the binary operators of multiplication and addition on the indeterminates. If the cost of taking a scalar multiple is $0$, and the cost of using addition and multiplication is $1$, then consider the following: Let $P=X_1+X_2$ (cost $1$). Let $Q=P \cdot P$ (cost $1$). Now at this point $Q$ is $(X1+X2) \cdot (X1+X2)$. Am I allowed to be done with the computation at this point? Or must I end with the explicit $X_1^2+ 2 X1 \cdot X2 +X_2^2$ in order to have computed $X_1^2+2X_1 \cdot X_2+ X_2^2$? (Which would take $5$ steps). - I don't understand. You say you can use multiplication and addition "on the indeterminates". a) I'm not sure what that means, and b) in $Q=P\cdot P$ you're using multiplication on "$P$", by which you may be denoting either the result of performing $X_1+X_2$ or the polynomial $X_1+X_2$, neither of which is an indeterminate. The only way I can make sense of the first part of the question is if you in fact mean that you can perform multiplication and addition on the values taken by the indeterminates; but in that case $(X_1+X_2)\cdot(X_1+X_2)$ is the same as $X_1^2+2X_1\cdot X_2+X_2^2$. – joriki Aug 23 '11 at 5:58 Sorry, I was a bit tired last night and couldn't find the answer to this anywhere. Anyways yes, I mean you can perform multiplication and addition on the values taken on by the indeterminates. And you answered my question. Thanks! – seagaia Aug 23 '11 at 13:13 ## 1 Answer Since you said my comment answered your question, I'm turning it into an answer so the question can be marked as answered: If multiplication and addition refer to the values taken by the indeterminates, then $(X_1+X_2)\cdot(X_1+X_2)$ is the same as $X_1^2+2X_1\cdot X_2+X_2^2$, so there's nothing left to do; the computation was carried out efficiently with cost $2$ instead of cost $5$. (By the way, under this interpretation of the question, presumably the coefficients of the polynomials are considered to be taken from some (relatively simple) subset of the possible values of the indeterminates? Else it wouldn't make sense to have scalar multiplication cost 0 and indeterminate multiplication cost 1, since they'd be the same operation.) -	{"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": 23, "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": 0.9522695541381836, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/4635/understanding-weight-on-an-inclined-plane	# Understanding weight on an inclined plane I'm trying to solve a problem where I have an object resting on an inclined plane, with the angle of the plan being alpha, and the weight being w. I'm having trouble figuring out how I can calculate the component of the weight parallel to the plane. I also want to find out the weight component perpendicular to the plane. I don't want an outright answer, more of an explanation to help me understand. Thanks! - John, what is Your school background, especially what about math? – Georg Feb 5 '11 at 19:33 I need a sketch tool that's really handy for questions like this. – dmckee♦ Feb 5 '11 at 20:49 @dmckee: try pencil and paper :) – Marek Feb 5 '11 at 21:29 @Marek: Thanks. I'm partial to white boards, myself. – dmckee♦ Feb 5 '11 at 21:32 1 @dmckee: but I really meant it. If you have a scanner (or even just a digital camera) at your disposal, I don't think there's any quicker way than drawing with hand. Though white board would do the trick too, I suppose. Personally, I like drawing stuff on my windows. It produces cool math/phys shadows too :) – Marek Feb 5 '11 at 22:16 show 1 more comment ## 2 Answers Here's a quick sketch: Gravity is the vector $u$. Its components in the plane and against the plane are $v$ and $w$ respectively. You want to find $v$. The angle between the plane and horizontal is the same as the angle between $w$ and $u$, which allows you to find a simple trig relation to solve the problem. - Imagine the inclined plane, and imagine an arrow pointing straight down, representing your weight (the force of gravity). Then imagine rotating that picture so that the plane is horizontal, with the arrow rotated sideways by alpha. Imagine now a horizontal x axis, and a vertical y axis, like they always draw in the textbooks. The x and y axes should cross (intersect) at the tail of the weight arrow. Now imagine a horizontal line that starts at the tip of the arrow, and ends at the (vertical) y axis. That line is the component of the weight that is parallel to the inclined plane. Then use what you know about sines and cosines (trigonometry) to calculate the amount of weight in the directions parallel and perpendicular to the incline. -	{"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": 0.9384157061576843, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/64219?sort=newest	## Indecomposable vector spaces and the axiom of choice ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) It is a known result by A. Blass that the axiom of choice is equivalent to the assertion that every vector space has a basis. (Rubin's Equivalents of the Axiom of Choice: form B) It is also known that the axiom of choice equivalent to the existence of a field $F$ such that for every vector space $V$ over $F$ has the property that if $S\subseteq V$ is a subspace of $V$ then there is $S'\subseteq V$ such that $S\oplus S'=V$. In this case, assuming the axiom of choice does not hold, we have a vector space without a basis somewhere in the model, and over every field there is a vector space $V$ that has a subspace without a complement. My question: Can we derive from the absence of choice that there will be a vector space (over some field? any field?) such that for any nontrivial subspace $S$ of $V$ there is no subspace $S'$ of $V$ such that $S\oplus S'=V$? My initial instinct was to try and take a vector space without a basis and start decomposing it, somehow deriving if we stop at a successor stage we have found such subspace, and if we happened to hit a limit stage with an empty subspace then we can construct a basis. I think I overlooked something because I couldn't finish my argument. Edit: Just to be clear, I am not looking whether or not this is consistent with ZF, but rather if the negation of choice implies there exists such vector space. - I think the tag linear-algebra would also be appropriate. – SNd May 7 2011 at 17:58 ## 2 Answers This is not meant to be a complete answer to the question (which I find very interesting), but rather an explanation for why the particular approach you described is bound to run into difficulties. I'm new here, and I'm not sure this belongs as an "answer;" please let me know if this is inappropriate. It is consistent with ZF that there is a vector space (over the field of two elements) which does not have a basis and yet every nontrivial subspace admits a decomposition into two complementary subspaces. In fact, one such example is quite familiar. Set $V = \mathcal{P}(\omega)/\mathrm{FIN}$, viewed as a vector space over $\mathbb{Z}/2\mathbb{Z}$ (with, of course, symmetric difference serving as the addition operation). A fairly straightforward Baire category argument shows that $V$ does not have a basis in a model of ZFDC + all sets of reals have the Baire property. However, any subspace $W \subseteq V$ with more than two elements can be decomposed as $S_0 \oplus S_1$: simply fix some element $A \in W$ which does not almost contain every element of $W$ and set $S_0 = \{B \in W : B \mbox{ is almost contained in A}\}$ and $S_1 = \{B \in W : B \mbox{ is almost disjoint from A}\}$. I'm sure the same argument actually works for $\mathcal{P}(\omega)$ itself, but I hate thinking about finite sets. Edit: OK, now I'm worried about offending finite sets. I don't hate thinking about you in general, but I'd rather not think about you in the particular Baire category argument I have in mind. - This example is very interesting as itself, and it does explain why my approach ran into some difficulties. Do note that this works for the vector space Francois mentioned, as every subspace is of finite dimension, and spaces of finite dimension can easily be decomposed even without the axiom of choice. – Asaf Karagila May 8 2011 at 11:27 Right. I suppose I should've emphasized that I was going for a simple example (working in ZFDC) with lots of subspaces. – Clinton Conley May 8 2011 at 11:41 ### 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. H. Läuchli [Auswahlaxiom in der Algebra, Comment. Math. Helv. 37 1962/1963, MR143705] constructed a permutation model wherein there is a vector space which is not finite dimensional and such that all of its proper subspaces are finite dimensional. This is an example of a vector space with the property you want. - This example appears in Jech's The Axiom of Choice. I am well aware of this construction, my question is whether or not this sort of behaviour (where you can replace "finite subspaces" by "subspaces of cardinality $<\kappa$" for a fixed $\kappa$) can always be found in the absence of choice. – Asaf Karagila May 7 2011 at 20:42 Ah! That's not the way I read your question. I will have to think about the situation you have in mind. – François G. Dorais♦ May 8 2011 at 1:34 I will clarify my question as well. Thank you! – Asaf Karagila May 8 2011 at 5:35	{"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": 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": 0.9559314250946045, "perplexity_flag": "head"}
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My question is: Is $\operatorname{rank}(A-B) \geq \operatorname{rank}(A) - \operatorname{rank}(B)$ true in general? Or maybe under certain assumptions?	{"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": 118, "mathjax_display_tex": 5, "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": 0.9021719098091125, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/256482/transition-functions-for-the-tautological-bundle	# Transition functions for the tautological bundle Define the tautological bundle over $CP^1$ to be $\tau = \{[a_1, a_2], (z_1, z_2) \in CP^1\times\mathbb{C}^2 \| \exists \lambda \in \mathbb{C} \;\text{such that} \;\lambda (z_1,z_2) = (a_1, a_2) \}.$ Then $\tau$ trivializes over open sets $U_i = \{[a_1,a_2] \| a_i \neq 0\}, i=1,2$ where $(\pi, \phi_i):\pi^{-1}(U_1) \rightarrow CP^1\times\mathbb{C}$ is given by $\phi_i([a_1,a_2]) = a_i.$ Note that $(\pi, \phi_1)^{-1}:U_1\times\mathbb{C} \rightarrow \pi^{-1}(U_1)$ is given by $([a_1,a_2], w) \rightarrow ([a_1, a_2], (w, \frac{a_2}{a_1}w)).$ Thus the transition function $T_{12}$ is given by $[a_1, a_2] \rightarrow \frac{a_2}{a_1}.$ Up to homotopy we need only look at the map on the equator of $CP^1 = S^2.$ Since we can identify $U_1$ with $S^2 \setminus \infty = \mathbb{C}$ by $[a_1,a_2] \rightarrow \frac{a_2}{a_1},$ the transition function $T_{12}$ appears to be a degree one map $S^1 \rightarrow S^1.$ Is this right? If so, why is this bundle always denoted $O_\mathbb{P}(-1)?$ Shouldn't it be called $O_\mathbb{P}(1)?$ -	{"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": 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": 0.9300082325935364, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?t=184615&page=2	Physics Forums ## A proton is released from rest at the positive plate of a parallel-plate capacitor. A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the negative plate with a speed of 50,000 m/s. What will be the proton's final speed if the experiment is repeated with double the amount of charge on each capacitor? I seem to be stuck on this one. I initially tried to use 1/2 m$$^{}_{i}$$v$$^{2}$$ + V$$^{}_{i}$$q = 1/2 m$$^{}_{f}$$v$$^{2}$$ + V$$^{}_{f}$$q but i ran into a wall. Help please. 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 Recognitions: Homework Help Science Advisor That's the wrong way to think of it. What is the same about the capacitor? (Hint, it's the same capacitor) What does doubling the charge of a capacitor do to the voltage across the capacitor? What effect does this voltage increase have on the electric potential energy of the electron? the capacitor plates are still the same distance apart. same size. same signs(one is positive and one is still negative) by doubling the the charge of the plates it would seem that the voltage would increase by a factor of 2 as well. increasing the voltage would increase the electrical potential of the proton. Are these asumtions correct? Recognitions: Homework Help Science Advisor ## A proton is released from rest at the positive plate of a parallel-plate capacitor. Quote by Xaspire88 the capacitor plates are still the same distance apart. same size. same signs(one is positive and one is still negative) by doubling the the charge of the plates it would seem that the voltage would increase by a factor of 2 as well. increasing the voltage would increase the electrical potential of the proton. Are these asumtions correct? Doubling the voltage will increase the potential energy of the proton by how much? And then, how much more KE will the proton have when it reaches the other plate? the potential energy = Vq so if the voltage increases by a factor of two then the potential energy will increase by a factor of two. Which would then make sense to me that the kenetic energy would have to double to keep conservation of energy? Admin Quote by Xaspire88 the potential energy = Vq so if the voltage increases by a factor of two then the potential energy will increase by a factor of two. Which would then make sense to me that the kenetic energy would have to double to keep conservation of energy? Correct. The proton (initially at rest, so KE = 0) is accelerated across the potential difference. The electric potential energy is converted to the proton's KE. So if initially the proton had a final velocity of 50,000m/s, once the charge on the plates was doubled and the potential energy was doubled and the kinetic was therefor doulbed would the resulting final velocity be double that of the original , 100,000m/s? if Ke= 1/2mv^2. the mass is the same but since the kinetic energy doubled the velocity would as well? Correct line of thought? Admin Careful. What is the relationship between magnitude of velocity and kinetic energy. Take V - and the kinetic energy, KE = 1/2 mV2, now let V become 2V and KE = 1/2 m (2V)2. Find the expression for V in terms of KE from the first expression. If we solve for v in terms of KE from the first expresion v= $$\sqrt{(2KE)/m}$$ now that is without using 2v using KE= 1/2m2v^2 v= $$\sqrt{KE/m}$$ or since you have it as KE = 1/2 m (2V)^2 when you solve for V in terms of KE would v= 1/2 $$\sqrt{(2KE)/m}$$ my work: KE = 1/2 m (2V)^2 2KE=m(2V)^2 (2KE)/m = (2V)^2 squareroot((2KE)/m)= 2V 1/2 squareroot((2KE)/m) = V Recognitions: Homework Help So if the energy doubles... how does the speed change? rather than have (2V) wouldn't i have (2KE) since we have determined that we have double the PE which become double the KE? so the expression would then be 2KE = 1/2 mv^2 which would mean that V in terms of KE would equal v = $$\sqrt{(4KE)/m}$$ which simplified would be v = 2 $$\sqrt{KE/m}$$? would make sense to me that the increase in v due to the increase in KE would be a factor of 2? or should it only be an increase of $$\sqrt{2}$$? without the increase v= $$\sqrt{(2KE)/m}$$ after v = $$\sqrt{(4KE)/m}$$ since the only difference is from 2 to 4....... the increase would be by a factor of $$\sqrt{2}$$? which would then make the velocity after doubling the charge on the plates v= 70,710.7m/s. Recognitions: Homework Help Quote by Xaspire88 $$\sqrt{2}$$? Yes, that's the answer... if you are unsure, just take the ratio of the new velocity to the old: $$\sqrt{(2KE)/m}$$ This is correct for v... so v1 = $$\sqrt{(2KE_1)/m}$$ v2 = $$\sqrt{(2KE_2)/m}$$ What does v2/v1 come out to? Recognitions: Homework Help Quote by Xaspire88 which would then make the velocity after doubling the charge on the plates v= 70,710.7m/s. yes. well the ratio of the new velocity to the old is going to be $$\sqrt{2}$$ (70,710.7m/s)/(50,000m/s) = $$\sqrt{2}$$ which doesn't really confirm anything for me because i came up with Vfinal by multiplying 50,000m/s by $$\sqrt{2}$$ so it only make since that working backwards would give me that? Maybe i missed the point of your question. I'm sorry if this is frustrating. I really appreciate the time and effort from all of you. Recognitions: Homework Help Quote by Xaspire88 well the ratio of the new velocity to the old is going to be $$\sqrt{2}$$ (70,710.7m/s)/(50,000m/s) = $$\sqrt{2}$$ which doesn't really confirm anything for me because i came up with Vfinal by multiplying 50,000m/s by $$\sqrt{2}$$ so it only make since that working backwards would give me that? Maybe i missed the point of your question. I'm sorry if this is frustrating. I really appreciate the time and effort from all of you. Oops... sorry. I understand now... I didn't mean take the ratio of the numbers... I gave expressions for v2 and v1 in terms of E2 and E1... take the ratio of v2/v1 using those two expressions... oh ok now i see that by simplifying $$\sqrt{(2(2KE))/m}$$ / $$\sqrt{(2KE)/m}$$ = $$\sqrt{2}$$ which confirms that it would increase by a factor of $$\sqrt{2}$$. Thank you guys. Recognitions: Homework Help Quote by Xaspire88 oh ok now i see that by simplifying $$\sqrt{(2(2KE))/m}$$/$$\sqrt{(2KE)/m}$$ = $$\sqrt{2}$$ which confirms that it would increase by a factor of $$\sqrt{2}$$. Thank you guys. yup. you're welcome. 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http://mathoverflow.net/revisions/16706/list	Return to Answer 3 added references The reason that in stochastic calculus the left-hand and right-hand sums give different integrals really all boils down to quadratic variations. Processes such as Brownian motion have non-zero quadratic variation. Suppose that you are integrating a process X with respect to some other process Y, then choosing a partition 0=t0≤...≤tn=t the approximations using left and right hand sums respectively are, $$\int_0^t X\ dY\approx \sum_{k=1}^nX_{t_{k-1}}(Y_{t_k}-Y_{t_{k-1}})$$ $$\int_0^t X\ \overleftarrow{d}Y\approx \sum_{k=1}^nX_{t_{k}}(Y_{t_k}-Y_{t_{k-1}})$$ The difference between these can be bounded as follows $$\sum_{k=1}^n(X_{t_k}-X_{t_{k-1}})(Y_{t_k}-Y_{t_{k-1}}) \le\max_k\vert X_{t_k}-X_{t_{k-1}}\vert\sum_k\vert Y_{t_k}-Y_{t_{k-1}}\vert$$ The final term on the right hand side converges to the variation of Y as the mesh of the partition goes to zero and, if X is continuous, the first term goes to zero. So, in standard calculus where integration is always with respect to finite variation functions, it makes no difference whether the left hand or right hand sums are used. Alternatively, the Cauchy-Schwarz inequality can be applied to get the following bound. $$\sum_{k=1}^n(X_{t_k}-X_{t_{k-1}})(Y_{t_k}-Y_{t_{k-1}}) \le\sqrt{\sum_{k=1}^n(X_{t_k}-X_{t_{k-1}})^2\sum_{k=1}^n(Y_{t_k}-Y_{t_{k-1}})^2}.$$ As the mesh of the partition goes to zero, the terms inside the square root converge to the quadratic variations of X and Y respectively, denoted by [X] and [Y]. Again, in standard calculus, we use (continuous) finite variation functions, which have zero quadratic variation. However, in stochastic calculus, processes such as Brownian motion have non-zero quadratic variation. Convergence to the quadratic variation along partitions occurs in the sense of convergence in probability - it does not have to converge in the usual sense with any positive probability. A Brownian motion B has [B]t=t, so the left and right hand sums can converge to different numbers. With the Stratonovich integration the correct thing to use is the average of the left and right hand sums, not the mid-point. For Ito processes, which are integrals with respect to Brownian motion and time, it makes no difference. This is because their quadratic variations are absolutely continuous. However, for general continuous semimartingales, the mid-point sums don't actually have to converge to anything. (See Sur quelques approximations d'intégrales stochastiques by Marc Yor). So, Stratonovich integration uses the average of the left and right hand sums, and the difference between this and the Ito integral is precisely half of what you get using the right-hand sums. As to the question of whether this difference shows up in standard (non-stochastic) calculus, the answer is, as far as I know, hardly ever. In fact, given any continuous functions then you can choose a sequence of partitions along which the quadratic variation vanishes as the mesh goes to zero (I'll leave this as an exercise!). So, given any continuous function with non-zero quadratic variation with respect to some sequence of partitions, so that the left and right hand sums converge to different numbers then, there will be other partitions along which the quadratic variation vanishes. So the integral isn't really defined at all in this case. However, there is one case I have seen where left and right hand sums for deterministic functions converge to different numbers. For this to happen, you have to fix some sequence of partitions with mesh going to zero, and stick to using these to define the integral. Different partitions could lead to different results. Hans Follmer has published some papers a paper using this idea (in the 70s, I think, I'll come back with some references later)1981 (Calcul d'Ito Sans Probabilities). There aren't any natural (and useful) cases that I know of where this occurs, but you can construct some examples. Given a Brownian motion, the quadratic variation along a sequence of partitions where each is a refinement of the previous one will converge with probability one. So, selecting a Brownian motion sample path at random, you can then define integrals using these partitions in a pathwise sense, rather than using the machinery of stochastic integration. They will converge to the same thing though, so this is a bit of a cheat.Alternatively, you could construct a continuous function along the lines of the Weierstrass function while forcing the quadratic variation to converge to a nonzero number along a given sequence of partitions. Then, left and right hand Riemann sums along these partitions will converge to a different answer. For example, let s(t) be the 'sawtooth' function s(t) = 1-\|1-2{t/2}\| ({t}=fractional part of t). Then, define the following function on the unit interval, $$f(t) = s(t)+\sum_{n=1}^\infty 2^{-(n+1)/2}s(2^nt).$$ This has quadratic variation 1, calculated along dyadic partitions. The following left-hand, right-hand and `Stratonovich' integrals are easily verified, $$\int_0^1 f df = (f(1)^2-f(0)^2-[f]_1)/2.$$ $$\int_0^1 f \overleftarrow{d}f = (f(1)^2-f(0)^2+[f]_1)/2.$$ $$\int_0^1 f \partial f = (f(1)^2-f(0)^2)/2.$$ 2 added toy example Given a Brownian motion, the quadratic variation along a sequence of partitions where each is a refinement of the previous one will converge with probability one. So, selecting a Brownian motion sample path at random, you can then define integrals using these partitions in a pathwise sense, rather than using the machinery of stochastic integration. They will converge to the same thing though, so this is a bit of a cheat.Alernativelycheat.Alternatively, you could construct a continuous function along the lines of the Weierstrass function while forcing the quadratic variation to converge to a nonzero number along a given sequence of partitions. Then, left and right hand Riemann sums along these partitions will converge to a different answer. For example, let s(t) be the 'sawtooth' function s(t) = 1-\|1-2{t/2}\| ({t}=fractional part of t). Then, define the following function on the unit interval,f(t) = s(t)+\sum_{n=1}^\infty 2^{-(n+1)/2}s(2^nt).This has quadratic variation 1, calculated along dyadic partitions. The following left-hand, right-hand and `Stratonovich' integrals are easily verified, $$\int_0^1 f df = (f(1)^2-f(0)^2-[f]_1)/2.$$ $$\int_0^1 f \overleftarrow{d}f = (f(1)^2-f(0)^2+[f]_1)/2.$$ $$\int_0^1 f \partial f = (f(1)^2-f(0)^2)/2.$$ 1 The reason that in stochastic calculus the left-hand and right-hand sums give different integrals really all boils down to quadratic variations. Processes such as Brownian motion have non-zero quadratic variation. Suppose that you are integrating a process X with respect to some other process Y, then choosing a partition 0=t0≤...≤tn the approximations using left and right hand sums respectively are, $$\int_0^t X\ dY\approx \sum_{k=1}^nX_{t_{k-1}}(Y_{t_k}-Y_{t_{k-1}})$$ $$\int_0^t X\ \overleftarrow{d}Y\approx \sum_{k=1}^nX_{t_{k}}(Y_{t_k}-Y_{t_{k-1}})$$ The difference between these can be bounded as follows $$\sum_{k=1}^n(X_{t_k}-X_{t_{k-1}})(Y_{t_k}-Y_{t_{k-1}}) \le\max_k\vert X_{t_k}-X_{t_{k-1}}\vert\sum_k\vert Y_{t_k}-Y_{t_{k-1}}\vert$$ The final term on the right hand side converges to the variation of Y as the mesh of the partition goes to zero and, if X is continuous, the first term goes to zero. So, in standard calculus where integration is always with respect to finite variation functions, it makes no difference whether the left hand or right hand sums are used. Alternatively, the Cauchy-Schwarz inequality can be applied to get the following bound. $$\sum_{k=1}^n(X_{t_k}-X_{t_{k-1}})(Y_{t_k}-Y_{t_{k-1}}) \le\sqrt{\sum_{k=1}^n(X_{t_k}-X_{t_{k-1}})^2\sum_{k=1}^n(Y_{t_k}-Y_{t_{k-1}})^2}.$$ As the mesh of the partition goes to zero, the terms inside the square root converge to the quadratic variations of X and Y respectively, denoted by [X] and [Y]. Again, in standard calculus, we use (continuous) finite variation functions, which have zero quadratic variation. However, in stochastic calculus, processes such as Brownian motion have non-zero quadratic variation. Convergence to the quadratic variation along partitions occurs in the sense of convergence in probability - it does not have to converge in the usual sense with any positive probability. A Brownian motion B has [B]t=t, so the left and right hand sums can converge to different numbers. With the Stratonovich integration the correct thing to use is the average of the left and right hand sums, not the mid-point. For Ito processes, which are integrals with respect to Brownian motion and time, it makes no difference. This is because their quadratic variations are absolutely continuous. However, for general continuous semimartingales, the mid-point sums don't actually have to converge to anything. So, Stratonovich integration uses the average of the left and right hand sums, and the difference between this and the Ito integral is precisely half of what you get using the right-hand sums. As to the question of whether this difference shows up in standard (non-stochastic) calculus, the answer is, as far as I know, hardly ever. In fact, given any continuous functions then you can choose a sequence of partitions along which the quadratic variation vanishes as the mesh goes to zero (I'll leave this as an exercise!). So, given any continuous function with non-zero quadratic variation with respect to some sequence of partitions, so that the left and right hand sums converge to different numbers then, there will be other partitions along which the quadratic variation vanishes. So the integral isn't really defined at all in this case. However, there is one case I have seen where left and right hand sums for deterministic functions converge to different numbers. For this to happen, you have to fix some sequence of partitions with mesh going to zero, and stick to using these to define the integral. Different partitions could lead to different results. Hans Follmer has published some papers using this idea (in the 70s, I think, I'll come back with some references later). There aren't any natural (and useful) cases that I know of where this occurs, but you can construct some examples. Given a Brownian motion, the quadratic variation along a sequence of partitions where each is a refinement of the previous one will converge with probability one. So, selecting a Brownian motion sample path at random, you can then define integrals using these partitions in a pathwise sense, rather than using the machinery of stochastic integration. They will converge to the same thing though, so this is a bit of a cheat.Alernatively, you could construct a continuous function along the lines of the Weierstrass function while forcing the quadratic variation to converge to a nonzero number along a given sequence of partitions. Then, left and right hand Riemann sums along these partitions will converge to a different answer.	{"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": 15, "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": 0.9255994558334351, "perplexity_flag": "head"}
http://mathoverflow.net/questions/56182/almost-complex-integrability-and-algebraic-varieties	## Almost Complex Integrability and Algebraic Varieties ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $J$ be an almost complex structure on an algebraic variety $V$. As we all know, $J$ comes from a complex structure if the Nijenhuis tensor of $J$ vanishes. What I would like to know is if there exists a simpler characterisation of integrability than this for varieties (as opposed to general manifolds). - 4 When you say that $V$ is an algebraic variety, are you assuming that it is complex-algebraic (i.e., already has a complex structure), and that $J$ is an additional almost-complex structure? – S. Carnahan♦ Feb 21 2011 at 16:48 1 @Janos: If you want equivalent conditions to the Nijenhuis tensor vanishing then one is that the induced $\bar \partial$ operator defines a complex, i.e. that $\bar \partial^2 = 0$. Another one is that the exterior derivative decomposes as $d = \partial + \bar \partial$. If you want explicit examples of almost complex manifolds that are not complex, that's going to be more difficult, see the answers to tinyurl.com/4zrkar6 – Gunnar Magnusson Feb 21 2011 at 18:03 1 No, wait, do you want to hit two birds with one stone and get integrability for the almost complex structure and projectivity of the resulting complex manifold in one swoop? I'm not sure that's going to be doable... just look at two-dimensional tori. You can realize any complex torus as an integrable complex structure on $M = \mathbb R^2/\mathbb Z^2$, but not all tori are projective. Given an arbitrary integrable complex structure on $M$, I don't know how to link its properties to the question of projectivity of the torus. – Gunnar Magnusson Feb 21 2011 at 18:10 1 There is a simpler criterion that does integrability and Kählerness in one step: if there is both a Riemann metric $g$ and an almost complex structure $J$ and $\nabla J =0$ (i.e., $J$ is parallel w.r.t. the Levi-Civita connection of the Riemann metric), then $J$ is automatically integrable and $g$ is a kähler metric. – Johannes Ebert Feb 23 2011 at 22:13 1 @Gunnar: complex tori of complex dimension $1$ are automatically projective (they are smooth elliptic curves). In general, compact Riemann surfaces are smooth projective algebraic curves... – Qfwfq Feb 23 2011 at 23:50 show 10 more comments ## 1 Answer If you want equivalent conditions to the Nijenhuis tensor vanishing then one is that the induced $\bar \partial$ operator defines a complex, i.e. that $\bar \partial^2 = 0$. Another one is that the exterior derivative decomposes as $d = \partial + \bar \partial$. If you want explicit examples of almost complex manifolds that are not complex, that's going to be more difficult, see the answers to tinyurl.com/4zrkar6 To find the $\bar \partial$ operator associated to an almost complex structure $J$ on a smooth manifold $M$, one needs to note that $J$ induces a splitting $T_M \otimes \mathbb C = T^{1,0} \oplus T^{0,1}$ of the tangent bundle into $i$ and $-i$ eigenvectors (the eigenspaces being marked with $(1,0)$ and $(0,1)$, respectively). The same thing happens on the level of 1-forms (and indeed on the level of $k$-forms): they split into $(p,q)$-forms like on complex manifolds. If $\pi^{p,q} : \bigwedge^k T_M \to \bigwedge^{p,q} T_M$ is the projection onto the space of $(p,q)$-forms, then the $\bar \partial : \bigwedge^{p,q} T_M \to \bigwedge^{p,q+1} T_M$ operator associated to $J$ is $\bar \partial_J = \pi^{p,q+1} \circ d$. Once one does the calculations this comes out to $$\bar \partial \alpha = \frac 1 2 \left( d \alpha + i d J \alpha \right)$$ for a $(p,q)$-form $\alpha$. A similar formula holds for the $\partial$ operator, you just have to change $i$ to $-i$. (I may have confounded the signs here.) A very good reference for the linear algebra parts (i.e. most of this) is Chapter 2 of Huybrecht's "Complex geometry". For the conditions equivalent to the vanishing of the Nijenhuis tensor I seem to remember the first chapter of http://tinyurl.com/4cqspu7 Moroianu's notes on Kahler geometry being very helpful when I went through this a couple of months ago. Finally, on why the vanishing of the Nijenhuis tensor implies that we indeed have a complex structure I recommend Demailly's book - Chapter 8, section 11 (page 396) has all the details: http://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/agbook.pdf - Great, thanks a lot! – Janos Erdmann Feb 21 2011 at 19:09	{"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": 42, "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": 0.9127153754234314, "perplexity_flag": "head"}
http://mathhelpforum.com/number-theory/6849-prove-question.html	# Thread: 1. ## a prove question How should I prove this question? Prove that any two integers that are associates have the same norm. 2. Originally Posted by beta12 How should I prove this question? Prove that any two integers that are associates have the same norm. Two elements a, b that are associates have the following property: a\|b and b\|a If we are speaking of integers then the only possibilities are that $b = \pm a$ Thus \|b\| = \|a\|. -Dan 3. Originally Posted by beta12 How should I prove this question? Prove that any two integers that are associates have the same norm. What does it mean associates? According to PlanetMath and Wikipedia associates are the units of the ring with unity. But what ring are you in? Surly (CaptainBlank) word not in $\mathbb{Z}$ where the associates are -1 and 1. Are you in the ring of quadradic integers? 4. Thanks topsquark. Hi perfecthacker, Yes, in quadratic integers. Here is the definition of associate : Two quadratic intergers alpha and beta are associates provided that the quotient alpha/beta is a unit. That mean N(alpha/beta) = 1 or -1 5. Originally Posted by beta12 Thanks topsquark. Hi perfecthacker, Yes, in quadratic integers. Here is the definition of associate : Two quadratic intergers alpha and beta are associates provided that the quotient alpha/beta is a unit. That mean N(alpha/beta) = 1 or -1 Interesting. My book does talk about associates being defined on an integer ring (which is why I was thinking there might be a typo in your question...that you weren't talking about integers, but two elements of an integer ring), but my book doesn't define them as being units. The reason I didn't mention my thoughts about the "typo" is that two associates (as defined in my book) don't necessarily have the same norm. -Dan 6. The reason I didn't mention my thoughts about the "typo" is that two associates (as defined in my book) don't necessarily have the same norm. -Dan[/QUOTE] How does your book define associates? That is interesting issue. The conclusion which I gave are according to my book and lectures. I am sure no " typo " in it. =============== Also, perfecthacker, I am so sorry for the double post. I did really forget. 7. Originally Posted by beta12 How does your book define associates? The book is "Algebra" by Thomas Hungerford, Chapter 3, section 3, pg. 135. "A nonzero element a of a commutative ring R is said to divide an element $b \in R$ (notation: a\|b) if there exists $x \in R$ such that ax = b. Elements a, b of R are said to be associates if a\|b and b\|a." In an earlier section (Chapter 3, section 1, pg. 116) it defines a unit: "An element a in a ring R with identity is said to be left invertible if there exists $c \in R$ such that $ca = 1_R$. The element c is called a left inverse of a. An element $a \in R$ that is both left and right invertible is said to be invertible or to be a unit." I edited out the right invertible definition to make this easier to read. You can supply that extension easily enough. Clearly my book is making a distinction between the definitions of unit and associate. And I note that in $\mathbb{Z}_6$ that 2 and 4 are associates, but clearly not units. (According to the above definitions.) However these are general definitions and not specific to quadratic rings. (To which I've forgotten the definition of, and my book doesn't mention.) Perhaps that makes the difference. -Dan 8. I see.	{"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": 0.9495059847831726, "perplexity_flag": "middle"}
http://mathhelpforum.com/algebra/120112-tricky-algebraic-manipulation.html	Thread: 1. tricky algebraic manipulation I need this $\frac{-18477 s-2.82843\times 10^8}{s^2+15307 s+4.\times 10^8}$ to be in this format : $\frac{s+a}{(s+a)^2+w^2}$ I can easily factor the top to get the "s+a", $-18477\frac{s+15307 }{s^2+15307s+4*10^8}$ Now $(s+15307)^2=s^2+30614 s+234304249$ but how can I factor the bottom in such a way to get the required form? Can someone help please? 2. Have you tried completing the square? Do the numbers have to be integers? $s^2+15307s+4\times10^8 = (s^2 + 15307s + 7653.5^2) - 7653.5 ^2 + 4 \times 10^8$ $(s+ 7653.5)^2 + (4 \times 10^8 - 7653.5^2)$ 3. Originally Posted by Gusbob Have you tried completing the square? Do the numbers have to be integers? $s^2+15307s+4\times10^8 = (s^2 + 15307s + 7653.5^2) - 7653.5 ^2 + 4 \times 10^8$ $(s+ 7653.5)^2 + (4 \times 10^8 - 7653.5^2)$ Good idea. So I get: $-18477\frac{s+15307 }{(s+7653)^2+18477^2}$ which is tantalizingly close, but not in the required form. 4. Sorry I didn't read the question carefully. Didn't realise the square on the bottom has to be the square of the numerator. In this case, I don't think it will be possible to make it into the format you wish, unless 'w' can have 's' in it. This is because we cannot make the bottom square $(s + 15307)^2$ without introducing more 's' into the denominator, which would have to be part of 'w'. 5. It's ok, thanks for the help anyway. Yeah I was beginning to think it's not possible too. Oh well. Search Tags View Tag Cloud Copyright © 2005-2013 Math Help Forum. All rights reserved.	{"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": 10, "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": 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": 0.9525319933891296, "perplexity_flag": "middle"}
http://pvpmc.org/modeling-steps/shading-soiling-and-reflection-losses/incident-angle-reflection-losses/	PV Performance Modeling Collaborative A Collaborative Effort Between Sandia National Laboratories and Industry Aimed at Improving Photovoltaic Performance Modeling # Incident Angle Reflection Losses Unless the PV array is mounted on a 2-axis tracker, the incident angle for the direct component of the solar radiation will not be normal except for a few rare instances, depending on the orientation. When the angle of incidence is greater than zero, there are optical losses due to increased reflections from the module materials that need to be quantified. To define the loss factor due to angle of incidence we define a “angle of incidence modifier”, IAM, which in the case of the beam component of incident irradiance is defined as: $IAM_{B}=\frac{\tau&space;(\theta&space;)}{\tau&space;(0))}=\frac{R(\theta)-A(\theta&space;)}{R(0)-A(0)}$ where $\tau$ is the weighted transmittance, R is the weighted reflectance, and A is the weighted absorptance as a function of angle. The weighting is done with the product of the spectral response of the PV module and the spectral distribution of the radiation. Subsections: ## Email Sign Up Brought to you by Sandia National Laboratories Exceptional service in the national interest © 2012 Sandia Corporation \| Questions & 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": 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": 0.18, "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": 0.8523405194282532, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/118389/pullback-and-pushforward	# Pullback and pushforward Say I have a scheme $X$, irreducible and of finite type over a field $k$, and a closed subscheme $Y$ of $X$ with associated closed immersion $i: Y \to X$. Consider a sheaf $F$ on $X$ (for the étale topology). How should I think about the sheaf $i_* i^* F$ and the kernel of $F \to i_* i^* F$ in a more or less "concrete" way? Is it possible to visualize these sheaves somehow, or at least to compare them with $F$ in an imaginative way? -	{"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": 0.947855532169342, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/85102/list	Return to Answer 1 [made Community Wiki] This is profoundly less interesting to me than the other two answers already posted, but I think when I was in high school it would have been otherwise. Especially after having seen a bit of calculus. I recall being amazed that there were formulas --- polynomial formulas! --- for sums like $$s(n) := \sum_{i=1}^n i^k.$$ The first values, $k=0,1$ made sense. For $k=2$, things seemed a bit lucky, and for higher $k$ miraculous indeed. Here's the proof/formula I have in mind. First, define $\Delta f(x) := f(x+1)-f(x)$ and the falling factorial $x^{\underline 0} := 1, x^{\underline n} := (x-n+1) x^{\underline{n-1}}$. Then $\Delta x^{\underline n} = n x^{\underline{n-1}}$ (for $n\geq 1$). Next, prove that if $\Delta^{k+1} f(x) = 0$ and $\Delta^{k+1} g(x) =0$, then $f(x)-g(x)= c$, and from this get that if $\Delta^{k+1} f(x)=0$, then $f(x)$ is a degree $k$ polynomial. Finally, we can get the formula by using only $s(0),s(1),s(2),\dots,s(k),s(k+1)$ and Lagrange Interpolation: $$s(n) = \sum_{i=0}^{k+1} s(i) \frac{n(n-1)(n-2)\cdots (n-k-1)}{n-i} \cdot \frac{i-i}{i(i-1)(i-2)\cdots (i-k-1)}$$ with the need to cancel before multiplying. That's ugly enough that I hesitate to press the "Post Your Answer" button, but it has so very many ingredients that would have rocked my world in high school. And the final formula is easy enough to remember.	{"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": 14, "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": false, "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": 0.965505063533783, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-statistics/72769-conditional-probability-problem-makes-no-sense-me-intuitively.html	# Thread: 1. ## This conditional probability problem makes no sense to me intuitively "The king comes from a family of 2 children. What is the probability that the other child is his sister?" I know if I work through this problem using conditional probability concepts, the answer is 2/3. However, could someone explain why this is the answer? I don't see why this question is any different from the following: "A coin is flipped 2 times. The first flip lands on heads. What is the probability that the second flip lands on tails?" Yet the answer to the second question seems to obviously be 1/2. 2. Originally Posted by paulrb "The king comes from a family of 2 children. What is the probability that the other child is his sister?" I know if I work through this problem using conditional probability concepts, the answer is 2/3. However, could someone explain why this is the answer? I don't see why this question is any different from the following: "A coin is flipped 2 times. The first flip lands on heads. What is the probability that the second flip lands on tails?" Yet the answer to the second question seems to obviously be 1/2. King question: The king is a boy. The sample space for the King question is therefore BG, GB, BB where BG means the king has an older sister and GB means that the king has a younger sister. Two of the three outcomes are favourable to the king having a sister. Coin question: "A coin is flipped 2 times. The first flip lands on heads. What is the probability that the second flip lands on tails?" The sample space is HT, HH. One of the two outcomes are favourable to the second toss being a tail. Alternate coin question (King question in disgiuse): "A coin is flipped 2 times. One of the flips lands on heads. What is the probability that the other flip landed on tails?" The sample space is HT, HH, TH. Two of the three outcmes are favourable to the other flip being a tail .... 3. Thank you, that helps, but I don't understand why I'm incorrect using this reasoning: We know the king has a sibling. The sibling may be: 1) An older sister 2) A younger sister 3) An older brother 4) A younger brother 2 of those possibilities involve the other sibling being a sister, and 2 of the possibilities involve the other sibling being a brother. Assuming each possibility is equally likely, there is a 2/4 chance the sibling is a brother and a 2/4 chance the sibling is a sister. Or 1/2 for each. 4. Originally Posted by paulrb "The king comes from a family of 2 children. What is the probability that the other child is his sister?" I know if I work through this problem using conditional probability concepts, the answer is 2/3. However, could someone explain why this is the answer? I don't see why this question is any different from the following: "A coin is flipped 2 times. The first flip lands on heads. What is the probability that the second flip lands on tails?" Yet the answer to the second question seems to obviously be 1/2. Hi Paulrb, I'm going to disagree with your first answer. The probability that the other child is a girl is 1/2, for the reasons stated in your analysis of the coin flip problem. Don't confuse your king-sibling problem with the classic problem where we are told that at least one child is a boy and we are asked to find the probability that both children are boys. In your problem we have more information: we know the king is male and we are asked about the probability that his sibling is female (or male, it makes no difference.) This is different than being told that at least one child is a boy. Maybe this will be clearer if we employ some symbols. Let $X_i = 1$ if the ith child is a boy, zero otherwise. In the king-sibling problem we are told $X_1 = 1$ and we are asked to find $P(X_2 = 1)$. In the "at least one boy" version we are told $X_1 + X_2 \geq 1$ and we are asked to find $P(X_1 = 1 \text{ and } X_2 =1)$. These are two different problems. It's very difficult to come up with a real-world situation where all you know is that at least one child is a boy. Usually you have seen one child, and that child was a boy; that's different. That is one of the reasons why the "at least one..." problem gives people so much trouble. 5. I may be beating that ole dead horse, but here goes. GAME: I toss two coins out of your sight. Then cover the outcome, only you can remove the covering. You then declare if both coins show the same side or if both coins show different sides. You then remove the cover to see if you win. Now you absolutely know the possible outcomes. $\begin{array}{cc} H & H \\ H & T \\ T & H \\ T & T \\ \end{array}$ So you can easily see that either choice is equally likely. So you choose either. But now I condition the game by telling you that one coin is heads. Now you also absolutely know the possible outcomes. $\begin{array}{cc} H & H \\ H & T \\ T & H \\ \end{array}$ Would you now choose the option that both are the same? Of course not! Choosing the option that “they are different” is clearly best. Conditional probabilities has the effect of narrowing the possible outcomes. 6. Is this problem different though? The siblings aren't identical objects; they're distinct people who were born at different times. It may be more like flipping a penny and a quarter. We are told one of the coins lands on heads, without being told whether that coin is a penny or a quarter. Therefore, the other coin may be: 1) A quarter that landed on heads 2) A quarter that landed on tails 3) A nickel that landed on heads 4) A nickel that landed on tails. Meaning, a 1/2 probability of landing on heads if each of the above possibilities is equally likely. We cannot rule out any of the above possibilities because we were only given half the information required to do so. We need to know the coin's denomination as well as whether it landed on heads or tails, but we are not told the coin's denomination, only that it landed on heads. Similarly, we were not told if the king was born first or second. To awkward - I said 2/3 because that's what the solutions manual for my book says. I typed the problem verbatim. Also, if you search Google for this problem, other people have worked solutions for it and arrived at 2/3. To be honest, I am still not fully convinced that is the correct answer, but then again I don't think so many people would be incorrect. EDIT: Shortly after typing this I think I figured it out. Since we are told one of the coins landed on heads, there is a 1/2 chance the quarter may NOT land on tails, since it landed on heads. There is also a 1/2 chance the penny may NOT land on tails, since it landed heads. Therefore, each of those possibilities should only count half. 1 * quarter lands on heads (there is no way that possibility can be ruled out) .5 * quarter lands on tails (there is a 1/2 chance that possibility can be ruled out) 1 * penny lands on heads (there is no way that possibility can be ruled out) .5 * penny lands on tails (there is a 1/2 chance that possibility can be ruled out) So with the "weighted" possibilities, there's a 2/3 chance the other coin lands on heads and a 1/3 chance it lands on tails. The same logic applies to my problem. Except...wait, did I get something backwards? In the king problem there was a 2/3 chance the other sibling was a girl (opposite sex of the other) but in this case there's a 2/3 chance the other coin lands on heads (same side as the other) 7. Yes, having thought the problem over, I can see that the "at least one boy" interpretation is probably (no pun intended) the intent of the author of the problem. In that case the probability that the other child is a girl is 2/3, as others have pointed out. However, I think there are other interpretations possible, and quite likely in the real world. When the problem says "the king comes from a family of two children", what does that mean? If it means that someone saw one of the two children and that one happened to be male, then the probability that the other child is female is 1/2. But if it means no more and no less than that at least one of the children is male, then the probability that the other child is female is 2/3. Altogether, I think the problem statement is sufficiently ambiguous to allow for either possibility. 8. Originally Posted by paulrb Is this problem different though? The siblings aren't identical objects; they're distinct people who were born at different times. It may be more like flipping a penny and a quarter. EDIT: Shortly after typing this I think I figured it out. Since we are told one of the coins landed on heads, there is a 1/2 chance the quarter may NOT land on tails, since it landed on heads. There is also a 1/2 chance the penny may NOT land on tails, since it landed heads. Therefore, each of those possibilities should only count half. No, I think you are confusing yourself. The point that you are missing is this: even though siblings are distinct, they are still animals and as such they are male or female. Likewise, even though coins are distinct, they still have only two sides, heads or tails. It is binary, if we know that one of two outcomes is a 1, then there are just three possible outcomes. 9. Ok... I understand what you're saying. Still, what I said from my previous post is also true: Do you agree with this: We know the king has a sibling. The sibling may be: 1) An older sister 2) A younger sister 3) An older brother 4) A younger brother Even if this is a more complicated way to do the problem, I would like to do it this way to convince myself that the result is the same. Is there a way to show that these 4 possibilities are not equally likely, and that they are skewed so that there is a 2/3 chance the other sibling is a girl? Edit: Perhaps this is the reason why the problem specifically stated "the king?" Most people know that it's the firstborn who becomes king, not the second born. Therefore, the probability that the sibling is an older brother is 0. Then there are 3 possibilities, giving a 2/3 chance that the other sibling is his sister. Maybe that's a coincidence, though. Still, it makes all the 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": 7, "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": 0.9739049077033997, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/41307/how-does-one-calculate-the-volume-of-a-nucleus-and-the-volume-of-an-atom-in-thi?answertab=active	How does one calculate the volume of a nucleus and the volume of an atom (in this case hydrogen)? The hydrogen atom contains 1 proton and 1 electron. The radius of the proton is approximately 1.0 fm (femtometers), and the radius of the hydrogen atom is approximately 53 pm (picometers). - 1 Do you mean, how do you find the volume from the given radii, or how these radii are obtained? The former is straightforward (see answer below), the latter is a bit more complicated. – Benji Remez Oct 20 '12 at 22:05 2 Answers The Size of a proton is an extremely difficult calculation, It cannot be done by hand(so far). It requires an in depth understanding of Strong forces(Non-Abelian Gauge theory) and super-computers(See Lattice QCD) The radius of the Hydrogen atom is relatively straight forward.(See Bohr Radius) On can derive it from old quantum theory thought angular-momentum Quantization. you will find the derivations here. http://en.wikipedia.org/wiki/Bohr_model http://en.wikipedia.org/wiki/Bohr_radius Radius to volume is easy $$V=\frac{4}{3}\pi r3$$ - Indeed the proton calculation is so hard that the right thing to do is measure the size. But even then there is some ambiguity concerning what you mean by radius. You can choose the RMS charge radius, half the FWHM charge radius, the strong nuclear interaction hard-core repulsion radius, or various figures taken from the parton distribution functions in the forward scattering limit. Most of the choices fall somewhere between 0.5 and 1.5 fm, and it is common to see people pick 0.9 or 1.0 fm to use for BoTE calculations. – dmckee♦ Oct 20 '12 at 22:41 Model them as spheres. The volume of a sphere is $$V = \frac{4}{3}\pi r^3.$$ This should straightforwardly give you the answer. - 1 What's the old joke? First, assume the sheep is a sphere ... – dbaseman Oct 21 '12 at 18:28	{"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": 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": false, "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": 0.884412407875061, "perplexity_flag": "middle"}
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My quantum computation instructor keeps referring to the vector space in which he is using Dirac's bra-ket notation as an "inner product space", but doesn't it need additional properties to use 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": 124, "mathjax_display_tex": 12, "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": 0.9269253611564636, "perplexity_flag": "middle"}
http://nrich.maths.org/253/note	What is the limit as $x\to \infty$ of $x^n$?	{"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": 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": 0.18, "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": 0.9968705177307129, "perplexity_flag": "head"}
http://mathematica.stackexchange.com/questions/4702/how-can-i-solve-a-difference-differential-equation	# How can I solve a difference-differential equation? How do I ask Mathematica to try to solve a recursive relation that defines a sequence of functions? For example, suppose I know that $g_n(x) = g_{n-1}'(x)$ for $n > 0$ and that $g_0(x) = e^{2x}$. How can I ask Mathematica to find a closed form for $g_n(x)$? (This is just a placeholder equation to highlight my question; I know the answer of $g_n(x) = 2^n e^{2x}$, $n\geq 0$.) A less trivial instance of the problem would be the Hermite polynomial recursion, $$H_{n+1}(x) = 2xH_n(x) - H_n'(x)$$ I don't see how to convince either `DSolve` or `RSolve` to solve it for me. `DSolve` is unhappy because $n-1$ is used on the RHS: ````DSolve[{g[n, x] == 2D[g[n - 1, x], x]}, g, {n, x}] ```` `RSolve` just echoes my input: ````RSolve[{g[n, x] == 2D[g[n - 1, x], x], g[0, x] == Exp[2x]}, g, {n, x}] ```` I know finding a closed-form solution is going to be hopeless in most instances, but it seems like some cases like the above $g_n(x)$ should be doable. I have been unable to find any examples in the Mathematica documentation addressing this type of problem. - Are you looking for a closed form solution in terms of $n$, or you just want to compute the result for each $n$ successively? – Szabolcs Apr 24 '12 at 15:38 1 – Artes Apr 24 '12 at 15:42 I tried ```Solve[{Series[g[n, x], {x, 0, 5}, {n, 0, 5}] == Series[2D[g[n - 1, x], x], {x, 0, 5}, {n, 0, 5}]}, g, {n, x}]``` but didn't found the solution. I think you will try to search a closed form asymptotically. – GarouDan Apr 24 '12 at 16:33 @Szabolcs I'm looking for closed forms. I know that g=Table[0,{t,0,10}]; g[[1]]=Exp[2x]; Table[g[[i+1]]=D[g[[i]],x],{i,1,9}] will compute the result (I'm sure there are more elegant ways). Artes Thanks. I'm more interested in how to deal with this paradigm rather than Hermite polynomials in particular. – UVW Apr 24 '12 at 17:01 ## 2 Answers For the simple example in the question, `FindSequenceFunction` can be used to infer the general form: ````g[0]=Exp[2x]; g[n_]:=g[n]=Expand[D[g[n-1],x]] FindSequenceFunction[g/@Range[5],n] Out[3]= 2^n E^(2 x) ```` - +1 Really nice. The sentence For the simple example in the question should not be taken lightly, as FindInstance is not able to find many "easy" sequences. – belisarius Apr 24 '12 at 21:53 Thanks, @Simon. It was probably unreasonable of me to hope for more. This looks like a reasonable approach. – UVW Apr 25 '12 at 16:35 This is recursion, not solution-finding. That makes it fast and straightforward. For instance, the Hermite polynomial example, with memoization of the function (not just of its values at previous arguments of $x$), might look like this (although I'm sure the real experts can find a more elegant way to accomplish the same thing): ````ClearAll[g]; g[n_Integer, x_] := g[n][x]; g[0] = Function[{x}, 0]; g[1] = Function[{x}, 1]; g[n_Integer][x_] := With[{}, g[n] = Function[{y}, Evaluate@ Expand[2 y g[n - 1, y] - D[g[n - 1, y], y]]]; g[n][x] ] ```` After executing, say, ````In[2]:= g[5,x] Out[2]= 12 - 48 x^2 + 16 x^4 ```` the definition of `g` will be ````? g g[n_Integer][x_]:=With[{},g[n]=Function[{y},Evaluate[Expand[2 y g[n-1,y]-\!$ \SubscriptBox[\(\[PartialD]$, $y$]$g[n - 1, y]$\)]]];g[n][x]] g[0]=Function[{x},0] g[1]=Function[{x},1] g[2]=Function[{y$},2 y$] g[3]=Function[{y$},-2+4 y$^2] g[4]=Function[{y$},-12 y$+8 y$^3] g[5]=Function[{y$},12-48 y$^2+16 y$^4] g[n_Integer,x_]:=g[n][x] ```` Those are just closed-forms for particular values of $n$. I thought the O.P. wanted a closed-form expression as a function of $n$ (and $x$, of course). – murray Apr 24 '12 at 19:34 OK, thanks for the clarification @UVW. Nevertheless, often a list of solutions can be a good start: you can then pick out the coefficients of the components of the solutions and search for closed forms using `FindSequenceFunction` and their ilk. Sometimes it works! – whuber Apr 24 '12 at 20:14	{"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": 14, "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": 0.8686196208000183, "perplexity_flag": "middle"}
http://stats.stackexchange.com/questions/12647/assuming-two-gaussian-distributions-of-equal-mean-and-variance-then-how-differe	# Assuming two Gaussian distributions of equal mean and variance, then how different can we expect the top X members of each group to be? Here's the thread I got the idea from: http://www.quora.com/Do-men-have-a-wider-variance-of-intelligence-than-women/answer/Ed-Yong Basically, this is a model that might be able to explain why there aren't more females in prestigious math/science competitions - it might be a statistical artifact arising from the simple fact that there are far more males than females in math/science. If this model applies, then we may not need to assume that male intelligence has higher variance than female intelligence. The question I'd like to see addressed: If we assume equal means and equal variances (but different sample sizes), then is the model in the paper still the best model when used for predicting, say, the gender composition of the team of the 5-10 best players? Rather than just the gender composition of the grandmaster? http://rspb.royalsocietypublishing.org/content/276/1659/1161.full#sec-3 has the diagram and use of the model They basically used pairing between the top 100 males and top 100 females. Is that a valid assumption to make though? It works for grandmasters - that's true - but would it work if we're trying to select the top 10 people in any field? It's entirely possible, after all, that the expected distributions would be different if we're trying to select from a random distribution of the top 5 players of each gender, rather than the n-th ranked player of each gender. As you increase the number of players you select for a "winning" team, for example, then maybe the distributions play out in a different way. I would expect the smaller group to have higher variance in mean than the larger group. We know that to be true when averaging over the entire population distribution (as a consequence of the central limit theorem). But what if we just want 10 people from each population instead? The fact is that a lot of "potentially" top people will end up dropping out because they would do something other than spend hours a day to practice for a "winning team" High variability of the extreme value though - that makes sense if we're talking about the very top. In a large population, the extreme value is going to be very consistent. Whereas in a small population, the extreme value is going to have A LOT of variability - but that extreme value spends far more time in the left part of the (mean of extreme values) as compared to the right part of it. So if you had a head-to-head match up most years, the population with the larger sample size will win. The thing is, what about a head-to-head matchup of the top 10 members of each distribution? It would be some sort of average between the model the paper used (1 to 1 matchups) and the model where we simply had matchups of the two entire populations with each other. - 1 Does the problem imply the assumption that the population (sample) sizes are very different $n_{male}>>n_{female}$? – Dmitrij Celov Jul 5 '11 at 8:18 I do not see a clear question here. "Equal means and variances" of what? What precisely is the "model" in the referenced paper (I don't find one)? – whuber♦ Jul 5 '11 at 13:28 Dmitrij: Yes, the problem implies the assumption that the sample sizes are very different. – InquilineKea Jul 5 '11 at 17:22 whuber: Basically, we have two Gaussian distributions whose means and variances are equal. The models are just Gaussian distributions of "talent" or "intelligence" – InquilineKea Jul 5 '11 at 17:23 @Inq But (again) what is the question? By the last paragraph you seem to conclude you don't have a question. The preceding paragraphs refer to a "head-to-head matchup of the top 10 members" in distinction to "1 to 1 matchups": what is the distinction? How do you "match" a group of 10 against another group of 10, if not on a one-to-one basis? – whuber♦ Jul 5 '11 at 18:32 show 1 more comment ## 2 Answers Let's look at the top 3 of 100 Gaussians vs. the top 3 of 1000. Real statisticians will give formulas for this and more; for the rest of us, here's a little Monte Carlo. The intent of the code is to give a rough idea of the distributions of $X_{(N-2)} X_{(N-1)} X_{(N)}$; running it gives ````# top 3 of 100 Gaussians, medians: [[ 2. 2.1 2.4]] # top 3 of 1000 Gaussians, medians: [[ 2.8 2.9 3.2]] ```` If someone could do this in R with rug plots, that would certainly be clearer. ````#!/usr/bin/env python # Monte Carlo the top 3 of 100 / of 1000 Gaussians # top 3 of 100 Gaussians, medians: [[ 2. 2.1 2.4]] # top 3 of 1000 Gaussians, medians: [[ 2.8 2.9 3.2]] # http://stats.stackexchange.com/questions/12647/assuming-two-gaussian-distributions-of-equal-mean-and-variance-then-how-differen # cf. Wikipedia World_record_progression_100_metres_men / women import sys import numpy as np top = 3 Nx = 100 Ny = 1000 nmonte = 100 percentiles = [50] seed = 1 exec "\n".join( sys.argv[1:] ) # run this.py top= ... np.set_printoptions( 1) # .1f np.random.seed(seed) print "Monte Carlo the top %d of many Gaussians:" % top # sample Nx / Ny Gaussians, nmonte times -- X = np.random.normal( size=(nmonte,Nx) ) Y = np.random.normal( size=(nmonte,Ny) ) # top 3 or so -- Xtop = np.sort( X, axis=1 )[:,-top:] Ytop = np.sort( Y, axis=1 )[:,-top:] # medians (any percentiles, but how display ?) -- Xp = np.array( np.percentile( Xtop, percentiles, axis=0 )) Yp = np.array( np.percentile( Ytop, percentiles, axis=0 )) print "top %d of %4d Gaussians, medians: %s" % (top, Nx, Xp) print "top %d of %4d Gaussians, medians: %s" % (top, Ny, Yp) ```` - What is this code supposed to be doing? Exactly what question does it purport to answer? – whuber♦ Jul 5 '11 at 13:28 Thank you for the clarification. If you get a chance, please insert that into your reply for the benefit of future readers. – whuber♦ Jul 5 '11 at 15:40 1 BTW the CDF for the max of $N$ iid variates whose CDF is $F$ is just $F^N$. This follows directly from the definitions of CDF and independence. Thus, for any $N$, you can replace the Monte-Carlo simulation with a single (numerical) integration. – whuber♦ Jul 5 '11 at 17:07 1 True -- but a) one can Monte Carlo any distribution given by data, e.g. times for the 100-metre dash. b) $F^N$ gives me no feel, no intuition, for the distributions for N = 100, 1000, and imho beginners have to build up intuition; rug plots anyone ? – Denis Jul 5 '11 at 17:24 2 (a) The formula I gave is completely general; (b) $F^N$ is incredibly simple to plot :-). The mathematics provides intuition far better than the results of a few simulations, because we can generalize from the math using pure reasoning, an approach that is not available from inspecting simulation results. – whuber♦ Jul 5 '11 at 18:25 The earlier answer doesn't address the question of the gender composition of the top $5$ or top $10$ players. The analytical answer is simple, and it doesn't depend on the underlying distribution (as long as it is the same for men and for women, and continuous, and each person's ability is assumed to be independent of anyone else's). Under these assumptions, the number of women among the top $k$ follows a hypergeometric distribution very close to a binomial distribution. If there are $10$ times as many men participating as women, then each spot has a $1/11$ chance to be occupied by a woman. For that question, the techniques used to produce the results in the paper you cite are not needed. If you want to find the expected value of the $i$th highest out of a sample of size $n$, this is the expected value of an order statistic, and this does depend (slightly) on the distribution. For a uniform distribution on $[0,1]$, the expected value of the $i$th highest value out of $n$ is $\frac{n+1-i}{n+1}$. I think the expected value for an order statistic of a normal distribution doesn't have a closed form in general, but there are good approximations which tell you how to adjust the naive guess of $\Phi^{-1}(\frac{n+1-i}{n+1})$ standard deviations above the mean. While it may be worth understanding these order statistics as a null hypothesis, I doubt that the distribution of ratings for chess players is so well approximated by the normal distribution that the top ratings out of many millions of players are properly predicted by the corresponding values for a normal distribution. Typically, when you use a normal approximation, you don't count on it working several standard deviations from the mean, and it certainly doesn't work in the other direction. -	{"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": 0.9177863597869873, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?t=580787	Physics Forums ## How to find type of conic section from its general equation? Given the general equation of conic section: Ax2+Bxy+Cy2+Dx+Ey+F=0 How can we find the type of conic section from the above equation? I've used method of rotating coodinate around the origin (from Oxy to Ox'y', Ox' makes an angle θ to Ox, counterclockwise) such that the symmetry axes of the conic curve are Ox' and Oy', respectively. The general equation of conic section in the new coodinate without coefficient B': A'x'2+C'y'2+D'x+E'y+F'=0 The new coefficients and parameters would be found from origin: tan2θ=B/(A-C) A', C' is the roots of quadratic equation: X2-(A+C)X+(4AC-B2)/4=0 D'=Dcosθ+Esinθ E'=-Dsinθ+Ecosθ F'=F. And then find the type of conic section normally. But when I applied this method, there was something strange. Consinder the relation: x2+2xy+y2-8x+8y=0 From above, we can find: θ=π/4 or 3π/4 (There're two orientations of Ox'y') A', C' are found from: X(X-2)=0. So A', C' can be alternative. The parabola has a paralleil to Ox' or Oy' is symmetric axis depends on which value of A' and C' we choose. So, we can find four cases of orientation of this parabola. But the graph I obtained using graph draw software just only gave one case. (The vertex of parabola direct toward the quad-corner II(upper left corner)) I do not understand at all. Could you explain me why does it ignore the other cases and do they actually exist? Thank you so much! :) Mentor Quote by truongson243 Given the general equation of conic section: Ax2+Bxy+Cy2+Dx+Ey+F=0 How can we find the type of conic section from the above equation? I've used method of rotating coodinate around the origin (from Oxy to Ox'y', Ox' makes an angle θ to Ox, counterclockwise) such that the symmetry axes of the conic curve are Ox' and Oy', respectively. The general equation of conic section in the new coodinate without coefficient B': A'x'2+C'y'2+D'x+E'y+F'=0 The new coefficients and parameters would be found from origin: tan2θ=B/(A-C) A', C' is the roots of quadratic equation: X2-(A+C)X+(4AC-B2)/4=0 D'=Dcosθ+Esinθ E'=-Dsinθ+Ecosθ F'=F. And then find the type of conic section normally. But when I applied this method, there was something strange. Consinder the relation: x2+2xy+y2-8x+8y=0 From above, we can find: θ=π/4 or 3π/4 (There're two orientations of Ox'y') A', C' are found from: X(X-2)=0. So A', C' can be alternative. The parabola has a paralleil to Ox' or Oy' is symmetric axis depends on which value of A' and C' we choose. So, we can find four cases of orientation of this parabola. This isn't a parabola - it's a circle. If A = C, the conic section is a circle. As you already know, the xy term indicates a rotation. Let me assume that B = 0 so that there is no rotation. If A = 0 and C $\neq$ 0, the conic is a parabola that opens left or right. If C = 0 and A $\neq$ 0, the conic is a parabola that opens up or down. Opening left vs right or up vs. down are determined by the agreement or disagreement in sign between the squared term and the linear term in the other variable. For example, y2 + x = 0 (A = 0, C = 1, D = 1) is a parabola that opens to the left. For another example, x2 - y = 0 (A = 1, C = 0, E = 1) is a parabola that opens up. If A and C are opposite in sign, the conic is a hyperbola. If A and C are the same sign, but not equal, the conic is an ellipse. If A = 0 and C = 0, the "conic" is a degenerate hyperbola (actually two straight lines). Quote by truongson243 But the graph I obtained using graph draw software just only gave one case. (The vertex of parabola direct toward the quad-corner II(upper left corner)) I do not understand at all. Could you explain me why does it ignore the other cases and do they actually exist? Thank you so much! :) Quote by Mark44 This isn't a parabola - it's a circle. If A = C, the conic section is a circle. Thank you so much but it's only a circle in case the term of xy is equal to zero. In this case, when I change the coordinate into which Ox' is incline 45° to Ox (horizontal), the new coefficients of general equation in new coordinate are A', C' (B' = 0 in the new coor. so that there is no rotation) satisfy: X2-2X=0, so I have to choose A' = 0 or C' = 0 and the parabola will open left or downward respectively. But, the computer only illustrated the case C=0, so we have the graph of function: y=-x2/4$\sqrt{2}$ (in new coordinate) Why didn't it choose A'=0 so we'll have the graph of x=-y2/4$\sqrt{2}$, the open-left parabola or doesn't it exist? Could you have a look at this link below? It's what my computer illutrated for the relation x2+2xy+y2-8x+8y=0 or, (x+y)2=8(x-y) http://s880.photobucket.com/albums/a...t=parabola.jpg ## How to find type of conic section from its general equation? The type of the conic section can be read from its discriminant D=b2-4ac. D>0: hyperbola D=0: parabola D<0: ellipse, incl circle as a special case when b=0, a=c. This assumes the conic section exists over R, is irreducible. (edit: and given by ax2+bxy+cy2+dx+ey+f=0) Thread Tools \| \| \| \| \|-----------------------------------------------------------------------------------\|----------------------------------\|---------\| \| Similar Threads for: How to find type of conic section from its general equation? \| \| \| \| Thread \| Forum \| Replies \| \| \| Calculus & Beyond Homework \| 5 \| \| \| General Math \| 4 \| \| \| General Physics \| 1 \| \| \| Precalculus Mathematics Homework \| 1 \| \| \| Linear & Abstract Algebra \| 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": 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": 0.3, "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": 0.9048700928688049, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?p=4151241	Physics Forums ## Expected value paradox? Not long ago I was surprised to learn that when trying to maximize the expected long-term growth rate of your money, it is sometimes necessary to bet on an outcome that has negative expected value (in addition to outcomes that have positive expectation). See http://www.physicsforums.com/showthread.php?t=637064. I think I get that now, although I still don't consider it as obvious as everyone else seemed to. I'm used to problems where you are trying to maximize the total expected value on one round of betting. Surely, I thought, you would never bet any amount of your money on an outcome with negative expectation if you were trying to maximize your total expected value. Can you think of an exception to this? Can you think of a situation where there are multiple outcomes to bet on, and in order to maximize your expected value (not expected rate of return or longterm growth rate) you must bet some fraction of your money on an outcome with negative EV, along with the other positive EV bets you make? 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 Recognitions: Homework Help Science Advisor Expectation values (for a single round) are additive. You cannot increase them with a bet with negative expectation value. I think I get that now, although I still don't consider it as obvious as everyone else seemed to. Here an example which might be clearer: Two fair dice is rolled. If they show "1,1", you get 35.9 times your bet. If they show anything else, you get 1.5 times your bet. The second one has a positive expectation value, the first one a negative one. You want to bet a lot of money on the second (80%+? Did not calculate it) and keep the rest. But then you risk losing much which throws you back by several rounds. Give ~2% of your money to the first one, and you lose at most 30%, while the money increase in the other case is just a little bit smaller. Quote by mfb Expectation values (for a single round) are additive. You cannot increase them with a bet with negative expectation value. If the odds you are getting are fixed, then you definitely can't increase your expected value by betting on a negative EV outcome. But what if the amount you bet affects the odds ("moves the line"), as is the case in a bet pool? Then your expected value is no longer linear in the amounts you bet on each outcome. Here an example which might be clearer: Two fair dice is rolled. If they show "1,1", you get 35.9 times your bet. If they show anything else, you get 1.5 times your bet. The second one has a positive expectation value, the first one a negative one. You want to bet a lot of money on the second (80%+? Did not calculate it) and keep the rest. But then you risk losing much which throws you back by several rounds. Give ~2% of your money to the first one, and you lose at most 30%, while the money increase in the other case is just a little bit smaller. That's a good example. I understand the reasoning behind it, but I still wouldn't consider it obvious for someone who hadn't spent some time thinking about it. In fact, I think someone with a little mathematical knowledge is even more likely to get that wrong. One of the first things you learn when you start learning probability is that smart gamblers look for bets with a postive EV and avoid those with a negative EV. Once you understand why that is, it is surprising to learn that it is not strictly true. Recognitions: Homework Help Science Advisor ## Expected value paradox? But what if the amount you bet affects the odds ("moves the line"), as is the case in a bet pool? Then your expected value is no longer linear in the amounts you bet on each outcome. Then you have to consider all correlated bets at the same time and find the ideal distribution. Quote by mfb Then you have to consider all correlated bets at the same time and find the ideal distribution. Yes. But would you be surprised to find yourself betting some of your money on an outcome with a negative expectation? It would seem strange to me. Recognitions: Homework Help Science Advisor If your bets are correlated, it is meaningless to talk about the expectation value of a single one. You can compare "expectation value with it" and "expectation value without" - the former one should be larger with an ideal strategy, so this bet (given all others) has a positive expectation value. Quote by mfb If your bets are correlated, it is meaningless to talk about the expectation value of a single one. You can compare "expectation value with it" and "expectation value without" - the former one should be larger with an ideal strategy, so this bet (given all others) has a positive expectation value. I see what you are saying, but I still think it is possible to talk about the expected value of each bet separately. Your total winnings is the sum of the winnings from each bet, so the expected value is the sum of the expected values of each bet. This is true even if the outcomes are correlated (even mutually exclusive, for an extreme case). This was also the case for the problem of maximizing longterm growth rate when the odds were fixed. The outcomes were mutually exclusive, but we still talked about some of them having positive EV and others having negative EV. Recognitions: Homework Help Science Advisor This is true even if the outcomes are correlated (even mutually exclusive, for an extreme case). No, correlations between the outcomes do not matter. Correlations between the expected money and other bets can be relevant, e. g. "if you bet 1 € on option A and win, you get 2 €. However, if you bet an additional 1 € on option B and option A wins, you get 10 €". If A has a positive expectation value, you would want to bet 1€ on B as well as it increases the expectation value - even if B itself and alone would have a negative expectation value. Okay, you convinced me. When your bet affects the odds you get on the various outcomes, it is somewhat arbitrary to try to assign expected values to the individual outcomes. So there's not much of a paradox here, just an oddity. Thank you, and good work :) I don't want to disappoint people who looked at this thread hoping for a puzzle though. If you want a challenge, try to figure out the optimal bets to make on mutually exclusive outcomes when you are the last person to place your bets in a bet pool. For example, say a group of people are betting on the outcome of a single die roll. Everyone has placed their bets except you, and the breakdown of the pool so far is $1110 on 1$1333 on 2 $1754 on 3$1792 on 4 $1961 on 5$2050 on 6 Assume that the die is fair and that you have a large enough budget to make any bet necessary. How much do you bet on each outcome to maximize your expected value? Recognitions: Gold Member Homework Help Science Advisor Are you talking about poker? I think that's the classic case of betting on an expected negative outcome (at least by the odds) in order to maximize your long term gain. The obvious case being bluffing to beat someone out of the pot. The subtler case being caught bluffing in an affordable pot to tempt others to challenge you when you do have a strong hand. Quote by BobG Are you talking about poker? I think that's the classic case of betting on an expected negative outcome (at least by the odds) in order to maximize your long term gain. The obvious case being bluffing to beat someone out of the pot. The subtler case being caught bluffing in an affordable pot to tempt others to challenge you when you do have a strong hand. Actually, I hadn't even thought about situations where psychology comes into play. That is a little tricky because you have to assign a subjective probability for you opponent to fold. It might be worth it if the pot odds you're getting are good. So that would really be positive EV. The subtler case you mentioned really would be negative EV, at least for that hand. That is very similar to mfb's example. You bet on a small, negative EV bet now for the opportunity to make a large +EV bet later. mfb convinced me that it is difficult to disentangle the amount of EV that comes from each individual bet in cases like that. So your answer is a good answer to my question. I originally thought this was strange enough to qualify as "parodoxical" in the informal sense (think Parrondo's paradox). But it appears to be much more common than I thought. I am used to thinking that smart gamblers make only positive EV bets, but there are so many exceptions to this rule that maybe I shouldn't have been surprised when I stumbled on a particular one. Tags expected value, parimutuel Thread Tools \| \| \| \| \|----------------------------------------------\|--------------------------------------------\|---------\| \| Similar Threads for: Expected value paradox? \| \| \| \| Thread \| Forum \| Replies \| \| \| Set Theory, Logic, Probability, Statistics \| 6 \| \| \| Calculus \| 1 \| \| \| General Math \| 0 \| \| \| General Discussion \| 18 \| \| \| Special & General Relativity \| 4 \|	{"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": 0.9601282477378845, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/245598/triple-integral-of-a-function	# Triple integral of a function Upon integration, $\int f(x) \implies$ area of curve. $\iint f(x) \implies$ volume under the curve. $\iiint f(x) \implies$ ? . We are expected to come up with something 4 dimensional? I simply know. $\iiint 1 \implies$ Volume. $\iiint \rho(x) \implies$ gives mass. - 1 What's the question? – Raskolnikov Nov 27 '12 at 10:53 2 Yes, it would be the hypervolume of a 4D object. This is a bit tough to imagine, but check out what James Stewart has to say in his Calculus text; in particular, look at his section on "Applications of Triple Integrals" (16.6 p.1031). – B.D Nov 27 '12 at 11:13 The question is $\int\int\int f(x)=?$ As for James Stewart, He does not say anything more than what you have already said and what I already wrote. Thank you, though. – ήλιος Nov 27 '12 at 11:33 ## 1 Answer We know that a continuous function is integrable over a closed bounded domain. In fact if $f(x,y,z)=1$ on the domain $D$, then you certainly know that the triple integral gives the volume od $D$: $$V=\iiint_D dV$$ Now if $f$ be a positive function, then $$\iiint_D f(x,y,z) dV$$ can be interpreted as the hypervolume (i.e. the 4 dimensional volume) of the region in 4-space (which as @B.D noted is a bit hard to imagine)having the set $D$ as its 3-dimentional base and having its top on the hypersurface $u=f(x,y,z)$. Of course and indeed, this is not specially useful interpretation but, many more useful cases arise in application which you are aware of them. -	{"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": 0.9463571906089783, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/164256/how-do-i-know-which-method-of-revolution-to-use-for-finding-volume-in-calculus/164260	# How do I know which method of revolution to use for finding volume in Calculus? Is there any easy way to decide which method I should use to find the volume of a revolution in Calculus? I'm currently in the middle of my second attempt at Calculus II, and I am getting tripped up once again by this concept, which seems like it should be rather straight forward, but I can't figure it out. If I have the option of the disk/washer method and the shell method, how do I know which one I should use? - After a while, if you do enough examples, you will in essence have "seen everything." – André Nicolas Jun 28 '12 at 17:26 ## 3 Answers The first thing to understand is that you don’t directly choose the method of integration: you determine what kind of integration will be easier, based on the shape of the region in question, and that determines which method you’ll use. Draw the region that’s being revolved. Then ask yourself: does it slice up nicely into vertical strips, or do horizontal strips work better? 1. If the region has boundaries of the form $y=f(x)$, $y=g(x)$, $x=a$, and $x=b$, the answer is almost always that vertical strips are simpler: for each $x$ from $a$ to $b$ you have a strip of length $f(x)-g(x)$ or $g(x)-f(x)$, depending on which of $f(x)$ and $g(x)$ is larger. 2. Similarly, if the region has boundaries of the form $x=f(y)$, $x=g(y)$, $y=a$, and $y=b$, the answer is almost always that vertical strips are simpler: for each $y$ from $a$ to $b$ you have a strip of length $f(y)-g(y)$ or $g(y)-f(y)$, depending on which of $f(y)$ and $g(y)$ is larger. 3. The case of a region bounded by $y=f(x)$ and $y=g(x)$, where you have to solve for the points of intersection of the two curves in order to find the horizontal extent of the region, is really just special case of (1). Similarly, if the region is bounded by $x=f(y)$ and $x=g(y)$, and you have to solve for the vertical extent of the region, you’re looking at a special case of (2). What you want is a way of slicing the region into vertical or horizontal strips whose endpoints are defined in the same way. Take, for instance, the region bounded by $y=x$ and $y=x(x-1)$. If you slice it into vertical strips, each strip runs from the parabola at the bottom to the straight line at the top, so the strip at each $x$ between $0$ and $2$ has its bottom end at $x(x-1)$ and its top end at $x$. If you were to slice it into horizontal strips, the ones between $y=0$ and $y=2$ would have their left ends on the straight line and their right ends on the parabola, but the ones between $y=-1/4$ and $y=0$ would have both ends on the parabola. Thus, you’d need a different calculation to handle the part of the region below the $x$-axis from the one that you’d need for the part above the $x$-axis. Whether the boundaries are given in the form $y=f(x)$ or the form $x=f(y)$ is often a good indication: the former tends to go with vertical slices and the latter with horizontal slices. It’s far from infallible, however, and in some problems some boundaries are given in one form and some in the other. You should always look at a picture of the region. You want slices whose endpoints are defined consistently, and you want slices that don’t have any gaps in them. Once you’ve decided which way to slice up the region, sketch in the axis of revolution. If it’s parallel to your slices, each slice will trace out a cylindrical shell as it revolves about the axis. If, on the other hand, it’s perpendicular to your slices, each slice will trace out a washer or disk as it revolves about the axis. In either case the proper method of integration has automatically been determined for you. - Use whichever method you find easiest to apply for each problem you encounter. They should give the same result (or something is wrong with the way you apply them). As a practical matter, just pick one of them at random until you work up enough experience to have a hunch about what is likely to work easiest for the kind of shape you're looking at. If your first choice gets you into integrals that are hard for you to solve, put it aside for a while and see if it the other way is easier going. Continue switching between the two methods, throwing progressively more effort at each until one of them yields. - 1 Also, setup a few problems using both methods to see how one way can be easier than the other. – dls Jun 28 '12 at 17:13 This is the method I used to use last time I took Calc II. If I didn't know which one is was, I would do both methods until each got pretty hard, then I would pick the easier looking equation to finish. But sometimes it would then end up as the one that looked easier in the middle was harder to finish. I guess it is kind of just random then. – Nick Anderegg Jun 28 '12 at 17:19 @Nick: It is absolutely not random. There is a straightforward way to choose that works in the overwhelming majority of problems; see my answer. – Brian M. Scott Jun 28 '12 at 17:23 @Brian: I assumed we were dealing with cases where both ways of slicing make sense in the first place. – Henning Makholm Jun 28 '12 at 17:29 @Henning: I’ve the advantage of having spent many years watching people struggle with volumes of revolution, so I knew that that wasn’t a safe assumption. Textbooks often contribute to the problem, both by giving the impression that one chooses the method first and by including exercises requiring students to use the ‘wrong’ method. – Brian M. Scott Jun 28 '12 at 17:35 I think that there are some standard formulae in integral calculus to calculate the Volume of revolution. These offer you a straight forward method ( I don't know anything about the dish washer method ) but you still have to know how to trace the curves to apply the following formulae on them:- $V=\pi\int_a^by^2dx$ where $y=f(x)$ is the equation of the curve and $x=a$ and $x=b$ are the limits of the revolution. $V=\pi\int_c^dx^2dy$ where $x=g(y)$ is the equation of the curve and $y=c$ and $y=d$ are the limits of the revolution. To calculate the volume of revolution when polar coordinated are given, you substitute $y=cos\theta$ and $x=sin\theta$ and calculate the limits accordingly. Also, you need to see that the given formulae are valid only when the revolution is occuring about the x-axis or y-axis.If the curve is revolving around x=constant or y=constant, then you need to change the origin and make suitable adjustements. If you stilll face any problem in solving problems on this topics ( I faced many myself ), I would be happy to help. You can even message me problems on:- suri.kunal007@gmail.com -	{"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": 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": 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": 0.9545893669128418, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?s=cd79260dce1bc057844bd82696bc1d59&p=3991780	Physics Forums ## kirchhoff voltage law hi! i understand how the krichoff current law works , but can someone explain me the derivation/proof of the voltage law . PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus It follows directly from law of conservation of energy.I haven't seen it's derivation anywhere. does anyone else have a mathematical or logical proof ? Recognitions: Science Advisor ## kirchhoff voltage law The Kirchhoff rules are special cases of Maxwell's equations for quasistationary problems. For AC it is simple to state: It's applicable as long as the geometrical extensions of the electric circuit are small compared to the wavelength of the electromagnetic fields, i.e., if $l \ll \lambda=c/f$, where $f$ is the frequency of your AC, because then the retardation of electromagnetic fields can be neglected. You can find a derivation in the marvelous (however somewhat a bit older) textbook A. Sommerfeld, Lectures on Theoretical Physics, Vol. 3 (Electromagnetism). It should follow conceptually from the idea that the voltage at any given point has only one value. If you go around a loop and come back to that point, the voltage should not be different--i.e. the sum of all voltage changes around that closed loop therefore must be zero. Muphrid is right. It it true of any situation in which each point in space has a particular value. It would even be true of a chart of random numbers. If you travel in any closed path, the sum of all positive transitions and negative transitions must be zero. But why does each point in space have a unique value? It is because the electric force is a conservative force. The work that an external agent does on a charge to make it go from point A to point B is equal in magnitude and opposite in sign to the work that the electric field does on the charge to make it go from point B to point A. In other words, such a thing as potential exists in the first place. muphrid and mikelepore , together your comments answered my question ... ty I know it is common these days to be clever and fashionable and teach Kirchoff's voltage law as an equation summing to zero. However that was not the original formulation and I would not recommend it as it can actually lead to incorrect results if used in that way. For instance connect a 12 volt battery in parallel with a 6 volt battery and a load and try performing a KVL analysis. Or try to perform a KVL on a circuit containing a current source. The original formulations was in line with the proinciple of conservation of energy as mentioned earlier. The sum of the EMFs equals the algebraic sum of the IR products in a closed loop Used in this form KVL will not betray you. Recognitions: Science Advisor Even better is to remember that Kirchhoff's Law is an integrated form of Faraday's Law stating $$\vec{\nabla} \times \vec{E} = -\partial_t \vec{B}.$$ This is integrated along the wires and compact resistors, capacitors, and inductances. This also explains the sign of currents and emf's as being defined according to the right-hand rule, implemented in the definition of the rotation (curl) operator on the left-hand side via Stoke's theorem for infinitesimal surfaces and their boundaries, where by convention the relative orientation of the area-normal vectors and the boundary curve is according to the right-hand rule. Are you sure Kirchoff knew all that? Recognitions: Science Advisor I'm not so sure historically. Of course, we don't need to bother know what as been known then. The good thing with natural sciences is that you don't need to learn too much of outdated stuff but you can start with the best knowledge one has, and in the context of this topic that's Maxwell's electrodynamics. Yes if you are at University, but I think Sam is in high school,although he is asking some penetrating questions. Quote by Studiot Yes if you are at University, but I think Sam is in high school,although he is asking some penetrating questions. yes im still in high school Not 'still', Sam, you are doing well. Keep questioning as you have done and you will go far. Quote by Studiot For instance connect a 12 volt battery in parallel with a 6 volt battery and a load and try performing a KVL analysis. When designing with ideal circuit elements, that connection is undefined, like a math lesson about never dividing by zero. You placed 12-6= 6 volts across the zero resistance of the interconnecting wires, and power v^2/R went to infinity. But does anyone know what happens if you do it in real lfe? Do the wires get hot and perhaps melt? Or do the chemical reactions in the batteries fail to deliver the advertised voltage? Recognitions: Science Advisor I didn't know that "sambarbarian" is at high school, and for sure I can only also encourage you to ask such questions. These are really the right ones to ask concerning science! I really thought you are at undergrad university level, and of course I do not expect that you know Maxwell's equations at high school! Sorry for my misunderstanding. Thread Tools \| \| \| \| \|--------------------------------------------\|----------------------------------------------\|---------\| \| Similar Threads for: kirchhoff voltage law \| \| \| \| Thread \| Forum \| Replies \| \| \| Classical Physics \| 1 \| \| \| General Physics \| 4 \| \| \| Engineering, Comp Sci, & Technology Homework \| 1 \| \| \| Introductory Physics Homework \| 9 \| \| \| Introductory Physics Homework \| 9 \|	{"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": 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": 0.9467270374298096, "perplexity_flag": "middle"}
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http://mathoverflow.net/questions/123739/is-there-a-conceptual-reason-why-topological-spaces-have-quotient-structures-whil/123753	## Is there a conceptual reason why topological spaces have quotient structures while metric spaces don’t? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Of the mathematical objects that I am familiar with, it is normally the case that the product of 2 objects is an object of the same type and that an equivalence relation on an object induces a quotient object of the same type. I think I have some understanding as to why the product of 2 fields is not a field, because a field is not an algebra in the universal algebra sense. But I don't see a reason as to why an equivalence relation on a metric space fails to induce a quotient structure, apart from the fact that it just doesn't work. - 5 Are you thinking about the category $\mathcal{C}$ of metric spaces and continuous maps, or the category $\mathcal{D}$ of metric spaces and metric-preserving maps (ie isometric embeddings)? In $\mathcal{C}$ (which is essentially a full subcategory of topological spaces) I don't think that there is anything interesting to say about your question. Category $\mathcal{D}$ is much more natural if you want to make analogies with algebraic categories. – Neil Strickland Mar 6 at 11:53 2 I'm not sure whether this really answers your question, but the reason why limits and colimits of topological spaces exist is that on a given set there are always the coarsest and the finest topology satisfying some condition. On the other hand it doesn't seem sensible to talk about "coarsest" and "finest" metrics. – Karol Szumiło Mar 6 at 12:47 2 On the other hand, a quotient norm is often seen for normed spaces. – Gerald Edgar Mar 6 at 13:22 5 I'm dubious of the motivation in the first paragraph. Plenty of structures require some conditions for a quotient to be an object of the same type. For example, modding out a group by an equivalence relation, even an equivalence relation arising from a subgroup, does not in general give a group; you need normality. Even in the case of topological spaces, where you can always define quotients (as per Karol's comment), you need conditions for the resulting space to be nice. So your question could just as easily ask about Hausdorff spaces, or many other things, instead of metric spaces. – Noah Stein Mar 6 at 13:44 2 @Noah: I agree in principle, but note that quotients do exist in Hausdorff spaces, as Hausdorff spaces are a reflective subcategory of all topological spaces, hence cocomplete. – David Carchedi Mar 6 at 16:12 show 2 more comments ## 2 Answers You can define a (pseudo)metric on a quotient of a metric space. Let $X$ be a metric space with metric $d$ and an equivalence relation $\sim$. Say that a chain between two points $x,y\in X$ is a sequence of points $x=a_0\sim b_0$, $a_1\sim b_1$, $\ldots$ $a_n\sim b_n=y$, and define the length of such a chain to be $\sum d(b_i,a_{i+1})$. We can now define the distance $d([x],[y])$ between two equivalence classes to be the infimum of all lengths of chains from $x$ to $y$. It's easy to see that this is a pseudometric on $X/\sim$ (a metric where the distance between two points might be $0$). This descends to a true metric on the quotient $Y=X/\sim'$, where $x\sim' y$ if $d([x],[y])=0$. Furthermore, $Y$ can be characterized by the following universal property: distance-decreasing maps from $Y$ to a metric space $Z$ are naturally in bijection with distance-decreasing maps from $f:X\to Z$ such that $f(x)=f(y)$ whenever $x\sim y$. More generally, a similar construction shows that the category of metric spaces and distance-decreasing maps has all connected colimits (colimits over connected diagrams). If you generalize metrics to allow the distance between two points to be infinite, you can construct all colimits, and also all limits (use the sup metric on products). - 1 This kind of limit/quotient is important in the theory of group actions on $\mathbb{R}$-trees, for instance the paper of Levitt and Paulin, "Geometric group actions on trees", MR1428059. – Lee Mosher Mar 6 at 13:24 Eric, do you mean 1-Lipschitz (non-expansive) maps, or is it strictly distance-decreasing and/or strictly preserving (isometries)? It seems that what you're saying is easy to prove for 1-Lipshitz maps, but anyway are there any good references for this sort of questions? – Sergey Melikhov Mar 7 at 4:10 See also ncatlab.org/nlab/show/… . – Qiaochu Yuan Mar 7 at 4:50 ### 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. One way to get around the lack of quotients is to work in the category of Uniform Spaces. This is more general than a metric space, but still gives you a way to talk about how "close" two points are. Uniform spaces are also cool because they simultaneously generalize metric spaces and topological groups. I believe their study was popular back in the 60s and 70s for instance because this category of uniform spaces has much nicer categorical properties than either of the categories it generalizes. For instance, with uniform maps it's complete and cocomplete, as can be seen here. Even if your maps are just continuous maps, you still get quotients as can be seen here. Isbell's book is a great reference for this subject. Side note: I like Eric Wofsey's answer a lot and I wonder if it's secretly related to this answer. The wikipedia page points out the connection between uniform spaces and pseudo-metrics but for categorical things it's probably easy to work with the former than the latter. - Yes, your answer is related to Eric's. The category of metric spaces and uniformly continuous maps has many quotients (though not all) defined by the formula in Eric's answer. (In other words, that formula gives a well-defined metric, in many cases, if the original metric is changed without changing the uniform structure.) For example, you can define mapping cylinder and join in this category (which you can't do for metric spaces and continuous maps). This is done in arxiv.org/abs/1106.3249 – Sergey Melikhov Mar 7 at 3:30	{"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": 28, "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": 0.9307101368904114, "perplexity_flag": "head"}
http://mathhelpforum.com/algebra/105713-sequences-series.html	Thread: 1. Sequences, Series 4. Prove that the terms of the sequence $U_n = n^{2}-10n+27$ are positive. For what value of $n$ is $U_n$ smallest? How do I do this? 2. Well first, n = 0 for $U_n$ to be the lowest, because any negative numbers squared would turn out positive, 0 is smaller than any positive number, so to answer your question n = 0. Im not sure how to prove that they are positive, but I think it has something to do with squaring, anytime you square a negative it becomse positive, and -10 times any negative number will become positive. 3. Originally Posted by Viral 4. Prove that the terms of the sequence $U_n = n^{2}-10n+27$ are positive. For what value of $n$ is $U_n$ smallest? How do I do this? First we need to complet the square $n^2-10n+27=n^2-10n+25+2=(n-5)^2+2$ Since $(n-5)^2 \ge 0$ we know that $(n-5)^2+2\ge 0+2=2>0$ and since this is a parabola its mininum occurs at the vertex when n=5 you get 2. 4. Can you expand on: $<br /> <br /> (n-5)^2+2\ge 0+2=2>0<br />$ a bit please. I understood the completing the square bit fine. 5. Originally Posted by Viral Can you expand on: $<br /> <br /> (n-5)^2+2\ge 0+2=2>0<br />$ a bit please. I understood the completing the square bit fine. If you square any real number it is always non negative so $(n-5)^2$ is always bigger than or equal to zero. i.e the smallest it can ever be is 0 and that occurs when n=5 so the smallest value your squence can take it 2. 6. Thanks I got that now. How would I work out the first part of the question though? (the proof)	{"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": 13, "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": 0.941517174243927, "perplexity_flag": "middle"}
http://jeremykun.com/tag/geometric-transformations/	# Double Angle Trigonometric Formulas Posted on May 19, 2012 by Problem: Derive the double angle identities $\sin(2\theta) = 2\sin(\theta)\cos(\theta)\\ \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)$ Solution: Recall from linear algebra how one rotates a point in the plane. The matrix of rotation (derived by seeing where $(1,0)$ and $(0,1)$ go under a rotation by $\theta$, and writing those coordinates in the columns) is $A = \begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix}$ Next, note that to rotate a point twice by $\theta$, we simply multiply the point (as a vector) by $A$ twice. That is, multiply by $A^2$: $AAv = A^2v$ Computing $A^2$ gives the following matrix: $A^2 = \begin{pmatrix} \cos^2(\theta) - \sin^2(\theta) & -2\sin(\theta)\cos(\theta) \\ 2\sin(\theta)\cos(\theta) & \cos^2(\theta) - \sin^2(\theta) \end{pmatrix}$ But rotating twice by $\theta$ is the same as rotating once by $2\theta$, so we have the equality: $\begin{pmatrix} \cos(2\theta) & -\sin(2\theta) \\ \sin(2\theta) & \cos(2\theta) \end{pmatrix} = \begin{pmatrix} \cos^2(\theta) - \sin^2(\theta) & -2\sin(\theta)\cos(\theta) \\ 2\sin(\theta)\cos(\theta) & \cos^2(\theta) - \sin^2(\theta) \end{pmatrix}$ The matrices are equal, so they must be equal entrywise, giving the identities we desire. $\square$ Discussion: There are (painful, messy) ways to derive these identities by drawing triangles on the unit circle and cultishly chanting “soh-cah-toa.” The key idea in this proof that one might study geometric transformations, and it is a truly mature viewpoint of mathematics. Specifically, over the last two hundred years the field of mathematics has changed focus from the study of mathematical “things” to the study of transformations of mathematical things. This proof is an elementary example of the power such perspective can provide. If you want to be really high-brow, start asking about transformations of transformations of things, and transformations of those transformations, and recurse until you’re doing something original. ### Like this: Posted in Linear Algebra, Proof Gallery \| \| Cancel	{"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": 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": 0.902401328086853, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/33883/spectral-decomposition-of-a-normal-matrix?answertab=votes	# Spectral decomposition of a normal matrix I'd like to find the spectral decomposition of $A$: $$A = \begin{pmatrix} 2-i & -1 & 0\\ -1 & 1-i & 1\\ 0 & 1 & 2-i \end{pmatrix}$$ i.e. $A=\sum_{i}\lambda_i P_i$ where $P_i$ are the coordinate matrices (in the standard basis) of the corresponding orthogonal transformations in the spectral decomposition of $T_A$ and $\lambda_i$ are the eigenvalues. I started off by showing that $A$ is normal, piece of cake. Then found the eigenvalues of $A$, those are: $\lambda_1 = 2-i, \lambda_2 = 3-i, \lambda_3 = -i$. I tried using these known facts from the spectral theorem: • $A=(2-i)P_1+(3-i)P_2-iP_3$ • $I=P_1+P_2+P_3$ • $\forall i\neq j, P_i P_j=0$ • $P^_i=P_i$ The only example I have in my book uses these but I couldn't get it to work here. The terms don't cancel out it seems. What else can I try? - 1 The $P_i$ are the orthogonal projections onto the Eigenspaces, so you need to find the eigenspaces of the eigenvectors. Find a nonzero eigenvector for each eigenvalue; because $A$ is normal, the eigenvectors will be mutually orthogonal. Then you just need to normalize them; the $P_i$ are given by the orthogonal projections onto the subspaces spanned by these vectors. – Arturo Magidin Apr 19 '11 at 19:30 Shouldn't $P_i$ be the projection onto the $i$-th eigenspace? Also, shouldn't they sum to the identity instead of to 0? – lhf Apr 19 '11 at 19:30 @Arturo: I'm stuck after finding the eigenspaces. I got $$V_{\lambda_1}=sp{(\frac{1}{\sqrt 2}, 0, \frac{1}{\sqrt 2})}, V_{\lambda_2}=sp{(\frac{-1}{\sqrt 3}, \frac{1}{\sqrt 3}, \frac{1}{\sqrt 3})}, V_{\lambda_3}=sp{(\frac{-1}{\sqrt 6}, \frac{-2}{\sqrt 6}, \frac{1}{\sqrt 6})}$$ And I know that if $v=v_1+v_2+v_3, v_i \in V_{\lambda_i}$ then $P_i = v_i$. But how do I actually find $[P_i]$? I'm confused. – daniel.jackson Apr 20 '11 at 11:10 1 The $P_i$ are the orthogonal projections onto the spans of the eigenspaces. The eigenvectors you have are already an orthonormal basis, and the matrices of $P_i$ relative to that basis are very easy: for example, if you order your basis as $\beta=[v_{\lambda_1},v_{\lambda_2},v_{\lambda_3}]$, then $$P_1 = \left(\begin{array}{ccc}1&0&0\\0&0&0\\0&0&0\end{array}\right).$$ To write $[P_i]_{\beta}$ in terms of the standard orthonormal basis of $\mathbb{C}^3$, just take the change of basis matrix $M$, and compute $M[P_i]_{\beta}M^{-1}$ (cont) – Arturo Magidin Apr 20 '11 at 12:53 Here, $M$ changes from the $\beta$ basis to the standard basis, so the columns of $M$ are the vectors of $\beta$. – Arturo Magidin Apr 20 '11 at 12:56 show 1 more comment ## 3 Answers The following result is useful "A matrix is normal if and only if it is unitarily similar to a diagonal matrix, and therefore any matrix A satisfying the equation $A^{}A=AA^{}$ is diagonalizable". That is $$\mathbf{A} = \mathbf{U} \mathbf{\Lambda} \mathbf{U}^ .$$ The next step is to find the eigenvectors $v_i$. Once that done, then you can get the matrices $P_i$ such that $$P_i = v_iv^{}_i,$$ which gives, $$A=(2-i)P_1+(3-i)P_2-iP_3 = (2-i)v_1v^{}_1+ (3-i)v_2v^{}_2-i v_3v^{}_3.$$ Note: You should check that $P_i,i=1,2,3$ satisfy the properties you listed above. Eigenvectors: I computed the eigenvectors by Maple which corresponds to the eigenvalues that you already computed $2-i,3-i,-i$ respectively $$v_1= \left[ \begin {array}{c} 1\\0 \\1 \end {array} \right ], v_2=\left[ \begin {array}{c} -1\\1 \\1 \end {array} \right ], v_3=\left[ \begin {array}{c} -1\\-2 \\1 \end {array} \right ]$$ - Using the primary decomposition theorem (PDT): Find the minimal polynomial of $A$. Clearly that would be $m_A(x)=(x-\lambda_1)(x-\lambda_2)(x-\lambda_3)$. Define $f_i(x)=\frac{m_A(x)}{(x-\lambda_i)}$. Observe that $f_1,...,f_3$ are co-prime (i.e. $gcd(f_1,f_2,f_3)=1$). Hence you can find polynomials $g_1,g_2,g_3$ such that $g_1f_1+g_2g_2+g_3f_3=1$. \ Finally, define $P_i=g_i(A)f_i(A)$. Check why does it work! - it is also possible to compute the Projection matrices without finding each eigen vector (corresponding to each eigen-value) of the matrix A. The trick is to use the Caley-Hamilton Theorem. First find the partial fraction decomposition of 1/charpoly(A,x) where charpoly(A,x) is the characteristics polynomial of the matrix A factored in the complex field as linear factorr. -	{"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": 44, "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": 0.893899142742157, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2009/05/08/	# The Unapologetic Mathematician ## DeMorgan’s Laws And here’s the post I wrote today: Today, I want to prove two equations that hold in any orthocomplemented lattice. They are the famous DeMorgan’s laws: $\displaystyle\neg(x\vee y)=\neg x\wedge\neg y$ $\displaystyle\neg(x\wedge y)=\neg x\vee\neg y$ First, we note that $x\leq x\vee y$ by definition. Since our complementation reverses order, we find $\neg(x\vee y)\leq\neg x$. Similarly, $\neg(x\vee y)\leq\neg y$. And thus we conclude that $\neg(x\vee y)\leq\neg x\wedge\neg y$. On the other hand, $\neg x\wedge\neg y\leq\neg x$ by definition. Then we find $x=\neg\neg x\leq\neg(\neg x\wedge\neg y)$ by invoking the involutive property of our complement. Similarly, $y\leq\neg(\neg x\wedge\neg y)$, and so $x\vee y\leq\neg(\neg x\wedge\neg y)$. And thus we conclude $\neg x\wedge\neg y\leq\neg(x\vee y)$. Putting this together with the other inequality, we get the first of DeMorgan’s laws. To get the other, just invoke the first law on the objects $\neg x$ and $\neg y$. We find $\displaystyle\begin{aligned}\neg x\vee\neg y&=\neg\neg(\neg x\vee\neg y)\\&=\neg(\neg\neg x\wedge\neg\neg y)\\&=\neg(x\wedge y)\end{aligned}$ Similarly, the first of DeMorgan’s laws follows from the second. Interestingly, DeMorgan’s laws aren’t just a consequence of order-reversal. It turns out that they’re equivalent to order-reversal. Now if $x\leq y$ then $x=x\wedge y$. So $\neg x=\neg(x\wedge y)=\neg x\vee\neg y$. And thus $\neg y\leq\neg x$. Posted by John Armstrong \| Algebra, Lattices \| 8 Comments ## Upper-Triangular Matrices and Orthonormal Bases I just noticed in my drafts this post which I’d written last Friday never went up. Let’s say we have a real or complex vector space $V$ of finite dimension $d$ with an inner product, and let $T:V\rightarrow V$ be a linear map from $V$ to itself. Further, let $\left\{v_i\right\}_{i=1}^d$ be a basis with respect to which the matrix of $T$ is upper-triangular. It turns out that we can also find an orthonormal basis which also gives us an upper-triangular matrix. And of course, we’ll use Gram-Schmidt to do it. What it rests on is that an upper-triangular matrix means we have a nested sequence of invariant subspaces. If we define $U_k$ to be the span of $\left\{v_i\right\}_{i=1}^k$ then clearly we have a chain $\displaystyle U_1\subseteq\dots\subseteq U_{d-1}\subseteq U_d=V$ Further, the fact that the matrix of $T$ is upper-triangular means that $T(v_i)\in U_i$. And so the whole subspace is invariant: $T(U_i)\subseteq U_i$. Now let’s apply Gram-Schmidt to the basis $\left\{v_i\right\}_{i=1}^d$ and get an orthonormal basis $\left\{e_i\right\}_{i=1}^d$. As a bonus, the span of $\left\{e_i\right\}_{i=1}^k$ is the same as the span of $\left\{e_i\right\}_{i=1}^k$, which is $U_k$. So we have exactly the same chain of invariant subspaces, and the matrix of $T$ with respect to the new orthonormal basis is still upper-triangular. In particular, since every complex linear transformation has an upper-triangular matrix with respect to some basis, there must exist an orthonormal basis giving an upper-triangular matrix. For real transformations, of course, it’s possible that there isn’t any upper-triangular matrix at all. It’s also worth pointing out here that there’s no guarantee that we can push forward and get an orthonormal Jordan basis. Posted by John Armstrong \| Algebra, Linear Algebra \| 1 Comment ## 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": 36, "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": 0.8916230201721191, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/45264/algebraic-proof-of-a-trig-matrix-identity	# Algebraic proof of a trig matrix identity? I'll put the question first, and then the background, because I'm not sure that the background is necessary to answer the question: I have a geometric proof, but is there an elegant algebraic proof that $$\left[ \begin{matrix}-1 & 1 + \cos\frac{\pi}{2n} \\ -2 & 1 + 2\cos\frac{\pi}{2n}\end{matrix} \right] ^n \left[ \begin{matrix}0 \\ 1\end{matrix} \right] = \left[ \begin{matrix}\cot \frac{\pi}{4n} \\ \cot \frac{\pi}{4n}\end{matrix} \right]$$ Background: This is motivated by the latest maths puzzle from Spanish newspaper El País (problem statement there in Spanish, obviously). NB the deadline for submitting solutions to win the prize has passed, so don't worry about cheating. The relevant part of the problem is this: We have two straight lines (in Euclidean geometry) and we wish to draw a zigzag between them. We start at the intersection point and draw a straight line segment of length $r$ along one of the lines (which we shall call the "horizontal line"). We then draw a straight line segment of length $r$ from the end-point to the other ("non-horizontal") line. We alternate between the two with straight line segments of the same length, without overlapping or doubling back. The 20th such line segment is perpendicular to the horizontal line. What is the angle between the lines? There's a simple geometric solution (which I shan't state here, in case anyone wants to solve it himself - although I expect that an answer may include spoilers, so if you do want to solve it yourself, look away now), but before finding it I went down an algebraic approach which led me to an equation involving a matrix similar to the one above raised to the 10th power. Basically the equation above is the generalisation to the case with $2n$ line segments and the answer (derived from the geometric proof) substituted for the unknown angle. The question is motivated by little more than curiosity, because I already have a proof. I've tried to prove it myself, but the best I've got so far is a messy expression in terms of some matrices whose powers do have a relatively nice closed form. Filling in some gaps: if the lines intersect at the origin, the horizontal line is along the x-axis, the non-horizontal line is in the first quadrant, and the angle between the lines is $\alpha$, then the points are $P_{2n} = (x_{2n}, x_{2n} \tan \alpha)$ and $P_{2n+1} = (x_{2n+1}, 0)$. Given that the distance between each pair of consecutive points is $r$ we find that $x_{2n}$ and $x_{2n+2}$ are the two roots of $$(z - x_{2n+1})^2 + z^2 \tan^2 \alpha = r^2$$, and we can use the properties of the quadratic equation to get $$x_{2n+2} = \frac{2}{1 + \tan^2 \alpha}x_{2n+1} - x_{2n}$$ By using the symmetry of the triangle formed by $P_{2n+1}$, $P_{2n+2}$, $P_{2n+3}$, we get $$x_{2n+3} = 2x_{2n+2} - x_{2n+1}$$ Putting this all together we can get a recurrence which leads to an expression for $(x_{2n}, x_{2n+1})$ in terms of the nth power of a matrix applied to $(x_{0} = 0, x_{1} = 1)$. Putting any more details risks spoiling the value of $\alpha$. The RHS of the original equation comes from observing that if the $(2n)^{th}$ line segment is perpendicular to the horizontal line, $x_{2n-1} = x_{2n} = x_{2n+1}$ and $\tan \alpha = r / x_{2n}$. The furthest I've got so far with the matrix is to split it out as $$\left( \cos\frac{\pi}{2n} \left[ \begin{matrix}0 & 1 \\ 0 & 2\end{matrix} \right] + \left[ \begin{matrix}-1 & 1 \\ -2 & 1\end{matrix} \right] \right)^n = \sum_{k=0}^{n} \left( \begin{matrix}n \\ k\end{matrix} \right) \cos^k\frac{\pi}{2n} \left[ \begin{matrix}0 & 1 \\ 0 & 2\end{matrix} \right]^k \left[ \begin{matrix}-1 & 1 \\ -2 & 1\end{matrix} \right]^{n-k}$$ where both of the matrix powers on the RHS have closed forms - the markup here doesn't seem to like the URLs and I've taken too long to write this up and need to run, but http://www.wolframalpha.com/input/?i=[[0,1],[0,2]]^k and http://www.wolframalpha.com/input/?i=[[-1,1],[-2,1]]^(r-k) - The binomial theorem can only be applied if the matrices in question commute. Do they? – Gerry Myerson Jun 14 '11 at 6:40 2 The usual way to raise a matrix to a power is to diagonalize it first, that is, to find $A^n$, first find an invertible matrix $P$ and a diagonal matrix $D$ such that $P^{-1}AP=D$ - this you do by letting the columns of $P$ be the eigenvectors of $A$, and the diagonal entries of $D$, the eigenvalues of $A$. Then $A^n=PD^nP^{-1}$, which is good because calculating $D^n$ is trivial. – Gerry Myerson Jun 14 '11 at 6:42 ## 2 Answers [moved from comment to answer] Let $\alpha=\pi/2n$ and let's call Peter's 2x2 matrix $$A=\left(\begin{array}{rr}-1&1+\cos\alpha\\-2&1+2\cos\alpha\end{array}\right).$$ Luboš's answer tells us that the eigenvalues of $A$ are $e^{\pm i\alpha}$, so we expect the matrix $A$ to be similar to the rotation matrix $$R=\left(\begin{array}{rr}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{array}\right).$$ We want to find a $2\times2$ matrix $P$ such that $PA=RP$. This is a homogeneous linear system in the unknown entries of $P$. As $R$ commutes with all the rotation matrices, we are at liberty to multiply $P$ from the left with one. Therefore we can assume that (for example) the bottom left entry of $P$ is equal to zero. The resulting system is not hard to solve, and let's pick among the solutions $$P=\left(\begin{array}{rc}2&-1-\cos\alpha\\0&\sin\alpha\end{array}\right).$$ Then $A=P^{-1}RP$, so $A^n=P^{-1}R^nP$. Here we benefit from the fact that $$R^n=\left(\begin{array}{rr}\cos n\alpha&\sin n\alpha\\-\sin n\alpha&\cos n\alpha\end{array}\right)=\left(\begin{array}{rr}0&1\\-1&0\end{array}\right).$$ Therefore $$A^n\left(\begin{array}{r}1\\0\end{array}\right)=P^{-1}\left(\begin{array}{rr}0&1\\-1&0\end{array}\right)\left(\begin{array}{c}-1-\cos\alpha\\\sin\alpha\end{array}\right)= \frac{1}{\sin\alpha}\left(\begin{array}{r}1+\cos\alpha\\1+\cos\alpha\end{array}\right).$$ The identities $\sin\alpha=2\sin(\alpha/2)\cos(\alpha/2)$ and $1+\cos\alpha=2\cos^2(\alpha/2)$ then give Peter's claim as now $(1+\cos\alpha)/\sin\alpha=\cot(\alpha/2)=\cot(\pi/4n).$ - You may calculate the eigenvalues of the matrix on the left hand side (which is exponentiated to the $n$-th power) which are simply $$\lambda_\pm = + \exp(\pm i\pi / 2n)$$ So the matrix $A$ is written as $A=CDC^{-1}$ where $D={\rm diag}(\lambda_+,\lambda_-)$. The corresponding eigenvectors are $$(\frac{1}{2}(1 +\exp(\pm i\pi/2n),1)^T$$ which are written as columns of $C$, in the order "plus minus" again. Given the form of your vector $(0,1)^T$ you added, only the second column of $C^{-1}$ is important. It is $$\mp\frac{1}{2\sin (\pi/2n)} (1 + \exp(\mp \pi i/2n))$$ well, I mean $(n_{1},n_{2})^T$ where $n_{1,2}$ are the numbers above with the signs "upper one, lower one". Clearly, $D^n$ is just ${\rm diag}(i,-i)$ because $\exp(i\pi /2)=i$ and so on. And I hope that the rest of $C D C^{-1} v$ is already easily calculated. Note that we had $1/\sin(\pi/2n)$ over there which is $2\sin(\pi/4n)\cos(\pi/4n)$ and the cosine cancels, leaving the sign which combines to the cotangent after some work. - Thanks, Luboš, for the general approach. @Jyrki, would you mind posting your comment as an answer? It seems to be more elegant, probably because it exploits the specific structure of the matrix in question, and is certainly worth an up-vote. – Peter Taylor Jun 14 '11 at 16:28 Thanks, @Jyrki, complexophobia is a legitimate concern! :-) If you have time and energy, it would be good if you could write a full-fledged answer. @Peter, thanks to you, too. – Luboš Motl Jun 14 '11 at 17:06 @Peter, @Luboš: Since you ask so nicely! – Jyrki Lahtonen Jun 14 '11 at 18:58	{"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": 59, "mathjax_display_tex": 13, "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": 0.946037769317627, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/112024/determine-the-percentage-needed-to-subtract-to-get-the-base-value/112034	# Determine the percentage needed to subtract to get the base value Given the value $N$ and the percentage $p$, where $N = G + G \times (p/100)$, $d$ is the percentage you need to subtract of $N$ to get the base value $G$. • Is there a general formula for determining $d$? • Is there a formula independent of $G$ and $N$, where $d$ can be calculated only from $p$? - This sounds like a homework question. Write an equation that describes how $d$ relates to $N$ and $G$… – Seamus Feb 22 '12 at 13:32 ## 2 Answers $$N=G(1+\frac{p}{100})$$ $$N(1-\frac{d}{100})=G$$ Putting $G$ of the second equation in the first equation you have: $$(1+\frac{p}{100})(1+\frac{d}{100})=1$$ So:$$d=\frac{100p}{100+p}$$ - So if I understood you correctly, for any numbers $N$, $G$ and $p$ for which $N=G+G\frac{p}{100}$ you want to compute $d$ such that $G=N-N\frac{d}{100}$. This should be achievable quite easily by substituting the formula for $G$ into that of $N$ (or vice versa) and eliminating terms: $$N = N - N\frac{d}{100} + N\frac{p}{100} - N\frac{dp}{10000}$$ $$1 = 1 - \frac{d}{100} + \frac{p}{100} - \frac{dp}{10000}$$ $$d + \frac{dp}{100} = p$$ $$d = \frac{p}{1+\frac{p}{100}}$$ -	{"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": 23, "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": 0.8965088129043579, "perplexity_flag": "head"}
http://mathforum.org/mathimages/index.php?title=Transformation_Matrix&oldid=26052	# Transformation Matrix ### From Math Images Revision as of 14:15, 19 July 2011 by 144.118.94.68 (Talk) (diff) ←Older revision \| Current revision (diff) \| Newer revision→ (diff) A transformation matrix is a special matrix that can describe 2d and 3d transformations. Transformations are frequently used in linear algebra and computer graphics, since transformations can be easily represented, combined and computed. ## Contents ### Computing Transformations If you have a transformation matrix you can evaluate the transformation that would be performed by multiplying the transformation matrix by the original array of points. For example in 2d suppose you had a transformation matrix of $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ then the transfomations of the points $\begin{bmatrix} x \\ y \end{bmatrix}$ would be $\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}$. Similarly, to perform 3d transformation $\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$ on the points $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ you would use $\begin{bmatrix} x' \\ y' \\ z' \end{bmatrix} = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ ### Examples in 2D Graphics In 2D graphics Linear transformations can be represented by 2x2 matrices. Most common transformations such as rotation, scaling, shearing, and reflection are linear transformations and can be represented in the 2x2 matrix. Other affine transformations can be represented in a 3x3 matrix. #### Rotation For rotation by an angle θ clockwise about the origin, the functional form is x' = xcosθ + ysinθ and y' = − xsinθ + ycosθ. Written in matrix form, this becomes: $\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} \cos \theta & \sin\theta \\ -\sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$ Similarly, for a rotation counterclockwise about the origin, the functional form is $x' = x \cos \theta - y \sin \theta$ and $y' = x \sin \theta + y \cos \theta$ and the matrix form is: $\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} \cos \theta & -\sin\theta \\ \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$ #### Scaling For scaling we have $x' = s_x \cdot x$ and $y' = s_y \cdot y$. The matrix form is: $\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} s_x & 0 \\ 0 & s_y \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$ #### Shearing For shear mapping (visually similar to slanting), there are two possibilities. For a shear parallel to the x axis has $x' = x + ky$ and $y' = y$; the shear matrix, applied to column vectors, is: $\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$ A shear parallel to the y axis has $x' = x$ and $y' = y + kx$, which has matrix form: $\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ k & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$ ### 2D Affine Transformations Affine transformations are represented by transformation matrices that are one higher dimension than the regular transformations. For example a 2d shear transformation $\begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix}$ could be represented by the affine transformation matrix $\begin{bmatrix} 1 & k & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$. The transformation performed by the affine transformation matrix can be found in the same manner as a regular transformation matrix with 1 extra dimension added on the matrix and vector or orignal points. Instead of performing the transformation on the points $\begin{bmatrix} x \\ y \end{bmatrix}$ you would perform the transformation on the points $\begin{bmatrix} x \\ y \\ 1 \end{bmatrix}$. If you calculate the transformation you end up with $\begin{bmatrix} x \\ y \\ w \end{bmatrix}$ where w is a value that you can discard if you are only interested in the (x,y) transformation. #### Translation Affine transformations are typically used instead because only affine transformations allow translations. The matrix form of x and y translation is: $\begin{bmatrix} 1 & 0 & T_x \\ 0 & 1 & T_y \\ 0 & 0 & 1 \end{bmatrix}$ ### 2D Transformation applet If you can see this message, you do not have the Java software required to view the applet. ### Examples in 3D Graphics Objects in three dimensions can be transformed using transformation matrices in the same way as two dimensional objects. Three dimensional transformation matrices are 3x3 matrices. Three dimensional affine transformation matrices are 4x4 matrices. #### Scale For scaling we have $x' = s_x \cdot x$, $y' = s_y \cdot y$ and $z' = s_z \cdot z$. The matrix form is: $\begin{bmatrix} s_x & 0 & 0 \\ 0 & s_y & 0 \\ 0 & 0 & s_z \end{bmatrix}$ #### Rotation There are three different sets of rotation in the three dimenstional transformation matrix, one for each axis that can be rotated around. X axis rotation: $\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos \theta & -\sin \theta \\ 0 & \sin \theta & \cos \theta \end{bmatrix}$ Y axis rotation: $\begin{bmatrix} \cos \theta & 0 & \sin \theta \\ 0 & 1 & 0 \\ -\sin \theta & 0 & \cos \theta \end{bmatrix}$ Z axis rotation: $\begin{bmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix}$ ### 3d Transformation Applet Matrix Transformation Applet ### Composing transformations The ability to compose multiple transformation matrix into one matrix is very convenient when you are to calculate many transformations. You can take any number of individual transformations and combine them into a single transformation matrix by multiplying the matrices together. It is important to remember that the order in which you multiply the matrices together is significant.	{"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": 32, "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": 0.9010453820228577, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/211291/show-that-if-l-1-1-and-l-2-2-are-both-modular-lattices-then-so-is	# Show that if $(L_1;≤_1)$ and $(L_2;≤_2)$ are both modular lattices then so is $(L_1 \times L_2;≤)$ [duplicate] Possible Duplicate: Cross Product of Partial Orders Suppose that $(L_1;≤_1)$ and $(L_2;≤_2)$ are partially ordered sets. We define a partial order $≤$ on the set $L_1 \times L_2$ in the most obvious way - we say $(a,b)≤(c,d)$ if and only if $a≤_1 c$ and $b≤_2 d$. a) Show that if $(L_1;≤_1)$ and $(L_2;≤_2)$ are both lattices, then so is $(L_1 \times L_2;≤)$. b) Show that if $(L_1;≤_1)$ and $(L_2;≤_2)$ are both modular lattices, then so is $(L_1 \times L_2;≤)$. c) Show that if $(L_1;≤_1)$ and $(L_2;≤_2)$ are both distributive lattices, then so is $(L_1 \times L_2;≤)$. d) Show that if $(L_1;≤_1)$ and $(L_2;≤_2)$ are both Boolean algebras, then so is $(L_1 \times L_2;≤)$. I am having trouble solving this problem. Any help would be appreciated. - Well, show us what you did on (a) and where you got stuck. For example, what is the definition of "lattice". – GEdgar Oct 11 '12 at 21:14 ## marked as duplicate by Henning Makholm, BenjaLim, tomasz, Thomas, NorbertOct 13 '12 at 17:40 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 I've been trying to attempt this solution. Here is where I get stuck: c) $L_1$ and $L_2$ are modular, so $$L_1: a_1 < a_3 \rightarrow a_1 \vee (a_2 \wedge a_3) = (a_1 \vee a_2) \wedge {a_3}$$ $$L_2: b_1 < b_3 \rightarrow b_1 \vee (b_2 \wedge b_3) = (b_1 \vee b_2) \wedge {b_3}$$ $$L_1\times L_2: (a_1,b_1) < (a_3,b_3) \rightarrow (a_1,b_1) \vee ((a_2,b_2) \wedge (a_3,b_3)) = ((a_1,b_1) \vee (a_2,b_2)) \wedge (a_3,b_3)$$ For $L_1\times L_2$ we have $$(a_1,b_1) < (a_3,b_3) = a_1 < a_3 \text{ and } b_1 < b_3$$ I'm not too sure how to use all this information to prove that $L_1\times L_2$ is a modular lattice. Any guidance would be greatly appreciated. -	{"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": 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": 0.9477140307426453, "perplexity_flag": "head"}
http://quant.stackexchange.com/questions/4529/why-is-the-capm-securities-market-line-straight/4531	# Why is the CAPM securities market line straight? Let $\gamma$ be the expected return, in terms of its exponential growth rate, of the market asset. If we set $\gamma=\mu-\sigma^2/2$ as explained by the Doléans-Dade exponential, then the expected return of a balanced portfolio with fraction $\beta$ invested in the market asset, and the remainder lent or borrowed at risk-free rate, is $$R = r_f + \beta(\mu-r_f) - \beta^2\sigma^2/2.$$ I have plotted $R$ against $\beta$ in the following chart, where for purposes of the chart $r_f=0.04$, $\gamma=0.13$, and $\sigma=0.2$. I know this effect is not my imagination, because Fernholz and others have quantified the "excess returns" of a balanced portfolio (where the green line lies above the red line) in their framework of "stochastic portfolio theory", and I myself have noticed this and alluded to it in my answer to How to calculate compound returns of leveraged ETFs? Risk aversion notwithstanding, I find it absurd to think that unlimited expected gains are available simply by being highly leveraged in the market. So why does the CAPM use a straight line as if this were the case? - ## 3 Answers The efficient frontier should be expressed in terms of arithmetic returns since only these returns can account for cross-sectional aggregation. Hence, if you assume the log returns of the risky portfolio are $X_{p} \sim N(\mu,\sigma^{2})$, then you first have to convert it to log-normal moments before combining it with the risk-free rate, $r_{f}$. However, it should be noted that while the median equals the mean for the normal distribution, that is not the case for the log normal distribution. Hence, the median arithmetic return will not show the same linear relationship against the standard deviation that the mean does. Similarly, if using CVaR as a measure of risk, then the risk will increase by even more as leverage increases due to the non-normality of the log normal distribution. - Thank you! You've explained it very well for me. Sorry, not registered can't upvote for now. – justin-- Nov 15 '12 at 20:09 The CAPM is just a model, not the truth nobody believes it. So you shouldn't apply it blindly. As a model it is useful but it has larger defects than this. - It is straight line because it is defined by y=mx+c equation. No other reason. It is not adequate and certainly something that you can to gain infinite returns. You want to make a model that accounts for convexity like you have , or use a multi factor model like Fama French or your own model which is evolving and is a N factor polynomial function ,please do so . Only reason why a linear approximation is used is because it is simple and sort of works. You can read about fama french model here . - 1 Fama French has more factors, but it's still a straight-line model. – justin-- Nov 15 '12 at 17:50 My point , you want a curve add higher powers it you can justify. Everyone knows CAPM is not sufficient Risk vs Return gauge. – Ash Nov 15 '12 at 17:54	{"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": 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": 0.9493443369865417, "perplexity_flag": "middle"}
http://www.nag.com/numeric/cl/nagdoc_cl23/html/S/s17avc.html	# NAG Library Function Documentnag_airy_bi_vector (s17avc) ## 1 Purpose nag_airy_bi_vector (s17avc) returns an array of values of the Airy function, $\mathrm{Bi}\left(x\right)$. ## 2 Specification #include <nag.h> #include <nags.h> void nag_airy_bi_vector (Integer n, const double x[], double f[], Integer ivalid[], NagError fail) ## 3 Description nag_airy_bi_vector (s17avc) evaluates an approximation to the Airy function $\mathrm{Bi}\left({x}_{i}\right)$ for an array of arguments ${x}_{\mathit{i}}$, for $\mathit{i}=1,2,\dots ,n$. It is based on a number of Chebyshev expansions. For $x<-5$, $Bix=atcos⁡z+btsin⁡z-x1/4,$ where $z=\frac{\pi }{4}+\frac{2}{3}\sqrt{-{x}^{3}}$ and $a\left(t\right)$ and $b\left(t\right)$ are expansions in the variable $t=-2{\left(\frac{5}{x}\right)}^{3}-1$. For $-5\le x\le 0$, $Bix=3ft+xgt,$ where $f$ and $g$ are expansions in $t=-2{\left(\frac{x}{5}\right)}^{3}-1$. For $0<x<4.5$, $Bix=e11x/8yt,$ where $y$ is an expansion in $t=4x/9-1$. For $4.5\le x<9$, $Bix=e5x/2vt,$ where $v$ is an expansion in $t=4x/9-3$. For $x\ge 9$, $Bix=ezutx1/4,$ where $z=\frac{2}{3}\sqrt{{x}^{3}}$ and $u$ is an expansion in $t=2\left(\frac{18}{z}\right)-1$. For , the result is set directly to $\mathrm{Bi}\left(0\right)$. This both saves time and avoids possible intermediate underflows. For large negative arguments, it becomes impossible to calculate the phase of the oscillating function with any accuracy so the function must fail. This occurs if $x<-{\left(\frac{3}{2\epsilon }\right)}^{2/3}$, where $\epsilon $ is the machine precision. For large positive arguments, there is a danger of causing overflow since Bi grows in an essentially exponential manner, so the function must fail. ## 4 References Abramowitz M and Stegun I A (1972) Handbook of Mathematical Functions (3rd Edition) Dover Publications ## 5 Arguments 1: n – IntegerInput On entry: $n$, the number of points. Constraint: ${\mathbf{n}}\ge 0$. 2: x[n] – const doubleInput On entry: the argument ${x}_{\mathit{i}}$ of the function, for $\mathit{i}=1,2,\dots ,{\mathbf{n}}$. 3: f[n] – doubleOutput On exit: $\mathrm{Bi}\left({x}_{i}\right)$, the function values. 4: ivalid[n] – IntegerOutput On exit: ${\mathbf{ivalid}}\left[\mathit{i}-1\right]$ contains the error code for ${x}_{\mathit{i}}$, for $\mathit{i}=1,2,\dots ,{\mathbf{n}}$. ${\mathbf{ivalid}}\left[i-1\right]=0$ No error. ${\mathbf{ivalid}}\left[i-1\right]=1$ ${x}_{i}$ is too large and positive. ${\mathbf{f}}\left[\mathit{i}-1\right]$ contains zero. The threshold value is the same as for NE_REAL_ARG_GT in nag_airy_bi (s17ahc), as defined in the Users' Note for your implementation. ${\mathbf{ivalid}}\left[i-1\right]=2$ ${x}_{i}$ is too large and negative. ${\mathbf{f}}\left[\mathit{i}-1\right]$ contains zero. The threshold value is the same as for NE_REAL_ARG_LT in nag_airy_bi (s17ahc), as defined in the Users' Note for your implementation. 5: fail – NagError Input/Output The NAG error argument (see Section 3.6 in the Essential Introduction). ## 6 Error Indicators and Warnings NE_BAD_PARAM On entry, argument $〈\mathit{\text{value}}〉$ had an illegal value. NE_INT On entry, ${\mathbf{n}}=〈\mathit{\text{value}}〉$. Constraint: ${\mathbf{n}}\ge 0$. 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. NW_IVALID On entry, at least one value of x was invalid. ## 7 Accuracy For negative arguments the function is oscillatory and hence absolute error is the appropriate measure. In the positive region the function is essentially exponential-like and here relative error is appropriate. The absolute error, $E$, and the relative error, $\epsilon $, are related in principle to the relative error in the argument, $\delta $, by $E≃ x Bi′x δ,ε≃ x Bi′x Bix δ.$ In practice, approximate equality is the best that can be expected. When $\delta $, $\epsilon $ or $E$ is of the order of the machine precision, the errors in the result will be somewhat larger. For small $x$, errors are strongly damped and hence will be bounded essentially by the machine precision. For moderate to large negative $x$, the error behaviour is clearly oscillatory but the amplitude of the error grows like amplitude $\left(\frac{E}{\delta }\right)\sim \frac{{\left\|x\right\|}^{5/4}}{\sqrt{\pi }}$. However the phase error will be growing roughly as $\frac{2}{3}\sqrt{{\left\|x\right\|}^{3}}$ and hence all accuracy will be lost for large negative arguments. This is due to the impossibility of calculating sin and cos to any accuracy if $\frac{2}{3}\sqrt{{\left\|x\right\|}^{3}}>\frac{1}{\delta }$. For large positive arguments, the relative error amplification is considerable: $εδ∼x3.$ This means a loss of roughly two decimal places accuracy for arguments in the region of $20$. However very large arguments are not possible due to the danger of causing overflow and errors are therefore limited in practice. None. ## 9 Example This example reads values of x from a file, evaluates the function at each value of ${x}_{i}$ and prints the results. ### 9.1 Program Text Program Text (s17avce.c) ### 9.2 Program Data Program Data (s17avce.d) ### 9.3 Program Results Program Results (s17avce.r)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 64, "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": 0.6800842881202698, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/70132/list	## Return to Answer 3 deleted 1 characters in body A bit belatedly, but perhaps marginally usefully: first, yes, indeed, all you needed was that $f\in C^o_c(G)$ gave a compact operator, not even necessarily Hilbert-Schmidt or trace-class, tho' the latter does lead to further interesting things. Second, about integrals, Gelfand and Pettis (c. 1928) effectively created a very nice notion of "integral" which works wonderfully for continuous, compactly-supported functions with values in any locally-convex, quasi-complete tvs. (These integrals are sometimes called "weak", but this is misleading in several ways.) The characterization is "weak" in the sense that $\int_X f\in V$ is uniquely determined by the fact that, for every continuous linear functional $\lambda$ on $V$, $\lambda(\int_X f)=\int_X \lambda\circ f$, where the scalar-valued integral of continuous, compactly-supported is unambiguous. One also proves that the integral is in the closure of the convex hull of the image $f(X)$, which gives a grip on "estimates". One immediate corollary is that for any continuous linear $T:V\rightarrow W$, the integral of $Tf$ is $T$ of the integral of $f$, which leads to justification of differentiation under integrals, and such. Further, giving operators on a Hilbert space the "strong operator topology" (not norm...) by $p(T)=\sup_{\|x\|\le 1} \|Tx\|$, which is what makes $G\times V\rightarrow V$ continuous, actually the operator-valued integral $\int_G f(g)\,T(g)\,dg$ makes sense, for $f\in C^o_c(G)$. E.g., see here . Various natural continuations of this, such as vector-valued holomorphic functions, were treated by Schwartz and Grothendieck. E.g., see here . Very handy on occasion. Third, about repns "decomposing discretely, with finite multiplicities". I think for practical purposes it's not so clear what a "decomposition" would mean outside the Hilbert space context, although notions of compact or trace-class or nuclear operators have senses in larger contexts. I've heard gossip about concerted efforts at Yale in the 1950s to be able to talk about "spectral theory" in more general contexts, but it seems that it just doesn't work very well beyond Hilbert spaces, and the Fredholm alternative for special operators on Banach spaces. That such a decomposition succeeds for $L^2(compact-quotient)$ was arguably known to several people in the 1950s already: probably Gelfand et alia, but also Selberg and others, and certainly Langlands by the early 1960s. Probably those people would say that the compact quotient case was "obvious", and that the issue of serious interest was the non-compact quotient case, where one has to do some serious work to prove that the operators are still trace-class on cuspforms. Probably Gelfand-PS gave the first more-or-less proof of that, although the reader has pretty substantial responsibilities there. The "continuous" decompositions we know for general unitary repns of type I groups are not very useful, unfortunately, in that they give no particular information. Indeed, one can execute the proof that Eisenstein series span various bits of continuous spectra... literally decomposing $L^2$ functions... without even formulating the general notion, somewhat like we can prove Fourier inversion without explaining how to view $L^2(\mathbb R)$ as the Hilbert direct integral of one-dimensional repns... Edit: after @pm's commentanswer, I realized I was not clear in what I wrote, at best: First, yes, I was thinking only of $G$ unimodular. Second, for many applications one wants to take trace, and trace class is a proper subset of Hilbert-Schmidt is a proper subset of "compact", altho' the composition of two Hilbert-Schmidt is trace class (perhaps by definition). Again, the most difficult issue is proving trace class _on_cuspforms_ for integral operators attached to not-necessarily very smooth functions on a Lie group (e.g.), which Langlands did in the 1960s, I think roughly the same time as Gelfand-PS, tho' perhaps much worse documented. Adele-group-or-not is not the key point, I think. I am not aware of any systematic approach to not-co-compact $\Gamma$ in general topological groups $G$, only for reductive Lie or adele or similar. Re: chronology, one should note that Harish-Chandra proved sharp versions of admissibility of unitaries of reductive Lie groups in the 1950s, tho' I do not know whether his results explicitly mentioned "trace class" issues. And, yes, since the composition of two Hilbert-Schmidt ops is trace-class, the Cartier/Dixmier-Malliavin result that all smooth functions are finite sums of convolutions of smooth functions certainly crushes certain issues. Many interesting things here! :) 2 added 1343 characters in body Edit: after @pm's comment, I realized I was not clear in what I wrote, at best: First, yes, I was thinking only of $G$ unimodular. Second, for many applications one wants to take trace, and trace class is a proper subset of Hilbert-Schmidt is a proper subset of "compact", altho' the composition of two Hilbert-Schmidt is trace class (perhaps by definition). Again, the most difficult issue is proving trace class _on_cuspforms_ for integral operators attached to not-necessarily very smooth functions on a Lie group (e.g.), which Langlands did in the 1960s, I think roughly the same time as Gelfand-PS, tho' perhaps much worse documented. Adele-group-or-not is not the key point, I think. I am not aware of any systematic approach to not-co-compact $\Gamma$ in general topological groups $G$, only for reductive Lie or adele or similar. Re: chronology, one should note that Harish-Chandra proved sharp versions of admissibility of unitaries of reductive Lie groups in the 1950s, tho' I do not know whether his results explicitly mentioned "trace class" issues. And, yes, since the composition of two Hilbert-Schmidt ops is trace-class, the Cartier/Dixmier-Malliavin result that all smooth functions are finite sums of convolutions of smooth functions certainly crushes certain issues. Many interesting things here! :) 1 A bit belatedly, but perhaps marginally usefully: first, yes, indeed, all you needed was that $f\in C^o_c(G)$ gave a compact operator, not even necessarily Hilbert-Schmidt or trace-class, tho' the latter does lead to further interesting things. Second, about integrals, Gelfand and Pettis (c. 1928) effectively created a very nice notion of "integral" which works wonderfully for continuous, compactly-supported functions with values in any locally-convex, quasi-complete tvs. (These integrals are sometimes called "weak", but this is misleading in several ways.) The characterization is "weak" in the sense that $\int_X f\in V$ is uniquely determined by the fact that, for every continuous linear functional $\lambda$ on $V$, $\lambda(\int_X f)=\int_X \lambda\circ f$, where the scalar-valued integral of continuous, compactly-supported is unambiguous. One also proves that the integral is in the closure of the convex hull of the image $f(X)$, which gives a grip on "estimates". One immediate corollary is that for any continuous linear $T:V\rightarrow W$, the integral of $Tf$ is $T$ of the integral of $f$, which leads to justification of differentiation under integrals, and such. Further, giving operators on a Hilbert space the "strong operator topology" (not norm...) by $p(T)=\sup_{\|x\|\le 1} \|Tx\|$, which is what makes $G\times V\rightarrow V$ continuous, actually the operator-valued integral $\int_G f(g)\,T(g)\,dg$ makes sense, for $f\in C^o_c(G)$. E.g., see here . Various natural continuations of this, such as vector-valued holomorphic functions, were treated by Schwartz and Grothendieck. E.g., see here . Very handy on occasion. Third, about repns "decomposing discretely, with finite multiplicities". I think for practical purposes it's not so clear what a "decomposition" would mean outside the Hilbert space context, although notions of compact or trace-class or nuclear operators have senses in larger contexts. I've heard gossip about concerted efforts at Yale in the 1950s to be able to talk about "spectral theory" in more general contexts, but it seems that it just doesn't work very well beyond Hilbert spaces, and the Fredholm alternative for special operators on Banach spaces. That such a decomposition succeeds for $L^2(compact-quotient)$ was arguably known to several people in the 1950s already: probably Gelfand et alia, but also Selberg and others, and certainly Langlands by the early 1960s. Probably those people would say that the compact quotient case was "obvious", and that the issue of serious interest was the non-compact quotient case, where one has to do some serious work to prove that the operators are still trace-class on cuspforms. Probably Gelfand-PS gave the first more-or-less proof of that, although the reader has pretty substantial responsibilities there. The "continuous" decompositions we know for general unitary repns of type I groups are not very useful, unfortunately, in that they give no particular information. Indeed, one can execute the proof that Eisenstein series span various bits of continuous spectra... literally decomposing $L^2$ functions... without even formulating the general notion, somewhat like we can prove Fourier inversion without explaining how to view $L^2(\mathbb R)$ as the Hilbert direct integral of one-dimensional repns...	{"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": 40, "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": 0.9597620964050293, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/20089-subgroups-normal-subgroups.html	# Thread: 1. ## Subgroups and normal subgroups Hey I have a question out of my algebra text (Hungerford, p.92 #13) It asks to show that if G is a group containing a proper subgroup H of finite index, then it contains a proper normal subgroup of finite index. It's easy to show the fact if G is finite or abelian, but I'm left with the last case if G is nonabelian and infinite. I'm trying to work with the center, which I know is normal, but I don't know how to manipulate it to create a resultant normal group of finite index, and obviously I need to incorporate the given subgroup H somehow. Letting C be the center of G, I tried to consider the group HC = { hc \| h in H, c in C} but I get no success trying to show this is either normal or has finite index. I am terrible with conjugation in groups in general, and considering this exercise involves a degree of counting cosets, I'm suspecting conjugation is going to play a part, which is even more intimidating. Any assistance would be appreciated. 2. Could you check the problem statement? As it stands, the statement is false. For example, the Icosahedral group is a finite simple group. 3. Originally Posted by mathisfun1 Could you check the problem statement? As it stands, the statement is false. For example, the Icosahedral group is a finite simple group. The proposition holds as stated, even for simple groups. The proper normal subgroup induced doesn't have to be nontrivial. The group generated by the identity is normal and has index equal to the order of the group, which is finite for finite groups. In the meantime, I sought help through my professor, and we established a homomorphism between G and the symmetric group of H's left cosets. The kernel of this is the subgroup I've been looking for: it's obviously normal, and it has finite index because G/Ker can't have more than n! elements, where n is the index of H. Thanks anyway. I'm sure I'll have more. 4. How about $A_5$? All it s subgroups are of finite index. Yet it contains no proper non-trivial normal subgroup. 5. Originally Posted by ThePerfectHacker How about $A_5$? All it s subgroups are of finite index. Yet it contains no proper non-trivial normal subgroup. Right, the proposition doesn't guarantee the induced normal subgroup is necessarily nontrivial. In $A_5$, the subgroup generated by the identity permutation is normal and isn't equal to $A_5$, so it satisfies the conclusion. The fact that it's trivial doesn't contradict the proposition. For finite groups (with at least two elements), the proposition itself is easy because the trivial subgroup is always normal and has finite index. The challenge I had was showing the statement for nonabelian infinite groups, in which the trivial subgroup has infinite index so I had to find another one. 6. Here is what your problem comes down to: "If G is infinite and has a proper group of finite index it cannot be simple." Look at the contrapositive, "If G is infinite and has a proper subgroup of finite index it cannot have be simple". Let us prove the contrapositive by contradiction. Let G be and infinite group and H < G where (G:H) is finite. We can view G as an G-set by defining a group action on H as "left-translation on the cosets of H". Let X be the left cosets of H then this creates a homomorphism f between G and S_X (symettric group). Thus, there is a homomorphic map f: G --> S_X.** This mapping cannot possibly be one-to-one because G is an infinite group and \|S_n\|=n!. So kernel(f) cannot be trivial. Since G is simple it means kernel(f) = G. But H contains kernerl(f), thus H=G. We have shown that if (G:H) is finite it means H=G, i.e. G has no proper subgroups of finite index. Q.E.D. )It has not proper non-trivial normal subgroups. )Define the operation g(aH) = (ga)H, this maps left cosets of H into left cosets of H and satisfies the conditions for being a G-set. **)This should be known to you. If not define f(g) = i_g(x) where i_g(x) is the map i_g(x) = gx. Now show it is a homomorphism. I can retype this in LaTeX if you want, but group theory is often were simple on notation. 7. I tried to find a way to create normal larger subgroups from given groups. But I was not able to (what you were trying). Neither was I able to find a construction on the internet. I am convinced there is no nice construction that we can use here. 8. Originally Posted by ThePerfectHacker I tried to find a way to create normal larger subgroups from given groups. But I was not able to (what you were trying). Neither was I able to find a construction on the internet. I am convinced there is no nice construction that we can use here. I agree, after a lot of playing around, I don't think there is without knowing more about what the group itself looks like. Your proof which constructs a finite indexed kernel is the one I ended up working out with the assistance of my professor. Thank you TPH. 9. In the beginning I tried to use a the normalizer subgroup. So given $H<G$ define $N[H] = \{ g\in G \| ghg^{-1} \in H \}$. If this normalizer subgroup is not normal repeat the contruction .... But this does not work too well, because it depends on the group structure we are dealing with.	{"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": 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": 0.9500495791435242, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/159637-finding-stationary-points.html	# Thread: 1. ## Finding stationary points. Can somebody help me find the stationary points of this: \displaystyle{Q(x) = \frac{4x^2 + 20x + 25}{e^x}} and if the above doesnt display right it is: 4x^2+20x+25/e^x and give the value of the points? Also say whether points are a local max or min and explain why to me? Thank you in advance! 2. What work have you done so far? 3. I did the quotient rule and got it down to a quadratic over e^x -4x^2-12x-5/e^2 what do I do from there? or is that right? lol 4. You need to be more careful with your parentheses. Technically, you wrote -4x^2-12x-(5/e^2). Also, I think you forgot the 2x in the exponent. Anyway, aside from notation issues, I think you have the general idea: $\displaystyle Q'(x)=-\frac{4x^{2}+12x+5}{e^{2x}}.$ What now? How do you find stationary points? 5. e^2x on the bottom? are you sure? Well you have to find out which values make the equation equal zero after the first derivative I'm fairly certain. Then you put that point back into the equation to see what value you get for that point. THEN get the second derivative and and judging by what THAT point is you can tell if you have a local max or min.... but I def need some help lol 6. e^2x on the bottom? are you sure? Quotient rule: low dee high minus high dee low over the square of what's below, or $\displaystyle\left(\frac{f}{g}\right)'=\frac{gf'-fg'}{g^{2}}.$ Since you have an $e^{x}$ in the denominator of the original fraction, you'd better have an $(e^{x})^{2}=e^{2x}$ in the denominator afterwards, right? You're not going to lose the variable x, at any rate. So what do you get when you set the derivative equal to zero? 7. Do I set the -4X^2-12x-5 to zero and forget about the bottom right now? In that case I get x=-5 and x=-3? -4x^2-12x-5=0 -4x^2-12x=5 -x(4x-12)=5 x=-5 & x=-3? 8. You are correct in forgetting about the denominator, since it will never contribute towards a zero. However, you are not solving the quadratic correctly. You cannot assume that if ab = c, where c is not zero, that a=c, or b=c. Just take 2 x 3 = 6. It is not true that 2 = 6 or 3 = 6. Complete the square, or use the quadratic formula. What do you get? 9. Okay. So I used quadratic formula instead. I got x=-2.5 and x=-0.5 10. So those are the x-coordinates for your stationary points. You need to find the y-coordinates, I think. What else are you going to need to do, do you think? 11. Okay, so now I put the x values back into the function to get the y values. which should tell me if I have a local max or min? 12. So I put the values back into the function and ignored the e^x or e^2x.... right thing to do? I got x=-62.5 and x=14	{"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": 0.9050585031509399, "perplexity_flag": "middle"}

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