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http://mathhelpforum.com/discrete-math/143065-another-probability-question.html
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Thread:
1. Another probability question
A colony of rare spherical cocci bacteria lives deep under lunar ice. After thousands of years of slowly reproducing (a process that often results in abject failure and death) in this harsh climate, only one lonely bacteria named John remains.
Suppose that the probability of an individual bacteria dividing successfully into two bacteria is 3/4, and that a bacteria must attempt the division process within two months of its immediately prior division experience.
John's time is up. It is time for him to attempt a division. What is the probability that poor John will be the proud predecessor of never-ending successive generations of his bacterial species?
2. Originally Posted by chickeneaterguy
A colony of rare spherical cocci bacteria lives deep under lunar ice. After thousands of years of slowly reproducing (a process that often results in abject failure and death) in this harsh climate, only one lonely bacteria named John remains.
Suppose that the probability of an individual bacteria dividing successfully into two bacteria is 3/4, and that a bacteria must attempt the division process within two months of its immediately prior division experience.
John's time is up. It is time for him to attempt a division. What is the probability that poor John will be the proud predecessor of never-ending successive generations of his bacterial species?
I think the general way to approach this is to take 1 and subtract the probability that the race goes extinct.
For the first generation, the probability of extinction is 1/4.
In the second generation, the probability is (3/4)(1/4)(1/4), which is the probability that both children fail given that the first generation was successful.
For the third generation, it is (3/4)^3*(1/4)^4 + (3/4)(1/4)(3/4) * (1/4)^2 + (3/4)(3/4)(1/4) * (1/4)^2.
Since it's getting messy, there's probably a better way to organize things, and possibly you can get an exact result somehow. (I'm reminded of those infinite sums that turn out to have $\pi^2$ in the answer, although I have no idea what it would be in this case.) But if you only need an approximation, you can get a pretty small error quickly this way, just on paper. And with a computer program you can get to a high number of generations quickly using dynamic programming.
Oh and since the bacteria reproduce asexually, we don't need to consider the possibility that they take varying times to decide to reproduce. Whether they wait a minute or two months, the overall result is the same. But I am assuming that the bacteria have limited lifespans, and so they all attempt to reproduce.
3. Originally Posted by undefined
I think the general way to approach this is to take 1 and subtract the probability that the race goes extinct.
For the first generation, the probability of extinction is 1/4.
In the second generation, the probability is (3/4)(1/4)(1/4), which is the probability that both children fail given that the first generation was successful.
For the third generation, it is (3/4)^3*(1/4)^4 + (3/4)(1/4)(3/4) * (1/4)^2 + (3/4)(3/4)(1/4) * (1/4)^2.
I assume the 2nd generation the 3/4 is coming from the chance that the other survived and the other 2 could die?
I don't understand the 3rd gen one.
4. I chose to double-post instead of editing my first post, so that people will get email notification if desired. If this is not the proper etiquette, please let me know.
I think it's possible to get an exact answer using Markov chain theory. Some literature is here (Chapter 11 of Introduction to Probability by Grinstead and Snell, 2nd ed, freely distributable under a GNU license.) Extinction can be set up as a sole absorbing state, and then the matrix inverse can be found as described in the .pdf file. I haven't actually used this method before, but I believe it applies here.
5. Originally Posted by chickeneaterguy
I assume the 2nd generation the 3/4 is coming from the chance that the other survived and the other 2 could die?
I don't understand the 3rd gen one.
Oh, we posted at almost the same time, haha.
Yes, the second generation probability I gave is p(B|A) where A = the first reproduction was successful, and B = the race goes extinct at that generation.
For the 3rd generation, we consider cases.
Case 1: John was successful, and both of his two children were successful. So there are 4 bacteria, and the probability of extinction right now is (1/4)^4 times the probability of getting to this state, or (3/4)(3/4)(3/4).
Case 2: John was successful, and Child A failed, but Child B succeeded. There are two bacteria now, so we have (1/4)^2 times the probability of getting there, (3/4)(1/4)(3/4).
Case 3: Same as case 2 but with Child A and Child B reversed.
6. Originally Posted by undefined
Oh, we posted at almost the same time, haha.
Yes, the second generation probability I gave is p(B|A) where A = the first reproduction was successful, and B = the race goes extinct at that generation.
For the 3rd generation, we consider cases.
Case 1: John was successful, and both of his two children were successful. So there are 4 bacteria, and the probability of extinction right now is (1/4)^4 times the probability of getting to this state, or (3/4)(3/4)(3/4).
Case 2: John was successful, and Child A failed, but Child B succeeded. There are two bacteria now, so we have (1/4)^2 times the probability of getting there, (3/4)(1/4)(3/4).
Case 3: Same as case 2 but with Child A and Child B reversed.
Okay, this makes a whole lot of sense. I'll try it out. Thanks.
7. Actually, I think there should only be one child out of the first split. I think it it splitting into two, and not creating two extra.
8. Originally Posted by chickeneaterguy
Actually, I think there should only be one child out of the first split. I think it it splitting into two, and not creating two extra.
My understanding is that after a successful division, the parent no longer exists, and two children exist in its place.
For unsuccessful division, the parent dies and there are no children.
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http://mathoverflow.net/questions/tagged/projective
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## Tagged Questions
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### Projective spaces with nonconstant regular functions
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### What is the Birkhoff norm of a Perron vector?
Let $A$ be a positive matrix. What is known about the Birkhoff norm of its Perron vector? By the Birkhoff norm of a vector $x$ I refer to the quantity $\frac{\max{x}}{\min{x}}$. …
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http://physics.stackexchange.com/questions/53986/component-of-force-field-tangent-to-a-curve
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# Component of force field tangent to a curve
I recently had to solve the following problem:
A force $F=10\hat x- 8\hat y$ is applied to an object that is constrained to travel towards increasing values of x along the path defined by $y=x^2$ and $z=0$. Find the component of $F$ that is tangent to this path at the point $(2,4,0)$.
Attempt: I used Pythagorean's theorem to $F=12.81 N$ since the $x$ and $y$ components are given. I just wasn't sure if this is what the questions is asking for, or is there something else I have do do?
Edit. More generally, suppose that one is given a curve and a force field on the plane. How would one go about computing the component of such a force field in the direction tangent to the curve at a given point?
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1
@Qmechanic I added an edit to make the question more broadly conceptually applicable; does the edit suffice to reopen this question? – joshphysics Feb 16 at 0:17
## 1 Answer
At the specified point, there is a certain tangent line to the path. Also at that point, the vector field $\mathbf F$ has certain components. We want to know the component of $\mathbf F$ in the direction of the tangent line to the path at the point given.
Here's how I would approach this.
Let's restrict to the $x$-$y$ plane. The curve on which the object is traveling can be written as a parameterized curve in 2D as $$\mathbf x(s) = (x(s), y(s)) = (s, s^2)$$ The unit tangent vector to this path at parameter value $s$ is $$\mathbf T(s) = \frac{\mathbf x'(s)}{|\mathbf x(s)|} = \frac{(1, 2s)}{\sqrt{1+4s^2}}$$ The tangential component of a vector field $\mathbf F(\mathbf x)$ at a given point along the curve corresponding to parameter value $s$ is therefore $$F_T(s) = \mathbf F(\mathbf x(s))\cdot\mathbf T(s)$$ I'll let you try to figure out the rest.
Cheers!
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http://crypto.stackexchange.com/questions/5122/impact-of-algorithms-for-factoring-using-elliptic-curves-over-mathbbq?answertab=oldest
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# Impact of algorithms for factoring using elliptic curves over $\mathbb{Q}$
Recently a few papers have appeared that describe a new approach to factoring, using elliptic curves over $\mathbb{Q}$. See, e.g.,
• Factoring integers and computing elliptic curve rational points, Iftikhar A. Burhanuddin and Ming-Deh A. Huang.
• Factoring integers using elliptic curves over Q, Xiumei Li, Jinxiang Zeng, arXiv:1207.0274v2.
However, these papers don't seem to describe the complexity of these novel methods. What is the running time of these methods? Are these methods faster than the standard known algorithms for factoring? What is the impact of these new ideas on the security of RSA and other factoring-based cryptosystems? Should we be concerned?
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## 1 Answer
I don't think there's anything to worry about here. Remarks 4.2 and 4.3 in the second paper point out that these approaches need to first find a point on a curve with a very large conductor/discriminant, and that seems to be very hard. (Harder than factoring the integer using NFS.)
There are many ways to compute factorings and discrete logarithms where you first do some very hard number theoretic computations (e.g. related to an elliptic curve over the rationals), then easily factor or compute a d.log. I designed such an algorithm myself once, and it was utterly useless.
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http://mathoverflow.net/revisions/25986/list
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## Return to Answer
2 Corrected typo
Etimology
Etymology may help. Compact, from the Latin Compactus, past participle of the verb Compingere: "to pack together closely and firmly". You may have an idea of how strong is the hypothesis of compactness if you look at what happens when even total boundedness is lacking.
Just consider e.g. the possibly most familiar infinite dimensional object, the separable Hilbert space H=l2. Its unit closed ball is not compact. A continuous real valued function on it may be unbounded, or bounded without minimum and maximum value. A linear form on H may be everywhere discontinuous and locally unbounded. There is a continuous transformation of the unit closed ball with no fixed points. The ball itself retracts on its boundary. Infinitely many disjoint unit open balls may be packed within a ball of radius $1+ \sqrt{2}$. The linear group GL(H) is connected. There are injective linear continuous transformations of H that are not surjective....
1
Etimology may help. Compact, from the Latin Compactus, past participle of the verb Compingere: "to pack together closely and firmly". You may have an idea of how strong is the hypothesis of compactness if you look at what happens when even total boundedness is lacking.
Just consider e.g. the possibly most familiar infinite dimensional object, the separable Hilbert space H=l2. Its unit closed ball is not compact. A continuous real valued function on it may be unbounded, or bounded without minimum and maximum value. A linear form on H may be everywhere discontinuous and locally unbounded. There is a continuous transformation of the unit closed ball with no fixed points. The ball itself retracts on its boundary. Infinitely many disjoint unit open balls may be packed within a ball of radius $1+ \sqrt{2}$. The linear group GL(H) is connected. There are injective linear continuous transformations of H that are not surjective....
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http://physics.stackexchange.com/questions/53184/would-a-magnetic-rod-through-mars-shield-it-from-solar-wind
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# Would a magnetic rod through Mars shield it from solar wind?
Would a magnetic rod going all the way through Mars shield it from solar wind?
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## 2 Answers
Theoretically yes.
It would have to produce a magnetic field of sufficient strength of course.
The Earth's magnetic field at the Earth's surface ranges from 25 to 65 µT. Given that Mars is approximately 50% smaller than the Earth a smaller field would probably be sufficient.
The benefits of the shielding would be:
1. The level of radiation at the surface would be greatly reduced thus making Mars a safer place to live.
2. Mars would be able to hold a thicker atmosphere than it has now. One of the reasons that Mar's atmosphere is so thin is due to the solar wind actually stripping away the atmospheric gases.
If such a rod could be built and installed it would be a constant strength and would do a better job of protecting Mars than Earth's field does of protecting the Earth as that fluctuates and is "due" for one of it's periodic reversals.
However, it's probably beyond our current technologies to build such a rod.
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You needn't necessarily run a rod all the way through. A wire wrapped around the circumference of Mars would do the job as well.
Since a magnetic dipole (which is basically what a bar magnet is) can be thought of as a current loop, wrapping a wire around Mars would achieve the same thing.
I have to point out though, that this solution is not very feasible at all. The magnetic field produced by a current through a loop is given by $$B = \frac{\mu _0 I }{2 \pi r} \text{Tesla}$$ Therefore for a magnetic field of ~ $1 \mu T$ and the radius of Mars being $3397$ $km$, we'd need a current of about $100000000$ $A$ ($10^8$ $A$) which is a massive amount of current to generate and sustain.
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http://mathoverflow.net/questions/94513/tracking-spectral-sequence-differentials
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## Tracking spectral sequence differentials
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I read a number of posts here on MO, but haven't quite found an answer to the question of where the differentials in a spectral sequence come from.
I came across a differential $d^{0,1}$ on the $E_2$-page from a 2-dim to a 3-dim vector space (in an application of the cohomological Lyndon-Hochschild-Serre s.s.). I know that $E^{0,1}_{\infty}=\mathrm{ker} d^{0,1}$.
A priori, the kernel has dimension 0, 1 or 2. How can I determine the dimension?
In general, my question is, whether there is a way of tracking down differentials in a s.s. explicitly. Where do they come from in the first place? In the books I have read, differentials seem to be given... Still, I imagine, somewhere hidden in the proof must be a construction that proves they exist and are determined. After all, in my calculation, even if 0, the kernel of my differential does not get lost, but does get promoted to star in the second cohomology group, rather than the first. And this surely has to be prevented by a properly converging s.s.
(And just out of curiosity: How would such an explicit construction be possible in the Grothendieck s.s.?)
Thank you for any insights.
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In the Grothendieck spectral sequence the differentials come from the injective (or projective) resolutions involved (the construction can be found in XX.9 of Lang: Algebra): They give rise to a double complex. By filtering the double complex on obtains a filtered complex and a filtered complex always yields a spectral sequence with $(E_r,d_r)$ given explicitly in terms of the filtered complex (see Prop. XX.9.1 in Lang's book). – Ralph Apr 19 2012 at 11:35
There is a difference between defining what a spectral sequence is and constructing one. In the definition of `spectral sequence' the differentials are given. When constructing a spectral sequence like the Hochschild-Serre spectral sequence one must construct the differentials. This is of course exactly what Hochschild and Serre did. – Wilberd van der Kallen Apr 20 2012 at 7:36
## 2 Answers
This answer by Ralph gives an explicit description of the differential in the LHSSS you mention.
Another thought: presumably you have a group extension $1\to K\to G\to G/K\to 1$. If you are using field coefficients (as you seem to imply) then there are no extension problems and $$H^1(G)\cong E_{\infty}^{0,1}\oplus E^{1,0}_\infty \cong \ker d^{0,1}\oplus E^{1,0}_2.$$
Presumably you can calculate the first cohomologies of $G$ and $G/K$, in which case a simple dimension count should give you the dimension of the kernel of $d^{0,1}$.
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That has it somewhat backwards. I want to calculate the dimension of $H^1(G,M)$, with twisted coefficients from $K$ and $G/K$, which happen to be abelian. I thought the best way of getting at the cohomology of $G$ would be via the s.s. Indeed, the terms are easy to calculate, but the result is obscured by the differential. Maybe it would have been wiser to bite the bullet and calculate $H^1(G,M)$ directly via the bar resolution...? – s.barmeier Apr 19 2012 at 12:10
2
$H^1(G;M)$ is just the quotient of the derivations $Der(G,M)$ by the subgroup $Ider(G,M)$ of inner derivations. Also note that the seven-term-exact sequence yields the exact sequence $0 \to H^1(G/K;M) \to H^1(G;M) \to H^1(K;M)^{G/K} \to H^2(G/K;M^K)$ that could be of help. – Ralph Apr 19 2012 at 12:32
### 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.
Without meaning to be snarky, I think there's some confusion here about what spectral sequences are for. In particular, the sense in which a SpSeq (or even a long exact sequence!) serves as a computational tool isn't that it tells you how to calculate things --- particularly, not how to calculate the things in it. Rather, it gives you a way to organize things you already know how to calculate; maybe it suggests a helpful summary of your calculations; and it will certainly remind you of what things you don't already know --- it will provide helpful questions to consider --- if you try to squeeze modules out of it.
As a small example, the usual snake-shapped diagram chase proving the isomorphism of, e.g. de-Rham and real Čech cohomology, is neatly summarized by noting that there's a related double complex which is very nearly exact in every way, and thus all the later pages are mostly zero, while the transgression is an isomorphism. We get a proof, because we already know what enough of the modules and differentials are.
If you really want to ask what the pages and differentials of some SpSeq are, there are algorithmic tools (i.e., programmes --- usually not suitable for human consumption!), provided you can properly specify what your particular spectral sequence is about. See e.g. (Ana Romero's thesis)1, building on homological perturbation lemmata (start at 2).
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http://mathhelpforum.com/pre-calculus/154058-multiplying-imaginary-factors-polynomial.html
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# Thread:
1. ## [Solved] Multiplying imaginary factors of a polynomial
The problem:
$(x - e^{\frac{i\pi}{4}})(x - e^{\frac{-i\pi}{4}})$
My attempt:
I used the distributive rule to expand it, and got:
$x^2 - x(e^{\frac{i\pi}{4}} + e^{\frac{-i\pi}{4}}) + 1$
I recognise that $cos(\frac{\pi}{4}) = \frac{1}{2}(e^{\frac{i\pi}{4}} + e^{\frac{-i\pi}{4}})$
So I get $x^2 - x(2cos(\frac{\pi}{4})) + 1$
= $x^2 - \frac{2x}{\sqrt{2}} + 1$
= $x^2 - \sqrt{2}x + 1$
I'm unsure if this is correct, since I don't have a solution. Any assistance would be greatly appreciated.
2. Originally Posted by Glitch
The problem:
$(x - e^{\frac{\pi}{4}})(x - e^{\frac{-\pi}{4}})$
My attempt:
I used the distributive rule to expand it, and got:
$x^2 - x(e^{\frac{\pi}{4}} + e^{\frac{-\pi}{4}}) + 1$
I recognise that $cos(\frac{\pi}{4}) = \frac{1}{2}(e^{\frac{\pi}{4}} + e^{\frac{-\pi}{4}})$
So I get $x^2 - x(2cos(\frac{\pi}{4})) + 1$
= $x^2 - \frac{2x}{\sqrt{2}} + 1$
= $x^2 - \sqrt{2}x + 1$
I'm unsure if this is correct, since I don't have a solution. Any assistance would be greatly appreciated.
$\cos x = \frac12(e^{ix} + e^{-ix})$ (notice the i's in the exponents). So either the problem should read $(x - e^{\frac{\pi i}{4}})(x - e^{\frac{-\pi i}{4}})$, or the solution should have cosh instead of cos. If the problem should have had i's in the exponents then the solution is correct.
3. Whoops, yeah, there's supposed to be 'i'. Thanks.
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http://math.stackexchange.com/questions/98065/how-many-ways-can-b-balls-be-distributed-in-c-containers-with-no-more-than/98356
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# How many ways can $b$ balls be distributed in $c$ containers with no more than $n$ balls in any given container?
I think there are $\binom{b+ c - 1}{c-1}$ ways to distribute $b$ balls in $c$ containers. (Please correct me if that's a mistake.) How does this change if I am not allowed to place more than $n$ balls in any given container?
If we call this number $N(b,c,n)$ then I can come up with
$$N(b,c,n) = \sum_{k=0}^n N(b-k, c-1, n)$$
with the base cases
$$N(b,c,n) = \binom{b+ c - 1}{c-1}$$
for $n\geq b$, and
$$N(b,c,n) = 0$$
for $n<b/c$.
Is there some better way of writing this than that recursion relation?
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## 5 Answers
If you carry out the inclusion-exclusion argument mentioned by true blue anil in general, you get
$$N(b,c,n)=\sum_i(-1)^i\binom{c}i\binom{b+c-1-i(n+1)}{c-1}\;;$$
whether this is actually any more useful than the recurrence is another question.
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You could use inclusion-exclusion.
I'll illustrate with a concrete case, you can then work out the formula.
Place 10 identical balls in 5 distinct containers with no more than 4 balls in any container.
We see that if left unconstrained, 1 or 2 containers may violate the restriction, so we allot 5 balls each to 1 or 2 containers, and correct for such configurations to get
C(14,4) - C(5,1)*C(9,4) + C(5,2)*C(4,4) = 381
Now generalise.
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This is the number of solutions to $x_1+x_2+ \cdots + x_c = b$ where $0 \leq x_i \leq n$ for $i = 1, \dots, c$.
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As Thomas pointed out, you can use generating functions to help with this problem. Also, Brian rightly notes you could use PIE and also he observes which method is more useful computationally is a different question altogether.
Below, I will produce a standard (and in my opinion still a beautiful and more expressive) argument to prove why is $N(b,c) = {{b+c-1}\choose{c-1}}$.
So, let us assume that you have got $b$ bananas you want to distribute among $c$ children. Some children can get $0$ bananas and you want to give away all $b$ bananas. Now let us imagine that you have $c-1$ "fake bananas" which are added to your initial stock giving a total of $b+c-1$ bananas. You can imagine that the bananas left to the first fake banana $c_1$ are reserved for child $C_1$. The bananas between fakes $c_i$ and $c_{i+1}$ are reserved for child $C_i$ for all $i < c-1$ and the last child gets all bananas to the right of fake banana $c_{c-1}$. Thus, the number of ways you can distribute these bananas is precisely equal to the places where you can have $c-1$ fake bananas among $b+c-1$ bananas. And the number of bananas, therefore is ${{b+c-1}\choose{c-1}}$.
I know this does not really answer your question but it is good to know about this argument in case you do not know it already.
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Thank you. Yes, that was how I got that identity. – Mark Eichenlaub Jan 12 '12 at 14:37
For $n\gt 2$ you can reduce the number of additions in the recursion if you replace
$$N(b,c,n) = \sum_{k=0}^n N(b-k, c-1, n)$$
by
$$N(b,c,n) = N(b-1, c, n)+N(b, c-1, n)-N(b-k-1, c-1, n).$$
It may not be a saving, but you can avoid multiplication and division if you replace the base cases by $N(0,c,n)=1$ and $N(b,c,n)=0$ for $b\lt 0$.
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http://mathhelpforum.com/math/195681-test-preparation-print.html
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# Test preparation
Printable View
• March 7th 2012, 12:34 AM
KristianS
Test preparation
Hey guys
Im new here and its good to see that these forums are active and full of friendly people. I was wondering if someone could help me with some test preparation. My unit that I am revising on consists of Linear algebra/functions and I need help getting some videos to cover certain topics.
• Solve equations "Y"
• Determine if an equation is linear "X"
• Tell the difference between a function and a relation "Y"
• State the domain and range of a function or relation "Y"
• State if a function is discrete or continuous "X"
• Find the gradient of a linear function, graphically and algebraically "Y"
• Find the x and y intercepts of a linear function "X"
• Find the equation of a linear function "X"
• Find a function that is parallel or perpendicular to another function "Y"
• Determine if points are collinear "Y"
• Find the distance between two points "Y"
• Find the midpoint between two points "Y"
That is everything I need to know, if it has an "X" on it then I need help, if it has a "Y" on it then I don't. I would greatly appreciate if anyone could recommend some videos for me to watch that will cover the topics with "X's" on them. I am currently doing year 11 maths B in Australia (Hardest math besides extension). I really want to do well and I'm going to be studying for this to insure I get a good result. If someone could post up some videos that cover the topics that I need help with it would be greatly appreciated. Sorry if this is in the wrong section, I'm new here :).
Thankyou
• March 16th 2012, 11:21 PM
Chris11
Re: Test preparation
Hey. Just so you know, this really shouldn't be in the advanced algebra section. Anyways, I don't have any videos, but I suppose that making some would be a good idea, allthough I don't really have the time. Regardless, there's an oranization called Kahn Academy that you might want to check out. He probally has videos on this sort of thing on his website, and these videos are free as Kahn Academy is a nonprofit. Anyways, here's a quick run down of the sorts of things you do would need to check out above.
1. Solve equations "Y" I'm assuming by this you mean that you have a system of linear equations. In this case, then what you do is the following. Bring any constant terms in any equantion to one side. Call the variables $x_1,...,x_n$. Then, make sure that you have 0s for the xs that don't appear in the
• August 5th 2012, 08:37 PM
henrysy231
Re: Test preparation
Quote:
Im new here and its good to see that these forums are active and full of friendly people. I was wondering if someone could help me with some test preparation. My unit that I am revising on consists of Linear algebra/functions and I need help getting some videos to cover certain topics.
Math tests are, to most people, tedious and bland. However, if you have a record of failed math tests, or feel that you can't understand math no matter how hard you try, math tests are terror and tedium rolled into one. Yet, math is doable by anyone once you have the basic approaches sorted. Indeed, John Louis von Neumann once said: If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is. Honestly, it's doable!
When your next math test is coming up, and you have absolutely no idea how to study, ace it, or even pass, don't panic. Relax, take these tips along with you, and you will ace that math test.
__________________________
Even Giants Born Small
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http://math.stackexchange.com/questions/63020/eigenvalues-of-some-peculiar-matrices
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# Eigenvalues of some peculiar matrices
While I was toying around with matrices, I chanced upon a family of tridiagonal matrices $M_n$ that take the following form: the superdiagonal entries are all $1$'s, the diagonal entries take the form $m_{j,j}=4 j (2 n + 3) - 8 j^2 - 6 n - 5$, and the subdiagonal entries take the form $m_{j+1,j}=4 j (2 j - 1) (2 n - 2 j + 1) (n - j)$.
For example, the $4\times 4$ member of this family looks like this:
$$M_4=\begin{pmatrix} 7 & 1 & 0 & 0 \\ 84 & 27 & 1 & 0 \\ 0 & 240 & 31 & 1 \\ 0 & 0 & 180 & 19\end{pmatrix}$$
I checked the eigenvalues of members of this family and I found that each member has the squares of the first few odd integers as eigenvalues. (For example, the eigenvalues of $M_4$ are $1,9,25,49$.) I couldn't find a way to prove this though.
I wish someone would help me! Thanks!
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7
You... you were just... toying around when you found this? :o Anyway, the proof of this fact (assuming it's true) will likely go the route of explicitly constructing the eigenvectors. We should try to find a pattern in them and that should lead to the answer. Either that or more or less determine the characteristic polynomial and show its roots are squares of odds. – anon Sep 9 '11 at 4:56
2
Off-topic, but I could not resist reading this in Tara Strong's voice. – Harry Stern Sep 9 '11 at 4:56
3
well, I can easily check that the trace is correct :) And I'm pretty sure if you successively (starting from i=1) subtract the proper multiples of row i from row i+1 (to make the (i+1, i) entry 0), then the diagonal entries become 1*(2n-1), 3*(2n-3), etc. So the determinant is correct too. – Ted Sep 9 '11 at 6:15
2
The way you came up with this matrix might suggest some useful ways to transform your matrix to a simpler form. – user1551 Sep 9 '11 at 10:30
2
Not sure if this is relevant, but the diagonal terms of $M$ can be written as $m_{jj} = 2(2j - 3/2)[2(n-j)+3/2] - 1/2$ and $m_{j,j+1} = 4 j (n-j)(2j - 1)[2(n-j) + 1]$, which contains a lot of terms of the form $(Aj - B)(A(n-j) + B)$. So there's some structure to the madness `;-)`. – Willie Wong♦ Sep 9 '11 at 14:29
## 3 Answers
I think I have something. My solution's a bit convoluted, and I'd be glad to see a shorter path to do this.
Since symmetric matrices are "easier" to handle, we apply a symmetrizing diagonal similarity transformation $\mathbf D^{-1}\mathbf M\mathbf D$ to the matrix $\mathbf M$, where $\mathbf D$ has the diagonal elements
$$d_{k,k}=\left(\prod_{j=1}^{k-1} \left(4j(2j-1)(2n-2j+1)(n-j)\right)\right)^\frac12$$
(This transformation of an unsymmetric tridiagonal matrix to a symmetric one is due to Jim Wilkinson.) The new matrix, call it $\mathbf W$, has diagonal entries that are identical to that of $\mathbf M$, while the sub- and superdiagonal entries are the square roots of the subdiagonal entries of $\mathbf M$. For instance, here is $\mathbf W_4$:
$$\begin{pmatrix} 7 & 2 \sqrt{21} & 0 & 0 \\ 2 \sqrt{21} & 27 & 4 \sqrt{15} & 0 \\ 0 & 4 \sqrt{15} & 31 & 6 \sqrt{5} \\ 0 & 0 & 6 \sqrt{5} & 19 \end{pmatrix}$$
I've found that $\mathbf W$ is symmetric positive definite; it should thus have a Cholesky decomposition $\mathbf W=\mathbf C^\top\mathbf C$, where the Cholesky triangle $\mathbf C$ is an upper bidiagonal matrix. Luckily, the entries of $\mathbf C$ take a (somewhat) simple(r) form:
$$\begin{align*}c_{k,k}&=\sqrt{(2k-1)(2n-2k+1)}\\c_{k,k+1}&=2\sqrt{k(n-k)}\end{align*}$$
Here's $\mathbf C_4$ for instance:
$$\begin{pmatrix} \sqrt{7} & 2 \sqrt{3} & 0 & 0 \\ 0 & \sqrt{15} & 4 & 0 \\ 0 & 0 & \sqrt{15} & 2 \sqrt{3} \\ 0 & 0 & 0 & \sqrt{7} \end{pmatrix}$$
Why look at the Cholesky decomposition when it's the eigenvalues that we're interested in? It is sometimes expedient to compute the eigenvalues of a symmetric positive definite matrix by considering the singular values of its Cholesky triangle. More precisely, if $\sigma_1,\dots,\sigma_n$ are the singular values of $\mathbf C$, then the eigenvalues of $\mathbf W$ are $\sigma_1^2,\dots,\sigma_n^2$.
Here's where it clicked for me. On a hunch, I decided to consider the Golub-Kahan tridiagonals corresponding to $\mathbf C$.
It is part of the theory of the singular value decomposition that if $\mathbf C$ has the singular values $\sigma_1,\dots,\sigma_n$, then the $2n\times 2n$ block matrix
$$\mathbf K=\left(\begin{array}{c|c}\mathbf 0&\mathbf C^\top \\\hline \mathbf C&\mathbf 0\end{array}\right)$$
has the eigenvalues $\pm\sigma_1,\dots,\pm\sigma_n$. It is also known that there exists a permutation matrix $\mathbf P$, such that this matrix realizes a similarity transformation between $\mathbf K$ and a special tridiagonal matrix $\mathbf T$:
$$\mathbf T=\mathbf P\mathbf K\mathbf P^\top$$
and $\mathbf T$ looks like ($n=4$):
$$\begin{pmatrix} 0 & \sqrt{7} & 0 & 0 & 0 & 0 & 0 & 0 \\ \sqrt{7} & 0 & 2 \sqrt{3} & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 \sqrt{3} & 0 & \sqrt{15} & 0 & 0 & 0 & 0 \\ 0 & 0 & \sqrt{15} & 0 & 4 & 0 & 0 & 0 \\ 0 & 0 & 0 & 4 & 0 & \sqrt{15} & 0 & 0 \\ 0 & 0 & 0 & 0 & \sqrt{15} & 0 & 2 \sqrt{3} & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 \sqrt{3} & 0 & \sqrt{7} \\ 0 & 0 & 0 & 0 & 0 & 0 & \sqrt{7} & 0 \end{pmatrix}$$
Note the structure of $\mathbf T$: the diagonal entries are zero, and the off-diagonal entries are the diagonal and superdiagonal entries of $\mathbf C$ "riffled" together. $\mathbf T$ is what is referred to as a Golub-Kahan tridiagonal matrix.
As it turns out, a further diagonal similarity transformation $\mathbf T^\prime=\mathbf F\mathbf T\mathbf F^{-1}$, with diagonal entries $f_{k,k}=\sqrt{\binom{2n-1}{k-1}}$ turns $\mathbf T$ to a rather famous set of matrices. Here's the $2n\times 2n$ matrix $\mathbf T^\prime$, for $n=4$:
$$\begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 7 & 0 & 2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 6 & 0 & 3 & 0 & 0 & 0 & 0 \\ 0 & 0 & 5 & 0 & 4 & 0 & 0 & 0 \\ 0 & 0 & 0 & 4 & 0 & 5 & 0 & 0 \\ 0 & 0 & 0 & 0 & 3 & 0 & 6 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 & 0 & 7 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{pmatrix}$$
$\mathbf T^\prime$ is what is known as the Clement-Kac(-Sylvester) matrix. It is well-known (see here or here, for instance) that the $2n\times 2n$ Clement-Kac(-Sylvester) matrix has the eigenvalues $\pm1,\dots,\pm(2n-1)$ (and thus these are the eigenvalues of $\mathbf T$ and $\mathbf K$ as well). From this, we find that the singular values of $\mathbf C$ are $1,\dots,2n-1$ (the first few odd numbers), and thus the eigenvalues of $\mathbf W=\mathbf C^\top\mathbf C$ and the original matrix $\mathbf M$ are $1,9,\dots,(2n-1)^2$.
Whew!
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I have yet to comprehend what you have revealed here, but I'm pretty amazed at your persistence in solving this problem! – user1551 Nov 10 '11 at 13:47
OK, finished reading it. This is ... ingenious. – user1551 Nov 10 '11 at 17:58
I'm still hoping somebody can outdo my answer, though. :) – J. M. Nov 10 '11 at 18:07
Here is the eigendecomposition $M_4= VDV^{-1}$ confirming @Craig's observation:
$$\begin{pmatrix} 1 &1 &1 &1\\ 42 &18 &2 &-6\\ 840 &-120 &-120 &72\\ 5040 &-3600 &2160 &-720 \end{pmatrix} \begin{pmatrix}49\\&25\\&&9\\&&&1\end{pmatrix}\begin{pmatrix} 1 &1 &1 &1\\ 42 &18 &2 &-6\\ 840 &-120 &-120 &72\\ 5040 &-3600 &2160 &-720 \end{pmatrix}^{-1}$$ If you can provide a little hint on how you came up with this, then we can probably understand why $V$ has this special structure. Also, you can distribute the squares to the left and right matrices and obtain an even weirder situation.
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If I'm not mistaken the largest eigenvector is of the form
$\begin{pmatrix} 1 \\ (2n-1)!/(2n-3)! \\ (2n-1)!/(2n-5)! \\ \vdots \\ (2n-1)!/1! \end{pmatrix}$.
This shouldn't be too difficult to prove, and I think the smallest eigenvector has a similarly compact form. My suspicion is that there is a simple rotation or pair of rotations you can do to see the eigenvalues explicitly. I would look for something of the form $ULDL^{-1}U^{-1}$ or $LUDU^{-1}L^{-1}$ where $L$ is non-zero except on the diagonal and subdiagonal and $U$ is zero except on the diagonal and superdiagonal.
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http://www.physicsforums.com/showthread.php?p=3732237
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Physics Forums
Simple problem: Gain after n trials given success rate and profibility
I'm stuck on what I'm sure is a very simple problem.
I'm trying to calculate predicted output for my share trading model.
For example, for a strategy the probability of it winning on each trade may be 0.4 and hence a loss is 0.6.
But for each trade it wins, my account balance is increased by 5% and each loss it loses 2%.
How do I calculate how much I would have won or lost after n amount of trades?
Can you break it down to on average each trade its a win or loss of x% so I can just use a simple compound interest formula to calculate expected profit after n amount of trades?
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Quote by qwirky64 I'm stuck on what I'm sure is a very simple problem. I'm trying to calculate predicted output for my share trading model. For example, for a strategy the probability of it winning on each trade may be 0.4 and hence a loss is 0.6. But for each trade it wins, my account balance is increased by 5% and each loss it loses 2%. How do I calculate how much I would have won or lost after n amount of trades? Can you break it down to on average each trade its a win or loss of x% so I can just use a simple compound interest formula to calculate expected profit after n amount of trades?
Your net gain/loss n trades is $\delta_G np - \delta_L n(1-p)$ so for 10 trades you have: 0.05 (10)0.4 -0.02(10)0.6= 0.08 This is a net gain per trade of 0.008 (0.8%). The formula for compound interest is $N(t)=N_{0}e^{rt}$ where r is the rate and t is the time in units over which the rate is calculated.
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Quote by qwirky64 Can you break it down to on average each trade its a win or loss of x%
What is the basis for the percentage? Is it x% of your original investment? Or is it x% of your current investment when the trade is made?
Simple problem: Gain after n trials given success rate and profibility
Quote by Stephen Tashi What is the basis for the percentage? Is it x% of your original investment? Or is it x% of your current investment when the trade is made?
I think he's working from a trading account where the number of trades is a surrogate for time. So the average net rate of gain per trade is 0.008 and t=n where n is the number of trades. His "stake" N(0) can be any reasonable amount to sustain extended trading assuming his assumptions hold. He would be carrying his gains forward from trade to trade for this calculation to be valid. If he takes any money out, he would need to redefine his base.
EDIT: Note that for the ten trades I described above, I did not add the compounding which would be quite small. N(t)= 1.00 exp{(.008)10}= 1.08329 vs 1.08000 without compounding. Note I'm just working from the fixed assumptions of the model described and not saying anything about actual trading in stocks.
Yes its from a trading account, so if I start with $5000 and I have a win and earn 5% then I end up with$5250, then If I win again I win 5% of the $5250 ($5512.5) and so on... I'm confused with the formula because lets say I start with 1000. and my probability of winning or losing is 50/50. and each win or loss results in a 10% gain or loss of my account. If I win first I end up with 1100, then I lose I end up with 990. Of if I lose first I end up with 900 and then if I win I end up with 990. (same result) But if I put these values into the formula I end up with a value of 1000.
Quote by qwirky64 Yes its from a trading account, so if I start with $5000 and I have a win and earn 5% then I end up with$5250, then If I win again I win 5% of the $5250 ($5512.5) and so on... I'm confused with the formula because lets say I start with 1000. and my probability of winning or losing is 50/50. and each win or loss results in a 10% gain or loss of my account. If I win first I end up with 1100, then I lose I end up with 990. Of if I lose first I end up with 900 and then if I win I end up with 990. (same result) But if I put these values into the formula I end up with a value of 1000.
That's the answer you should get. The only way I know how to estimate your gains or losses with a simple formula is to establish your long term net gain/loss probability and use that in the compounding formula (see my last post). If you want real time trade to trade compounding, you're better off with a simple computer program. But it won't be predictive of the long term outcome given your assumptions.
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If I win first I end up with 1100, then I lose I end up with 990. Of if I lose first I end up with 900 and then if I win I end up with 990. (same result) But if I put these values into the formula I end up with a value of 1000.
Which formula? Also, there are 4 possible combinations of winning or losing in a sequence of two trades. You only considered two.
Depending on whether you define the random variable in terms of percent of your current stake or percent of your initial stake, you get different problems to solve.
To consider it as a percentage of your current stake, let's get rid of the idea of "percentage" completely , since it is a source of ambiguity in practical discussions:
Define the random variable X as:
P(X = 1.05) = 0.4
P(X = 0.98) = 0.6
Let {X1,X2,.... Xn} be an independent sequence of n random variables, each with the same distribution as X.
Let the random variable R equal the product (K)(X1)(X2)...(Xn) where K is you initial stake.
The term "gain" is also ambiguous. (e.g. is its R - K? R/K ? (R-K)/K etc.). Let's look at the expected value E(R) of R.
E(R) =K E( (X1)(X2)...Xn)). Since the Xi are independent random variables, you can compute this as E(R) =K E(X1) E(X2)...E(Xn) = K (E(X))^n
Quote by Stephen Tashi ine the random variable X as: P(X = 1.05) = 0.4 P(X = 0.98) = 0.6 Let {X1,X2,.... Xn} be an independent sequence of n random variables, each with the same distribution as X. Let the random variable R equal the product (K)(X1)(X2)...(Xn) where K is you initial stake. The term "gain" is also ambiguous. (e.g. is its R - K? R/K ? (R-K)/K etc.). Let's look at the expected value E(R) of R. E(R) =K E( (X1)(X2)...Xn)). Since the Xi are independent random variables, you can compute this as E(R) =K E(X1) E(X2)...E(Xn) = K (E(X))^n
OK. I'm not quite sure how to apply this product formula since you have two independent values X= 1.05, Y=0.98. I applied it as $E(R)= (E(X)^k)(E(Y)^{n-k})$ Using this get 1.076 after ten trades (from a base of 1) where k=4, n=10. My original result is was 1.08 without compounding and 1.08329 with compounding for 10 trades.
I'm sure I'm missing something obvious. Could you show how to apply your formula for say 11 trades using P(X)=0.4, X=1.05, and Y=0.98?
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Quote by SW VandeCarr OK. I'm not quite sure how to apply this product formula since you have two independent values X= 1.05, Y=0.98.
I'm viewing the result of a trade as one random variable with two possible outcomes.
It's expected value is (1.05)(0.4) + (0.98)(0.6). So I'd raise that quantity to the 11 th power to get the expected multiplier for 11 trades.
Quote by Stephen Tashi I'm viewing the result of a trade as one random variable with two possible outcomes. It's expected value is (1.05)(0.4) + (0.98)(0.6). So I'd raise that quantity to the 11 th power to get the expected multiplier for 11 trades.
Interesting. That formula looks something like mine except I multiplied both additive terms by the delta terms specified by the OP in order to get an average rate estimate for each trade to use in the compound interest formula (as requested by the OP). That formula requires the delta term (in this case an average +0.008 per trade). When I computed this for 11 trades I got 1.0920 vs 1.0916 when raising to powers, so both ways give very similar results. In fact for 100 trades I get 2.226 with the compound interest formula (using x=.008 x100, base = 1) and 2.218 raising to powers.
So are you saying. If I have a 0.5 chance winning and hence 05. losing, and each win or loss is +10% or -10% respectively. I would write it as... P(X = 1.1) = 0.5 P(X = 0.9) = 0.5 and that after 5 trades I would have ((1.1)(0.5) + (0.9)(0.5))^5=1 Does this mean on average after 5 trades with those odds I would end up with the same amount of money?
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Quote by SW VandeCarr Interesting. That formula looks something like mine except I multiplied both additive terms by the delta terms specified by the OP in order to get an average rate estimate for each trade to use in the compound interest formula as requested by the OP
I had a different interpretation of the mention of the compound interest formula. I didn't interpret it as a request to determine an equivalent interest rate for the result of the trading or to give the trader credit for having his funds invested while not trading. I took it merely as a question about whether the formula could be somehow be used to find the result of N trades without considering time at all.
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Quote by qwirky64 So are you saying. If I have a 0.5 chance winning and hence 05. losing, and each win or loss is +10% or -10% respectively. I would write it as... P(X = 1.1) = 0.5 P(X = 0.9) = 0.5 and that after 5 trades I would have ((1.1)(0.5) + (0.9)(0.5))^5=1 Does this mean on average after 5 trades with those odds I would end up with the same amount of money?
Yes.
In any practical application, there will be some minimum stake that the trader needs in order to make a trade. The above formula doesn't account for the possibility of "ruin" when the traders stake falls below that minimum.
Quote by Stephen Tashi Yes. In any practical application, there will be some minimum stake that the trader needs in order to make a trade. The above formula doesn't account for the possibility of "ruin" when the traders stake falls below that minimum.
You could put in some kind of a safety feature into the program if the size of the stake drops below some fraction of its original size, say 50%.
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http://mathoverflow.net/questions/34390/on-proving-that-a-certain-set-is-not-empty-by-proving-that-it-is-actually-large/34410
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## On proving that a certain set is not empty by proving that it is actually large
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It happens occasionally that one can prove that a given set is not empty by proving that it is actually large. The word "large" here may refer to different properties.
For example, one can prove that a certain set is not empty by proving that its cardinality is big, as in the proof that there exist transcendental numbers : The set of algebraic numbers is countable, but the set of real numbers is uncountable, so there is uncountably many transcendental numbers.
One could also prove that a certain set is not empty by proving, for example, that it has positive measure, that it is dense, etc.
What are some good examples of such proofs?
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Some of these proofs can be categorized as the probabilistic method, and “The Probabilistic Method” by Noga Alon and Joel H. Spencer (Wiley) seems to be a very good book on it. – Tsuyoshi Ito Aug 3 2010 at 14:50
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Community wiki? – Dylan Wilson Aug 3 2010 at 18:47
## 12 Answers
Many existence proofs which exploit the idea of Baire category.
For instance, existence of a metrically transitive automorphism of the closed unit square was first obtained by the category method (see "Measure-preserving homeomorphisms and metrical transitivity" by Oxtoby and Ulam) . Another classical example is due to Banach who proved that every function from a residual subset of $C[0,1]$ is nowhere differentiable.
A nice and elementary book by Oxtoby discusses these and many other applications of the category method.
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One of my favorites in this regard is the existence of a first-order linear PDE with no solution. One writes down a PDE of a certain form with one of the coefficients $f$ left undetermined, finds some conditions on $f$ necessary for the existence of a solution, and then shows the set of $f$ satisfying these conditions is meager in $L^\infty$. Fritz John's PDE book has the details. – Nate Eldredge Aug 4 2010 at 22:32
That's an amazing example! Thanks. – Andrey Rekalo Aug 4 2010 at 22:42
Nice example! I also like the proof of the existence of many continuous functions $f$ on the unit circle whose fourier series diverge at a given point, using Baire's theorem. – Malik Younsi Aug 5 2010 at 12:17
### 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.
Frequently in arguments in algebraic number theory, one has to choose a prime that satisfies some list of conditions, and is bigger than some given bound. (A simple example of the kind of condition that I am thinking of would be that the prime should lie in a given congruence class modulo some number $n$.) One then interprets the conditions in such a way as to be able to apply the Cebotarev density theorem to conclude that there are infinitely many primes satisfying the given conditions, and so in particular a suitable prime can be found that lies above the desired bound. (In the simple example, one would use Dirichlet's theorem on the existence of infinitely many primes in arithmetic progression, which from this point of view is a special case of Cebotarev density.)
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Sard's lemma is an example - the set of regular values is non-empty since it has positive measure.
The following example also answers your question: Recently I proved the following lemma.
Let $f\colon M\to N$ be a smooth map, $M$ is a non-empty paracompact manifold. Let $k$ be a maximal rank of a differential $df(x)$ over $x\in M$. Then there exists a point $y$ in $f(M)$ such that the rank of the differential $df$ is maximal for all points of $f^{-1}(y)$.
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This page of the Tricki describes exactly the technique you are looking for.
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Virtually any existence proof in analytic number theory goes by giving a quantitative lower bound (or asymptotic formula) for the number of objects of interest. For example, there are infinitely many primes of the form $x^2 + y^4$ was shown by Friedlander and Iwaniec by finding an asymptotic formula for the counting function of such primes.
When I was in graduate school, a paper appeared on the arxiv purporting to prove the twin prime conjecture by establishing an asymptotic formula for the number of twin primes up to $x$. I remarked to my advisor, Iwaniec, that surely there was a mistake since the conclusion was so strong. He responded by saying that giving the asymptotic formula was actually a good sign. (It should go without saying that there was a mistake in the paper).
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There are non-Borel sets that are Lebesgue measurable. This is proved in the following way. First show that the Borel sigma algebra for the real line is uncountable with cardinality of the real line. On the other hand, you have the Cantor set which is uncountable(cardinality = $\mathbb R$) and is of Lebesgue measure zero. Since Lebesgue measure is complete, every subset of the Cantor set belongs to the Lebesgue sigma algebra and therefore the Lebesgue sigma algebra has cardinality of the power set of the reals.
Construction of explicit examples would require axiom of choice.
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When you say that the Borel sigma algebra is "uncountable," do you mean to say that its cardinality is the same as the cardinality of the reals? – Charles Staats Aug 3 2010 at 15:15
@Charles Staats : Yes, that is what I mean. I have edited to incorporate this. – Anweshi Aug 3 2010 at 15:25
Your "require Axiom of Choice" is, I think, wrong. As your link shows, without AC it is consistent that every set of reals is Borel, so even your cardinality proof fails. – Gerald Edgar Aug 3 2010 at 17:24
@Gerald Edgar: What I had in mind when I linked, was the example of Hamel basis which I mentioned there. Yes, it is true that without AC the proof that the Borel sigma algebra has cardinality that of the real line, would fail. – Anweshi Aug 3 2010 at 17:36
Explicit examples of sets that are Lebesgue measurable but not Borel are known. But, of course, not as easy as your cardinality proof. I mean here constructions of analytic sets that are not Borel, which are quite explicit. (Unlike the Hamel basis examples, which are not.) – Gerald Edgar Aug 4 2010 at 13:37
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C.Cornut and V.Vatsal's proof (as in e.g Inventiones mathematicae 148) of the non-triviality of CM points on zero and one-dimensional quaternionic Shimura varieties as one goes up an anticylcotomic $\mathbb Z_{p}$-extension $K_{\infty}$ is by showing that reduction modulo $\ell$ of CM points is actually onto points modulo $\ell$ for infinitely many $\ell$. In particular, their proof actually implies the much stronger statement that the $\mathbb Z_{p}[[\Gamma]]$-module generated by the image of the norm-coherent CM points under the Kummer map has trivial $\mu$-invariant (here $\Gamma$ is the Galois group of $K_{\infty}/K$).
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As Tsuyoshi Ito commented, the probabilistic method in combinatorics is an example. You use a probability measure on the space of possibilities, and show that the set with the desired probabilities has positive measure, hence is nonempty.
The classic example of this is the result by Erdős (1947) that the Ramsey number $R(t,t)$ grows at least exponentially with $t$. If you consider a random coloring of the edges of the complete graph on $n$ vertices, then the probability that a particular complete subgraph on $t$ vertices is monochromatic is $2^{1-{t \choose 2}}$. If ${n\choose t} 2^{1-{t \choose 2}} \lt 1$, then a random coloring has no monochromatic subgraph with positive probability. This is the case for $n = \sqrt 2^t$, $t\gt 2$, so $R(t,t) \ge \sqrt2^t$ for $t \gt 2$.
For slightly smaller $n$, most random colorings of the complete graph on $n$ vertices have no monochromatic subgraph of size $t$, but finding a construction has been an open problem.
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A kind of analogue of Sard's Lemma in Algebraic Geometry is Bertini's Theorem:
"Given a linear sistem $|L|$ on a smooth projective variety X, its general element is smooth outside the base points".
In particular, if $|L|$ is base-point free then the set of smooth elements in $|L|$ is dense, in particular non-empty.
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The theme is prevalent in combinatorial number theory and ergodic theory. Consider the ergodic Szemerédi theorem, for instance. It says that if $(X,\mathcal{B},\mu,T)$ is a measure-preserving system and $A \in \mathcal{B}$ has $\mu (A) >0$, then $\forall k \in \mathbb{N}$ $\exists n \in \mathbb{N}$ so that $\mu (A \cap T^{-n}A \cap T^{-2n}A \cap \cdots \cap T^{-kn}A)>0$. It turns out that it's easier to prove that measure of the intersection is positive for infinitely many $n$. In fact, Furstenberg's original proof showed that $\liminf_{N \to \infty} \frac{1}{N} \sum_{n=1}^{N} \mu (A \cap T^{-n}A \cap T^{-2n}A \cap \cdots \cap T^{-kn}A) > 0$.
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Not quite an answer to the question but related (large modulo $2$):
Zagier's beautiful proof (Amer. Math. Monthly 97 (1990), no. 2, 144) that every prime congruent to $1$ modulo $4$ is a sum of two squares is based on the fact that the cardinality of a certain finite set is odd and the set is thus non-empty.
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5 = 1 (mod 4). Is 5 a square? – John Bentin Aug 3 2010 at 17:24
Is any prime a square? – Kevin Ventullo Aug 3 2010 at 18:24
Any prime in an empty set is a square. So first you have to establish that the set isn't empty. – John Bentin Aug 3 2010 at 20:15
I do not understand these comments. The Theorem does not state that $5$ is a square (a prime is never a square by definition) but that $5$ is a sum of two squares: $5=2^2+1^1$. – Roland Bacher Aug 6 2010 at 9:19
I just heard this in a talk by Jan Krajicek: It is known that there are Boolean functions that require a "large" circuit to compute them, but the proof is probabilistic. (I.e., the measure of the set of such Boolean function (say of arity n) is known to be nonzero.) No explicit construction of such functions is known.
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http://physics.stackexchange.com/questions/16262/why-is-the-principle-of-superposition-true-in-em-does-it-hold-more-generally/16269
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# Why is the Principle of Superposition true in EM? Does it hold more generally?
• In the theory of electromagnetism (EM), why is the principle of superposition true? Can we read it off from Maxwell's equations directly?
• Does it have any limit of applicability or is it a fundamental property of nature?
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## 4 Answers
The principle of superposition comes from the fact that equations you solve most of the time are made of Linear operators (just like the derivative). So as long as you are using these operators you can write something like
$$\mathcal L\cdot \psi = 0$$
where $\mathcal L$ is a linear operator and, let say, $\psi$ is a function that depends on coordinates that $\mathcal L$ is acting on. The superposition principle is the same that saying this
$$\mathcal L \cdot \left(\sum_i \psi_i \right) = \mathcal L\cdot\psi_1 + \mathcal L\cdot\psi_2 + ... = 0$$
holds. An example when it doesn't would be, for example...
$$\mathcal L^2 \cdot\left(\sum_i \psi_i\right) \neq \mathcal L^2\cdot\psi_1 + \mathcal L^2\cdot\psi_2 + ...$$
in general (for the Laplacian in Euclidean space it is equal to). So then, the question is if Maxwell equations are linear. And they are because they are made up with this kind of operators. For instance, Gauss law for two different electric fields can be written as one
$$\nabla \cdot \vec{E}_1 = \rho_1/\varepsilon \quad; \quad \nabla \cdot \vec{E}_2 = \rho_2/\varepsilon$$
$$\nabla \cdot \overbrace{\left(\vec{E}_1 + \vec{E}_2\right)}^{\vec{E}} = \frac{\overbrace{\rho_1 + \rho_2}^{\rho_T}}{\varepsilon} \Rightarrow \boxed{\nabla \cdot\vec E = \frac{\rho_T}{\varepsilon}}$$
just because $\nabla$ is a linear operator.
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It is true up to very high filed strengths. For too high strengths the field itself is not stable, it can create real pairs. It is like a limit on a field strength in a capacitor. The capacitor dielectric can break.
EDIT: Classical Maxwell equations are linear indeed so the principle of superposition is implemented into them. But break of a dielectric can be introduced too as a resistance depending on the field strength. Thus one can make the Maxwell equation non-linear starting from some threshold strength.
In fact, the dielectric break or capacitor discharge due to cold electron emission (classical non linearity) occurs "well before" creating electron-positron pair in vacuum.
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No explanation of WHY or how it can be seen in Maxwell's equations. – FrankH Oct 27 '11 at 17:00
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It is explained in each and every textbook. – Vladimir Kalitvianski Oct 27 '11 at 17:07
Within the realm of Maxwell's equations, the principle of superposition is exactly true because Maxwell's equations are linear in both the sources and the fields. So if you have two solutions to Maxwell's equations for two different sets of sources then the sum of those two solutions will be a solution to the case where you add together the two sets of sources.
This will only break down when Maxwell's equations break down, for example, when the field strengths are so high that pair production becomes significant. In that case the quantum field theory of Quantum Electrodynamics (QED) would be required. Now, quantum theories are also linear, at least as far as the quantum wave function is concerned, however the probabilities that quantum theories predict depend on the magnitude of the wave function, so in that sense they are nonlinear, and therefore superposition will not apply to the results.
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-1 You are repeating Vladimir's answer, and further, you are muddling things--- the field equations are nonlinear in the usual sense in strong field electromagnetism, because of pair creation. – Ron Maimon Oct 27 '11 at 16:46
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@ron, I am not repeating Vladamir's answer, the question was WHY, all Vladimir said was it is true without any explanation as to Why. Furthermore, the question was "Can we read it off of Maxwell's equations?" which Vladimir also did not address. – FrankH Oct 27 '11 at 16:59
Static field strength has nothing to do with the wave function. It is an external field. – Vladimir Kalitvianski Oct 27 '11 at 17:09
The question on why is layish, trying to answer it even more. This principle is plainly a observation of experiments. Maxwells equations came later, and if Maxwell tried some nonlinear ansatz, he would have realized soon that this is wrong. "Truth" is not a useful expression in physics. – Georg Oct 28 '11 at 11:02
You are right. "Truth" is not useful expression in physics. On the other hand, I was not asking why the p. of superposition is true in an absolute sense... just why physicists believe it is valid, in which situation we can use it, whether it is related to some more general principle (are other fundamental forces also linear?). You seem to suggest that there were mainly empirical reasons to accept it and then Maxwell embodied the principle in his theory. This is fine and answers the question. To me, it seems that the my question is perfectly admissible. – quark1245 Oct 30 '11 at 21:28
While the first part of the question has been answered satisfactorily, everybody seems to confuse the unconditional linearity of the Maxwell equations with the often observed linearity of the constitutive relations for the material law. The field of nonlinear optics is concerned with the behavior of light in nonlinear media where the constitutive relations are no longer linear.
However, the superposition principle is already violated if even a single electron gets accelerated by the field. So nonlinear media are nothing exotic, even if most media are well described by linear constitutive relations for small field strengths.
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http://math.stackexchange.com/questions/184660/divisibility-for-numbers-like-13-17-and-19-compartmentalization-method?answertab=oldest
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# divisibility for numbers like 13,17 and 19 - Compartmentalization method
For denominators like 13, 17 i often see my professor use a method to test whether a given number is divisible or not. The method is not the following : Ex for 17 : subtract 5 times the last digit from the original number, the resultant number should be divisible by 17 etc...
The method is similar to divisibility of 11. He calls it as compartmentalization method. Here it goes.
rule For 17 :
take 8 digits at a time(sun of digits at odd places taken 8 at a time - sum of digits at even places taken 8 at a time)
For Ex : $9876543298765432..... 80$digits - test this is divisible by 17 or not.
There will be equal number of groups (of 8 digits taken at a time) at odd and even places. Therefore the given number is divisible by 17- Explanation.
The number 8 above differs based on the denominator he is considering.
I am not able to understand the method and logic both. Kindly clarify.
Also for other numbers like $13$ and $19$, what is the number of digits i should take at a time? In case my question is not clear, please let me know.
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I cannot figure out the rule. What are the "blocks of 8 digits at even/odd places"? For example, given the number 1234567890123456, what would the two blocks look like? – celtschk Aug 20 '12 at 15:27
I think he means the two 8 digit numbers formed by every other digit, in this case $13579135$ and $24680246$. – axblount Aug 20 '12 at 15:38
@celtschk: you split into blocks of 8 starting from the ones digit, so the first block would be $90123456$, the second would be $12345678$. Then we have $1234567890123456 \equiv 90123456-12345678 \pmod {17}$ – Ross Millikan Aug 20 '12 at 15:49
## 4 Answers
You quote two different rules with different results. When testing for divisibility by 17 by subtracting 5 times the last digit from the orignal number without its last digit, you are using the fact that $51$ is divisible by $17$, so $10a+b \equiv 10a-50b \pmod {17}$, then the fact that $10(a-5b)$ is a multiple of $17$ if and only if $(a-5b)$ is. Unless you do further computation, you lose the remainder if the original number is not a multiple.
When you take blocks of 8 digits, you use the fact that $10^8+1 \equiv 0 \pmod {17}$, so $10^8a+b \equiv b-a \pmod {17}$ You retain the remainder in this case. For 13, you need half the period of its repeating decimal, which is 6, so you use blocks of 3. Note that $10^3+1=1001 \equiv 0 \pmod {13}$
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Your professor is using the fact that $100000001=10^8+1$ is divisible by $17$. Given for example your $80$-digit number, you can subtract $98765432\cdot 100000001=9876543298765432$, which will leave zeros in the last $16$ places. Slash the zeros, and repeat. After $5$ times you are left with the number $0$, which is divisible by $17$, and hence your $80$-digit number must also be divisible by $17$.
When checking for divisibility by $17$, you can also subtract multiples of $102=6\cdot 17$ in the same way.
For divisibility by $7$, $11$, or $13$, you can subtract any multiple of the number $1001=7\cdot 11\cdot 13$ without affecting divisibility by these three numbers. For example, $6017-6\cdot 1001=11$, so $6017$ is divisible by $11$, but not by $7$ or $13$.
For divisibility by $19$, you can use the number $1000000001=10^9+1=7\cdot 11\cdot 13\cdot 19\cdot 52579$. By subtracting multiples of this number, you will be left with a number of at most $9$ digits, which you can test for divisibility by $19$ by performing the division.
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The whole process can be explained by writing
A^x/B = C where C can be +1 or -1 or 0. Here C is the remainder got on division of A^x by B.
Now we operate on base 10. So A takes a value of 10. B is the divisor for which you are attempting to set up a rule for. This method is called compartmentalisation. It is used to check on divisibility / remainder for numbers as you had stated .. not for the normal smal numbers. There are much effective rules for the same.
Suppose you talk of divisibility by 9: Consider any 2 digit number AB. This can be represented in terms of der place values as
AB = 10A+B = 9A+A+B.
Now if you were to check divisibility by 9.
[9A + (A+B)]/9. Then the first term 9A is divisible. Now if the whole number is to be divisible, we would want the terms (A+B) to also be divisible. So the divisibility of the number depends on (A+B) which is nothing but the digit sum. Which is why you see that the divisibility rule for 9 is "The digit sum of the number must be divisible by 9" Also you got to note that: 10^x/9 = +1 for whatever be the value of x. This results in another rule: 10^x/3 = +1 which is the rule for divisibility by 3.
Now consider the RULE FOR 7 A^x/B= C
Here A=10 (base 10), B = 7. Now we have to choose values for x such that it gives a remainder of C= +1 or -1.
we see the following pattern: 10^0/7 = +1
10^3/7 = -1
10^6/7 = +1
So we see that +1 and -1 remainders alternate every three powers of ten. This gives our compartmentalisation rule. The given number from left to right has to be grouped in threes and the rule of +1 and -1 for every triplet has to be applied. You will see it :
ABCPQRXYZ be a number. To check the divisibility by 7, we write it as:
[10^6 ABC]/7 + [10^3 PQR]/7 + [10^0 XYZ]/7
Now 10^6/7=+1 , 10^3/7 = -1 and so on.. This reduces to
(1xABC) (-1XPQR) (1*XYZ)
RULE : Sum of triplets at odd places - sum of triplets at even places
eg: 100200140240 /7 . What would be the remainder?
100 | 200 | 140 | 240 |
-1 | +1 | -1 | +1
• 100- 140 = -240
200 + 240 = 440
You get 440-240 = 200
200/7 gives a R= 4
Now consider the RULE FOR 11 10^0 / 11 = +1 10^1/ 11 = -1 10^2/ 11 = +1 10^3/11 = - 1
FOr 11, you can do it in 2 ways. One is that you notice that every alternate power of 10 divided by 11 gives an alternating pattern of +1 and -1. So the compartmentalisation will now be for every digit. So you do a
Sum of digits at odd places - Sum of digits at even places.
But if you did notice the fact that 10^0/11 = +1 and 10^3/11 = -1 Then you will notice that 11, same as 7, has +1 and -1 alternating for triplets. So, you can proceed compartmentalising the number into triplets and doing the same thing as we did for 7.
Sum of triplets at odd places - Sum of triplets at even places.
So we see that so far : 1. 10^x / B = +1 will hold good for three cases i guess (not sure it may be more ... ) They are 3, 9 , 37.
Even 10^3/37 = +1 . So here there are no alternate +1 and -1 as 10^6/37 will also be +1. Note that (a^m)^n = a^mn
**Rule for 37 would be : Group the number into triplets and apply a +1 to each triplet.
1. 10^x / B = +1 or -1 will hold good for 7, 11, 17 Rule for 17 (thus) : 10^8/17 = -1. So, 10^16/17 would be +1. So it becomes like (Sum of digits at odd places taken 8 at a time - sum of digits at even places taken 8 at a time. )**
3. Also 10^x/B = 0 (gives a remainder of 0 )when B are powers of 2 or powers of 5.
I still dont know if it is possibile to cover all primes (atleast 2 digits) in this method. Perhaps you may have to use other techniques like Chinese Remainder Theorem or the Basic Remainder Theorem and so on.. That is because I still dont see any patterns for 23 and many more numbers. If anybody could add anything more to this .. it would be great.
Hope this post helps.
Cheers :)
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The Similar can also be said for divisibility by $13$
$10^3/13$ gives a remainder of $-1$
$10^6/13$ gievs a remainder of $+1$
So again the rule for $13$, will be same as the rule for $7, 11$. Group the numbers into triplets as we see alternate changes of $-1$ to $+1$ between every three powers of $10$. So the rule for $13$ will also be
(Sum of triplets at odd places - Sum of triplets at even places) or Sum of digits at odd places taken $3$ at a time - Sum of digits at even places taken three at a time)
As i said, all this can be summarised as:
$10^x/B = +1$ holds for divisors (values of B) like $3,9, 37$
$10^x/B = -1$ and $+1$ for divisiors (values of B) like $7, 11, 13, 17$
$10^x/B = 0$ for divisors (values of B) like powers of $2$ and $5$
where $+1,0,-1$ are the remainders of the respective divisions
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# Tagged Questions
2answers
652 views
### What is a non linear $\sigma$ model?
What exactly is a non linear $\sigma$ model? In many books one can view many different types of non linear $\sigma$ models but I don't understand what is the link between all of them and why it is ...
1answer
1k views
### Differentiating Propagator, Greens function, Correlation function, etc
For the following quantities respectively, could someone write down the common definitions, their meaning, the field of study in which one would typically find these under their actual name, and most ...
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http://alanrendall.wordpress.com/2011/07/
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# Hydrobates
A mathematician thinks aloud
## Archive for July, 2011
### Cytokine dynamics, part 2
July 28, 2011
In a previous post I discussed a paper of Lev Bar-Or about the dynamics of the interactions between T cells and macrophages via the cytokines they produce. The model of that paper is a four-dimensional dynamical system with eighteen parameters. In that post I also discussed a paper of mine where I proved some mathematical results about this system. The features I captured were that for certain values of the parameters there is one stationary solution which is a global attractor while for other special values of the parameters there are three stationary solutions, two of which are stable. Both these cases were already seen in phase portraits included in the original paper. Some numerical simulations I did at that time did not reveal any more complicated behaviour.
More recently experiments I did using Mathematica led me to discover some new phenomena. For certain values of the parameters it is possible to extract a two-dimensional model system with just two parameters $A$ and $C$ and this already displays interesting dynamical phenomena. The parameter $C$ encodes the effect of antigen presentation. A two-dimensional system has the advantage that its qualitative properties can be displayed effectively in the plane. I found the command ‘Manipulate’ in Mathematica to be very helpful in searching through parameter space. First I used it on a phase portrait. This did not get me very far for the following reasons. I found it quite hard to recognize the important dynamical features in the phase portrait and I found that searching in parameter space quickly became tiring. Next I tried displaying nullclines as a method to look for stationary solutions. There it was easier to see what was going on since it is only necessary to see where two curves intersect. My first attempt with this approach also gave nothing new but the fact that it was less tiring meant that I was encouraged to try again. The second attempt revealed parameter values where there are seven stationary solutions. Once I had found these parameter values I could go back and compute the corresponding phase portrait. It shows that four of the seven steady states are sinks while the other two are saddle points.
Having seen these pictures I had the ambition to prove the existence of seven stationary solutions, four of which are stable, under conditions on the parameters which are as general as possible. I succeeded in doing this. In addition I was able to obtain generalizations of these results to the original four-dimensional system. These theorems and their proofs are contained in a new paper. I will now describe a key technique which was used. I first thought that this method was quite different from anything I had used before but I later saw that it does bear some resemblance to the method of Fuchsian equations. I leave it to the interested reader to think about what the relations are. The idea for proving the existence of stationary solutions is as follows. Often there is an obvious way of defining a mapping $\phi$ whose fixed points are in one-to-one correspondence with stationary solutions of the dynamical system. Suppose now that for a given system it is possible to guess that there is a stationary solution near some point with coordinates $x_*$. Consider a small box of size $\epsilon$ containing this point. If it is possible to show that this box is invariant under $\phi$ then the Brouwer fixed point theorem implies the existence of a stationary solution inside that box. If it can further be shown that the restriction of $\phi$ to the box is a contraction the stationary solution is the only one in the box. In the example the necessary estimates are obtained with the help of a parameter which has the property that the stationary solution tends to $x_*$ as it gets large. The procedure is then to choose the parameter large and the size of the box to be neither too small nor too large. With the obvious mapping $\phi$ this is enough to capture the four stable stationary solutions. The other three are more difficult and it is necessary to define a different mapping $\psi$ which has the same fixed points as $\phi$ but has other good mapping properties in addition. This more refined technique can be used to get the other three stationary solutions, at least in certain cases.
On the biological side these results suggest a scenario where the alternative states of the immune system are not just distinguished by the dominance of Th1 or Th2 cytokines but also different discrete possibilities for the strength of the dominance. Changing from one of these steady states to the other leads to a 50% increase in the total cytokine concentration and so it is by no means a small effect. I am not aware that this kind of phenomenon has been observed experimentally. One thing I will take away from this project is that in the future I plan to try to combine analytical and numerical approaches in an increasingly sophisticated way when trying to investigate the properties of dynamical systems.
Posted in dynamical systems, immunology, mathematical biology | Leave a Comment »
### Fuchsian equations
July 21, 2011
Fuchsian equations are a class of differential equations which have played a big role in my research and which have now found a niche in the mathematical relativity community. In fact they have a very wide range of applications. The story begins with Lazarus Fuchs in the mid nineteenth century. He was interested in the case of a higher order linear scalar ODE for a complex-valued function whose coefficients are analytic except for isolated singularities. The cases which I have mainly been concerned with are first order nonlinear systems of PDE for real-valued functions whose coefficients may only be smooth. It sounds like almost everything is different in the two cases and so what is the similarity? The basic idea is that a Fuchsian problem concerns a system of PDE with a singularity in the coefficients where the desire is to describe the behaviour of solutions near the singularity. If certain structural conditions are satisfied then for any formal power series solution of the problem there is a corresponding actual solution. In the original situation of Fuchs the series converged and the solution was its limit. More generally there need be no convergence and the series is only asymptotic. A related task is to find singular solutions of a regular system. By suitable changes of variables this can sometimes be transformed to the problem of finding a regular solution of a singular system.
With hindsight my first contact with a problem of Fuchsian type was in some work I did with Bernd Schmidt (Class. Quantum Grav. 8, 985) where we proved the existence of certain models for spherically symmetry stars in general relativity. Writing the equations in polar coordinates leads to a system of ODE with a singularity corresponding to the centre of symmetry. We proved the existence near the centre of solutions satisfying the appropriate boundary conditions by doing a direct iteration. In later years I was interested in the problem of mathematical models of the big bang in general relativity. I spent the academic year 1994-95 at IHES (I have written about this in a previous post) and in that period I often went to PDE seminars in Orsay. On one occasion the speaker was Satyanad Kichenassamy and his subject was Fuchsian equations. This is a long term research theme of his and he has written about it extensively in his books ‘Nonlinear Wave Equations’ and ‘Fuchsian Reduction: Applications to Geometry, Cosmology and Mathematical Physics’. I found the talk very stimulating and after some time I realized that this technique might be useful for studying spacetime singularities. Kichenassamy and I cooperated on working out an application to Gowdy spacetimes and this resulted in a joint paper. A general theorem on Fuchsian systems proved there has since (together with some small later modifications) been a workhorse for investigating spacetime singularities in various classes of spacetimes.
One of the highlights of this development was the proof by Lars Andersson and myself (Commun. Math. Phys. 218, 479) that there are very general classes of solutions of the Einstein equations coupled to a massless scalar field whose initial singularities can be described in great detail. In later work with Thibault Damour, Marc Henneaux and Marsha Weaver (Ann. H. Poincare 3, 1049) we were able to generalize this considerably, in particular to the case of solutions of the vacuum Einstein equations in sufficiently high dimensions. For these results no symmetry assumptions were necessary. More recently these results were generalized in another direction by Mark Heinzle and Patrik Sandin (arXiv:1105.1643). Fuchsian systems have also been applied to the study of the late-time behaviour of cosmological models with positive cosmological constant. In the meantime there are more satisfactory results on this question useing other methods (see this post) but this example does show that the Fuchsian method can be applied to problems in general relativity which have nothing to do with the big bang. In general this method is a kind of machine for turning heuristic calculations into theorems.
The Fuchsian method works as follows. Suppose that a system of PDE is given and write it schematically as $F(u)=0$. I consider the case that the equation itself is regular and the aim is to find singular solutions. Let $u_0$ be an explicit function which satisfies the equation up to a certain order in an expansion parameter and which is singular on a hypersurface defined by $t=0$. Look for a solution of the form $u=u_0+t^\alpha v$ where $t^\alpha$ is less singular than $u_0$. The original equation can be rewritten as $G(v)=0$, where $G$ is singular. Now the aim is to show that there is a unique solution $v$ of the transformed equation which vanishes as $t\to 0$. A theorem of this kind was proved in the analytic case in the paper mentioned above which I wrote with Kichenassamy. Results on the smooth case are harder to prove and there are less of them known. A variant of the procedure is to define $u_0$ as a solution (not necessarily explicit) of an equation $F_0(u_0)=0$ which is a simplified version of the equation $F(u)=0$. In the cases of spacetime singularities which have been successfully handled the latter system is the so-called velocity-dominated system.
Posted in general relativity, partial differential equations | 2 Comments »
### When is a dynamical system of mass action type?
July 1, 2011
At the moment I am at the European Conference on Mathematical and Theoretical Biology in Krakow. This is a joint conference with the Society for Mathematical Biology and it is very large, with more than 900 participants. This leads to a huge number of parallel sessions and the need to choose very carefully in order to get the most profit from the conference. On one day, for instance, there were two cases with two sessions on immunology occurring simultaneously.
On Tuesday I went to a session on biochemical reaction networks. This included a talk by Gheorghe Craciun with a large expository component which I found enlightening. He raised the question of when a system of ODE with polynomial coefficients can be interpreted as coming from a system of chemical reactions with mass action kinetics. He mentioned a theorem about this and after asking him for details I was able to find a corresponding paper by Hars and Toth. This is in the Colloquia Mathematica Societatis Janos Bolyai, which is a priori not easily accessible. The paper is, however, available as a PDF file on the web page of Janos Toth. A chemical reaction network gives rise to a system of equations of the form $\dot x_i=p_i-x_iq_i$ where the $p_i$ and $q_i$ are polynomials with positive coefficients. They represent the contributions from reactions where the species with concentration $x_i$ is on the right and left side respectively. The result of Hars and Toth is that any system of this algebraic form can be obtained from a reaction network. It was pointed out by Craciun in his talk that this means that arbitrarily complicated dynamics can be incorporated into systems coming from reaction networks. If we have a system of the form $\dot x_i=f_i-g_i$ we can replace it by $\dot x_i=(x_1\ldots x_n)(f_i-g_i)$. This changes the system but does not change the orbits of solutions. If, for instance, we start with the Lorenz system with unknowns $x$, $y$ and $z$ we can simply translate the coordinates so as to move the interesting dynamics into the region where all coordinates are positive and then multiply the result by $xyz$. This preserves the strange attractor structure. This result may be compared with the Perelson-Wallwork theorem discussed in a previous post.
The construction of the reaction network reproducing the given equations is not very complicated. The main problem is keeping track of the notation. Suppose we start with a system of $n$ equations in $k$ variables $x_i$ which is polynomial and satisfies the necessary condition already mentioned. The reaction network can be constructed in the following way. (It is not at all unique.) Introduce one species $X_i$ for each $x_i$. The right hand side of each equation is a sum of terms of the form $Ax_1^{m_i}\ldots x_k^{m_k}$ and one reaction is introduced for each of these terms. To explain what it is suppose without generality that it belongs to the first equation. If $A>0$ hen the reaction transforms the complex $m_1X_1+m_2X_2+\ldots m_kX_k$ to the complex $(m_1+1)X_1+m_2X_2+\ldots m_kX_k$ with rate constant $A$. The only species where there is a net production is $X_1$ and so this reaction only contributes to the first equation. Moreover it does so with the desired term. On the other hand if $A<0$ the reaction transforms the complex $m_1X_1+m_2X_2+\ldots m_kX_k$ to the complex $(m_1-1)X_1+m_2X_2+\ldots m_kX_k$ with rate constant $-A$. The assumption on the system assures that $m_1-1\ge 0$.
Posted in dynamical systems, mathematical biology | 2 Comments »
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http://yehar.com/blog/?p=430
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# Marginal notes
These are miscellaneous notes and recipes.
## Mapping a pair of integers to an integer
2011-05-26
How to map from a pair of integers to an integer? Negative or positive, independent of the size of the two integers, you can convert them into balanced ternary to get rid of the signs and then interleave the digits starting from the least significant digits. Pad with leading zeros if the other integer is longer.
Interleaving the balanced ternary trits of a pair of integers gives a fractal structure.
When working with non-negative numbers or on a fixed binary bit depth, interleaving bits works fine (see Z-order curve).
## Logarithm of a sum of large numbers
2010-03-30
When dealing with large numbers, it can be useful, as an approximation, not to deal with the numbers themselves, but with their logarithms. This has the added benefit that multiplication of two numbers corresponds to taking the sum of their logarithms. But what if you want to take the sum of the numbers, $a$ and $b$, how can you avoid dealing with the possibly horrendously large numbers themselves? This problem may be encountered for example when computing the total unnormalized probability (a Boltzmann factor or partition function) of mutually exclusive events. Here's a way:
Presume that $a>0$, $b>0$, and $a \ge b$. If $a < b$ you can always swap the two numbers.
$\mathrm{ln}(a + b) = \mathrm{ln}\big(a(1+\frac{b}{a})\big) = \mathrm{ln}\,a + \mathrm{ln}(1+\frac{b}{a}) = \mathrm{ln}\, a + \mathrm{ln}\big(1 + \mathrm{e{x}p}(\mathrm{ln}\,b - \mathrm{ln}\,a)\big)$.
If $\mathrm{ln}\,b - \mathrm{ln}\,a$ is a very large negative number, then $\mathrm{ln}(a + b) \approx \mathrm{ln}\,a$.
For the logarithm of the difference of $a$ and $b$, we would proceed similarly:
Presume that $a>0$, $b>0$, and $a > b$. These conditions are necessarily satisfied if $\mathrm{log}\,a$, $\mathrm{log}\,b$, and $\mathrm{log}(a - b)$ can be calculated.
$\mathrm{ln}(a - b) = \mathrm{ln}\big(a(1-\frac{b}{a})\big) = \mathrm{ln}\,a + \mathrm{ln}(1-\frac{b}{a}) = \mathrm{ln}\, a + \mathrm{ln}\big(1 - \mathrm{e{x}p}(\mathrm{ln}\,b - \mathrm{ln}\,a)\big)$.
If $\mathrm{ln}\,b - \mathrm{ln}\,a$ is a very large negative number, then $\mathrm{ln}(a - b) \approx \mathrm{ln}\,a$.
$\mathrm{ln}\big(1 + \mathrm{e{x}p}(x)\big)$ and $\mathrm{ln}\big(1 - \mathrm{e{x}p}(x)\big)$ are called Gaussian logarithms, because Carl Friedrich Gauss was the first to publish printed tables of them.
One application for a Gaussian logarithm is in calculation of the transition energy to a state from its complement state that covers the remainder of the phase space or thermodynamic ensemble. When the Boltzmann factor of the state and the partition function of the system are known, but both expressed as energy ($E_{\mathrm{state}}$ and $E_{\mathrm{system}}$, respectively), the calculation for the transition energy $E_{\mathrm{transition}}$ turns out as: $E_{\mathrm{transition}} = E_{\mathrm{state}} - E_{\mathrm{system}} - k_{\beta} T\ \mathrm{ln}\big(1-\mathrm{e{x}p}(-\frac{E_{\mathrm{state}} - E_{\mathrm{system}}}{k_{\beta}T})\big)$, where $k_{\beta}$ is the Boltzmann constant and T is temperature. The transition the other way would simply change the sign of the transition energy.
## Newbie bug with C++ nested templates
2010-03-02
Say, you want to create a map from stings to vectors of strings. Why does this (with the appropriate STL includes):
typedef std::vector VectorOfStrings; typedef std::map StringToVectorOfStrings;
work, while this doesn't:
typedef std::map> StringToVectorOfStrings;
The latter works if you modify it a little bit, so that the consecutive >'s are not recognized as >>:
typedef std::map > StringToVectorOfStrings;
## Two IRC channels in different windows using EPIC IRC client
2009-10-15
Start your IRC session like this:
/window new /window show 1 /bind ^I parse_command {/window size 6} {/window last} {/window move 1}
Now you have two windows on top of each other, and you are doing things in the bottom one. You can switch between the windows by pressing tab.
The normal usage is to join on two different channels:
/join #firstchannel
Press tab.
/join #secondchannel
Whenever you want to view or talk on the other channel, press tab.
## Embedding a Java Applet in a WordPress 2.8.4 post (also works in 3.2.1)
2009-09-12
In the HTML editor:
`<div id="dyemixer">APPLET HERE<script type="text/javascript">// <![CDATA[ document.getElementById('dyemixer').innerHTML = '<applet width="550" height="420" code="DyeMixer.class" archive="http://yehar.com/blog/wp-content/uploads/2009/09/DyeMixer.jar" alt="To use this applet, you need a Java virtual machine (Java plug-in) for your web browser.">'; // ]]></script></div>`
This also survives editing the post in the visual editor. You can even center the applet like you would a normal paragraph. Note that the div's id and the argument to getElementById must be identical. Using a similar JavaScript script, you should be able to put also other arbitrary HTML in a WordPress post or page.
## Generation of random numbers from a truncated exponential distribution
2009-07-30
To generate a random number with a truncated exponential distribution:
1. Generate a random number $x$ from an exponential distribution with the rate parameter you want
2. Calculate $x$ modulo where-to-truncate. This would not be integer modulo but for example fmod() floating-point remainder.
Could be useful if you have exponentially distributed random numbers handy.
## Anti-imaged wavelet transform
2002-11-25
This is an idea of a wavelet-type audio processing scheme, where you decompose the audio into octave bands, each octave sampled at half the sampling frequency compared to the one-up octave band. Then you process the bands in whatever way you wish, and reconstruct the signal from the bands. The benefit of the method compared to usual wavelets is the low level of imaging noise when bands are processed separately. The original music-dsp posting has the details:
http://aulos.calarts.edu/pipermail/music-dsp/2002-November/018600.html
The processing scheme ("/2" means decimation by 2, "*2" means dilution by zeros to 2x samplerate)
Frequency responses of filters that could be used in the scheme
## Frequency shifting (audio effect)
2001-09-02
Frequency shifting can be an interesting audio effect, although it does not usually preserve harmonic relations. Here is how to do it the wrong and the right way:
Complex exponential modulation illustrated on z-plane unit circle
The ugly mixture. Complicated aliasing caused by discarding the imaginary part from a complex exponential modulated signal.
Anti-aliased frequency shifting illustrated on z-plane unit circle
## Daubechies wavelets 1-38 in a .h file
2000-08-10
I copied some daubechies wavelets from a published database into this C header file, confirmed the data to be error-free and wrote a little introduction into wavelets in the comments.
## 2 Comments »
1. I had no luck doing the same thing you did to embed a Java applet in WordPress 3.0.5.
So I created a plugin specifically to do that: http://huyz.us/2011/the-easiest-way-to-embed-java-in-wordpress/
Comment by huyz — 2011-07-01 @ 20:40
2. huyz: that's great! I may start using it if I run into problems again
Comment by Olli Niemitalo — 2011-07-02 @ 10:59
RSS feed for comments on this post. TrackBack URL
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http://math.stackexchange.com/questions/14421/encrypting-algorithm/14424
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# Encrypting algorithm
I know there are some encrypting algorithm that once the text (the message) is encrypted there is no way to decrypt it.
I can think one way of doing this: using as encryption key the current date at the encryption moment. But there is one problem wiht that: Each time you encrypt the same message the result will be different.
I think there must be some "math-algorithm" to encrypt without out the posibility to decrypt and that always the same message turns out with the same encrypted message, but I realy can't imagine how to do it.
Note: I'm posting this question in this forum and not in stackoverflow.com intentionally, it is not referred to "how to programm the algorithm", it is refered to "how is the math algorithm".
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– J. M. Dec 15 '10 at 16:19
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If you want it not to be able to be decrypted, just flip each bit randomly. But I'm not sure what the use is... – Ross Millikan Dec 15 '10 at 16:51
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Are you talking about cryptographic hash functions? – Incognito Dec 15 '10 at 20:20
No, you can easily decrypt by simply guessing the date of encryption. Besides, as Ross said, if you won't be able to decrypt it [not even by you], so what's the use? – muntoo Dec 15 '10 at 23:35
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@Diego Brute force decryptically beat up your file through all the possible dates. – muntoo Dec 16 '10 at 17:05
show 1 more comment
## 3 Answers
Such algorithms do exist. Which one you mean depends on what you mean by "no way to decrypt it". You refer to either:
• A function which takes an input and generates a unique output, with there being ideally no known way to derive the input from the output. These are called hashing algorithms; examples of which include MD5, SHA256. Those links provide more detail than I possibly can.
• A cryptosystem that is impossible for any other person but the keyholder to decrypt. This is the one time pad J.M. mentions. This is the only currently proven-secure algorithm.
Edited in because this is probably too long for comments:
Diego, no problem. They're not, as they're two different things and you're comparing apples and oranges, so to speak. A cryptosystem is designed to be hard to undo without the key, but easy if you have the key. By contrast, a hashing function should only require 1 input, be one way and give a unique output for each input. The idea is to be able to generate a unique representation of the output without being able to deduce what the output is. Why?
Well aside from verifying the unique nature of your input, a hashing function is used as part of a digital signature. In a public key cryptosystem, you may well be aware that Alice can send Bob an encrypted message using Bob's public key. However, anyone can send Bob a message using Bob's public key. If Eve has broken the cryptosystem, she could intercept Alice's message, decrypt it, change it, re-encrypt it and send it on to Bob.
How do we prevent this? Well, let $H: x\rightarrow y$ be a map from $x$ some input to $y$ some output. It is trivial to compute $H$ but impossible (ideally, the truth is just very very unlikely) to compute $H^{-1}$.
Then let $C:x,k$ be some assymetric cryptosystem Alice and Bob know, with Alice's keys $a_p$ and $a_s$ (public and secret) and likewise Bob's keys $b_p$ and $b_s$.
Now we know that as this cryptosystem is assymetric, that $C:x,b_p \rightarrow z$ where $z$ is some ciphertext (encrypted secret) and that $C:z,b_s \rightarrow x$ i.e. if you use Bob's public key to encrypt, Bob's private key using exactly the same algorithm will decrypt the process.
This we know already. However, as discussed, a man in the middle attack is possible if Eve knows Bob's secret. However, Alice can do two things:
• Compute $C(H(x), a_s)$
• Compute $C(x+H(x), b_p)$
And send the second one to Bob. What's the point in this? Well, Bob can reverse both "encryptions", because he holds the opposite keys (his own private and Alice's public. He is left with two things: the ciphertext and a unique hash of the ciphertext. He can also compute H(x) and compare results. If they differ, he knows the message has been tampered with.
Why? Well, consider Eve sat gleefully with Bob's keys. She intercept's Alices message and sees it contains the two parts (she can decrypt the outer layer, Bob's message). However, now she's left with an impossible situation. She can compute the hash of her newly inserted message, but she does not know Alice's private key (if she does, the cryptosystem is truly broken). Therefore she cannot encrypt the new hash. So, if she tries to send the new message onwards, Bob will immediately be aware the system has been tampered with.
This is called a digital signature or digitally signing and is used to identify the message author as who they truly say they are. After all, only they can encrypt with their private key (allegedly). The importance of hashing in this scenario is that it generates a unique output for each given output. However, not being reversible is also desired so that it is impossible to determine what a secret is given the output. This also makes it a highly used password-storage mechanism (because in theory, you're not storing the passwords, just a unique representation of them).
As always in cryptography, someone is always trying to break hash functions. Rainbow tables are the technique used to pre-compute parts of hashes to make hash reversing easier. Note however that the longer the input, the larger the tables need to be. Hence why passwords should always be greater than 6 characters...
Finally, you may be interested to know in Hash weaknesses. If two hashes with unique input produce the same output, this is called a collision and is a serious weakness, especially if someone can work out any pattern to said collisions. MD5 is thought to suffer from some, as is SHA1. SHA2 may, also. In any case, NIST has launched SHA-3, a competition to find future hash functions offering security now and in many years to come. The finalists are available to view here.
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I think I was looking for hashing algorithms, thanks!. Why would a "decryptable" algorithm be more secure than a non "decryptable"? – Diego Dec 16 '10 at 12:02
Excelent post! I wish I could +10 it :P. I must admit I didn't totally understand all the explanation (with a litle of shame). I'm now going to re-read it. Thank you so much! – Diego Dec 17 '10 at 13:09
"This is the only currently proven-secure algorithm and even that has been thrown into doubt." - The security proof of OTP is so stupidly simple that I honestly couldn't believe this when I read it. That paper and all its references appear to be written by the same person: his argument is that, if we know one of two messages is being sent, and we know the probability of one message to be 10%, then after receiving the cyphertext the probability goes up to 50/50. And since the probability has changed, OTP is insecure. (cont.) – BlueRaja - Danny Pflughoeft Jan 17 '11 at 22:12
(cont.) This is obviously bogus - the probability-space has not changed, only the observer's interpretation of the probability (which it really shouldn't - the probability is still 10/90 after receiving the cyphertext). This is equivalent to throwing a loaded die, rigged to roll '6' 99% of the time, then covering up the result and claiming there is a 1/6th chance that a '3' was rolled because there are 6 possibilities. Please remove the very misleading text in your post; you'll give people the wrong idea. – BlueRaja - Danny Pflughoeft Jan 17 '11 at 22:13
I've read it and am very much inclined to agree with you - I do apologise for inserting rubbish. I clearly didn't read it correctly when I was looking for articles on the subject... it's quite clearly snake oil, or inverted snake oil. – Antony Vennard Jan 17 '11 at 22:22
show 1 more comment
As J.M. commented above, the only totally secure cryptographic algorithm is the one-time pad. Because this relies on both the sender and receiver of a message to be able to have access to the same arbitrarily long random sequence, it is often not feasible. Cryptography originally relied on shared keys, from which simple ciphers, such as Vigenere, Caesar, and substitution, come from. Unfortunately, all of these algorithms are so simple that using frequency analysis can often be used to break them very quickly.
Modern cryptography is generally split into two core areas: symmetric and asymmetric algorithms. Symmetric seems to be what you are describing, so I will describe that first. In symmetric algorithms, both users must share a key, which they often pass to one another via asymmetric cryptography or using Diffie-Hellman key exchange. Once the key is known between the two parties, eavesdroppers are generally going to have a very difficult time breaking the encryption. As with all things, certain algorithms are stronger than others. Older encryption schemes like DES are not as secure as some of the newer ones like AES, and it's not infeasible to try to break the weaker ones with sufficient computing power. Asymmetric cryptography, or public key cryptography, eliminates the need to share keys between users by admitting a public key and private key for each user. To send something to someone else, you encrypt using their public key (which they publish) and only they are able to decrypt using their private key. RSA and ECC are two of the more common public key protocols.
-
+1 from me for assym/sym crypto. – Antony Vennard Dec 15 '10 at 16:51
In asymmetric algorithms the sender and reciever don't have to share a key, (beside the public one)?? wow.. Still, I'm looking for something wich can't be decrypted. – Diego Dec 16 '10 at 12:00
@Diego Yes, the sender and receiver don't have to share a key. Anyone is able to see the public one. If you are looking for something which cannot be decrypted at all, then it sounds like you are talking about cryptographic hash functions, which Ninefingers addressed above. – Brandon Carter Dec 16 '10 at 17:36
I can give an answer, which may seem like a joke. But you can map all text (the set of all codes) to the same result.
Say for example I map the codes ABC to 1 BDCD to 1 ABDGR to 1
there is no opportunity for decoding here. :)
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Isn't that also called "compressing" data? Not mapping everything, just patterns, and storing it in a separate file to remember what each pattern is mapped to. If you don't store it in a separate file, you can remember it. So, the remembered thing can now be called a "key". Only if you have the key, will you know what each pattern represents. Just an example, of course. – muntoo Dec 15 '10 at 23:39
You are correct as far as I can remember from coding theory class. – picakhu Dec 16 '10 at 4:16
yes, you are correct, that really answer the question I've made. I just asume it woudn't be necesary to say that I don't want the same result repeating for different texts. – Diego Dec 16 '10 at 12:05
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http://physics.stackexchange.com/questions/27632/about-the-definition-motivation-properties-of-the-twisted-chiral-superfield-in
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# About the definition/motivation/properties of the twisted chiral superfield in ${\cal N}=2$ theories in $1+1$ dimensions
The following is in the context of the ${\cal N}=2$ supersymmetry in $1+1$ dimensions - which is probably generically constructed as a reduction from the ${\cal N}=1$ case in $3+1$ dimensions.
• In the $\pm$ notation what is the definition of ${\cal D}_+$ and ${\cal D}_{-}$, which I understand from context to be the gauge covariant superderivatives. (..It would be great if someone can relate them to the usual definition in the notation of say Wess and Bagger..)
• So what is the meaning/motivation of defining a twisted chiral superfield as, $\Sigma = \{\bar{{\cal D}}_{+}, {\cal D}_{-}\}$ (..naively this looks like an operator and not a field - I guess there is some way of arguing that the derivative terms which are not evaluated on something actually go to zero..)
I am guessing that in the above context it will be helpful if someone can explain as to what is meant by the following decomposition/reduction of the gauge field from $3+1$ dimensions,
$\sum _ {\mu = 0}^3 A_\mu dx^\mu = \sum _{\mu =0} ^1 A_\mu dy^\mu + \sigma (dy^2-idy^3) + \bar{\sigma}(dy^2+idy^3)$ ?
• From the above (does it/how does it) follow that one can write $\Sigma$ as,
$\Sigma = \sigma + \theta\lambda + \theta \bar{\theta}(F+iD)$
(..where I am not sure if $F,D,\sigma$ are real or complex scalar fields...and $\lambda$ is a Weyl fermion..)
• What is the R-charge of this twisted chiral super field? (..from some consistency conditions I would think that its 2..but I am not sure..)
I guess that the R-symmetry transformations act as,
• The "right" R symmetry keeps $\theta^-$s invariant and maps, $\theta^+ \mapsto e^{i\alpha}\theta^+$, $\bar{\theta}^+ \mapsto e^{-i\alpha}\bar{\theta}^+$
• The "left" R-symmetry keeps $\theta^+$ invariant and maps, $\theta^- \mapsto e^{-i\alpha}\theta^-$, $\bar{\theta}^- \mapsto e^{i\alpha}\bar{\theta}^+$.
Though I am not sure and like to understand as to why one wants to think of these two different R-symmetry groups as having two different origins - one coming from the rotation symmetry of the two spatial dimensions of the original $\cal{N}=1$, $1+3$ theory and another coming from R-symmetry of the $\cal{N}=1$, $U(1)$ gauge theory.
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## 1 Answer
After dimensional reduction from 4 to 2 dimensions, it is convenient to simply label the last two remaining dimensions as $+$ and $-$ instead of 1 and 2. So, basically you have ${\cal D}_- = {\cal D}_1$ and ${\cal D}_+ = {\cal D}_2$.
As for a motivation for twisted chiral superfields, I'm going to quote Witten [http://arxiv.org/abs/hep-th/9301042]:
Sigma models containing both chiral and twisted chiral superfields are quite lovely. Since mirror symmetry turns chiral multiplets into twisted chiral multiplets, it is likely that consideration of appropriate models containing multiplets of both types is helpful for understanding mirror symmetry.
The introduction given by Witten on twisted chiral superfields in the above paper should cover most of your questions.
I am curious though, where did you find your equations? I'm a bit confused by the F-Term in your twisted chiral superfield, as I thought it was common practice to use the WZ-gauge for these types of fields?
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– user6818 Apr 5 '12 at 22:56
– user6818 Apr 5 '12 at 23:06
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http://physics.stackexchange.com/questions/22747/why-do-we-like-gauge-potentials-so-much
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# Why do we like gauge potentials so much?
Today I read articles and texts about Dirac monopoles and I have been wondering about the insistence on gauge potentials. Why do they seem (or why are they) so important to create a theory about magnetic monopoles?
And more generally, why do we like gauge potentials so much?
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## 2 Answers
1) Postponing for a moment the issue of magnetic monopoles, one conventional answer is, that the gauge potential $A_{\mu}$ (as opposed to, e.g., the electric and magnetic $\vec{E}$ and $\vec{B}$ fields) constitute the true fundamental variables and (the photon field) of QED.
At the classically level, by saying that $A_{\mu}$ are fundamental variables, we mean that the Maxwell action $S[A]$ depends on $A_{\mu}$, and that the corresponding Euler-Lagrange equations are the remaining Maxwell equations, namely Gauss' and (modified) Ampere's laws. (The obsolete Maxwell equations are Bianchi identities, which are rendered vacuous by the existence of the gauge potential $A_{\mu}$.)
Quantum mechanically, it seems appropriate to mention the Aharonov Bohm effect, which shows that the $A_{\mu}$ gauge field has physical consequences not already encoded in the gauge-invariant $\vec{E}$ and $\vec{B}$ fields.
2) Up until now we haven't discussed magnetic monopoles and Dirac strings.
At a magnetic monopole, the $A_{\mu}$ gauge potential is not well-defined. This is the main topic of, e.g., this, this, and this questions.
Although magnetic monopoles so far have not been experimentally detected, there are theoretical reasons to believe they do exist, see e.g. this, this, and this questions.
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Aharonov-Bohm effect can be used as an argument if (and only if) elementary particles are considered as point-like objects. This argument is based on the assumption that particle's field is LOCALIZED in the area where $A_{\mu}$ is non-zero, while $E$ and $B$ are zero. – Murod Abdukhakimov Mar 24 '12 at 9:16
@Qmechanic: Tell me if I am right: a non-gauge potential would give the same E and B as a gauge potential, but the action would not be then as simple to calculate; in that sense a gauge potential, though non unique, is more fundamental. Correct? – Isaac Mar 24 '12 at 10:25
It is the gauge potentials, not the fields, that determine the quantum motion of particles. In either the Schrodinger equation or the path integral, the gauge field appears, not the E and B, and for nonabelian theories, this is impossible to fix because you don't have a integral Stokes law relation.
The interaction with charged particles is that a particle moving along a path gets a phase equal to the integral of A along the path. If you want to replace the A with B, you need to use the fact that the integral of A along a closed loop is the magnetic flux enclosed by the loop, and this is a nonlocal condition. So you can't write local equations of motion for a quantum particle using E and B.
The integral relation for B states that if you make a circle, and you want the phase that the charged particle will get if it moves on this circle, you draw some surface whose boundary is the circle, and the magnetic flux through this surface is the phase.
The dirac condition is simply the statement that if you have a monopole and draw a circle around the monopole, the flux through the northern hemisphere is equal to the flux through the southern hemisphere, up to a multiple of 2pi, which is an undetectable phase change. This tells you that the magnetic charge times the electric charge must be an integer multiple of $2\pi$.
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http://www.physicsforums.com/showthread.php?p=3858901
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Physics Forums
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## how to calculate required hp for machine
I have estimated the total MMOI as 16.5 kg-m^2, and on that basis, Bob_S value of 41 watts appears to be a reasonable estimate for the total power required to accelerate the roll (I calculated slightly less). The total energy of the spinning roll at 100 rpm is 903 J.
As to realism, now, these numbers are so small that I rather think that they are close to meaningless. The friction losses in the gear train, the bearings, the windage, etc. will consume more than what we have computed for the change in the energy state of the rotating mass. Thus I think that considerably more power will be required to overcome the system losses than to power the roll. Therefore, I think that these calculations are not realistic for evaluating the power required.
Quote by OldEngr63 Realism means that you really believe that 41 watts will do this job. I don't. Look at your overall calculation and tell us that the whole scenario is realistic, if you believe it is.
First, you have to do the math.
If the stored rotational energy after 60 seconds is
$$E=\frac{1}{2}I\omega^2=\frac{1}{2}mr^{2}\omega^2 = 1234 \text{ Joules}$$ then you have to add energy at the rate of 1234/60 joules/sec = 21 watts for 60 seconds.
This does not include motor inefficiency, friction losses, not using the right gear ratios, etc. Please check my math.
@ Bob_S: I did do the math, in considerably more detail than it appears you have done. It appears that you are simply saying that I = MR2 = 17.956 kg-m2 Note that the radius value you used is to the outer diameter of the rubber layer, well past the last of the steel structure. This gives an excessively large estimate for R. By a more detailed reckoning, based on the drawing, I came up with the figure that I gave previously I = 16.5 kg-m2 I think that you have somewhat over estimated the MMOI by putting it all at the rubber layer, well outside the steel, considering that part of it is internal structure. N = 100 rpm → ω = 10.47 rad/s ω2 = 109.66 1/s2 Because of the difference in the estimates for I, we differ substantially in the total energy content of the rotating roll. This is where our differences lie. My comments about these values being minor compared to the losses in the system and therefore not useful for determining drive power requirements still stand.
Quote by OldEngr63 @ Bob_S: I did do the math, in considerably more detail than it appears you have done. It appears that you are simply saying that I = MR2 = 17.956 kg-m2
You are correct. I decided to put the maximum value in $I=mr^2$, which is exactly twice the moment of inertia for a solid cylinder. It adds a few % to the result, as you found out.
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http://math.stackexchange.com/questions/51111/calculating-point-on-a-circle-given-an-offset?answertab=oldest
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# Calculating point on a circle, given an offset?
I have what seemed like a very simple issue, but I just cannot figure it out. I have the following circles around a common point:
The Green and Blue circles represent circles that orbit the center point. I have been able to calculate the distance/radius from the point to the individual circles, but I am unable to plot the next point on either circle, given an angle from the center point. Presently, my calculation looks like the following:
The coordinates of one of my circles is:
````y1 = 152
x1 = 140.5
````
And my calculation for the next point, 1 degree from the starting point `(140.5,152)` is:
````distance = SQRT((160-x1)^2 + (240-y1)^2) = 90.13
new x = 160 - (distance x COS(1 degree x (PI / 180)))
new y = 240 - (distance x SIN(1 degree x (PI / 180)))
````
My new x and y give me crazy results, nothing even close to my circle.
I can't figure out how to calculate the new position, given the offset of 160, 240 being my center, and what I want to rotate around. Where am I going wrong?
Update:
I have implemented what I believe to be the correct formula, but I'm only getting a half circle, e.g.
````x1 = starting x coordinate, or updated coordinate
y1 = starting y coordinate, or updated y coordinate
cx = 100 (horizontal center)
cy = 100 (vertical center)
radius = SQRT((cx - x1)^2 + (cy - y1)^2)
arc = ATAN((y1 - cy) / (x1 - cx))
newX = cx + radius * COS(arc - PI - (PI / 180.0))
newY = cy + radius * SIN(arc - PI - (PI / 180.0))
Set the values so next iteration of drawing, x1 and y1 will be the new
base for the calculation.
x1 = newX
y1 = newY
````
The circle begins to draw at the correct coordinates, but once it hits 180 degrees, it jumps back up to zero degrees. The dot represents the starting point. Also, the coordinates are going counterclockwise, when they need to go clockwise. Any ideas?
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And by circle you mean dot? From the diagram I see two circles (orbits) and three dots. Is this correct? – ja72 Jul 12 '11 at 21:49
@George: I comment here because I received the message "Please avoid extended discussions in comments.". 1) the behavior you describe in your comment below is caused most likely by the values of $y=\arctan x$, which are usually $−\pi/2<y<\pi/2$. If the current value is $y=-\pi/2$ the next value instead of being $-\pi/2-\pi/180$ is $\pi/2+\pi/180$. 2) I think you should use COS(arc + PI - (PI / 180.0)) and SIN(arc + PI - (PI / 180.0)) instead of COS(arc - PI - (PI / 180.0)) and SIN(arc - PI - (PI / 180.0)). (see my formulas). – Américo Tavares Jul 13 '11 at 20:53
... 3) But I have no explanation for the counterclokwise movement. Since the point $(x_1-x_c,y_1-y_c)=(140.5-160,152-240)$ is in the 3rd quadrant I added $\pi$ to 1.3527. – Américo Tavares Jul 13 '11 at 20:55
2
ATAN has no information on which quadrant your angle is in and so puts it in 2 fixed quadrants. Most languages provide a function ATAN2 that takes 2 arguments to fix this problem. ATAN2(y, x) = ATAN(y/x) normally. Using ATAN2 should fix your semicircle problem. – gereeter Jul 13 '11 at 22:05
@George: I updated the answer with a plot of a new circle centered at $(100,100)$, where the movement of the point is clockwise as $t$ increases. – Américo Tavares Jul 13 '11 at 22:09
show 1 more comment
## 5 Answers
Update 2: Here is the graph I got for $(x_{1},y_{1})=( 78. 965,12. 354)$, for the parametric circle $(x(t),y(t))$ centered at $(100,100)$
$$x=100+90.135\cos \left( 1.3527+\pi -t\frac{\pi }{180}\right) ,$$
$$y=100+90.135\sin \left( 1.3527+\pi -t\frac{\pi }{180}\right) .$$
together with the 4 points $(x(t),y(t))$ for $t=0,90,180,270$
$$(x_{1},y_{1})=(x(0),y(0)),(x(90),y(90)),(x(180),y(180)),(x(270),y(270)).$$
You might use the following equations in a for loop with $k=0$ to $k=359$, step $1$:
$$x=100+90.135\cos \left( 1.3527+\pi -k\frac{\pi }{180}\right) ,$$
$$y=100+90.135\sin \left( 1.3527+\pi -k\frac{\pi }{180}\right) .$$
to draw the "orbit" with a 1 degree interval.
Update: corrected coordinates of $(x_{1},y_{1})=(140.5,152)$.
You need to consider the new angle and not only the $1{{}^\circ}$ change. The argument of $\cos$ and $\sin$ is this new angle and not $1{{}^\circ}$.
Let $(x_{c},y_{c})=(160,240)$ be the center of the set of circles and $(x_{1},y_{1})=(140.5,152)$. The radius $r$ is
$$\begin{eqnarray*} r &=&\sqrt{\left( x_{c}-x_{1}\right) ^{2}+\left( y_{c}-y_{1}\right) ^{2}} \\ &=&\sqrt{\left( 160-140.5\right) ^{2}+\left( 240-152\right) ^{2}} \\ &=&90.135 \end{eqnarray*}$$
Call $(x,y)$ the new coordinates of $(x_{1},y_{1})$ rotated by an angle of $-1{{}^\circ}=-\dfrac{\pi }{180}$ around $(x_{c},y_{c})$ with a radius $r$. The new angle is $\theta'=\theta -\frac{\pi }{180}$, $\theta$ being the initial angle. Then
$$\begin{eqnarray*} x &=&x_{c}+r\cos \left( \theta -\frac{\pi }{180}\right), \\ y &=&y_{c}+r\sin \left( \theta -\frac{\pi }{180}\right), \end{eqnarray*}$$ where $\theta$ is the angle $\theta =\arctan \dfrac{y_{1}-y_{c}}{x_{1}-x_{c}}:$
$$\begin{eqnarray*} \theta &=&\arctan \frac{152-240}{140.5-160}=1.3527+\pi \text{ rad.}\\ &=&\frac{1.3527\times 180{{}^\circ}}{\pi }+180{{}^\circ}=257. 5{{}^\circ}\end{eqnarray*}$$ Thus $$\begin{eqnarray*} x &=&160+90.135\cos \left( 1.3527+\pi -\frac{\pi }{180}\right)= 138. 96 \\y &=&240+90.135\sin \left( 1.3527+\pi -\frac{\pi }{180}\right) = 152. 35 \end{eqnarray*}$$
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I'm not seeing where the 1 degree rotation is coming into effect in your last equation. Can you clarify? :) – George Jul 12 '11 at 23:06
@George: it enters in the subtraction of the angle $\dfrac{\pi}{180}$ from $\theta=1.4906+\pi$. (1 degree=$\pi/180$ rad) – Américo Tavares Jul 12 '11 at 23:46
@George: It is not the same in terms of the changes in the coordinates to have a change of 1 degree starting at an angle of e.g. 265.41 degrees (to 264.41 degrees) or the same 1 degree change but starting at let's say 0 degrees (to -1 degrees). – Américo Tavares Jul 13 '11 at 0:07
@George: I corrected the coordinates of $(x_1,y_1)=(140.5,152)$, which corresponds to an angle $\theta=257.5$ degrees. The rotated point makes an angle of $257.5-1=256.5$ degrees with the line $y_c$. My point is that you need to consider the new angle and not only the 1 degree change. The argument of $\cos$ and $\sin$ is this new angle and not 1 degree. – Américo Tavares Jul 13 '11 at 0:39
1
Wow, thank you for the in-depth answer. I never expected that much! I'd upvote you 10x if I could! – George Jul 13 '11 at 2:41
show 2 more comments
How about: $$\begin{matrix} x_{green} = x_{center} + r_{green} \cos \theta \\ y_{green} = y_{center} + r_{green} \sin \theta \end{matrix}$$
then then just increment your angle $\theta$ by 1 degree. Make sure you convert degrees to radians before using the `sin()` and `cos()` functions.
Similarly for the blue dot.
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George, instead of subtracting for the offset, try adding, i.e.
````distance = SQRT((160-x_{1})^2 + (240-y_{1})^2) = 90.13
new x = 160 + (distance x COS(1 degree x (PI / 180)))
new y = 240 + (distance x SIN(1 degree x (PI / 180)))
````
The part, $x_{new}=(distance)(cos(pi/180))$ is assuming the distance is from the origin (0,0). Since you are starting your x value from positive 160, you need to add that, i.e. $x_{new}=x_{center} + (distance)(cos(pi/180))$ $x_{new}=160 + (distance)(cos(pi/180))$ And similarly for the new y value.
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A look at the calculation reveals the basic issue. The calculation of new $x$ and new $y$ pays no attention to the old $x$ and old $y$. It is therefore not surprising that the new point is nowhere near the old point.
The center and radius are known. For the sake of plotting, we presumably need to keep track of $(x,y)$. From this we can compute the angle, which we might as well compute from the the usual reference half-line, the positive $x$-axis. There is a slight complication that the rotation seems to be clockwise. So, if we follow the usual mathematical conventions, soon the angle will be negative.
We now have a problem. Shall we have negative angles built into the code, or shall we change the usual convention about signs of angles. Each choice has built-in problems. Negative angles are intuitively unpleasant, but changing from the usual convention means that we may have to adjust standard formulas. I will choose to use the ordinary convention that counterclockwise is positive. But you probably shouldn't.
It is a nuisance to recompute the angle each time from the current $x$ and $y$. (In the old days, it was also excessively computationally costly). So my feeling is that it is best to keep track of of the angle.
In changing the angle by $1^\circ$, we need to decrement the angle by $\pi/180$ radians. Let $R$ be the radius. We want $$\theta_{\rm new}=\theta_{\rm old} -\frac{\pi}{180}$$ $$x_{\rm new}=160+R\cos(\theta_{\rm new})$$ $$y_{\rm new}=240+R\sin(\theta_{\rm new})$$ Recall that angles will be negative most of the time. One can work with the positive version by incrementing, and using $-R\sin(\theta_{\rm new})$. The $+$ sign in front of the $R\cos$ term would remain unchanged.
An alternative way of doing things is to look up how to carry out rotations by multiplying by an appropriate matrix. You will probably need the idea sometime in the future, so might as well test both ways now. In that case, you don't need to keep track of angle. All that one does is multiply the vector $(x_{\rm old}, y_{\rm old})$ by the appropriate matrix that does rotation by $1^\circ$ to get $(x_{\rm new}, y_{\rm new})$ . The matrix is precomputed, so the calculations are blazing fast.
It is not quite as simple as that! Fairly quickly, because of accumulating roundoff errors, there will be degradation of quality: the spinning circle will make a lousy clock. So adjustment steps have to be built in, if long-term accuracy is desired.
The matrix approach is faster. The one I discussed at some length is essentially what you chose to do, adjusted to make things work. What to do?
Code both. At some time or other, you will have need of each idea.
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Using "compass" coordinates with zero at north and positive angles in the clockwise direction is just as consistent as "math" coordinates with zero to the right and positive angles in the counter clockwise direction. All the same identities hold. – phv3773 Jul 13 '11 at 14:40
@phv3773: The internal consistency could never be an issue. And interchanging roles of $x$ and $y$ turns one to the other. The only possible difficulty is with mechanical "borrowing" of formulas. – André Nicolas Jul 13 '11 at 17:37
In order to account for the rounding issues, I assume it would be as simple as grabbing the starting point, and after 360 degrees of rotation, reset to the starting point. Do you think that would be a viable option, or would the circle degrade within the initial 360 degree rotation, to the point where it would be noticable by the human eye? – George Jul 13 '11 at 20:17
@George: A very practical solution, which I am sure would be plenty good enough! – André Nicolas Jul 13 '11 at 21:18
We can modify 6312's suggestion a bit to reduce the trigonometric effort. The key idea is that the trigonometric functions satisfy a recurrence relation when integer multiples of angles are considered.
In particular, we have the relations
$$\cos(\phi-\epsilon)=\cos\,\phi-(\mu\cos\,\phi-\nu\sin\,\phi)$$
$$\sin(\phi-\epsilon)=\sin\,\phi-(\mu\sin\,\phi+\nu\cos\,\phi)$$
where $\mu=2\sin^2\frac{\epsilon}{2}$ and $\nu=\sin\,\epsilon$. (These are easily derived through complex exponentials...)
In any event, since you're moving by constant increments of $1^\circ$; you merely have to cache the values of $\mu=2\sin^2\frac{\pi}{360}\approx 1.523048436087608\times10^{-4}$ and $\nu=\sin\frac{\pi}{180}\approx 1.745240643728351\times 10^{-2}$ and apply the updating formulae I gave, where your starting point is $\cos\,\phi=\frac{140.5-160}{\sqrt{(140.5-160) ^2+(152-240)^2}}\approx-0.2163430618226664$ and $\sin\,\phi=\frac{152-240}{\sqrt{(140.5-160) ^2+(152-240)^2}}\approx-0.9763174071997252$
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http://mathandcode.com/
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# Wire Bead Toys with Mathematica
Posted on March 13, 2013 by
These are toys that I loved as a kid. It’s a dead simple toy, but it’s fun to flick a bead and watch it coast along the track. So what are the mathematics of this? What are some properties of it, and how can you simulate and animate it with Mathematica?
There are two solutions. There’s the smart solution, and there’s the brute force solution. The smart solution is easier to do and gives you more intuition towards the problem, but the brute force solution is easy to plug in to mathematica!
Continue reading →
Posted in Uncategorized |
# Running Khan Academy Javascript Code Offline
Posted on September 22, 2012 by
I’ve found that a common question on the Khan Academy javascript computer science tutorials is: “How can I run this code offline?!”
It turns out it’s pretty easy using a little bit of html and a little bit of extra javascript. I explain more in my youtube video on the subject, but for a quick how-to:
2. extract the files to a folder on your computer
3. modify myCode.js, making use of the extra commands “size” and “framerate”, and leaving the first and last line. (also, downloading and replacing any images you use)
4. right click processing.html and select open with and your browser.
This is the exact code that I used for my gravity app. Now you can switch between khanacademy computer science, and your own offline applications!
Posted in Uncategorized |
# The Three Body Problem
Posted on September 19, 2012 by
Find the simulation here: http://www.mathandcode.com/gravity/
The two-body problem consists of two objects under gravitational attraction. The attraction force is the newtonian force $$G \frac{M_1 M_2}{d^2}$$, where G is some constant, M is mass and distance is the distance between the two objects.
In the two-body problem, the path of each particle can be solved for explicitly. We can get circles or ellipses, like the earth’s path around the sun, or we can get escape trajectories where the two particles pass by and never collide with each other again, taking the form of hyperbolas and parabolas.
The three-body problem is more difficult though. Three-bodies can cause chaos, and you can’t solve it explicitly for hyperbolas/parabolas. The only way to solve the general case is to simulate the system over time.
It’s hard to get to grips with the chaos of the three body problem, but this program tries to depict it.
It starts out with three bodies with the same mass/radius and sends them off in different velocities. From this, you can calculate the future path of the particles. This program shows the path of three objects over the full simulation, so instead of viewing three circles (the particles current positions), you see three lines (the particles paths given the starting conditions). As the initial velocity varies over time, the paths vary, and you get chaotic looking results.
Posted in Uncategorized |
# Gravitational Simulation
Posted on September 10, 2012 by
http://mathandcode.com/programs/javagrav/
Gravitational simulations can use a lot of processing power. This simulation, written in Java, simplifies it a bit by using a structure called a quadtree. Check out the program page for more information!
The results of the program are pretty awesome. I can simulate about 13,500 particles with universal gravitational attraction in real time.
Posted in Uncategorized |
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http://en.wikipedia.org/wiki/Gaussian_beam
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# Gaussian beam
Instantaneous intensity of a Gaussian beam.
A 5 mW green laser pointer beam profile, showing the TEM00 profile
In optics, a Gaussian beam is a beam of electromagnetic radiation whose transverse electric field and intensity (irradiance) distributions are well approximated by Gaussian functions. Many lasers emit beams that approximate a Gaussian profile, in which case the laser is said to be operating on the fundamental transverse mode, or "TEM00 mode" of the laser's optical resonator. When refracted by a diffraction-limited lens, a Gaussian beam is transformed into another Gaussian beam (characterized by a different set of parameters), which explains why it is a convenient, widespread model in laser optics.
The mathematical function that describes the Gaussian beam is a solution to the paraxial form of the Helmholtz equation. The solution, in the form of a Gaussian function, represents the complex amplitude of the beam's electric field. The electric field and magnetic field together propagate as an electromagnetic wave. A description of just one of the two fields is sufficient to describe the properties of the beam.
The behavior of the field of a Gaussian beam as it propagates is described by a few parameters such as the spot size, the radius of curvature, and the Gouy phase.[1]
Other solutions to the paraxial form of the Helmholtz equation exist. Solving the equation in Cartesian coordinates leads to a family of solutions known as the Hermite–Gaussian modes, while solving the equation in cylindrical coordinates leads to the Laguerre–Gaussian modes.[2] For both families, the lowest-order solution describes a Gaussian beam, while higher-order solutions describe higher-order transverse modes in an optical resonator.
The top portion of the diagram shows the two-dimensional intensity profile of a Gaussian beam that is propagating out of the page. The blue curve, below, is a plot of the electric field amplitude as a function of distance from the center of the beam. The black curve is the corresponding intensity function.
## Mathematical form
The Gaussian beam is a transverse electromagnetic (TEM) mode.[3] A mathematical expression for its complex electric field amplitude can be found by solving the paraxial Helmholtz equation, yielding[1]
$E(r,z) = E_0 \frac{w_0}{w(z)} \exp \left( \frac{-r^2}{w^2(z)} -ikz -ik \frac{r^2}{2R(z)} +i \zeta(z) \right)\ ,$
where[1]
$r$ is the radial distance from the center axis of the beam,
$z$ is the axial distance from the beam's narrowest point (the "waist"),
$i$ is the imaginary unit (for which $i^2 = -1$),
$k = 2 \pi/\lambda$ is the wave number (in radians per meter),
$E_0 = |E(0,0)|$,
$w(z)$ is the radius at which the field amplitude and intensity drop to 1/e and 1/e2 of their axial values, respectively,
$w_0 = w(0)$ is the waist size,
$R(z)$ is the radius of curvature of the beam's wavefronts, and
$\zeta(z)$ is the Gouy phase shift, an extra contribution to the phase that is seen in Gaussian beams.
Additionally, the field has a time dependence factor $e^{i\omega t}$ that has been suppressed in the above expression.
The corresponding time-averaged intensity (or irradiance) distribution is
$I(r,z) = { |E(r,z)|^2 \over 2 \eta } = I_0 \left( \frac{w_0}{w(z)} \right)^2 \exp \left( \frac{-2r^2}{w^2(z)} \right)\ ,$
where $I_0 = I(0,0)$ is the intensity at the center of the beam at its waist. The constant $\eta \,$ is the characteristic impedance of the medium in which the beam is propagating. For free space, $\eta = \eta_0 = \sqrt{\mu_0/\varepsilon_0} = 1/(\varepsilon_0 c) \approx 376.7 \ \Omega$.
## Beam parameters
The geometry and behavior of a Gaussian beam are governed by a set of beam parameters, which are defined in the following sections.
### Beam width or spot size
See also: Beam diameter
Gaussian beam width w(z) as a function of the axial distance z. w0: beam waist; b: depth of focus; zR: Rayleigh range; $\Theta$: total angular spread
For a Gaussian beam propagating in free space, the spot size (radius) w(z) will be at a minimum value w0 at one place along the beam axis, known as the beam waist. For a beam of wavelength λ at a distance z along the beam from the beam waist, the variation of the spot size is given by[1]
$w(z) = w_0 \, \sqrt{ 1+ {\left( \frac{z}{z_\mathrm{R}} \right)}^2 } \ .$
where the origin of the z-axis is defined, without loss of generality, to coincide with the beam waist, and where[1]
$z_\mathrm{R} = \frac{\pi w_0^2}{\lambda}$
is called the Rayleigh range.
### Rayleigh range and confocal parameter
At a distance from the waist equal to the Rayleigh range zR, the width w of the beam is[1]
$w(\pm z_\mathrm{R}) = \sqrt{2} w_0.$
The distance between these two points is called the confocal parameter or depth of focus of the beam:
$b = 2 z_\mathrm{R} = \frac{2 \pi w_0^2}{\lambda}\,.$
### Radius of curvature
R(z) is the radius of curvature of the wavefronts comprising the beam. Its value as a function of position is[1]
$R(z) = z \left[{ 1+ {\left( \frac{z_\mathrm{R}}{z} \right)}^2 } \right] \ .$
### Beam divergence
The parameter $w(z)$ increases linearly with $z$ for $z \gg z_\mathrm{R}$. This means that far from the waist, the beam is cone-shaped. The angle between the straight line $r=w(z)$ and the central axis of the beam ($r=0$) is called the divergence of the beam. It is given by[1]
$\theta \simeq \frac{\lambda}{\pi w_0} \qquad (\theta \mathrm{\ in\ radians}).$
The total angular spread of the beam far from the waist is then given by
$\Theta = 2 \theta\ .$
Because the divergence is inversely proportional to the spot size, a Gaussian beam that is focused to a small spot spreads out rapidly as it propagates away from that spot. To keep a laser beam very well collimated, it must have a large diameter. This relationship between beam width and divergence is due to diffraction. Non-Gaussian beams also exhibit this effect, but a Gaussian beam is a special case where the product of width and divergence is the smallest possible.
Since the gaussian beam model uses the paraxial approximation, it fails when wavefronts are tilted by more than about 30° from the direction of propagation.[4] From the above expression for divergence, this means the Gaussian beam model is valid only for beams with waists larger than about $2\lambda/\pi$.
Laser beam quality is quantified by the beam parameter product (BPP). For a Gaussian beam, the BPP is the product of the beam's divergence and waist size $w_0$. The BPP of a real beam is obtained by measuring the beam's minimum diameter and far-field divergence, and taking their product. The ratio of the BPP of the real beam to that of an ideal Gaussian beam at the same wavelength is known as M2 ("M squared"). The M2 for a Gaussian beam is one. All real laser beams have M2 values greater than one, although very high quality beams can have values very close to one.
### Gouy phase
The longitudinal phase delay or Gouy phase of the beam is[1]
$\zeta(z) = \arctan \left( \frac{z}{z_\mathrm{R}} \right) \ .$
The Gouy phase indicates that as a Gaussian beam passes through a focus, it acquires an additional phase shift of π, in addition to the usual $e^{-ikz}$ phase shift that would be expected from a plane wave.[1]
### Complex beam parameter
Main article: Complex beam parameter
Information about the spot size and radius of curvature of a Gaussian beam can be encoded in the complex beam parameter, $q(z)$:[5]
$q(z) = z + q_0 = z + iz_\mathrm{R} \ .$
The reciprocal $1/q(z)$ shows the relationship between $q(z)$, $w(z)$, and $R(z)$ explicitly:[5]
${ 1 \over q(z) } = { 1 \over z + iz_\mathrm{R} } = { z \over z^2 + z_\mathrm{R}^2 } - i { z_\mathrm{R} \over z^2 + z_\mathrm{R}^2 } = {1 \over R(z) } - i { \lambda \over \pi w^2(z) }.$
The complex beam parameter plays a key role in the analysis of Gaussian beam propagation, and especially in the analysis of optical resonator cavities using ray transfer matrices.
In terms of the complex beam parameter ${q}$, a Gaussian field with one transverse dimension is proportional to
${u}(x,z) = \frac{1}{\sqrt{{q}_x(z)}} \exp\left(-i k \frac{x^2}{2 {q}_x(z)}\right).$
In two dimensions one can write the potentially elliptical or astigmatic beam as the product
${u}(x,y,z) = {u}(x,z)\, {u}(y,z),$
which for the common case of circular symmetry where ${q}_x = {q}_y = {q}$ and $x^2 + y^2 = r^2$ yields[6]
${u}(r,z) = \frac{1}{{q}(z)}\exp\left( -i k\frac{r^2}{2 {q}(z)}\right).$
## Power and intensity
### Power through an aperture
The power P passing through a circle of radius r in the transverse plane at position z is
$P(r,z) = P_0 \left[ 1 - e^{-2r^2 / w^2(z)} \right]\ ,$
where
$P_0 = { 1 \over 2 } \pi I_0 w_0^2$
is the total power transmitted by the beam.
For a circle of radius $r = w(z) \,$, the fraction of power transmitted through the circle is
${ P(z) \over P_0 } = 1 - e^{-2} \approx 0.865\ .$
Similarly, about 95 percent of the beam's power will flow through a circle of radius $r = 1.224\cdot w(z) \,$.
### Peak and average intensity
The peak intensity at an axial distance $z$ from the beam waist is calculated using L'Hôpital's rule as the limit of the enclosed power within a circle of radius $r$, divided by the area of the circle $\pi r^2$:
$I(0,z) =\lim_{r\to 0} \frac {P_0 \left[ 1 - e^{-2r^2 / w^2(z)} \right]} {\pi r^2} = \frac{P_0}{\pi} \lim_{r\to 0} \frac { \left[ -(-2)(2r) e^{-2r^2 / w^2(z)} \right]} {w^2(z)(2r)} = {2P_0 \over \pi w^2(z)}.$
The peak intensity is thus exactly twice the average intensity, obtained by dividing the total power by the area within the radius $w(z)$.
## Derivation
The Gaussian beam formalism begins with the wave equation for an electromagnetic field in free space or in a homogeneous dielectric medium:[7]
$\nabla^2 U = \frac{1}{c^2} \frac{\partial^2 U}{\partial t^2},$
where $U$ may stand for any one of the six field components $E_x$, $E_y$, $E_z$, $B_x$, $B_y$, or $B_z$. The Gaussian beam formalism proceeds by writing down a solution of the form[7]
$U(x,y,z,t) = u(x,y,z) e^{-i(kz-\omega t)},$
where it is assumed that the beam is sufficiently collimated along the $z$ axis that $\partial^2 u/\partial z^2$ may be neglected. Substituting this solution into the wave equation above yields the paraxial approximation to the wave equation:[7]
$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 2ik \frac{\partial u}{\partial z}.$
Solving this differential equation yields an infinite set of functions, of which the Gaussian beam is the lowest-order solution or mode.
## Higher-order modes
See also: Transverse mode
Gaussian beams are just one possible solution to the paraxial wave equation. Various other sets of orthogonal solutions are used for modelling laser beams. In the general case, if a complete basis set of solutions is chosen, any real laser beam can be described as a superposition of solutions from this set. The design of the laser determines which basis set of solutions is most useful. In some cases the output of a laser may closely approximate a single higher-order mode. Hermite-Gaussian modes are particularly common, since many laser systems have Cartesian reflection symmetry in the plane perpendicular to the beam's propagation direction.
### Hermite-Gaussian modes
Twelve Hermite-Gaussian modes
Hermite-Gaussian modes are a convenient description for the output of lasers whose cavity design is not radially symmetric, but rather has a distinction between horizontal and vertical. In terms of the previously defined complex $q$ parameter, the amplitude distribution in the $x$-plane is proportional to
${u}_n(x,z) = \left(\frac{2}{\pi}\right)^{1/4} \left(\frac{1}{2^n n! w_0}\right)^{1/2} \left( \frac{{q}_0}{{q}(z)}\right)^{1/2} \left[\frac{{q}_0}{{q}_0^\ast} \frac{{q}^\ast(z)}{{q}(z)}\right]^{n/2} H_n\left(\frac{\sqrt{2}x}{w(z)}\right) \exp\left[-i \frac{k x^2}{2 {q}(z)}\right]$
where the function $H_n(x)$ is the Hermite polynomial of order $n$ (physicists' form, i.e. $H_1(x)=2x\,$), and the asterisk indicates complex conjugation. For the case $n=0$ the equation yields a Gaussian transverse distribution.
For two-dimensional rectangular coordinates one constructs a function ${u}_{mn}(x,y,z)=u_m(x,z) u_n(y,z)$, where $u_n(y,z)$ has the same form as $u_m(x,z)$. Mathematically this property is due to the separation of variables applied to the paraxial Helmholtz equation for Cartesian coordinates.[8]
Hermite-Gaussian modes are typically designated "TEMmn", where m and n are the polynomial indices in the x and y directions. A Gaussian beam is thus TEM00.
### Laguerre-Gaussian modes
The intensity profiles of twelve Laguerre-Gaussian modes
If the problem is cylindrically symmetric, the natural solutions of the paraxial wave equation are Laguerre-Gaussian modes. They are written in cylindrical coordinates using Laguerre polynomials
${u}(r,\phi,z)=\frac{C^{LG}_{lp}}{w(z)}\left(\frac{r \sqrt{2}}{w(z)}\right)^{|l|}\exp\left(-\frac{r^2}{w^2(z)}\right)L_p^{|l|} \left(\frac{2r^2}{w^2(z)}\right) \exp\left( i k \frac{r^2}{2 R(z)}\right)\exp(i l \phi)\exp\left[-i(2p+|l|+1)\zeta(z)\right],$
where $L_p^l$ are the generalized Laguerre polynomials, the radial index $p\ge 0$ and the azimuthal index is $l$. $C^{LG}_{lp}$ is an appropriate normalization constant; $w(z)$, $R(z)$ and $\zeta(z)$ are beam parameters defined above.
### Ince-Gaussian modes
In elliptic coordinates, one can write the higher-order modes using Ince polynomials. The even and odd Ince-Gaussian modes are given by [9]
$u_\varepsilon \left( \xi ,\eta ,z\right) = \frac{w_{0}}{w\left( z\right) }\mathrm{C}_{p}^{m}\left( i\xi ,\varepsilon \right) \mathrm{C} _{p}^{m}\left( \eta ,\varepsilon \right) \exp \left[ -ik\frac{r^{2}}{ 2q\left( z\right) }-\left( p+1\right) \psi _{GS}\left( z\right) \right] ,$
where $\xi$ and $\eta$ are the radial and angular elliptic coordinates defined by
$x = \sqrt{\varepsilon /2}w\left( z\right) \cosh \xi \cos \eta ,$
$y = \sqrt{\varepsilon /2}w\left( z\right) \sinh \xi \sin \eta.$
${C}_{p}^{m}\left( \eta ,\epsilon \right)$ are the even Ince polynomials of order $p$ and degree $m$, $\varepsilon$ is the ellipticity parameter, and $\psi _{GS}\left( z\right) =\arctan \left( z/z_\mathrm{R}\right)$ is the Gouy phase. The Hermite-Gaussian and Laguerre-Gaussian modes are a special case of the Ince-Gaussian modes for $\varepsilon=\infty$ and $\varepsilon=0$ respectively.
### Hypergeometric-Gaussian modes
There is another important class of paraxial wave modes in polar coordinates in which the complex amplitude is proportional to a confluent hypergeometric function.
These modes have a singular phase profile and are eigenfunctions of the photon orbital angular momentum. The intensity profile is characterized by a single brilliant ring with a singularity at its center, where the field amplitude vanishes.[10]
$u_{pm}(\rho,\theta;\zeta)= \sqrt{\frac{2^{p+|m|+1}}{\pi\Gamma(p+|m|+1)}} \frac{\Gamma(1+|m|+\frac{p}{2})}{\Gamma(|m|+1)} \,\,i^{|m|+1}\zeta^{\frac{p}{2}}(\zeta+i)^{-(1+|m|+\frac{p}{2})}\rho^{|m|}e^{-\frac{i\rho^2}{(\zeta+i)}}e^{im\phi}{}_{1}F_{1}\left(-\frac{p}{2}, |m|+1;\frac{r^2}{\zeta(\zeta+i)}\right),$
where $m$ is integer, $p\ge-|m|$ is real valued, $\Gamma(x)$ is the gamma function and ${}_{1}F_{1}(a,b;x)$ is a confluent hypergeometric function.
Some subfamilies of hypergeometric-Gaussian (HyGG) modes can be listed as the modified Bessel-Gaussian modes, the modified exponential Gaussian modes, and the modified Laguerre–Gaussian modes.
The set of hypergeometric-Gaussian modes is overcomplete and is not an orthogonal set of modes. In spite of its complicated field profile, HyGG modes have a very simple profile at the pupil plane:
$u(\rho,\phi,0) \propto \rho^{p+|m|}e^{-\rho^2+im\phi}.$
See Optical vortex, which explains that the outcoming wave from a pitch-fork hologram is a sub-family of HyGG modes. The HyGG profile while beam propagates along $\zeta$ has a dramatic change and it is not a stable mode below the Rayleigh range.
## Notes
1. Svelto, pp. 153–5.
2. Siegman, p. 642.
3. Svelto, p. 158.
4. Siegman (1986) p. 630.
5. ^ a b Siegman, pp. 638–40.
6. See Siegman (1986) p. 639. Eq. 29
7. ^ a b c Svelto, pp. 148–9.
8. Siegman (1986), p645, eq. 54
9. Bandres and Gutierrez-Vega (2004)
10. Karimi et. al (2007)
## References
• Saleh, Bahaa E. A. and Teich, Malvin Carl (1991). Fundamentals of Photonics. New York: John Wiley & Sons. ISBN 0-471-83965-5. Chapter 3, "Beam Optics," pp. 80–107.
• Mandel, Leonard and Wolf, Emil (1995). Optical Coherence and Quantum Optics. Cambridge: Cambridge University Press. ISBN 0-521-41711-2. Chapter 5, "Optical Beams," pp. 267.
• Siegman, Anthony E. (1986). Lasers. University Science Books. ISBN 0-935702-11-3. Chapter 16.
• Svelto, Orazio (2010). Principles of Lasers (5th ed.).
• Yariv, Amnon (1989). Quantum Electronics (3rd ed.). Wiley. ISBN 0-471-60997-8.
• F. Pampaloni and J. Enderlein (2004). "Gaussian, Hermite-Gaussian, and Laguerre-Gaussian beams: A primer". arXiv:physics/0410021 [physics.optics].
• Miguel A. Bandres and Julio C. Gutierrez-Vega (2004). "Ince Gaussian beams". Opt. Lett. (OSA) 29 (2): 144–146. Bibcode:2004OptL...29..144B. doi:10.1364/OL.29.000144. PMID 14743992. More than one of `|number=` and `|issue=` specified (help)
• E. Karimi, G. Zito, B. Piccirillo, L. Marrucci, and E. Santamato (2007). "Hypergeometric-Gaussian beams". Opt. Lett. (OSA) 32 (21): 3053–3055. arXiv:0712.0782. Bibcode:2007OptL...32.3053K. doi:10.1364/OL.32.003053. PMID 17975594. More than one of `|number=` and `|issue=` specified (help)
• Gaussian Beam Propagation - CVI Melles Griot Technical Guide
• Gaussian Beam Optics Tutorial, Newport
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http://mathoverflow.net/questions/39620?sort=oldest
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## Morphisms of a simple sheaf over an algebra to its double dual
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Given a smooth and projective surface $S$ over an algebraically closed field $k$ and a sheaf of Azumaya algebras $R$, i.e. $R$ is a locally free $O_S$-module of finite rank. Let $M$ be a coherent and torsion free $O_S$-module, which is also a left $R$-module, such that generically $M_\eta$ is a simple $R_\eta$-module. Then we have $Hom_R(M,M)=k$.
Now $M^*:=Hom_{O_S}(M,O_S)$ is a right $R$-module and $M^{**}$ is a left $R$-module. We have the canonical map $\iota: M \rightarrow M^{**}$.
Is it true that $Hom_R(M,M^{**})$ just consists of the muliples of $\iota$, i.e. is it a one dimensional $k$-vector space?
I tried to use the sequence $0\rightarrow M\rightarrow M^{**} \rightarrow Q\rightarrow 0$. Since $M$ is torsion free $Q$ has support in codimension 2. Then apply $Hom_R(M, - )$, which is left exact, so we get, with $Hom_R(M,M)=k$: $0\rightarrow k\rightarrow Hom_R(M,M^{**}) \rightarrow Hom_R(M,Q)$. But here i am stuck.
Or is this assertion wrong, i.e. are there more morphisms? If it is right, can it be generalized to a bigger class of algebras $R$?
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1
It is better to apply $Hom_R(-,M^{**})$. Since $Q$ is in codimension 2 one has $Hom(Q,M^{**}) = Ext^1(Q,M^{**}) = 0$, so $Hom(M,M^{**}) = Hom(M^{**},M^{**})$. – Sasha Sep 22 2010 at 15:36
Okay, i see this long exact sequence. But why do these groups vanish? Just because $Q$ live in codimension 2? I cannot see this. We still have all $H^0$ groups, e.g. $H^0(\mathcal{E}xt^1 (Q,M^{**}))$ which shows up in the local to global spectral sequence for $Ext^1$. Or am i missing the point here? – TonyS Sep 22 2010 at 16:58
1
Local $Hom$ and local $Ext^1$ vanish because $Q$ lives in codimension 2. Then local-to-global spectral sequence shows that global $Hom$ and $Ext^1$ vanish as well. – Sasha Sep 22 2010 at 17:29
1
It may be not too obvious, but it is a standard fact. Usually it is proved using the notion of depth e.t.c. But an easy way to explain this is the following. First, it is clear that local $Hom$ is supported at the support of $Q$. On the other hand, local $Hom$ is torsion free. Hence it is zero. Now locally we can choose a pair of Cartier divisors $D_1,D_2$ such that $Q$ is supported on $Z = D_1 \cap D_2$ (scheme theoretically) and $codim Z = 2$. Further, locally we can choose a surjection $O_D^n \to Q$. Let $Q′$ be the kernel. Then local $Hom$ for $Q′$ vanishes by the same reason. – Sasha Sep 22 2010 at 18:25
1
Sorry, $O_D$ should be $O_Z$. Continue. Hence local $Ext^1$ for $Q$ injects into local $Ext^1$ for $O_Z^n$. The latter can be computed explicitly using the Koszul resolution $$0 \to O(-D_1 - D_2) \to O(-D_1) \oplus O(-D_2) \to O \to O_Z \to 0.$$ The result is zero. – Sasha Sep 22 2010 at 18:27
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## 1 Answer
Any $R$-homomorphism (in fact any $\mathcal O_S$-homomorphism) `$M \to M^{**}$` extends to a morphism `$M^{**}\to M^{**}$` (as $M$ is locally free in codimension $1$ and `$M^{**}$` is the maximal extension from outside codimension $2$. This gives what you want. as `$Hom_R(M^{**},M^{**})=k$` for the same reason as it is true of $M$.
-
Thanks for your answer, i have some questions: what does extension mean here? Something like: given $f: M\rightarrow M^{**}$, there is a unique $g: M^{**} \rightarrow M^{**}$ with $f=g \iota$. Since $End_R(M^{**})=k$ $g$ must be a multiple of the identity and so f is multiple of $\iota$? How do I find this g exactly? – TonyS Sep 22 2010 at 15:54
1
Your interpretation is correct. You can find $g$ either by finding a codimension $2$ subset outside of which $\iota$ is an isomorphism and then note that (local) sections of `$M^{**}$` defined outside of a codimension $2$ subset extend over the subset (algebraic version of Hartog's theorem) or you can look at `$g^{**}\colon M^{**}\to M^{****}=M^{**}$`. – Torsten Ekedahl Sep 22 2010 at 16:23
So we actually don't need $R$ to be an Azumaya algebra. Nice! – TonyS Sep 22 2010 at 16:54
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http://mathoverflow.net/questions/tagged/inequalities
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## Tagged Questions
3answers
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### Minimizing/Maximizing the tail of the convex combinations of Chi Squared i.i.d random variables
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http://en.wikipedia.org/wiki/Multivariate_Gaussian
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# Multivariate normal distribution
(Redirected from Multivariate Gaussian)
"MVN" redirects here. For the airport with that IATA code, see Mount Vernon Airport.
Notation
Probability density function Many samples from a multivariate normal distribution, shown along with the 3-sigma ellipse, the two marginal distributions, and the two 1-d histograms.
$\mathcal{N}(\boldsymbol\mu,\,\boldsymbol\Sigma)$
μ ∈ Rk — location Σ ∈ Rk×k — covariance (nonnegative-definite matrix)
x ∈ μ+span(Σ) ⊆ Rk
$|(2\pi)^{k}\boldsymbol\Sigma|^{-\frac{1}{2}}\, e^{ -\frac{1}{2}(\mathbf{x}-\boldsymbol\mu)'\boldsymbol\Sigma^{-1}(\mathbf{x}-\boldsymbol\mu) },$ exists only when Σ is positive-definite
(no analytic expression)
μ
μ
Σ
$\frac{1}{2}\ln |2\pi e\boldsymbol\Sigma|$
$\exp\!\Big( \boldsymbol\mu'\mathbf{t} + \tfrac{1}{2} \mathbf{t}'\boldsymbol\Sigma \mathbf{t}\Big)$
$\exp\!\Big( i\boldsymbol\mu'\mathbf{t} - \tfrac{1}{2} \mathbf{t}'\boldsymbol\Sigma \mathbf{t}\Big)$
In probability theory and statistics, the multivariate normal distribution or multivariate Gaussian distribution, is a generalization of the one-dimensional (univariate) normal distribution to higher dimensions. One possible definition is that a random vector is said to be k-variate normally distributed if every linear combination of its k components has a univariate normal distribution. However, its importance derives mainly from the multivariate central limit theorem. The multivariate normal distribution is often used to describe, at least approximately, any set of (possibly) correlated real-valued random variables each of which clusters around a mean value.
## Notation and parametrization
The multivariate normal distribution of a k-dimensional random vector x = [X1, X2, …, Xk] can be written in the following notation:
$\mathbf{x}\ \sim\ \mathcal{N}(\boldsymbol\mu,\, \boldsymbol\Sigma),$
or to make it explicitly known that X is k-dimensional,
$\mathbf{x}\ \sim\ \mathcal{N}_k(\boldsymbol\mu,\, \boldsymbol\Sigma).$
with k-dimensional mean vector
$\boldsymbol\mu = [ \operatorname{E}[X_1], \operatorname{E}[X_2], \ldots, \operatorname{E}[X_k]]$
and k x k covariance matrix
$\boldsymbol\Sigma = [\operatorname{Cov}[X_i, X_j]], i=1,2,\ldots,k; j=1,2,\ldots,k$
## Definition
A random vector x = (X1, …, Xk)' is said to have the multivariate normal distribution if it satisfies the following equivalent conditions.[1]
• Every linear combination of its components Y = a1X1 + … + akXk is normally distributed. That is, for any constant vector a ∈ Rk, the random variable Y = a′x has a univariate normal distribution.
• There exists a random ℓ-vector z, whose components are independent standard normal random variables, a k-vector μ, and a k×ℓ matrix A, such that x = Az + μ. Here ℓ is the rank of the covariance matrix Σ = AA′. Especially in the case of full rank, see the section below on Geometric interpretation.
• There is a k-vector μ and a symmetric, nonnegative-definite k×k matrix Σ, such that the characteristic function of x is
$\varphi_\mathbf{x}(\mathbf{u}) = \exp\Big( i\mathbf{u}'\boldsymbol\mu - \tfrac{1}{2} \mathbf{u}'\boldsymbol\Sigma \mathbf{u} \Big).$
The covariance matrix is allowed to be singular (in which case the corresponding distribution has no density). This case arises frequently in statistics; for example, in the distribution of the vector of residuals in the ordinary least squares regression. Note also that the Xi are in general not independent; they can be seen as the result of applying the matrix A to a collection of independent Gaussian variables z.
## Properties
### Density function
#### Non-degenerate case
The multivariate normal distribution is said to be "non-degenerate" when the covariance matrix $\boldsymbol\Sigma$ of the multivariate normal distribution is symmetric and positive definite. In this case the distribution has density
$f_{\mathbf x}(x_1,\ldots,x_k) = \frac{1}{\sqrt{(2\pi)^k|\boldsymbol\Sigma|}} \exp\left(-\frac{1}{2}({\mathbf x}-{\boldsymbol\mu})^T{\boldsymbol\Sigma}^{-1}({\mathbf x}-{\boldsymbol\mu}) \right),$
where $|\boldsymbol\Sigma|$ is the determinant of $\boldsymbol\Sigma$. Note how the equation above reduces to that of the univariate normal distribution if $\boldsymbol\Sigma$ is a $1 \times 1$ matrix (i.e. a real number).
Bivariate case
In the 2-dimensional nonsingular case (k = rank(Σ) = 2), the probability density function of a vector [X Y]′ is
$f(x,y) = \frac{1}{2 \pi \sigma_x \sigma_y \sqrt{1-\rho^2}} \exp\left( -\frac{1}{2(1-\rho^2)}\left[ \frac{(x-\mu_x)^2}{\sigma_x^2} + \frac{(y-\mu_y)^2}{\sigma_y^2} - \frac{2\rho(x-\mu_x)(y-\mu_y)}{\sigma_x \sigma_y} \right] \right),$
where ρ is the correlation between X and Y and where $\sigma_x>0$ and $\sigma_y>0$. In this case,
$\boldsymbol\mu = \begin{pmatrix} \mu_x \\ \mu_y \end{pmatrix}, \quad \boldsymbol\Sigma = \begin{pmatrix} \sigma_x^2 & \rho \sigma_x \sigma_y \\ \rho \sigma_x \sigma_y & \sigma_y^2 \end{pmatrix}.$
In the bivariate case, we also have a theorem that makes the first equivalent condition for multivariate normality less restrictive: it is sufficient to verify that countably many distinct linear combinations of X and Y are normal in order to conclude that the vector [X Y]′ is bivariate normal.[2]
When plotted in the x,y-plane the distribution appears to be squeezed to the line:
$y\left( x \right) = {\mathop{\rm sgn}} \left( {{\rho }} \right)\frac{{{\sigma _y}}}{{{\sigma _x}}}\left( {x - {\mu _x}} \right) + {\mu _y}$
as the correlation parameter ρ increases. This is because the above expression is the best linear unbiased prediction of Y given a value of X.[3]
#### Degenerate case
If the covariance matrix $\boldsymbol\Sigma$ is not full rank, then the multivariate normal distribution is degenerate and does not have a density. More precisely, it does not have a density with respect to k-dimensional Lebesgue measure (which is the usual measure assumed in calculus-level probability courses). Only random vectors whose distributions are absolutely continuous with respect to a measure are said to have densities (with respect to that measure). To talk about densities but avoid dealing with measure-theoretic complications it can be simpler to restrict attention to a subset of $\text{rank}(\boldsymbol\Sigma)$ of the coordinates of $\mathbf{x}$ such that covariance matrix for this subset is positive definite; then the other coordinates may be thought of as an affine function of the selected coordinates.[citation needed]
To talk about densities meaningfully in the singular case, then, we must select a different base measure. Using the disintegration theorem we can define a restriction of Lebesgue measure to the $\text{rank}(\boldsymbol\Sigma)$-dimensional affine subspace of $\mathbb{R}^k$ where the Gaussian distribution is supported, i.e. $\{\boldsymbol\mu+\boldsymbol{\Sigma ^{1/2}}\mathbf{v} : \mathbf{v} \in \mathbb{R}^k \}$. With respect to this probability measure the distribution has density:
$f(\mathbf{x})=(\text{det}^*(2\pi\boldsymbol\Sigma))^{-\frac{1}{2}}\, e^{ -\frac{1}{2}(\mathbf{x}-\boldsymbol\mu)'\boldsymbol\Sigma^+(\mathbf{x}-\boldsymbol\mu) }$
where $\boldsymbol\Sigma^+$ is the generalized inverse and det* is the pseudo-determinant.[4]
### Higher moments
Main article: Isserlis’ theorem
The kth-order moments of x are defined by
$\mu _{1,\dots,N}(\mathbf{x})\ \stackrel{\mathrm{def}}{=}\ \mu _{r_{1},\dots,r_{N}}(\mathbf{x})\ \stackrel{\mathrm{def}}{=}\ E\left[ \prod\limits_{j=1}^{N}x_j^{r_{j}}\right]$
where r1 + r2 + ⋯ + rN = k.
The central k-order central moments are given as follows
(a) If k is odd, μ1, …, N(x − μ) = 0.
(b) If k is even with k = 2λ, then
$\mu _{1,\dots,2\lambda }(\mathbf{x}-\boldsymbol\mu )=\sum \left( \Sigma _{ij}\Sigma _{k\ell}\cdots\Sigma _{XZ}\right)$
where the sum is taken over all allocations of the set $\left\{ 1,\dots,2\lambda \right\}$ into λ (unordered) pairs. That is, if you have a kth ( = 2λ = 6) central moment, you will be summing the products of λ = 3 covariances (the -μ notation has been dropped in the interests of parsimony):
$\begin{align} & {} E[x_1 x_2 x_3 x_4 x_5 x_6] \\ &{} = E[x_1 x_2 ]E[x_3 x_4 ]E[x_5 x_6 ] + E[x_1 x_2 ]E[x_3 x_5 ]E[x_4 x_6] + E[x_1 x_2 ]E[x_3 x_6 ]E[x_4 x_5] \\ &{} + E[x_1 x_3 ]E[x_2 x_4 ]E[x_5 x_6 ] + E[x_1 x_3 ]E[x_2 x_5 ]E[x_4 x_6 ] + E[x_1 x_3]E[x_2 x_6]E[x_4 x_5] \\ &+ E[x_1 x_4]E[x_2 x_3]E[x_5 x_6]+E[x_1 x_4]E[x_2 x_5]E[x_3 x_6]+E[x_1 x_4]E[x_2 x_6]E[x_3 x_5] \\ & + E[x_1 x_5]E[x_2 x_3]E[x_4 x_6]+E[x_1 x_5]E[x_2 x_4]E[x_3 x_6]+E[x_1 x_5]E[x_2 x_6]E[x_3 x_4] \\ & + E[x_1 x_6]E[x_2 x_3]E[x_4 x_5 ] + E[x_1 x_6]E[x_2 x_4 ]E[x_3 x_5] + E[x_1 x_6]E[x_2 x_5]E[x_3 x_4]. \end{align}$
This yields $(2\lambda -1)!/(2^{\lambda -1}(\lambda -1)!)$ terms in the sum (15 in the above case), each being the product of λ (in this case 3) covariances. For fourth order moments (four variables) there are three terms. For sixth-order moments there are 3 × 5 = 15 terms, and for eighth-order moments there are 3 × 5 × 7 = 105 terms.
The covariances are then determined by replacing the terms of the list $\left[ 1,\dots,2\lambda \right]$ by the corresponding terms of the list consisting of r1 ones, then r2 twos, etc.. To illustrate this, examine the following 4th-order central moment case:
$E\left[ x_i^4\right] = 3\Sigma _{ii}^2$
$E\left[ x_i^3 x_j\right] = 3\Sigma _{ii} \Sigma _{ij}$
$E\left[ x_i^2 x_j^2\right] = \Sigma _{ii}\Sigma_{jj}+2\left( \Sigma _{ij}\right) ^2$
$E\left[ x_i^2x_jx_k\right] = \Sigma _{ii}\Sigma _{jk}+2\Sigma _{ij}\Sigma _{ik}$
$E\left[ x_i x_j x_k x_n\right] = \Sigma _{ij}\Sigma _{kn}+\Sigma _{ik}\Sigma _{jn}+\Sigma _{in}\Sigma _{jk}.$
where $\Sigma_{ij}$ is the covariance of xi and xj. The idea with the above method is you first find the general case for a kth moment where you have k different x variables - $E\left[ x_i x_j x_k x_n\right]$ and then you can simplify this accordingly. Say, you have $E\left[ x_i^2 x_k x_n\right]$ then you simply let xi = xj and realise that $\Sigma _{ii}$ = σi2.
### Likelihood function
If the mean and variance matrix are unknown, a suitable log likelihood function for a single observation x would be:[citation needed]
$\ln(L)= -\frac{1}{2} \ln |2\pi^k\boldsymbol\Sigma|\, -\frac{1}{2}(\mathbf{x}-\boldsymbol\mu)^{\rm T}\boldsymbol\Sigma^{-1}(\mathbf{x}-\boldsymbol\mu)$
where x is a vector of real numbers. The complex case, where z is a vector of complex numbers, would be: $2\ln(L) = -\ln |2\pi\boldsymbol\Sigma|\, -(\mathbf{z}-\boldsymbol\mu)^\dagger\boldsymbol\Sigma^{-1}(\mathbf{z}-\boldsymbol\mu)$.
A similar notation is used for multiple linear regression.[5]
### Entropy
The differential entropy of the multivariate normal distribution is[6]
$\begin{align} h\left(f\right) & = -\int_{-\infty}^\infty \int_{-\infty}^\infty \cdots\int_{-\infty}^\infty f(\mathbf{x}) \ln f(\mathbf{x})\,d\mathbf{x},\\ & = \frac12 \ln\left|2\pi e \boldsymbol\Sigma \right|.\\ \end{align}$
### Kullback–Leibler divergence
The Kullback–Leibler divergence from $\mathcal{N}_0(\boldsymbol\mu_0, \boldsymbol\Sigma_0)$ to $\mathcal{N}_1(\boldsymbol\mu_1, \boldsymbol\Sigma_1)$, for non-singular matrices Σ0 and Σ1, is:[7]
$D_\text{KL}(\mathcal{N}_0 \| \mathcal{N}_1) = { 1 \over 2 } \left\{ \mathrm{tr} \left( \boldsymbol\Sigma_1^{-1} \boldsymbol\Sigma_0 \right) + \left( \boldsymbol\mu_1 - \boldsymbol\mu_0\right)^{\rm T} \boldsymbol\Sigma_1^{-1} ( \boldsymbol\mu_1 - \boldsymbol\mu_0 ) -\ln { | e \boldsymbol \Sigma_0 | \over | \boldsymbol\Sigma_1 | } \right\}.$
The logarithm must be taken to base e since the two terms following the logarithm are themselves base-e logarithms of expressions that are either factors of the density function or otherwise arise naturally. The equation therefore gives a result measured in nats. Dividing the entire expression above by loge 2 yields the divergence in bits.
### Cumulative distribution function
The cumulative distribution function (cdf) F(x0) of a random vector x is defined as the probability that all components of x are less than or equal to the corresponding values in the vector x0. Though there is no closed form for F(x), there are a number of algorithms that estimate it numerically.
### Tolerance region
The equivalent for the univariate Normal distribution tolerance intervals in the multivariate case would yield a tolerance region. Such region consists of those vectors x satisfying
$({\mathbf x}-{\boldsymbol\mu})^T{\boldsymbol\Sigma}^{-1}({\mathbf x}-{\boldsymbol\mu}) \leq \chi^2_k(p).$
Here ${\mathbf x}$ is a $k$-dimensional vector, ${\boldsymbol\mu}$ is the known $k$-dimensional mean vector, $\boldsymbol\Sigma$ is the known covariance matrix and $\chi^2_k(p)$ is the quantile function for probability $p$ of the chi-squared distribution with $k$ degrees of freedom.
When $k = 2,$ the expression defines the interior of an ellipse and the chi-squared distribution simplifies to an exponential distribution with mean equal to two.
## Joint normality
### Normally distributed and independent
If X and Y are normally distributed and independent, this implies they are "jointly normally distributed", i.e., the pair (X, Y) must have multivariate normal distribution. However, a pair of jointly normally distributed variables need not be independent (would only be so if uncorrelated, $\rho = 0$ ).
### Two normally distributed random variables need not be jointly bivariate normal
The fact that two random variables X and Y both have a normal distribution does not imply that the pair (X, Y) has a joint normal distribution. A simple example is one in which X has a normal distribution with expected value 0 and variance 1, and Y = X if |X| > c and Y = −X if |X| < c, where c > 0. There are similar counterexamples for more than two random variables.[citation needed] In general, they sum to a mixture model.
### Correlations and independence
In general, random variables may be uncorrelated but highly dependent. But if a random vector has a multivariate normal distribution then any two or more of its components that are uncorrelated are independent. This implies that any two or more of its components that are pairwise independent are independent.
But it is not true that two random variables that are (separately, marginally) normally distributed and uncorrelated are independent. Two random variables that are normally distributed may fail to be jointly normally distributed, i.e., the vector whose components they are may fail to have a multivariate normal distribution. For an example of two normally distributed random variables that are uncorrelated but not independent, see normally distributed and uncorrelated does not imply independent.
## Conditional distributions
If μ and Σ are partitioned as follows
$\boldsymbol\mu = \begin{bmatrix} \boldsymbol\mu_1 \\ \boldsymbol\mu_2 \end{bmatrix} \quad$ with sizes $\begin{bmatrix} q \times 1 \\ (N-q) \times 1 \end{bmatrix}$
$\boldsymbol\Sigma = \begin{bmatrix} \boldsymbol\Sigma_{11} & \boldsymbol\Sigma_{12} \\ \boldsymbol\Sigma_{21} & \boldsymbol\Sigma_{22} \end{bmatrix} \quad$ with sizes $\begin{bmatrix} q \times q & q \times (N-q) \\ (N-q) \times q & (N-q) \times (N-q) \end{bmatrix}$
then, the distribution of x1 conditional on x2 = a is multivariate normal (x1|x2 = a) ~ N(μ, Σ) where
$\bar{\boldsymbol\mu} = \boldsymbol\mu_1 + \boldsymbol\Sigma_{12} \boldsymbol\Sigma_{22}^{-1} \left( \mathbf{a} - \boldsymbol\mu_2 \right)$
and covariance matrix
$\overline{\boldsymbol\Sigma} = \boldsymbol\Sigma_{11} - \boldsymbol\Sigma_{12} \boldsymbol\Sigma_{22}^{-1} \boldsymbol\Sigma_{21}.$[8]
This matrix is the Schur complement of Σ22 in Σ. This means that to calculate the conditional covariance matrix, one inverts the overall covariance matrix, drops the rows and columns corresponding to the variables being conditioned upon, and then inverts back to get the conditional covariance matrix. Here $\boldsymbol\Sigma_{22}^{-1}$ is the generalized inverse of $\boldsymbol\Sigma_{22}$.
Note that knowing that x2 = a alters the variance, though the new variance does not depend on the specific value of a; perhaps more surprisingly, the mean is shifted by $\boldsymbol\Sigma_{12} \boldsymbol\Sigma_{22}^{-1} \left(\mathbf{a} - \boldsymbol\mu_2 \right)$; compare this with the situation of not knowing the value of a, in which case x1 would have distribution $\mathcal{N}_q \left(\boldsymbol\mu_1, \boldsymbol\Sigma_{11} \right)$.
An interesting fact derived in order to prove this result, is that the random vectors $\mathbf{x}_2$ and $\mathbf{y}_1=\mathbf{x}_1-\boldsymbol\Sigma_{12}\boldsymbol\Sigma_{22}^{-1}\mathbf{x}_2$ are independent.
The matrix Σ12Σ22−1 is known as the matrix of regression coefficients.
In the bivariate case where x is partitioned into X1 and X2, the conditional distribution of X1 given X2 is[9]
$X_1|X_2=a \ \sim\ \mathcal{N}\left(\mu_1+\frac{\sigma_1}{\sigma_2}\rho(a-\mu_2),\, (1-\rho^2)\sigma_1^2\right).$
where $\rho$ is the correlation coefficient between X1 and X2.
### Bivariate conditional expectation
In the case
$\begin{pmatrix} X_1 \\ X_2 \end{pmatrix} \sim \mathcal{N} \left( \begin{pmatrix} 0 \\ 0 \end{pmatrix} , \begin{pmatrix} 1 & \rho \\ \rho & 1 \end{pmatrix} \right)$
the following result holds[citation needed]
$\operatorname{E}(X_1 | X_2 > z) = \rho { \phi(z) \over \Phi(-z) } ,$
where the final ratio here is called the inverse Mills ratio.
## Marginal distributions
To obtain the marginal distribution over a subset of multivariate normal random variables, one only needs to drop the irrelevant variables (the variables that one wants to marginalize out) from the mean vector and the covariance matrix. The proof for this follows from the definitions of multivariate normal distributions and linear algebra.[10]
Example
Let x = [X1, X2, X3] be multivariate normal random variables with mean vector μ = [μ1, μ2, μ3] and covariance matrix Σ (standard parametrization for multivariate normal distributions). Then the joint distribution of x′ = [X1, X3] is multivariate normal with mean vector μ′ = [μ1, μ3] and covariance matrix $\boldsymbol\Sigma' = \begin{bmatrix} \boldsymbol\Sigma_{11} & \boldsymbol\Sigma_{13} \\ \boldsymbol\Sigma_{31} & \boldsymbol\Sigma_{33} \end{bmatrix}$.
## Affine transformation
If y = c + Bx is an affine transformation of $\mathbf{x}\ \sim \mathcal{N}(\boldsymbol\mu, \boldsymbol\Sigma),$ where c is an $M \times 1$ vector of constants and B is a constant $M \times N$ matrix, then y has a multivariate normal distribution with expected value c + Bμ and variance BΣBT i.e., $\mathbf{y} \sim \mathcal{N} \left(\mathbf{c} + \mathbf{B} \boldsymbol\mu, \mathbf{B} \boldsymbol\Sigma \mathbf{B}^{\rm T}\right)$. In particular, any subset of the xi has a marginal distribution that is also multivariate normal. To see this, consider the following example: to extract the subset (x1, x2, x4)T, use
$\mathbf{B} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & \ldots & 0 \\ 0 & 1 & 0 & 0 & 0 & \ldots & 0 \\ 0 & 0 & 0 & 1 & 0 & \ldots & 0 \end{bmatrix}$
which extracts the desired elements directly.
Another corollary is that the distribution of Z = b · x, where b is a constant vector of the same length as x and the dot indicates a vector product, is univariate Gaussian with $Z\sim\mathcal{N}\left(\mathbf{b}\cdot\boldsymbol\mu, \mathbf{b}^{\rm T}\boldsymbol\Sigma \mathbf{b}\right)$. This result follows by using
$\mathbf{B}=\begin{bmatrix} b_1 & b_2 & \ldots & b_n \end{bmatrix}.$
Observe how the positive-definiteness of Σ implies that the variance of the dot product must be positive.
An affine transformation of x such as 2x is not the same as the sum of two independent realisations of x.
## Geometric interpretation
The equidensity contours of a non-singular multivariate normal distribution are ellipsoids (i.e. linear transformations of hyperspheres) centered at the mean.[11] The directions of the principal axes of the ellipsoids are given by the eigenvectors of the covariance matrix Σ. The squared relative lengths of the principal axes are given by the corresponding eigenvalues.
If Σ = UΛUT = UΛ1/2(UΛ1/2)T is an eigendecomposition where the columns of U are unit eigenvectors and Λ is a diagonal matrix of the eigenvalues, then we have
$\mathbf{x}\ \sim \mathcal{N}(\boldsymbol\mu, \boldsymbol\Sigma) \iff \mathbf{x}\ \sim \boldsymbol\mu+\mathbf{U}\boldsymbol\Lambda^{1/2}\mathcal{N}(0, \mathbf{I}) \iff \mathbf{x}\ \sim \boldsymbol\mu+\mathbf{U}\mathcal{N}(0, \boldsymbol\Lambda).$
Moreover, U can be chosen to be a rotation matrix, as inverting an axis does not have any effect on N(0, Λ), but inverting a column changes the sign of U's determinant. The distribution N(μ, Σ) is in effect N(0, I) scaled by Λ1/2, rotated by U and translated by μ.
Conversely, any choice of μ, full rank matrix U, and positive diagonal entries Λi yields a non-singular multivariate normal distribution. If any Λi is zero and U is square, the resulting covariance matrix UΛUT is singular. Geometrically this means that every contour ellipsoid is infinitely thin and has zero volume in n-dimensional space, as at least one of the principal axes has length of zero.
## Estimation of parameters
The derivation of the maximum-likelihood estimator of the covariance matrix of a multivariate normal distribution is perhaps surprisingly subtle and elegant. See estimation of covariance matrices.
In short, the probability density function (pdf) of a multivariate normal is
$f(\mathbf{x})= \frac{1}{\sqrt { (2\pi)^k|\boldsymbol \Sigma| } } \exp\left(-{1 \over 2} (\mathbf{x}-\boldsymbol\mu)^{\rm T} \boldsymbol\Sigma^{-1} ({\mathbf x}-\boldsymbol\mu)\right)$
and the ML estimator of the covariance matrix from a sample of n observations is
$\widehat{\boldsymbol\Sigma} = {1 \over n}\sum_{i=1}^n ({\mathbf x}_i-\overline{\mathbf x})({\mathbf x}_i-\overline{\mathbf x})^T$
which is simply the sample covariance matrix. This is a biased estimator whose expectation is
$E[\widehat{\boldsymbol\Sigma}] = \frac{n-1}{n} \boldsymbol\Sigma.$
An unbiased sample covariance is
$\widehat{\boldsymbol\Sigma} = {1 \over n-1}\sum_{i=1}^n (\mathbf{x}_i-\overline{\mathbf{x}})(\mathbf{x}_i-\overline{\mathbf{x}})^{\rm T}.$
The Fisher information matrix for estimating the parameters of a multivariate normal distribution has a closed form expression. This can be used, for example, to compute the Cramér–Rao bound for parameter estimation in this setting. See Fisher information for more details.
## Bayesian inference
In Bayesian statistics, the conjugate prior of the mean vector is another multivariate normal distribution, and the conjugate prior of the covariance matrix is an inverse-Wishart distribution $\mathcal{W}^{-1}$ . Suppose then that n observations have been made
$\mathbf{X} = \{\mathbf{x}_1,\dots,\mathbf{x}_n\} \sim \mathcal{N}(\boldsymbol\mu,\boldsymbol\Sigma)$
and that a conjugate prior has been assigned, where
$p(\boldsymbol\mu,\boldsymbol\Sigma)=p(\boldsymbol\mu\mid\boldsymbol\Sigma)\ p(\boldsymbol\Sigma),$
where
$p(\boldsymbol\mu\mid\boldsymbol\Sigma) \sim\mathcal{N}(\boldsymbol\mu_0,m^{-1}\boldsymbol\Sigma) ,$
and
$p(\boldsymbol\Sigma) \sim \mathcal{W}^{-1}(\boldsymbol\Psi,n_0).$
Then,[citation needed]
$\begin{array}{rcl} p(\boldsymbol\mu\mid\boldsymbol\Sigma,\mathbf{X}) & \sim & \mathcal{N}\left(\frac{n\bar{\mathbf{x}} + m\boldsymbol\mu_0}{n+m},\frac{1}{n+m}\boldsymbol\Sigma\right),\\ p(\boldsymbol\Sigma\mid\mathbf{X}) & \sim & \mathcal{W}^{-1}\left(\boldsymbol\Psi+n\mathbf{S}+\frac{nm}{n+m}(\bar{\mathbf{x}}-\boldsymbol\mu_0)(\bar{\mathbf{x}}-\boldsymbol\mu_0)', n+n_0\right), \end{array}$
where
$\begin{array}{rcl} \bar{\mathbf{x}} & = & n^{-1}\sum_{i=1}^{n} \mathbf{x}_i ,\\ \mathbf{S} & = & n^{-1}\sum_{i=1}^{n} (\mathbf{x}_i - \bar{\mathbf{x}})(\mathbf{x}_i - \bar{\mathbf{x}})' . \end{array}$
## Multivariate normality tests
Multivariate normality tests check a given set of data for similarity to the multivariate normal distribution. The null hypothesis is that the data set is similar to the normal distribution, therefore a sufficiently small p-value indicates non-normal data. Multivariate normality tests include the Cox-Small test[12] and Smith and Jain's adaptation[13] of the Friedman-Rafsky test.[14]
Mardia's test[15] is based on multivariate extensions of skewness and kurtosis measures. For a sample {x1, ..., xn} of k-dimensional vectors we compute
$\begin{align} & \widehat{\boldsymbol\Sigma} = {1 \over n} \sum_{j=1}^n \left(\mathbf{x}_j - \bar{\mathbf{x}}\right)\left(\mathbf{x}_j - \bar{\mathbf{x}}\right)^T \\ & A = {1 \over 6n} \sum_{i=1}^n \sum_{j=1}^n \left[ (\mathbf{x}_i - \bar{\mathbf{x}})^T\;\widehat{\boldsymbol\Sigma}^{-1} (\mathbf{x}_j - \bar{\mathbf{x}}) \right]^3 \\ & B = \sqrt{\frac{n}{8k(k+2)}}\left\{{1 \over n} \sum_{i=1}^n \left[ (\mathbf{x}_i - \bar{\mathbf{x}})^T\;\widehat{\boldsymbol\Sigma}^{-1} (\mathbf{x}_i - \bar{\mathbf{x}}) \right]^2 - k(k+2) \right\} \end{align}$
Under the null hypothesis of multivariate normality, the statistic A will have approximately a chi-squared distribution with 1/6⋅k(k + 1)(k + 2) degrees of freedom, and B will be approximately standard normal N(0,1).
Mardia's kurtosis statistic is skewed and converges very slowly to the limiting normal distribution. For medium size samples $(50 \le n < 400)$, the parameters of the asymptotic distribution of the kurtosis statistic are modified[16] For small sample tests ($n<50$) empirical critical values are used. Tables of critical values for both statistics are given by Rencher[17] for k=2,3,4.
Mardia's tests are affine invariant but not consistent. For example, the multivariate skewness test is not consistent against symmetric non-normal alternatives.[18]
The BHEP test[19] computes the norm of the difference between the empirical characteristic function and the theoretical characteristic function of the normal distribution. Calculation of the norm is performed in the L2(μ) space of square-integrable functions with respect to the Gaussian weighting function $\scriptstyle \mu_\beta(\mathbf{t}) = (2\pi\beta^2)^{-k/2} e^{-|\mathbf{t}|^2/(2\beta^2)}$. The test statistic is
$\begin{align} T_\beta &= \int_{\mathbb{R}^k} \left| {1 \over n} \sum_{j=1}^n e^{i\mathbf{t}^T\widehat{\boldsymbol\Sigma}^{-1/2}(\mathbf{x}_j - \bar{\mathbf{x})}} - e^{-|\mathbf{t}|^2/2} \right|^2 \; \boldsymbol\mu_\beta(\mathbf{t}) d\mathbf{t} \\ &= {1 \over n^2} \sum_{i,j=1}^n e^{-{\beta \over 2}(\mathbf{x}_i-\mathbf{x}_j)'\widehat{\boldsymbol\Sigma}^{-1}(\mathbf{x}_i-\mathbf{x}_j)} - \frac{2}{n(1 + \beta^2)^{k/2}}\sum_{i=1}^n e^{ -\frac{\beta^2}{2(1+\beta^2)} (\mathbf{x}_i-\bar{\mathbf{x}})^T\widehat{\boldsymbol\Sigma}^{-1}(\mathbf{x}_i-\bar{\mathbf{x}})} + \frac{1}{(1 + 2\beta^2)^{k/2}} \end{align}$
The limiting distribution of this test statistic is a weighted sum of chi-squared random variables,[20] however in practice it is more convenient to compute the sample quantiles using the Monte-Carlo simulations.[citation needed]
A detailed survey of these and other test procedures is available.[21]
## Drawing values from the distribution
A widely used method for drawing a random vector x from the N-dimensional multivariate normal distribution with mean vector μ and covariance matrix Σ works as follows:[citation needed]
1. Find any real matrix A such that A AT = Σ. When Σ is positive-definite, the Cholesky decomposition is typically used, and the extended form of this decomposition can always be used (as the covariance matrix may be only positive semi-definite) in both cases a suitable matrix A is obtained.[citation needed] An alternative is to use the matrix A = UΛ½ obtained from a spectral decomposition Σ = UΛUT of Σ.[citation needed] The former approach is more computationally straightforward but the matrices A change for different orderings of the elements of the random vector, while the latter approach gives matrices that are related by simple re-orderings. In theory both approaches give equally good ways of determining a suitable matrix A, but there are differences in computation time.
2. Let z = (z1, …, zN)T be a vector whose components are N independent standard normal variates (which can be generated, for example, by using the Box–Muller transform).
3. Let x be μ + Az. This has the desired distribution due to the affine transformation property.
## See also
• Chi distribution, the pdf of the 2-norm (or Euclidean norm) of a multivariate normally distributed vector (centered at zero).
• Complex normal distribution, for the generalization to complex valued random variables.
• Multivariate stable distribution extension of the multivariate normal distribution, when the index (exponent in the characteristic function) is between zero to two.
• Mahalanobis distance
• Wishart distribution
## References
1. Gut, Allan (2009) An Intermediate Course in Probability, Springer. ISBN 9781441901613 (Chapter 5)
2. Hamedani, G. G.; Tata, M. N. (1975). "On the determination of the bivariate normal distribution from distributions of linear combinations of the variables". The American Mathematical Monthly 82 (9): 913–915. doi:10.2307/2318494.
3. Wyatt, John. "Linear least mean-squared error estimation". Lecture notes course on applied probability. Retrieved 23 January 2012.
4. Rao, C.R. (1973). Linear Statistical Inference and Its Applications. New York: Wiley. pp. 527–528.
5. Gokhale, DV; NA Ahmed, BC Res, NJ Piscataway (May 1989). "Entropy Expressions and Their Estimators for Multivariate Distributions". Information Theory, IEEE Transactions on 35 (3): 688–692. doi:10.1109/18.30996.
6. Eaton, Morris L. (1983). Multivariate Statistics: a Vector Space Approach. John Wiley and Sons. pp. 116–117. ISBN 0-471-02776-6.
7. Jensen, J (2000). Statistics for Petroleum Engineers and Geoscientists. Amsterdam: Elsevier. p. 207.
8.
9. Cox, D. R.; Small, N. J. H. (1978). "Testing multivariate normality". Biometrika 65 (2): 263. doi:10.1093/biomet/65.2.263.
10. Smith, S. P.; Jain, A. K. (1988). "A test to determine the multivariate normality of a data set". IEEE Transactions on Pattern Analysis and Machine Intelligence 10 (5): 757. doi:10.1109/34.6789.
11. Friedman, J. H.; Rafsky, L. C. (1979). "Multivariate Generalizations of the Wald-Wolfowitz and Smirnov Two-Sample Tests". The Annals of Statistics 7 (4): 697. doi:10.1214/aos/1176344722.
12. Mardia, K. V. (1970). "Measures of multivariate skewness and kurtosis with applications". Biometrika 57 (3): 519–530. doi:10.1093/biomet/57.3.519.
13. Rencher (1995), pages 112-113.
14. Rencher (1995), pages 493-495.
15. Baringhaus, L.; Henze, N. (1991). "Limit distributions for measures of multivariate skewness and kurtosis based on projections". Journal of Multivariate Analysis 38: 51. doi:10.1016/0047-259X(91)90031-V.
16. Epps, Lawrence B.; Pulley, Lawrence B. (1983). "A test for normality based on the empirical characteristic function". Biometrika 70 (3): 723–726. doi:10.1093/biomet/70.3.723.
17. Baringhaus, L.; Henze, N. (1988). "A consistent test for multivariate normality based on the empirical characteristic function". Metrika 35 (1): 339–348. doi:10.1007/BF02613322.
18. Henze, Norbert (2002). "Invariant tests for multivariate normality: a critical review". Statistical Papers 43 (4): 467–506. doi:10.1007/s00362-002-0119-6.
### Literature
• Rencher, A.C. (1995). Methods of Multivariate Analysis. New York: Wiley.
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http://stats.stackexchange.com/questions/10206/conditions-for-central-limit-theorem-for-dependent-sequences/10212
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# Conditions for Central Limit Theorem for dependent sequences
Cumbersome technical assumptions (e.g., mixing properties) are used in the literature to prove Central Limit Theorems for dependent sequences. I sketched a proof that does not require any of these technical assumptions. Can you help me figure out what is wrong with this proof? The proof is at: http://www.statlect.com/central_limit_theorem_for_correlated_sequences.htm. Thanks in advance to all those who will be so generous and patient to read it.
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One thing that is not explicitly stated is that $V$ needs to be finite, so you would have to impose some sort of conditions on the covariance structure. This is usually what those "technical conditions" are doing, making sure that the variance doesn't get "too big" – probabilityislogic May 1 '11 at 14:14
@user4422, hand-waving is good, but I did not see any proof. All the technical details usually come up, after you start to write rigorous proof. As the text stands now it is just an idea or strategy of proof, not a proof. The 2 things which stand out is the first claim that omiting some elements changes nothing, another thing is the claim that the remaining sequence is iid, it is not and assymptotic closeness to iid does not immediately allows application of CLT for iid variables. – mpiktas May 1 '11 at 14:38
OK. Right, we must ensure that V stays finite. Is there any counter-example where V goes to infinity for a stationary ergodic process? – user4422 May 1 '11 at 15:50
## 1 Answer
Additional conditions are needed. (A near-proof of this fact is that many incredibly smart individuals have been thinking deeply about these issues for over 100 years. It is highly unlikely that something like this would have escaped all of them.)
First of all, note that the formula for $V$ that you give is part of the conclusion of the associated central limit theorem. See, for example, Theorem 7.6 on pages 416–417 of R. Durrett, Probability: Theory and Examples, 3rd. ed., which based on your link, you appear to have access to.
At any rate, here is a simple counterexample to your claim.
Let $X_0$ equal $+1$ with probability $1/2$ and $-1$ with probability $1/2$. Define $X_n = (-1)^n X_0$. Then $\{X_n\}$ is a stationary ergodic process with mean 0 and variance 1, but the Central Limit Theorem fails.
The properties of stationarity and ergodicity should be pretty easy to see as we can construct this process by defining a function over the states of a two-state Markov chain with stationary probability measure $\pi(x) = 1/2$ for $x \in \{0,1\}$.
Observe that this process yields a sequence of the form $-X_0, X_0, -X_0, \ldots$ and so
1. Even without appealing to any notions about ergodicity, it is easy to see that $\newcommand{\e}{\mathbb{E}}\bar{X}_n \to \e X_0 = 0$ almost surely, and,
2. $\newcommand{\Var}{\mathbb{V}\mathrm{ar}}\Var(S_n) = 0$ if $n$ is even and $1$ if $n$ is odd.
This already is enough to conclude that there is no way that any rescaling of $S_n$ can make it converge in distribution to a normal random variable. In fact, for every function $f$ such that $f(n) \to \infty$, $S_n / f(n) \to 0$ almost surely no matter how slowly $f$ diverges.
Note also that this example should make it clear that the formula for $V$ is a conclusion of the theorem. Indeed, for the example above, $$V_n = 1 + 2 \sum_{i = 1}^n \e X_0 X_i = \left\{ \begin{array}{rl} -1, & n \text{ odd}, \\ 1, & n \text{ even}, \end{array} \right.$$ which, of course, (a) makes no sense as a variance, (b) does not have a limit, and (c) is not asymptotically equivalent to $\Var(S_n)$. (NB: I use a slightly different form for $V_n$ than you do where mine matches that given in Durrett.)
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First of all thanks for your long and careful answer, I really appreciate that. As for the example, it is very interesting, but I am struggling to understand why the process is ergodic: the notion of ergodicity I have in mind, coupled with stationarity, implies strong mixing, i.e. asymptotic independence of the terms of the sequence, which in this example seem perfectly dependent. – user4422 May 1 '11 at 22:56
Sorry, I was wrong again. Mixing is stronger. Then it might be ergodic even if it's not mixing. – user4422 May 1 '11 at 23:05
OK. Now I understood why it is ergodic. Trivially all shift invariant subsets have either probability zero or one. – user4422 May 1 '11 at 23:33
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http://math.stackexchange.com/questions/242724/monochromatic-squares-in-a-colored-plane
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# Monochromatic squares in a colored plane
Color every point in the real plane using the colors blue,yellow only. It can be shown that there exists a rectangle that has all vertices with the same color. Is it possible to show that there exists a square that has all vertices with the same color ? If it is not possible, please give me an example of a coloring of the real plane that does not have monochromatic squares.
To give the viewers an idea about other similar results (which they might find useful), for any coloring (2 colors) of the real plane:
1) There exists three collinear points having the same color, such that one of the points is the midpoint of the line segment that joins the other two.
2)For any two angles $\theta,\phi$ there exists a monochromatic triangle that has angles $\theta,\phi,180-(\theta+\phi)$
3)For any angle $\theta$, there exists a monochromatic parallelogram with angle $\theta$
Now its natural to ask if there are any monochromatic squares.
Thank you
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Are you talking about the real plane of the integer plane (grid points)? – Arthur Nov 22 '12 at 18:08
I am talking about the real plane. The stronger version is for grid points. If you have a solution to the grid points please post it as an answer. It ll also be a solution to my problem – Amr Nov 22 '12 at 18:10
I'm just trying to understand the problem. Does the square have to have sides paralell to the axes? – Arthur Nov 22 '12 at 18:14
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Do you think the way I stated it is not clear ? – Amr Nov 22 '12 at 18:15
Well, with the discrete-tag, and the fact that coloring problems are usually in the grid. No, I didn't think so. – Arthur Nov 22 '12 at 18:16
show 2 more comments
## 1 Answer
Molina, Oza, and Puttagunta, Sane bounds on Van der Waerden-type numbers, write,
The following is known: no matter how the lattice points of a coordinate plane are colored red and blue, there exists a square whose corners are the same color (a monochromatic square). In fact, using more than just two colors will still guarantee a monochromatic square (one whose vertices are the same color). So for all $c$ (the number of colors), there is a number $G(c)$ where all colorings with $c$ colors of the lattice points of a $G(c) \times G(c)$ grid will contain a monochromatic square. Unfortunately, the necessary number of points is unknown, but bounds are known. These bounds are enormous (roughly $G(c)\le2^{2^c}\times2^{2^{2^{2^{2c}}}}$).
The references they cite are
1. W. Gasarch, C. Kruskal, and A. Parrish. Van der Waerden’s theorem: Variants and applications. [But the Chapter that is supposed to be about "The Square Theorem" isn't there]
2. R. Graham, B. Rothchild, and J. Spencer. Ramsey Theory. Wiley, 1990.
3. B. Landmann and A. Robertson. Ramsey Theory over the integers. AMS, 2003.
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– Andres Caicedo Nov 23 '12 at 7:23
@ Gerry Myerson, thank you – Amr Nov 23 '12 at 11:03
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http://math.stackexchange.com/questions/308024/truth-functional-completness-for-quantifiers
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# Truth functional completness for quantifiers
There is a definition for truth-functional completness for a set of propositional connectives. Is there a definition for truth-functional completness of a set of quantifiers and propositional connectives ?
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## 1 Answer
There cannot be a direct analogue, because a formula with quantifiers in it can depend on the truth values of infinitely many atomic propositions. There are $2^{2^{\kappa}}$ different possible truth functions of $\kappa$ inputs, and when $\kappa$ is infinite this much larger than the number of possible formulas one can build from any finite set of connectives. So in this sense there is no complete finite set of quantifiers and connectives.
I suppose one could consider complete sets containing (uncountably) infinitely many different quantifiers, but that's not very practical.
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http://mathoverflow.net/questions/102763/the-use-of-parentheses-to-mean-i-wont-tell-you-this-again/103060
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## the use of parentheses to mean “I won’t tell you this again”
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
A reader of one of my drafts found fault with my use of parentheses; I put the word "bounded" in parentheses in a statement of a certain theorem, and he replied "But the statement isn't true if the assumption of boundedness is dropped!"
That reader seemed to be thinking that parentheses mark things that are in some way inessential (as is sometimes the case in non-mathematical prose). But, as I wrote to him:
Here I am using parentheses to mean "Of course the interval must be bounded! In case some of you are nodding off, I'll include the stipulation of boundedness, but I might not include it next time." I wonder if that use of parentheses has a name?
Does this use of parentheses have a name, or any sort of pedigree that might dignify it, within or beyond mathematical writing?
I have no idea how to tag this post; it's a question about the (possibly nonexistent) subfield of modern Rhetoric that is concerned with the ways mathematicians use language to communicate ideas to other mathematicians. I'll be grateful if someone will suggest appropriate tags and add them (and I'll make a note of what the tag is, in case I need it again).
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It seems to me that these two uses of parentheses are not necessarily different. Ideally before you use parentheses to indicate "I won't tell you this again" you will say something like "all widgets are henceforth assumed to be bounded" and then when you write "(bounded) widget" it is an inessential reminder of this global assumption. – Trevor Wilson Jul 20 at 19:16
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I agree with Trevor. The parentheses around "bounded" should indicate that the theorem is true without that word, probably because some earlier convention said that boundedness is always tacitly understood. The reason for including the redundant word in parentheses would usually be that the convention was stated so long ago that the reader might have forgotten it. – Andreas Blass Jul 20 at 19:24
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In short, be explicit, as explicit as you can without becoming painful: the seconds you save by not writing things out will be charged to your readers in terms of time and unease. – Mariano Suárez-Alvarez Jul 20 at 19:25
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A shriek, "(bounded!)", says "reminder" and removes the possibility of the "inessential" interpretation at the cost merely of a single extra character. – David Feldman Jul 20 at 22:13
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Is it really worth making a convention to be able to write "interval" instead of "bounded interval"? You should remember that most readers won't read your paper linearly, i.e. they may jump directly to section 3.14 since that's all they care about. Not being able to read section 3.14 without having read sections 1.1-3.13 is a BIG disservice to most readers. – ABayer Jul 21 at 16:42
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## 2 Answers
Re: "Does this use of parentheses have a name?",
preterition |ˌpretəˈri sh ən|
noun (...) the rhetorical technique of making summary mention of something by professing to omit it.
ORIGIN late 16th cent.: from late Latin praeteritio(n-), from praeterire ‘pass, go by.’
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That's similar but not quite it, is it? The Eco's artiluge mentioned at the end of that section, though, would make for a fun paper :) – Mariano Suárez-Alvarez Jul 20 at 20:06
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@Mariano: to me that's exactly it, i.e. unless I'm missing something his "(bounded)" is synonymous with "(I won't mention bounded)". – Francois Ziegler Jul 20 at 20:18
I think a more typical use is found in this gem from Cicero's Against Catiline (quoting from the 1856 trans. on perseus.tufts.edu): "What? when lately by the death of your former wife you had made your house empty and ready for a new bridal, did you not even add another incredible wickedness to this wickedness? But I pass that over, and willingly allow it to be buried in silence, that so horrible a crime may not be seen to have existed in this city..." Cicero draws attention to the crime making himself appear generous for doing so. – Adam Saltz Jul 20 at 20:43
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(This is on the verge of becoming offtopic but) I think that preterition is the artiluge of saying that one is not saying something in order to say it, but the «(bounded)» does not carry that intention. It would be different if the theorem were something like «Don't get me started on the fact that we are assuming that our intervals are bounded, and let us just say that continuous functions on an interval are integrable.» – Mariano Suárez-Alvarez Jul 20 at 20:50
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Isn't one of the standard examples of preterition "I come to bury Caesar, not to praise him...", followed by much praising? Or, for a more recent exmplar, Peter Cook's sketch cvillewords.com/2007/11/09/… – Yemon Choi Jul 21 at 8:49
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I think it might be beneficial to see the actual context in which the comments were made (by me; not as a referee, but just someone that Jim wrote to and asked for comments on his nice paper, which by the way, has a fair bit of its provenance in various MO threads).
The work in question is on the arxiv here. Various properties of an ordered field $R$ are being considered and compared. The last two are:
(17) The Shrinking Interval Property: suppose $I_1 \supset I_2 \supset \ldots$ are (bounded) closed intervals in $R$ with lengths decreasing to zero. Then the intersection of the $I_n$'s is nonempty.
and
(18) The Nested Interval Property: Suppose $I_1 \supset I_2 \supset \ldots$ are (bounded) closed intervals in $R$. Then the intersection of the $I_n$'s is nonempty.
I was not thrilled with the use of (bounded) in (17), but I let it go. I objected to the use of (bounded) in (18).
Note that "(bounded)" is playing different roles in the two statements. In (17), it is a superfluous hypothesis: if the lengths of the intervals are decreasing to zero then necessarily all but finitely many of them are bounded. In (18) it certainly isn't. I found this lack of parallelism especially confusing: so confusing that the first time I read it I honestly did arrive at the (ridiculous) conclusion that Jim Propp was unaware that e.g. $\bigcap_{n=1}^{\infty} [n,\infty) = \varnothing$.
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I agree with Pete's comment completely. (In my earlier email correspondence with him, I confused (17) with (18), and missed the salient difference between them: in the first case the boundedness follows from the other hypotheses, and in the second it doesn't.) The "praeteritional" ("praeterite"?) use of parentheses is allowable for (17), but not for (18). Anyway, the responses I've received to this question have convinced me that in mathematical writing it's best to avoid confusion by being more explicit (e.g. "for the rest of this proof, all intervals are assumed to be bounded"). Thanks! – James Propp Jul 25 at 2:57
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http://physics.stackexchange.com/questions/21168/do-particles-have-different-spins-in-different-frames-of-reference
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# Do particles have different spins in different frames of reference?
Let's say we have two photons, whose momentum vectors point to opposite directions. Also spin angular momentum vectors of the photons point to opposite directions. (Sum of spins is zero)
Now we change into a frame where the momentum vectors point into almost same direction. Do the spin vectors point into almost same direction in this frame? (Is the sum of spins not zero in this frame?)
(Photons were far away from us, moving to the left and to the right, then we accelerated towards the photons)
(Momentum and spin vectors are either parallel or anti-parallel in massless particles, according quantum-mechanics)
I mean: Can acceleration of an observer change spins of particles that the observer observes?
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## 3 Answers
Yes, what you are suggesting is exactly what is happening, but that is if you have an expression which transforms like an axial vector which you can identify with the the spin of the photon. The inherent spin property of photons ($1\hbar$) and electrons ($\tfrac12\hbar$) is of course reference frame independent.
Maybe without realizing you brought forward here the issue of "what is the expression which represents the spin of the electromagnetic field", This field can not be expressed in a gauge invariant way because it contains the vector potential $A^\mu$.
The spin density in the Lorentz gauge would be:
${\cal C}^\mu ~~=~~ \epsilon_o\,\tfrac12\varepsilon^{\,\mu\nu\alpha\beta} F_{\alpha\beta}A_\nu ~~=~~ \epsilon_o\,\varepsilon^{\,\mu\alpha\beta\gamma} A_\alpha\partial_\beta A_\gamma$
Which (in vacuum) is equal to.
${\cal C}^\mu ~~=~~ \left( \begin{array}{c c c c} ~ 0 &-\tfrac1c\,\mathsf{H}_x &-\tfrac1c\,\mathsf{H}_y &-\tfrac1c\,\mathsf{H}_z \\ \tfrac1c\,\mathsf{H}_x & ~~~ 0 & \ \ ~~\mathsf{D}_z & ~-\mathsf{D}_y \\ \tfrac1c\,\mathsf{H}_y & ~-\mathsf{D}_z & ~~~ 0 & \ \ ~~\mathsf{D}_x \\ \tfrac1c\,\mathsf{H}_z & \ \ ~~\mathsf{D}_y & ~-\mathsf{D}_x & ~~~ 0 \end{array} \right) \left( \begin{array}{c} \ \ A_0 \\ -A_x \\ -A_y \\ -A_z \end{array} \right)$
From this expression one can already see that it transforms like an axial vector. If you go through the trouble of calculating the $A^\mu$ field of a circulating charge using Liénard Wiechert (like i did here) then you get indeed the required $1\hbar$ ratio with the momentum density for circular polarized photons and $0\hbar$ for linear polarized photons.
The latter expression is equivalent to the electron's spin density found via the Gordon decomposition of the axial Dirac current of the electron. In this case the matrix is given by the Magnetization Polarization tensor of the Dirac field while the column vector is given by the dynamic momentum of the electron. (The phase change rates minus the phase induced by the $A^\mu$ field, $\partial_\mu-ieA_\mu$).
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By the way, @Hans, what enables us to associate axial current with spin density? – Murod Abdukhakimov Feb 28 '12 at 22:38
My understanding is that Spin $S$ can be defined as the residual angular momentum in the rest frame. So, you will measure a different angular momentum $J$.
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I don't know if what i'm saying something correct since my QM knowledge is limited.
Spin for a photon is a binary quantity associated to the particle via a tensor product. It's not merely and arrow pointing up or down: It is a new property that objects have but it is defined in a space completely different from the position space. In fact one have functions for angular momentum, like the one used for orbitals for atoms and molecules, but not for spin.
So when one make a change of reference it should be specified if it is in the angular momentum space or spin space: one change of reference does not necessary imply the other.
I understand that acceleration involves relativistic consideration that i did not make.
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http://crypto.stackexchange.com/questions/6392/could-use-an-explanation-of-the-notation-for-an-oracle-adversary/6394
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# Could use an explanation of the notation for an oracle adversary
In the definition below, what exactly does it mean for the adversary (not even sure that's the correct term?) to equal one?
$$\underset{K}{Pr} \left [A^{F_k(\cdot)} = 1 \right ]$$
Source (Page 22/35?) or Imgur Link
Thanks in advance!
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Asymptotic security definitions ARGHHH – CodesInChaos Feb 19 at 17:15
## 1 Answer
This represents the probability over all $K$ that $A$ given an oracle access to $F_K(\cdot)$ outputs 1. You usually compare that to the probability of A ouptputting 1 while having oracle access to a random function and the difference tells you how well $A$ does at telling $F_K$ from random
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http://mathoverflow.net/questions/19285?sort=newest
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## How do you axiomatize topology via nets?
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Let $X$ be a set and let ${\mathcal N}$ be a collection of nets on $X.$
I've been told by several different people that ${\mathcal N}$ is the collection of convergent nets on $X$ with respect to some topology if and only if it satisfies some axioms. I've also been told these axioms are not very pretty.
Once or twice I've tried to figure out what these axioms might be but never came up with anything very satisfying. Of course one could just recode the usual axioms regarding open sets as statements about nets and then claim to have done the job. But, come on, that's nothing to be proud of.
Has anyone seen topology axiomatized this way? Does anyone remember the rules?
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Axiomatizing via filters is a much more fruitful endeavor. For that I suggest Bourbaki's "General Topology". Filters have a number of really nice applications to logic, set theory, and even category theory (filtered limits and colimits!). – Harry Gindi Mar 25 2010 at 13:40
@fpqc: Filters are in some exact sense "dual" to nets, right? – Theo Johnson-Freyd Mar 25 2010 at 19:28
I'm having a hard time giving you an answer that is completely elementary, so don't take this as me using overly complicated examples just to be a jerk: I would say that a net is to a filter what a pseudofunctor is to a fibered category. The filter encodes all equivalent nets, and getting a net from a filter just requires you to make choices (similar to choosing a cleavage, for example). So they're not really dual, but rather, related by something similar to the grothendieck construction. – Harry Gindi Mar 26 2010 at 4:57
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@TJF: Section 6 of the notes referred to in my answer below makes explicit the "correspondence" between nets and filters. It is not a perfect correspondence, and fpqc's description [I have ignored the categorical jargon] seems correct to me: a net is fundamentally an extrinsic object to the set in question, whereas a filter is intrinsic to the set. The assignment net |-> filter is canonical; the other one requires choices. Note that extrinsic constructions have their merits: I discuss this in Section 7. In particular, I think Bourbaki uses "too many filters". – Pete L. Clark Mar 26 2010 at 5:06
+1 Thanks for clarifying my admittedly unclear comment! – Harry Gindi Mar 26 2010 at 9:54
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## 2 Answers
Yes. This is given in Kelley's General Topology. (Kelley was one of the main mathematicians who developed the theory of nets so that it would be useful in topology generally rather than just certain applications in analysis.)
In the section "Convergence Classes" at the end of Chapter 2 of his book, Kelley lists the following axioms for convergent nets in a topological space $X$
a) If $S$ is a net such that $S_n = s$ for each $n$ [i.e., a constant net], then $S$ converges to $s$.
b) If $S$ converges to $s$, so does each subnet.
c) If $S$ does not converge to $s$, then there is a subnet of $S$, no subnet of which converges to $s$.
d) (Theorem on iterated limits): Let $D$ be a directed set. For each $m \in D$, let $E_m$ be a directed set, let $F$ be the product $D \times \prod_{m \in D} E_m$ and for $(m,f)$ in $F$ let $R(m,f) = (m,f(m))$. If $S(m,n)$ is an element of $X$ for each $m \in D$ and $n \in E_m$ and $\lim_m \lim_n S(m,n) = s$, then $S \circ R$ converges to $s$.
He has previously shown that in any topological space, convergence of nets satisfies a) through d). (The first three are easy; part d) is, I believe, an original result of his.) In this section he proves the converse: given a set $S$ and a set $\mathcal{C}$ of pairs (net,point) satisfying the four axioms above, there exists a unique topology on $S$ such that a net $N$ converges to $s \in X$ iff $(N,s) \in \mathcal{C}$.
I have always found property d) to be unappealing bordering on completely opaque, but that's a purely personal statement.
Addendum: I would be very interested to know if anyone has ever put this characterization to any useful purpose. A couple of years ago I decided to relearn general topology and write notes this time. The flower of my efforts was an essay on convergence in topological spaces that seems to cover all the bases (especially, comparing nets and filters) more solidly than in any text I have seen.
http://math.uga.edu/~pete/convergence.pdf
But "even" in these notes I didn't talk about either the theorem on iterated limits or (consequently) Kelley's theorem above: I honestly just couldn't internalize it without putting a lot more thought into it. But I've always felt/worried that there must be some insight and content there...
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+1 for mentioning the oft-overlooked "General Topology". – Steve D Mar 25 2010 at 8:51
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An interesting addendum: Suppose we make a tiny change (So tiny that some have done it inadvertantly by mistake.) We allow a directed set to be empty. Keep these axioms. Define open set, closed set, etc., from convergence as usual. Almost everything is the same as usual. But the whole space need not be open. So we end up in another recent question at MO. – Gerald Edgar Mar 25 2010 at 12:22
The classic reference on generalized convergence-I still insist-is Robert Bartle's review article.My old analysis teacher Gerald Itzkowitz learned it from that paper and so did I. – Andrew L Mar 26 2010 at 4:24
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@Andrew L: could you be a little more specific in your reference? – Pete L. Clark Mar 26 2010 at 4:31
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The "Handbook of Analysis and its Foundations" by Eric Schechter contains an extremely extensive and thorough discussion of all notions of convergence, including the relationship between nets and filters, a classification of various notions of subnets and similar material. I would highly recommend the book to anyone interested in the connection between filters and nets. – Michael Greinecker Jul 11 2010 at 18:46
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(too long for a comment to Pete's answer)
Garrett Birkhoff was my Ph.D. advisor. Let me provide a few remarks of a historical nature.
From a 25-year-old Garrett Birkhoff we have: Abstract 355, "A new definition of limit" Bull. Amer. Math. Soc. 41 (1935) 636. (Received September 5, 1935)
According to the report of the meeting (Bull. Amer. Math. Soc. 42 (1936) 3) the paper was delivered at the AMS meeting in New York on October 26, 1935.
In the abstract we find what would nowadays be called convergence of a filter base. (See also Definition 4 in Birkhoff's 1937 paper.) Birkhoff remarked to me once that Bourbaki never acknowledged his (Birkhoff's) priority.
It seems that some time after Birkhoff's talk, his father (G. D. Birkhoff) remarked that it reminded him of a paper of Moore and Smith. So young Garrett read Moore and Smith, and in the end adopted their system for the subsequent paper, calling it "Moore-Smith convergence in general topology". Since that Annals of Mathematics paper was received April 27, 1936, one can only imagine young Garrett working furiously for 6 months converting his previous filter-base material into the Moore-Smith setting!
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Did Bourbaki acknowledge anyone 's priority? I remember his thanking tits in a footnote for the exercises on Coxeter groups &c. – Mariano Suárez-Alvarez Jul 11 2010 at 23:18
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http://math.stackexchange.com/questions/30477/intuitive-idea-of-expectation-of-random-variable
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# Intuitive idea of Expectation of random variable?
I'm studying probability for preparing myself for machine learning. I came across this notion of Expectation of (random variable) or (function of random variable) like $E[X]$ or $E[g(X)]$. Can anybody explain me the intuitive idea of the above notion (taking into consideration it's implications in machine learning, may be)
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## 3 Answers
Expected value means the average value of the random variable. It is a weighted average of the values of the function, weighted by the probabilities of taking those values.
Expected value is a linear map from some random variables to numbers. This map throws away all of the information about a random variable except one number. This number is the center of mass of the distribution.
In case the random variable is a statistic you can measure from a sample, you can imagine repeating independent samples many times. In that case, the Strong Law of Large Numbers (a theorem) says that the average of the results will be close to the expected value in that the limit of the averages will converge to the expected value. This does not mean that you need a statistic to come from a repeated experiment for the expected value to make sense. Expected value makes sense even if you can only observe a statistic once.
If your text doesn't spend a lot of time on the central and fundamental idea of expected value, it could be that the author assumes you have already read another text which has done this, and so you might look for a more elementary text.
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Thanks for the answer. So, in case of NORMAL mean we just add the elements (considering 100% probability of each of them, which's rarely the case in the real world) & divide them...etc, but in case of EXPECTATION, the probability of individual elements (contributing to average) counts, so that the the average makes a real-world (full of uncertain info.) sense. Is that so? Correct me if I'm wrong somewhere. – Amit L Apr 3 '11 at 5:02
Mean and expected value are interchangeable. Both are weighted averages. – Douglas Zare Apr 11 '11 at 10:18
Think of you operate a gaming machine in a casino and it has $0$ $10$ $20$ $30$ $40$ $50$ usd prizes with each prize having probability of $p_1,p_2,\ldots p_6$,
If X is random variable defining amount of the prize, each time a game playing person wins, then say $E[X] = 5$ means that, on average lets say $1000$ people played a game, casino would pay $5000$ from its cash.
And again assume that for playing a game each time you buy a ticket which is $5$ USD.
Now $X$ is the prize amount you win, $5$ USD ticket fee, now casino gives $X$ USD to you but you give $5$ USD to casino. Now casino loses $(5-X)$ amount of money each time if $x=0$ it wins $5$ USD if $x=50$ it loses $45$ USD. I as an owner of casino want to calcculate long term profit or loss I use $E[f(X)] = E[5-x]$, now if I already know $E[X]$ in some way it its easy to calculate $E[f(x)]$
Brief note: I'm not an expert in probability so some of my intuitions might be wrong anyone please correct if so.
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In calculating the average value of a random variable the underlying assumption is that all possible outcomes of the random variable are equally probable.
Let's say that random variable $X$ has $3$ possible outcomes, $x_1,x_2,x_3,$ and all three are equally probable.
The average value of variable $X$ is : $(x_1+x_2+x_3)/3$ or $0.333(x_1) + 0.333(x_2) + 0.333(x_3)$
Also the probability of each outcome is 0.3333. So the expectation is : $$E(X) = 0.333(x_1+x_2+x_3) = (x_1 +x_2 + x_3)/3$$
In the case where the outcomes were not equally probable and occurred with a probability of $p_1,p_2\text{ and } p_3$ respectively, then $$E(X) = p_1\cdot x_1 + p_2\cdot x_2 + p_3\cdot x_3$$ for the discrete random variable $X$.
Generalizing, we could write $$E(X) = \sum p_i\,x_i$$ where the sum is over all outcomes.
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Welcome to the site! I LaTeX-ified your answer. Please feel free to roll back my edits if you think I messed up your exposition. – Rick Decker Jun 20 '12 at 1:32
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http://mathhelpforum.com/calculus/70055-evaluate-integral.html
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# Thread:
1. ## Evaluate the integral.
$<br /> \int \limits_0^{\frac{\\pi}{2}} \frac{cosx}{1+sin^2(x)}dx<br />$
2. Originally Posted by saiyanmx89
$<br /> \int \limits_0^{\frac{\\pi}{2}} \frac{cosx}{1+sin^2(x)}dx<br />$
Let $u = \sin x, \;\;\; \text{so} \;\;\; du = \cos x \, dx$
so (note the change of limits)
$\int_0^1 \frac{1}{1 + u^2} \; du$
You can do it from here.
3. why does u=sinx? how did you get sinx?
4. $sin^2(x)$ is the same thing as $(sinx)^2$
So, (sinx) is what you are using for your u-substitution
5. thank you!!!
6. So, would the answer be:
arctan(sinx) +c ???
7. No!! This is a definite integral! It will not have a " +C" only indefinite integrals use the +C. You should get a real number answer after you integrate and plug in your limits.
Also, look at danny's post. When he did the u-substitution, he changed his limits based on "u" so you don't have to plug "sinx" back into the problem. You can just evaluate it by plugging your new limits in for "u"
Do you understand the u-substitution part and why the limits changed?
Ask if you need clarification on how to do that!
8. So, the answer would be found from:
arctan(1)-arctan(0)
??
I'm still new to changing the limits. I understand how to change them but, I usually don't know when they need to be changed...
9. Originally Posted by saiyanmx89
So, the answer would be found from:
arctan(1)-arctan(0)
??
[snip]
Yes.
10. No problem! u-substitution can be tricky to learn!
You have probably been taught that when you use u-substitution, that you plug the "u" value back in to your problem before you integrate it. It is okay to do that and you can get the right answer, but you can save yourself a lot of time and paper, by changing your limits of integration based on "u" and just integrating "u" instead of whatever "u" was.
Here are the steps one by one:
$\int\limits_{0}^{\frac{\pi}{2}} \frac{cos x}{1 + sin^2 x} dx$
$\int\limits_{0}^{\frac{\pi}{2}} \frac{cos x}{1 + (sin x)^2} dx$
Let u = sin x
du = cos x dx * This will get rid of the cos x on the top and also your dx
$\int\limits_{0}^{\frac{\pi}{2}} \frac{1}{1 + (u)^2} du$
**NOW** This is where you would usually replace the "u" with "sin x"
But, instead of doing that, you can create new limits based on "u." You do this by plugging the original limits into your u= sin x equation and solving for "u"
So,
u = sin x * Your original limits were 0 and $\frac{\pi}{2}$
Plug those in to "x" and solve for "u" Those values will be your new limits of integration
$u = sin(0) = 0$ *this is your new lower limit
$u = sin (\frac{\pi}{2}) = 1$ * this is your new upper limit
So now you rewrite your integral like this:
$\int\limits_{0}^{1} \frac{1}{1 + (u)^2} du$
Then you integrate as usual but in terms of du!!
$\int\limits_{0}^{1} \frac{1}{1 + u^2} du$
$= [\arctan(u)] ^1_0 = \arctan (1) - \arctan (0)$
Hope that helps you! Sorry it took me so long! I am SLOW at typing LaTex!
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http://physics.stackexchange.com/questions/tagged/mathematical-physics+quantum-mechanics
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# Tagged Questions
0answers
6 views
### what is the magnetic quadrupole operator?
To find magnetic or electrical moments in quantum theory we must calculate the expectation value of an appropriate operator. the dipoles operator are similar and is easy to find but the magnetic ...
0answers
5 views
### what is the magnetic quadrupole moment of a nuclei in cylenderical coordinates?
what is the magnetic quadruple moment of a nuclei in cylenderical coordinates? the quadrupole moment of a nuclei is zero in spherical coordinates but in the cylenderical coordinates it can't be ...
2answers
143 views
### In quantum mechanics(QM), can we define a high-dimensional “spin” angular momentum other than the ordinary 3D one?
Inspired by my previous question Questions about angular momentum and 3-dimensional(3D) space? and another relevant question How to define angular momentum in other than three dimensions? , now I get ...
0answers
80 views
### Quantum uncertainty can explain the Riemann Hypothesis?
In the recent paper "Riemann Hypothesis as an Uncertainty Relation" (http://arxiv.org/abs/1304.2435) the author claims that the presence of zeros out of the critical line may lead to the violation of ...
0answers
117 views
### Question about the HVZ theorem
In this paper1 the authors cite the HVZ theorem2 saying that it follows from the method used by M. Reed & B. Simon without modifications; I don't really understand this point. Is there anyone who ...
2answers
157 views
### A question on the existence of Dirac points in graphene?
As we know, there are two distinct Dirac points for the free electrons in graphene. Which means that the energy spectrum of the 2$\times$2 Hermitian matrix $H(k_x,k_y)$ has two degenerate points $K$ ...
2answers
67 views
### Schmidt basis: Entanglement
I do not understand how any state in Hilbert Space $\mathcal{H}=\mathcal{H}_A\otimes\mathcal{H}_B$ of dimension $\text{dim}(\mathcal{H}_A)\times\text{dim}(\mathcal{H}_B)$ can be decomposed in the ...
3answers
171 views
### Is a quantum system mandatory for generating true random sequence?
Is a quantum system necessary if we want to generate true random sequence? The mathematical framework used for classical mechanics doesn't involve any random value. But the mathematical framework of ...
1answer
54 views
### Can I prove boundedness of an operator without checking it for its whole domain?
(I don't have a direct reference so this is a little fishy and I'll delete it if nobody recognises what I'm talking about, but I though for starters I'll ask anyway) I've heard at university that if ...
1answer
110 views
### Physical Significance of Operator Norm/Spectral Norm of a Quantum Operator
Is there any physical significance of operator norm/spectral norm of a hermitian operator?
1answer
167 views
### “An operator is hermitian”. Implications?
Alastair Rae states that there are 4 postulates of Quantum Mechanics in his text on the subject matter. The first part of his second postulate can be stated as: Every dynamical variable may be ...
1answer
151 views
### What is the mathematical background needed for quantum physics? [duplicate]
I'm a computer scientist with a huge interest in mathematics. I have also recently started to develop some interest about quantum mechanics and quantum field theory. Assuming some knowledge in the ...
1answer
318 views
### Physical Significance of Fourier Transform and Uncertainty Relationships
What is the physical significance of a fourier transform? I am interested in knowing exactly how it works when crossing over from momentum space to co ordinate space and also how we arrive at the ...
1answer
370 views
### Intuitive meaning of Hilbert Space formalism
I am totally confused about the Hilbert Space formalism of Quantum Mechanics. Can somebody please elaborate on the following points: The observables are given by self-adjoint operators on the ...
2answers
234 views
### Intuition for Path Integrals and How to Evaluate Them
I'm just starting to come across path integrals in quantum field theory, and want to get the right intuition for the them from the start. The amplitude for propagation from $x_a$ to $x_b$ is typically ...
1answer
143 views
### Equivalent Representations of Clifford Algebra
I'm reviewing David Tong's excellent QFT lecture notes here and am a little confused by something he writes on page 94. We've considered the standard chiral representation of the Clifford Algebra, ...
1answer
169 views
### Expectation value calculation for a weird operator
In the paper Fundamental monopoles and multimonopole solutions for arbitrary simple gauge groups.- E weinberg I am not being able to see one of the calculation. The author states (eqn 3.26) \langle ...
2answers
201 views
### Angular Momentum Operators Non-Degenerate
Typically one writes simultaneous eigenstates of the angular momentum operators $J_3$ and $J^2$ as $|j,m\rangle$, where $$J^2|j,m\rangle = \hbar^2 j(j+1)|j,m\rangle$$ J_3 |j,m\rangle = \hbar ...
2answers
243 views
### WKB method of approximation
Would it be legitimate to use the WKB approximation for a particle in a spherically symmetric Gaussian potential? $$V(r)~=~V_0(1-e^{-r^2/a^2}).$$ I'm not sure when to use which approximation ...
2answers
357 views
### Non-Degeneracy of Eigenvalues of Number Operator for Simple Harmonic Oscillator [duplicate]
Possible Duplicate: Proof that the One-Dimensional Simple Harmonic Oscillator is Non-Degenerate? I'm trying to convince myself that the eigenvalues $n$ of the number operator ...
3answers
544 views
### Canonical Commutation Relations
Is it logically sound to accept the canonical commutation relation (CCR) $$[x,p]~=~i\hbar$$ as a postulate of quantum mechanics? Or is it more correct to derive it given some form for $p$ in the ...
4answers
279 views
### Generalized quantum mechanics
I wonder if it's possible to discover another version of quantum theory that doesn't depend on complex numbers. We may discover a formulation of quantum mechanics using p-adic numbers, quaternions or ...
3answers
250 views
### Takhatajan's mathematical formulation of quantum mechanics
So I began skimming L. Takhatajan's Quantum Mechanics For Mathematicians, and saw the mathematical formulation of QM that he uses (page 51). (The PDF file is available here.) I've only taken a basic ...
0answers
111 views
### Research in Quantum Physics [closed]
I am Suvankar, a student of engienering. My branch is Electrical & Electronics Engineering. Although this is a Physics oriented website, I want to know whether it is possible to do M.Sc. after ...
3answers
305 views
### Why we use $L_2$ Space In QM?
I asked this question for many people/professors without getting a sufficient answer, why in QM Lebesgue spaces of second degree are assumed to be the one that corresponds to the Hilbert vector space ...
1answer
164 views
### State normalization in Dirac's formulation of quantum mechanics
Let us divide the time $T$ into $N$ segments each lasting $δt = T/N$. Then we write $\langle q_F | e^{−iHT} |q_I \rangle = \langle q_F | e^{−iHδt} e^{−iHδt} . . . e^{−iHδt} |q_I \rangle$ Our ...
1answer
141 views
### Problem in Hamiltonian
I need to elaborate the equation ,and need to know what is the physical significance and how matrices will manipulate in the equation \hat{H} = (\hat{\tau_3}+i\hat{\tau_2})\frac{\hat{p}^2}{2m_0}+ ...
1answer
187 views
### Question on Sakurai's treatment of the Harmonic Oscillator:
In Section 2.3 of the second edition of Modern Quantum Mechanics (which discusses the harmonic oscillator), Sakurai derives the relation $$Na\left|n\right> = (n-1)a\left|n\right>,$$ and states ...
4answers
565 views
### Reason for the discreteness arising in quantum mechanics?
What is the most essential reason that actually leads to the quantization. I am reading the book on quantum mechanics by Griffiths. The quanta in the infinite potential well for e.g. arise due to the ...
2answers
130 views
### Quantum Mechanics in terms of *-algebras
I'm currently trying to find my way into the geometric description of Quantum Mechanics. I therefor started reading: Geometry of state spaces. In: Entanglement and Decoherence (A. Buchleitner et ...
3answers
352 views
### Can we have discontinuous wavefunctions in the Infinite Square well?
The energy eigenstates of the infinite square well problem look like the Fourier basis of L2 on the interval of the well. So then we should be able to for example make square waves that are an ...
2answers
240 views
### What are the solution spaces of Nonlinear Schrödinger equations?
As we know, the solution space of Schrödinger equation is a Hilbert space, however, what about it of Nonlinear Schrödinger equations such as i\partial_t\psi=-{1\over ...
3answers
248 views
### If I go to the church of the greater Hilbert space, can I have Unitary Collapse?
Actually, unitary pseudo-collapse? Von Neuman said quantum mechanics proceeds by two processes: unitary evolution and nonunitary reduction, also now called projection, collapse and splitting. ...
1answer
267 views
### The Hermiticity of the Laplacian (and other operators)
Is the Laplacian operator, $\nabla^{2}$, a Hermitian operator? Alternatively: is the matrix representation of the Laplacian Hermitian? i.e. \langle \nabla^{2} x | y \rangle = \langle x | ...
2answers
185 views
### Quantum mechanics on Cantor set?
Has quantum mechanics been studied on highly singular and/or discrete spaces? The particular space that I have in mind is (usual) Cantor set. What is the right way to formulate QM of a particle on a ...
0answers
54 views
### Is it possible to use the properties of quantum mechnics to develop a computer that develop mathmatical theory? [closed]
Is it possible to use the properties of quantum mechnics to develop a computer that develop mathmatical theory? I want some reference please, because i want to get into very detailed. Thanks in ...
2answers
266 views
### Is there an analogue of configuration space in quantum mechanics?
In classical mechanics coordinates are something a bit secondary. Having a configuration space $Q$ (manifold), coordinates enter as a mapping to $\mathbb R^n$, $q_i : Q \to \mathbb R$. The primary ...
1answer
311 views
### Quantum mechanic newbie: why complex amplitudes, why Hilbert space?
I'm just starting learning quantum mechanics by myself (2 "lectures" so far) and I was wondering why we need to define quantum states in a complex vector space rater than a real one? Also I was ...
3answers
375 views
### Existence of creation and annihilation operators
In a multiple particle Hilbert space (any space of any multi-particle system), is it sufficient to define creation and annihilation operators by their action (e.g. mapping an n-particle state to an ...
1answer
331 views
### Schwinger representation of operators for n-particle 2-mode symmetric states
A bosonic (i.e. permutation-symmetric) state of $n$ particles in $2$ modes can be written as a homogenous polynomial in the creation operators, that is \left(c_0 \hat{a}^{\dagger n} + c_1 ...
1answer
78 views
### When does the “norm of quasi-eigenvectors” matter in calculations? For which physical results are these even used?
Which physical system in nonrelativistic quantum mechanics is actually described by a model, where the norm of the "position eigenstate" (i.e. the delta distribution as limit of vectors in the ...
1answer
183 views
### interpretation of Green function
Is there a physical interpretation of the existence of poles for a Green function? In particular how can we interpret the fact that a pole is purely real or purely imaginary? It's a general question ...
1answer
120 views
### Variational wavefunctions and “spread” of potential in quantum mechanics
A particle in a box has an energy that decreases with the size of the box. In the general case, it is often said that a variational solution for a "narrow and deep" potential is higher in energy than ...
3answers
405 views
### What areas of physics should a mathematician study to understand TQFT?
I am studying topological quantum field theory from the view point of mathematics.(axiomatic treatise) So it has no explanation about physics. I would like to know physic background of TQFT. But I ...
2answers
574 views
### Basic Question - Green's Functions in Quantum Mechanics
I am trying to learn about Green's functions as part of my graduate studies and have a rather basic question about them: In my maths textbooks and a lot of places online, the basic Greens function G ...
2answers
949 views
### Proof that the One-Dimensional Simple Harmonic Oscillator is Non-Degenerate?
The standard treatment of the one-dimensional quantum simple harmonic oscillator (SHO) using the raising and lowering operators arrives at the countable basis of eigenstates \$\{\vert n \rangle\}_{n = ...
2answers
696 views
### A book on quantum mechanics supported by the high-level mathematics
I'm interested in quantum mechanics book that uses high level mathematics (not only the usual functional analysis and the theory of generalised functions but the theory of pseudodifferential operators ...
1answer
476 views
### General procedure for Clebsch-Gordan expansions
I'm wondering if the Clebsch-Gordan series generalize to any orthonormal set of basis functions? If so, how would one go about deriving an expression for an arbitrary set of basis functions (perhaps ...
1answer
34 views
### Tip of a spreading wave-packet: asymptotics beyond all orders of a saddle point expansion
This is a technical question coming from mapping of an unrelated problem onto dynamics of a non-relativistic massive particle in 1+1 dimensions. This issue is with asymptotics dominated by a term ...
1answer
302 views
### Bra space and adjoint vectors
If I'm not wrong, a bra, $\langle \phi_n |$, can be thought as a linear functional that when applied to a ket vector, $| \phi_m \rangle$, returns a complex number; that is, the inner product it's a ...
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http://math.stackexchange.com/questions/tagged/contest-math?sort=faq&pagesize=15
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# Tagged Questions
Many people have their first exposure to real math in the form of a mathematics competition. Math contests show the competitors that math is a subject that makes great use of creativity. For those already initiated, math contests are a source of fantastic problems that often lead to extended ...
3answers
485 views
### How many rationals of the form $\large \frac{2^n+1}{n^2}$ are integers?
How many rationals of the form $\large \frac{2^n+1}{n^2},$ $(n \in \mathbb{N} )$ are integers? The possible values of $n$ that i am able to find is $n=1$ and $n=3$, so there are two solutions ...
2answers
407 views
### Euclidean Geometry Intersection of Circles
Two circles intersect in the Cartesian Coordinate system at points $A$ and $B$. Point $A$ lies on the line $y=3$. Point $B$ lies on the line $y=12$. These two circles are also tangent to the x-axis at ...
6answers
1k views
### Contest problem about convergent series
The following is probably a math contest problem. I have been unable to locate the original source. Suppose that $\{a_i\}$ is a set of positive real numbers and the series \sum_{n = 1}^\infty ...
1answer
270 views
### Prove that $\sum_{k=1}^n \frac{2k+1}{a_1+a_2+…+a_k}<4\sum_{k=1}^n\frac1{a_k}.$
Prove that for $a_k>0,k=1,2,\dots,n$, $$\sum_{k=1}^n \frac{2k+1}{a_1+a_2+\ldots+a_k}<4\sum_{k=1}^n\frac1{a_k}\;.$$
4answers
519 views
### Seemingly invalid step in the proof of $\frac{a^2+b^2}{ab+1}$ is a perfect square?
Recall the famous IMO 1988 question 6: Suppose that $\displaystyle\frac{a^2+b^2}{ab+1}=k\in\mathbb{N}$ for some $a,b\in\mathbb{N}$. Show that $k$ is a perfect square. Solutions can be found: ...
3answers
2k views
### The easy(?) part of IMO 2011 Problem 3
Let $f : \mathbb R \to \mathbb R$ be a real-valued function defined on the set of real numbers that satisfies $$f(x + y) \leq yf(x) + f(f(x))$$ for all real numbers $x$ and $y$. How can I prove that ...
4answers
503 views
### Irrational painting device
Part a) of the following problem appeared in one of the Putnam Exams (sorry, don't know which year exactly). If you want to solve Part a) don't read Part b). You have a painting device, which given ...
1answer
215 views
### Mathematical problem with square numbers in the decimal system
Moderator Note: this is a question from the Federal Mathematics Competition 2013. Good morning, here's another (pretty difficult) mathematical problem... The task may sound a little strange (I'm ...
1answer
276 views
### Is it possible for the number created by ordering $1$ to $n$ where $n > 1$ be a palindrome?
Is it possible for the number created by the consecutive numbers $1$ to $n$ where $n > 1$ be a palindrome eg. $1234567\ldots n$? I believe this is a contest problem, but how would one solve ...
1answer
166 views
### Smallest possible value on Fibonacci Function
Moderator Note: this is an open problem on brilliant.org Suppose $f$ is a polynomial with integer coefficients, such that for all non-negative integers $n$ the $n$-th Fibonacci number $u_n$ ...
1answer
195 views
### Applying a Function to Square Matrices
Moderator Note: This question is from a contest which ended 1 Dec 2012. Consider a polynomial $f$ with complex coefficients. Call such $f$ broken if we can find a square matrix $M$ such that \$M ...
3answers
228 views
### Expected Number of Coin Tosses to Get Five Consecutive Heads
A fair coin is tossed repeatedly until 5 consecutive heads occurs. What is the expected number of coin tosses?
1answer
86 views
### Uniformly distributed probability problem
May you have an idea for the following exercise I found from some olympiad. Each day you have to bring home one full can of water. To do so you go to the next well and make the can full. On the way ...
1answer
480 views
### Pennies on a checkerboard.
Here is a question on pennies on checkerboard. It isnt a homework question. I saw it in a book. ...
1answer
257 views
### How to find all rational numbers satisfy this equation?
Find all rational number $a,b,c$ satisfy: $$a+b+c=abc$$ I try to change this in different forms like $(ab-1)c = a+b$, $(ac-1)b = a+c$, $(cb-1)a = b+c$ etc but it won't help...
1answer
170 views
### Game Theory Matching a Deck of Cards
Moderator Note: This question is from a contest which ended 1 Dec 2012. Suppose we have a deck of cards labeled from $1$ to $52$. Let them be shuffled in a random configuration, then made ...
1answer
215 views
### What is shortcut to this contest algebra problem about polynomial?
The polynomial $P(x)=x^4 + ax^3 + bx^2 +cx + d$ has the property that $p(k)=11k$ for $k=1,2,3,4$. Compute $c$. The answer is $-39$.
1answer
530 views
### Does there exist a sequence $\{a_n\}_{n\ge1}$ with $a_n < a_{n+1}+a_{n^2}$ such that $\sum_{n=1}^{\infty}a_n$ converges?
Does there exist a sequence $\{a_n\}_{n\ge1}$ with $a_n < a_{n+1}+a_{n^2}$ such that $\sum_{n=1}^{\infty}a_n$ converges? Does there exist a sequence with the same property but with each term ...
4answers
584 views
### Solving for the implicit function $f\left(f(x)y+\frac{x}{y}\right)=xyf\left(x^2+y^2\right)$ and $f(1)=1$
How can I find all functions $f:\mathbb{R}\to\mathbb{R}$ such that $f(1)=1$ and $$f\left(f(x)y+\frac{x}{y}\right)=xyf\left(x^2+y^2\right)$$ for all real numbers $x$ and $y$ with $y\neq0$? PS. This is ...
6answers
689 views
### Find all polynomials $P$ such that $P(x^2+1)=P(x)^2+1$
Find all polynomials $P$ such that $P(x^2+1)=P(x)^2+1$
3answers
320 views
### Finding all integer solutions of $5^x+7^y=2^z$
Find all integers $x,y,z$ such that $5^x+7^y=2^z$. This one comes from an online contest that I arranged some years ago, and I can assure that a completely elementary solution exists.
2answers
1k views
### Olympiad calculus problem
This problem is from a qualifying round in a Colombian math Olympiad, I thought some time about it but didn't make any progress. It is as follows. Given a continuous function \$f : [0,1] \to ...
2answers
504 views
### Find a way from 2011 to 2 in four steps using a special movement
USAMTS 6/2/22 states: The roving rational robot rolls along the rational number line. On each turn, if the robot is at $\frac{p}{q}$, he selects a positive integer $n$ and rolls to ...
5answers
766 views
### Probability of random integer's digits summing to 12
What is the probability that a random integer between 1 and 9999 will have digits that sum to 12? As a user suggested, I could make a spreadsheet and count them, but is there a quicker way to do ...
2answers
471 views
### Translations of Kolmogorov Student Olympiads in Probability Theory
I am deeply interested in Kolmogorov's probability contest whose tests could be found in English for the five first years but there is no English translation to its problems from round 6 onward. I ...
1answer
448 views
### Contest problems with connections to deeper mathematics
We all know that problems from for example the IMO and the Putnam competition can sometimes have lovely connections to "deeper parts of mathematics". I would want to see such problems here which you ...
2answers
477 views
### How do people come up with difficult math Olympiad questions?
The problems that appear in difficult math competitions such as the IMO or the Putnam exam are usually very difficult and require some ingenuity to solve. They also usually don't look like they can be ...
3answers
682 views
### Olympiad Inequality Problem
Consider three positive reals $x,y,z$ such that $xyz=1$. How would one go about proving: $$\frac{x^5y^5}{x^2+y^2}+\frac{y^5z^5}{y^2+z^2}+\frac{x^5z^5}{x^2+z^2}\ge \frac{3}{2}$$ I really dont know ...
1answer
220 views
### Maximizing the volume of a rectangular prism
A rectangular prism has a surface area of $300$ square inches. What whole number dimensions give the prism the greatest volume? This is a math olympiad problem. It involves the volume and surface ...
2answers
476 views
### Proving there are an infinite number of pairs of positive integers $(m,n)$ such that $\frac{m+1}{n}+\frac{n+1}{m}$ is a positive integer
The question is: Show that there are an infinite number of pairs $(m,n): m, n \in \mathbb{Z}^{+}$, such that: $$\frac{m+1}{n}+\frac{n+1}{m} \in \mathbb{Z}^{+}$$ I started off approaching this ...
3answers
145 views
### Is there a non-constant function $f:\mathbb{R}^2 \to \mathbb{Z}/2\mathbb{Z}$ that sums to 0 on corners of squares?
A problem in the 2009 Putnam asks about functions $f:\mathbb{R}^2 \to \mathbb{R}$ such that whenever $A,B,C,D$ are corners of some square we have $f(A)+f(B)+f(C)+f(D)=0$. Without spoiling the problem ...
3answers
379 views
### Expected value uniform decreasing function
Moderator Note: This is part of a current contest. We are given a function $f(n,k)$ as for(i=0;i < k;i++) n = rand(n); return n; rand is defined as a ...
3answers
164 views
### Evaluate $\lim_{x\to\infty}\left(1+\frac{\ln x}{f(x)}\right)^{\displaystyle\frac{f(x)}{x}}$
Let's consider the function $f:\mathbb{R}\rightarrow(0,\infty)$, with $f(x)\cdot \ln f(x)=e^x$, $\forall x \in \mathbb{R}$. Then compute \lim_{x\to\infty}\left(1+\frac{\ln ...
1answer
183 views
### (Olympiad) Minimum number of pairs of friends.
I gave up, my approaches didn't work (induction, pigeon-hole, parity; though obviously there's a good chance I didn't use them cleverly): In a group of 12 people, every pair of them has a common ...
2answers
125 views
### Functional Equation: a little tricky
Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $f[f(x)^2+f(y)]=xf(x)+y$ for all real numbers $x$ and $y$. Clearly $f(x)=x$ is a solution, check by substitution. I'm at a loss as ...
1answer
228 views
### Proving or disproving $f(n)-f(n-1)\le n, \forall n \gt 1$, for a recursive function with floors.
The Olympiad-style question I was given was as follows: A function $f:\mathbb{N}\to\mathbb{N}$ is defined by $f(1)=1$ and for $n>1$, by: ...
2answers
692 views
### Let a; b; c and d be non-negative numbers such that a+b+c+d = 4. Prove that 4/(abcd) ≥ a/b + b/c + c/d + d/a
How would I approach this using only the AM - GM inequality? Are there any other methods that does not involve the AM-GM inequality?
0answers
210 views
### What is a way to do this combinatorics problem that could generalize to do any of problems similar to this but with more path?
A bug travels from $A$ to $B$ along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never ...
1answer
70 views
### Sequence $a_k=1-\frac{\lambda^2}{4a_{k-1}},\ k=2,3,\ldots,n$.
Consider the sequence $a_1, a_2,\ldots,a_n$ with $a_1=1$ and defined recursively by $$a_k=1-\frac{\lambda^2}{4a_{k-1}},\ k=2,3,\ldots,n.$$ Find $\lambda>1$ such that $a_n=0$. The answer is ...
3answers
416 views
### Finding a diagonal in a trapezoid given the other diagonal and three sides
The figure below is a trapezoid, what is the length of the red line? Thank you very much in advance!
2answers
185 views
### A question with the sequence $e_{n}=\left(1+\frac{1}{n}\right)^{n}$
Prove that $a$) the following sequence is increasing $$e_{n}=\left(1+\frac{1}{n}\right)^{n},\quad n\ge1;$$ $b$) the inequality below holds $$e_{n} \leq3,\quad n\ge1.$$
1answer
234 views
### Multiplication Table with a frame and picture of equal sum
Is there an $n \times n$ multiplication table such that if you form a border of width $k$ ("the frame") and sum its elements, the total will equal the sum of the remaining elements ("the picture")? ...
1answer
115 views
### Estimate variance given a sample of size one (7th Kolmogorov Student Olympiad)
This is problem 10 of the seventh Kolmogorov Student Olympiad in Probability Theory as translated by Jonathan Christensen in this thread. Given a sample of size one from the random variable \$\xi ...
1answer
212 views
### Putnam A-6: 1978: Upper bound on number of unit distances
Let n distinct points in the plane be given.prove that fewer than $2n^\frac{3}{2}$ pairs of them are at unit distance apart
2answers
370 views
### Exploring Properties of Pascal's Triangle $\pmod 2$
Moderator Note: This question is from a contest which ended 1 Dec 2012. Consider Pascal's Triangle taken $\pmod 2$: For simplicity, we will call a finite string of 0's and 1's proper if it ...
5answers
485 views
### Least wasteful use of stamps to achieve a given postage
You have sheets of 42-cent stamps and 29-cent stamps, but you need at least \$3.20 to mail a package. What is the least amount you can make with the 42- and 29-cent stamps that is ...
2answers
117 views
### Resource for Vieta root jumping
I can't seem to find a good resource on Vieta's root jumping, which would explain various scenarios that suggest using the technique. Does anyone have a suggestion for a reference? P.S. Not certain ...
3answers
216 views
### prove that $\text{rank}(AB)\ge\text{rank}(A)+\text{rank}(B)-n.$
If $A$ is a $m \times n$ matrix and $B$ a $n \times k$ matrix, prove that $$\text{rank}(AB)\ge\text{rank}(A)+\text{rank}(B)-n.$$ Also show when equality occurs.
1answer
95 views
### A quick question on general mathematics
I have the following question that I am currently unable to satisfactorily answer myself. My question is: Does the inequality \frac{a}{b} + \frac{b}{a} < \frac{f(a)}{f(b)} + ...
2answers
150 views
### Understanding Less Frequent Form of Induction? (Putnam and Beyond)
I won't paste the question here since my problem is not a technical one but a conceptual one. Book is here: (Page 22 of the pdf) I do not understand why it is necessarily to induct $2^{k}$ to show ...
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http://mathoverflow.net/questions/100693
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## Can we extend a continuous function with keeping Hausdorff dimension?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let X be a compact subset of R^d, and K be a compact subset of X, such that Dim_H(X)=Dim_H(K). Let F be a continuous function on K, Can we extend F from K to X, with keeping the continuous and the Hausdorff dimension of the gragh.
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What is the range of the function $F$? Is it also Euclidean space? Without specifying the range, the "Hausdorff dimension of the graph" makes no sense. – Lee Mosher Jun 26 at 16:30
Yes,I am sorrry, I mean taht F is a real-valued continuous function. – luka Jun 27 at 3:36
## 1 Answer
Yes, because you can extend any continuous mapping defined on $K$ to the whole space $\mathbb R^d$ so that it is locally Lipschitz outside $K$. Now the graph of $F$ on $X \setminus K$ has the same dimension as $X \setminus K$.
### Overly complicated way to construct the extension:
First take a Whitney decomposition of $Q\setminus K$, where $Q \subset \mathbb R^d$ is some dyadic cube containing $K$. Then enumerate the decomposition cubes $Q_i$ so that the diameter of $Q_i$ is decreasing. Next iteratively define $F$ on $Q_i$ as follows: For each corner $x$ of the cube $Q_i$ define $F(x)$ to be the value of $F$ at one of the points on $$K \cup \bigcup_{j < i} Q_j$$ which is closest to $x$ and then extend $F$ piecewise linearly to $Q_i$.
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(Please note that one has to be careful with the above construction: for example one should not select cubes to the decomposition which agree with the boundary of $K$ at other points besides the corners, one has to make sure that the extension agree on faces etc. All these things should be doable, but I left such details out.) – Tapio Rajala Jun 27 at 6:35
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http://physics.stackexchange.com/questions/2696/do-all-massive-bodies-emit-hawking-radiation
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# Do all massive bodies emit Hawking radiation?
It is known that any accelerated observer is subject to a heat bath due to Unruh radiation. The principle of equivalence suggests that any stationary observer on the surface of a massive body should also experience heat bath. Indeed by substituting surface gravity g into the formula for Unruh radiation one can find the black body temperature for a hypothetical hyper-cool planet:
$$T = \frac{\hbar g}{2\pi c k}$$
which is $3.9766×10^{-20}\ K$ for Earth
one even can find the time which it will take for Earth to evaporate: $5.69×10^{50}$ years.
Since the heat in the super-cold Earth cannot come out of nothing one should assume that it will come from decay of particles due to a certain mechanism.
Sometimes I heared an argument that an event horizon is needed for Hawking radiation to exist. But this can be countered by assumption of possibility of decay due to quantum virtual black holes (which inevitably should appear due to uncertainty principle, and the more massive and dense body is the greather concentration of virtual black holes inside it will be, eventually becoming similar to the concentration of bobbles inside a body of boiling water). Or just suggest that any massive body due to uncertainty principle can quantum tunnel into a black hole state so to emit Hawking radiation.
So what is the conclusion here?
• Can we say that all massive bodies are surrounded by the atmosphere of heated vacuum?
This is a weaker preposition: thermal state of surrounding vacuum does not mean energy transfer if the system is in thermodynamic equilibrium.
• Any body gradually evaporates, i.e. transfers its energy to the surrounding vacuum until completely vanishes?
This is a stronger preposition and suggests emission of radiation al loss of mass.
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I would warn you against extrapolating black hole temperature to estimating evaporation times. the back reaction problem (quantum effects create matter, which then perturbs the spacetime, which then change the quantum effects) in semi-classical gravity is extremely non-trivial, and once the star has radiated a large portion of its mass, the back-reaction effects will not be negligible. The premise of this problem, where the quantum radiation interacts with a matter distribution, is even more complicated. – Jerry Schirmer Jan 11 '11 at 2:08
Actually this estimation is mostly inaccurate because the temperature will decrease as the body evaporates (due to decrease of surface acceleration), unlike a BH which evaporates at accelerating rate. Thus one should better estimate the "half-evaporation" time for Earth rather than the expected time of complete evaporation which is infinite. The figure in the question is just for illustration, it is calculated out of presumption that the rate of evaporation does not change. One can better think of it as of an approximate time of evaporation of Earth mass from Saturn. – Anixx Aug 20 '11 at 8:41
## 6 Answers
The answer is clearly no, but it is interesting to see what goes wrong with the argument. I think the problems lies in the distinction between Unruh’s effect and Hawking’s. In Unruh’s case one has to be careful what is and what is not implied: an accelerated detector in empty flat space will behave as if immersed in a heat bath, in that it detects particles distributed thermally. However, this does not imply there is radiation, in the sense of energy flow from one place to another, empty space is empty, even in Rindler coordinates. It is only for a real black hole that you have real radiation emanating from the horizon.
(I vaguely remember a decent discussion of this in Birrell and Davies).
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Also presuming that black hole radiates and a body with identical mass but not a black hole does not, suggests that there are two different kinds or vacuum around them, an absurd idea. – Anixx Jan 11 '11 at 7:01
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1. Hawking formula has to do with the horizon, e.g the temperature has to do with the surface gravity, etc. It does not apply to objects with no horizons. – user566 Jan 11 '11 at 7:39
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2. Body as dense as a black hole is a black hole. Body with identical mass but much larger, is a different object which does not radiate. – user566 Jan 11 '11 at 7:40
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3. Virtual black holes are an hypothetical process, which if exists has some interesting consequence, but I don’t think that radiation from any massive object whatsoever is one of them. Placing the vacuum in a gravitational gradient is not sufficient to create Hawking radiation. – user566 Jan 11 '11 at 7:46
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To repeat the answer to the “two kinds of vacuum” comment - the details of a body consists of more than its mass, there are many many possible mass distributions with identical total mass, and the vacuum around them, as well as many other details, depend on the mass distribution. There is nothing absurd by saying that Hawking radiation depends on more than that one number, the total mass. It manifestly does, as the radiation is zero without an horizon. – user566 Jan 11 '11 at 15:50
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My previous answer is beside the point now that the question has been edited. There is a simpler example of a question of this type which has been analyzed in great detail in the literature, and that involves an electric charge which is stationary in a gravitational field. Since the power radiated is nonzero if a charge is accelerated one might, by the equivalence principle, expect such a charge to radiate. I believe the answer is that it does not, but that the analysis is subtle. You could start with the article D. Boulware, "Radiation from a uniformly accelerated charge", Annals of Physics 124 (1980), 169-187 and work your way forward from there by looking at citations of that paper.
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OK Then imagine a star which consists only of electrons. It becomes a black hole, it evaporates... – Anixx Jan 11 '11 at 0:56
– Sklivvz♦ Jan 11 '11 at 1:05
@ Jeff Harvey You did not understand my question. My question was not whether electron is stable (it probably is), but whether an observer on the surface of a massive planet will observe Hawking radiation. – Anixx Jan 11 '11 at 1:20
I rewrote the question. But if a planet consists only of cold hydrogen at near-zero temperature, one cannot suggest other source of energy other than proton decay. – Anixx Jan 11 '11 at 1:44
@Anixx: gravitational contraction of the planet would be another possible source of energy. – Jerry Schirmer Jan 11 '11 at 2:05
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There's a simple argument that massive bodies that are not black holes cannot emit Hawking radiation. Consider a single proton in its ground state. It does not emit radiation because if it were, it would have to decay to some lower-energy state. This eventually may happen when the proton decays, but the radiation is not black-body radiation, so it can't be considered Hawking radiation.
Now, let's consider a very large crystal of diamond in its lowest-energy state. Similarly, it cannot emit any radiation until one of its protons decays, in which case it will probably emit some high-energy gamma rays. So any radiation it is emitting is not black-body Hawking radiation.
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This answer neglects insanely improbable p+e to two photons which is gravitationally allowed. – Ron Maimon Jan 1 '12 at 17:08
@Ron: you're absolutely right. That's possibly more probable than proton decay, and needless to say it also will not produce black-body Hawking radiation. – Peter Shor Jan 1 '12 at 17:13
This answer only considers a black-body radiation but if a proton decays due to virtual black hole formation, the resulting photons will not belong to a black body spectrum, but still technically this can be considered Hawking radiation. That is if the black hole is too small its Hawking radiation is not necessarily following the black body distribution. – Anixx Jan 1 '12 at 17:42
@Anixx: a single proton is too small to count as a black hole (although if it decays via a virtual black hole, maybe you could consider it Hawking radiation). A massive crystal of diamond--the lowest energy state of a large number of carbon atoms--will be too big for the processes of proton decay or electron capture to look anything like Hawking radiation. – Peter Shor Jan 2 '12 at 14:59
The answer is no, because there is no horizon. A time-independent gravitational field emits no particles, because energy is conserved, just like a stationary charge distribution does not radiate. This is consistent with the equivalence principle, as explained below.
In the Unruh effect, the particle emmission is from the "black wall" acceleration horizon behind the observer. If you place a refrigerated baffle between the black wall and the observer, the radiation will be absorbed by the baffle, and will not be seen by the observer. An observer standing on a planet is indistinguishable from an accelerating observer with a planet between him and the acceleration horizon, and such an observer sees no emissions from the horizon, because the horizon is invisible.
You might object that the horizon is infinite, and the planet is finite, so shouldn't you be able to see the horizon far enough out? But if you go far enough out, you notice the curvature caused by the planet, and this will push the horizon out to infinity, because there is no horizon after all, you're just standing on a planet.
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It seems you did not read the question. Your stand that "no horizon=no radiation" has been already addressed by a conjecture that the horizon can emerge virtually. – Anixx Aug 15 '11 at 12:02
I didn't say no horizon means no radiation--- I said if there is stuff between you and the horizon, you see the stuff, not the emissions from the far-away virtual horizon. The virtual black hole business is silly--- a virtual black holes can pop in and pop out, but it conserves energy and doesn't lead to any detectible particle production unless there is a horizon visible. Every static solution has a conserved energy. – Ron Maimon Aug 18 '11 at 6:12
"a virtual black holes can pop in and pop out, but it conserves energy" - why it should conserve energy? why matter cannot decay through VBH fourmation? – Anixx Aug 19 '11 at 7:29
I would think that you can have some evaporation for massive objects without horizons. The usual way to think of black hole evaporation (see wikipedia) is that you get spontaneous particle pair creation near the horizon. Then particle '1' get sucked in and particle '2' escapes the black hole. Energy in total is conserved, so the black hole loses mass.
Now think of this planet. We have no horizon, but we do have a gravitational field. So when a spontaneous pair gets created (say outside the atmosphere), there is a (hyper small) chance that one will get sucked into the planet, while the other flies free, lowering the mass of the earth. I think that this is what you are talking about with your 'virtual black holes'.
So what would be the rate? Perhaps less than one particle per lifetime of the universe. The type of particle involved may also need to have special properties, e.g.: be extremely heavy/light and be charge free.
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As the question is currently phrased, the answer is a straightforward one. Hawking radiation does not occur with any object except one with a horizon, i.e. a black hole. The arguments leading to Hawking radiation are subtle, but, conceptually, the idea (via Susskind in The Black Hole War) is that quantum jitters become thermal jitters at the event horizon. (Susskind's explanation is much better, of course.)
To be a bit more complete and concrete, consider a standard quantum fluctuation of energy arising from the uncertainty relation between energy and time. Fluctuations on sufficiently small time scales will be large enough to result in particle-antiparticle pair creation. At the horizon of a black hole, the pair can be separated so that one particle falls into the black hole, while the other is shot out. This transforms a virtual particle (pair) into a real particle, which is then able to interact with other matter.
Now, one could say that this process is possible in any system with non-zero mass. However, the detailed and subtle mathematics of the process require an event horizon in order to occur. As a thought experiment, imagine that any mass could produce this effect. It would result in a wide-scale violation of the Uncertainty Principle because we would see changes in energy that exceeded those allowable over the associated time frame.
One other objection is that black hole temperature, which is tied to Hawking radiation, increases as the mass of the black hole decreases. If "normal" mass were to behave in a similar way, elementary particles would have tremendous "black-hole-like" temperatures.
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"It would result in a wide-scale violation of the Uncertainty Principle because we would see changes in energy that exceeded those allowable over the associated time frame." - can you please elaborate on this? It is known that particles decay due to energy fluctuations indeed occurs (with nuclear forces, for example), even if total energy of the potential barrier. It is not evident why this process cannot happen with a massive object. – Anixx Jan 12 '11 at 2:30
Regarding your second comment that this would lead to any particle to have very large temperature, you are wrong. Yes, the temperature of a black hole rises as its radius decreases, but only because the surface gravitational acceleration also increases. The Hawking temperature is proportional to surface gravity. As any particle has very tiny mass, its surface gravity is also very small and as such it cannot emit much Hawking radiation. That said, the radius does not matter, what matters is the surface gravity. – Anixx Jan 12 '11 at 2:34
@Anixx My reference to violation of the Uncertainty Principle is not with regard to particle decay, but particle-antiparticle pair creation. If pairs did not annihilate and return their borrowed energy to the vacuum, we would see quantum fluctuations in energy that are not allowable. – Mitchell Jan 12 '11 at 2:45
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@Anixx: how much quantitative general relativity or quantum field theory do you know? – Jerry Schirmer Jan 12 '11 at 3:30
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The qualitative answer depends on quantitative physics. – Jerry Schirmer Jan 12 '11 at 18:10
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http://physics.stackexchange.com/questions/13638/transformation-of-angles-in-special-relativity/13644
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# Transformation of angles in special relativity
My question is about the problem below
Depicted are two space ships (the USS Voyager and the USS Enterprise), each with velocity $v=c/2$ relative to the space station (Babylon 5). At the exact moment the two space ships are closest together (at a distance d) the USS Enterprise fires off a shuttle with velocity $u=3c/4$ relative to the space station.
The question is: Under what angle $\alpha'$ (as measured by USS Enterprise) must the shuttle be fired off in order to meet the USS Voyager?
I see two possible ways to approach this:
1. Calculate the angle $\alpha$ measured by the space station Babylon 5 and figure out how angles transform when we change our frame of reference.
2. Figure out the velocity in x-/y-direction in the frame of reference of the USS Enterprise and deduce the angle $\alpha'$ from that.
Now, if both ways are correct they should give the same answer. But they don't seem to, so what is wrong?
Here is what I have tried:
Let $u_x$, $u_y$ be the velocities (in the corresponding directions) the shuttle needs to have - as observed by Babylon 5. Let $u_x'$, $u_y'$ be the velocities of the shuttle observed by USS Enterprise.
Then by the rules for adding velocities:
\begin{eqnarray*} u_y' &=& \frac{u_y + v}{1 + \frac{vu_y}{c^2}} \\ u_x' &=& \frac{u_x }{1 + \frac{vu_y}{c^2}}\cdot \sqrt{1-\frac{v^2}{c^2}} \end{eqnarray*}
So this gives
$$\tan(\alpha') = \frac{u_y'}{u_x'} = \frac{u_y + v}{u_x} \cdot\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$
But on the other hand, we could imagine the guys on Babylon 5 drawing a big triangle to show the crew on USS Enterprise the trajectory the shuttle needs to take.
The people on Babylon 5 will then draw a triangle with angle $\alpha$ satisfying $$\tan(\alpha) = \frac {u_y}{u_x}$$
So, if side $a$ of this triangle is parallel to the x-axis and $b$ is parallel to the y-axis, then we have $$\frac ba = \tan(\alpha) = \frac{u_y}{u_x}$$
Since for the USS Enterprise $a$ is the same, but side $b$ is contracted, they will see a traingle with side $b' = b/\gamma$ and $a' = a$. Therefore
$$\tan(\alpha') = \frac{b'}{a'} = \frac ba \sqrt{1-\frac{v^2}{c^2}} = \frac{u_y}{u_x}\sqrt{1-\frac{v^2}{c^2}}$$
These are two different results in general. I cannot figure out what is wrong with either of them...
Thanks for reading, help will be greatly appreciated! :)
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O.k. I think that the second result is wrong: If $\alpha=0$ (i.e. $u_y=0$) then the shuttle must be fired off with a component in the direction opposite to the direction the USS Enterprise is moving. This is exactly what the first result says, but it is not at all what the second result says. I'm still unsure as to what exactly is the problem with the approach leading to the second formula but I think it has something to do with not taking the movement of USS Enterprise into account somehow. – Sam Aug 16 '11 at 17:52
## 2 Answers
Let's look at the Gallilean case first (to which this problem must reduce in non-relativistic limit anyway). Your first approach works the same way but we can get rid of unnecessarily complicated terms. In particular, we have $\tan(\alpha) = \frac{v}{u_x}$ and $\tan(\alpha') = \frac{2v}{u_x}$ (I've used that $u_y = v$ which is true also relativistically). You should convince yourself that this is the correct transformation.
Now, for the second approach we get that $\alpha = \alpha'$ since there are no contractions in Gallilean case. The conflict with the first approach rests on the fact that since the triangle is formed using the velocity vector, it will change when boosted. Therefore it's not valid to deduce that the triangle only transforms by contraction in one direction (which reduces to identity in non-relativistic limit). In fact, this contracting effect is quite negligible when compared to the deformation of the triangle caused by the boost.
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"... since the triangle is formed using the velocity vector, it will change when boosted." I think this is exactly right. Thanks! =) – Sam Aug 16 '11 at 20:10
There's an inconsistency with the way the velocity composition formulas are consistent with the axis labeled as in the diagram, but your $\tan$ equations are consistent with the axis labeled the other way around. I'll stick with the unconventional way the vertical axis is labeled $x$, the horizontal $y$ in your diagram, so
$\tan(\alpha') = \frac{u_y'}{u_x'} = \frac{u_y + v}{u_x} \cdot\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$
should be
$\tan(\alpha') = \frac{u_x'}{u_y'} = \frac{u_x}{u_y + v} \cdot\sqrt{1-\frac{v^2}{c^2}}$
Now let's look at your second method. The Lorentz contraction (LC) formula is a relationship between space intervals for the same pair of events in different frames, but simultaneous in one. I'll leave you to prove their time interval in the other frame $$\Delta t' = \gamma \frac V {c^2} \Delta x$$ Therefore, you need to be very clear in how you're using the LC formula to relate the space intervals of which pair of events.
A better way would be to find the coordinates of the event of the shuttle reaching USS voyager in the USS Enterprise frame using the Lorentz transformation$$y'=\gamma(y + Vt) = \gamma(b + Vt) = \gamma(u_yt + Vt),\qquad x'= x = a = u_xt$$ giving $$tan\alpha' = \frac {x'} {y'} = \frac{u_x}{\gamma(u_y + v)} \cdot$$ So the two methods are consistent with one another, as expected.
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http://mathoverflow.net/questions/81176/hyperbolicity-of-a-fundamental-group
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## Hyperbolicity of a fundamental group
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let G be the fundamental group of a compact 3-manifold which supports on its interior a complete non positively curved Riemannian metric and is a cilinder near de metric. Is G hyperbolic?
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What is the last part of the first sentence supposed to say? At any rate, the fundamental group of a flat 3-torus is not hyperbolic. – Richard Kent Nov 17 2011 at 14:01
You are right, then change nonpositively by negatively – Luis Jorge Nov 17 2011 at 14:24
It is a standard result in Riemannian geometry that the fundamental group of a compact Riemannian manifold with strictly negative curvature has hyperbolic fundamental group since the universal cover is $CAT(k)$ for some $k < 0$. Does that answer your question, or are you asking about something else? – Paul Siegel Nov 17 2011 at 14:32
Actually I want to know if groups of that form (the way I difined in the first question) satisfy the k.theoretic farrell jones conjecture. I have another question. In the case of zero curvature what kind of groups may occur? – Luis Jorge Nov 17 2011 at 14:43
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If you only require a negatively curved metric on the interior then $G$ need not be hyperbolic. For example, the figure eight knot complement with the hyperbolic metric (a) has constant negative curvature and (b) is the interior of a compact three-manifold. But $G$ is not hyperbolic because the peripheral group is ${\mathbb{Z}}^2$. – Sam Nead Nov 17 2011 at 18:11
## 1 Answer
Yes, this follows from the geometrization theorem. A negatively curved complete manifold is atoroidal, and the manifold is irreducible, so there is a complete hyperbolic metric on the interior by geometrization.
In your comments, you ask about zero curvature. Then it is just a Euclidean manifold, with fundamental group a Bieberbach group.
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I think he means word-hyperbolic (I didn't say the question made any sense...) – Igor Rivin Nov 17 2011 at 17:56
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I see Igor - I always confuse 3-manifolds and their fundamental groups. – Agol Nov 17 2011 at 19:34
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http://math.stackexchange.com/questions/85830/how-to-use-the-extended-euclidean-algorithm-manually/252923
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# How to use the Extended Euclidean Algorithm manually?
I've only found a recursive algorithm of the extended Euclidean algorithm. I'd like to know how to use it by hand. Any idea?
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I'm not sure what you mean. The Extended Euclidean Algorithm is inherently recursive. When you use it by hand, you use it recursively. – Jim Belk Nov 26 '11 at 18:13
– Martin Sleziak Nov 26 '11 at 18:14
– J. M. Nov 26 '11 at 18:16
33% accept rate is not much of admiring in my opinion. – Quixotic Nov 26 '11 at 18:29
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@Andrew, thanks for taking the time to accept answers in your old questions. – Jim Belk Nov 26 '11 at 20:06
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## 3 Answers
Perhaps the easiest way to do it by hand is in analogy to Gaussian elimination or triangularization, except that, since the coefficient ring is not a field, one has to use the division / Euclidean algorithm to iteratively descrease the coefficients till zero. In order to compute both $\rm\,gcd(a,b)\,$ and its Bezout linear representation $\rm\,j\,a+k\,b,\,$ we keep track of such linear representations for each remainder in the Euclidean algorithm, starting with the trivial representation of the gcd arguments, e.g. $\rm\: a = 1\cdot a + 0\cdot b.\:$ In matrix terms, this is achieved by augmenting (appending) an identity matrix that accumulates the effect of the elementary row operations. Below is an example from one of my old posts. It computes the Bezout representation for $\rm\:gcd(80,62) = 2\$ viz. $\ 7\cdot 80\: -\: 9\cdot 62\ =\ 2\:.\:$
````For example, to solve m x + n y = gcd(m,n) one begins with
two rows [m 1 0], [n 0 1], representing the two
equations m = 1m + 0n, n = 0m + 1n. Then one executes
the Euclidean algorithm on the numbers in the first column,
doing the same operations in parallel on the other columns,
Here is an example: d = x(80) + y(62) proceeds as:
in equation form | in row form
---------------------+------------
80 = 1(80) + 0(62) | 80 1 0
62 = 0(80) + 1(62) | 62 0 1
row1 - row2 -> 18 = 1(80) - 1(62) | 18 1 -1
row2 - 3 row3 -> 8 = -3(80) + 4(62) | 8 -3 4
row3 - 2 row4 -> 2 = 7(80) - 9(62) | 2 7 -9
row4 - 4 row5 -> 0 = -31(80) -40(62) | 0 -31 40
The row operations above are those resulting from applying
the Euclidean algorithm to the numbers in the first column,
row1 row2 row3 row4 row5
namely: 80, 62, 18, 8, 2 = Euclidean remainder sequence
| |
for example 62-3(18) = 8, the 2nd step in Euclidean algorithm
becomes: row2 -3 row3 = row4 when extended to all columns.
````
````In effect we have row-reduced the first two rows to the last two.
The matrix effecting the reduction is in the bottom right corner.
It starts as 1, and is multiplied by each elementary row operation,
hence it accumulates the product of all the row operations, namely:
````
$$\left[ \begin{array}{ccc} 7 & -9\\ -31 & 40\end{array}\right ] \left[ \begin{array}{ccc} 80 & 1 & 0\\ 62 & 0 & 1\end{array}\right ] \ =\ \left[ \begin{array}{ccc} 2 & 7 & -9\\ 0 & -31 & 40\end{array}\right ]$$
````Notice row 1 is the particular solution 2 = 7(80) - 9(62)
Notice row 2 is the homogeneous solution 0 = -31(80) + 40(62),
so the general solution is any linear combination of the two:
n row1 + m row2 -> 2n = (7n-31m) 80 + (40m-9n) 62
The same row/column reduction techniques tackle arbitrary
systems of linear Diophantine equations. Such techniques
generalize easily to similar coefficient rings possessing a
Euclidean algorithm, e.g. polynomial rings F[x] over a field,
Gaussian integers Z[i]. There are many analogous interesting
methods, e.g. search on keywords: Hermite / Smith normal form,
invariant factors, lattice basis reduction, continued fractions,
Farey fractions / mediants, Stern-Brocot tree / diatomic sequence.
````
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1
– Gone Jun 28 '12 at 20:24
You may like to check this and this.
Also, there is a well known table method which is very easy and fast for the manual solution.
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This is more a comment on the method explained by Bill Dubuque then a proper answer in itself, but I think there is an remarque so obvious that I dont understand that it is hardly ever made in texts discussing the extended Euclidean algorithm. This is the observation that you can save youself half of the work by computing only one of the Bezout coefficients. In orther words, instead of recording for every new remainder $r_i$ a pair of coefficients $k_i,l_i$ so that $r_i=k_ia+l_ib$, you need to record only $k_i$ such that $r_i\equiv k_ia\pmod b$. Once you will have found $d=\gcd(a,b)$ and $k$ such that $d\equiv ka\pmod b$, you can then simply put $l=(d-ka)/b$ to get the other Bezout coefficient. This simplification is possible because the relation that gives the next pair of intermediate coefficients is perfectly independent for the two coefficients: say you have $$\begin{aligned} r_i&=k_ia+l_ib\\ r_{i+1}&=k_{i+1}a+l_{i+1}b\end{aligned}$$ and Euclidean division gives $r_i=qr_{i+1}+r_{i+2}$, then in order to get $$r_{i+2}=k_{i+2}a+l_{i+2}b$$ one can take $k_{i+2}=k_i-qk_{i+1}$ and $l_{i+2}=l_i-ql_{i+1}$, where the equation for $k_{i+2}$ does not need $l_i$ or $l_{i+1}$, so you can just forget about the $l$'s. In matrix form, the passage is from $$\begin{pmatrix} r_i&k_i&l_i\\ r_{i+1}&k_{i+1}&l_{i+1}\end{pmatrix} \quad\text{to}\quad \begin{pmatrix} r_{i+2}&k_{i+2}&l_{i+2}\\ r_{i+1}&k_{i+1}&l_{i+1}\end{pmatrix}$$ by subtracting the second row $q$ times from the first, and it is clear that the last two columns are independent, and one might as well just keep the $r$'s and the $k$'s, passing from $$\begin{pmatrix} r_i&k_i\\ r_{i+1}&k_{i+1}\end{pmatrix} \quad\text{to}\quad \begin{pmatrix} r_{i+2}&k_{i+2}\\ r_{i+1}&k_{i+1}\end{pmatrix}$$ instead.
A very minor drawback is that the relation $r_i=k_ia+l_ib$ that should hold for every row is maybe a wee bit easier to check by inspection than $r_i\equiv k_ia\pmod b$, so that computational errors could slip in a bit more easily. But really, I think that with some practice this method is just as safe and faster than computing both coefficients. Certainly when programming this on a computer there is no reason at all to keep track of both coefficients.
A final bonus it that in many cases where you apply the extended Euclidean algorithm you are only interested in one of the Bezout coefficients in the first place, which saves you the final step of computing the other one. One example is computing inverse modulo a prime number $p$: if you take $b=p$, and $a$ is not divisible by it, then you know beforehand that you will find $d=1$, and the coefficient $k$ such that $d\equiv ka\pmod p$ is just the inverse of $a$ modulo $p$ that you were after.
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http://mathoverflow.net/questions/46279/if-q-is-a-subset-of-the-plane-of-size-less-than-continuum-then-does-every-closed
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## If Q is a subset of the plane of size less than continuum, then does every closed F in Q extend to a closed connected G in the plane with the same trace on Q? (Or is this independent of ZFC?)
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This question arises in connection with this MO question and especially with Sergei Ivanov's wonderful answer, which showed that for any countable set $Q\subset\mathbb{R}^2$ and every closed set $F\subset Q$, there is a closed connected $G\subset\mathbb{R}^2$ with $G\cap Q=F$. (In fact, he makes $G$ path-connected.)
My question is about the extent to which this phenomenon might generalize to higher cardinals, when the Continuum Hypothesis fails. For example, if the continuum $2^\omega$ is very large, then can we hope to handle uncountable sets $Q$ in the way Sergei handled the countable sets, provided that they have size less than the continuum? Or perhaps the best possible is always just the countable sets? Or is this independent of ZFC?
It seems sensible to introduce what seems to be a new cardinal characteristic here. Specifically, let $\kappa$ be the size of the smallest counterexample, that is, the smallest cardinal size of a set $Q\subset\mathbb{R}^2$ having a closed subset $F\subset Q$ for which there is no closed connected $G\subset\mathbb{R}^2$ with $G\cap Q=F$.
Sergei proved that this cardinal $\kappa$ is uncountable, and obviously $\kappa$ is at most the continuum (it is easy to make counterexamples of size continuum), and so $$\omega_1\leq\kappa\leq 2^\omega.$$ So the question is, what can we say about $\kappa$ in ZFC?
If the Continuum Hypothesis holds, of course, then the two endpoints above are identical and so $\kappa=2^\omega$. But is it consistent with ZFC that $$\omega_1\leq \kappa\lt 2^\omega?$$ Perhaps one can achieve particular values of $\kappa$ by forcing? Is the cardinal $\kappa$ related to other well-known cardinal characteristics? Perhaps the value of $\kappa$ is pushed up to the continuum $2^\omega$ by some of the standard forcing axioms?
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For clarity, when I refer to a closed set $F\subset Q$, what I mean is that $F$ is closed in the subspace $Q$; equivalently, $\bar F\cap Q=F$. So the issue is whether every $\bar F$ extends to a closed connected $G$ with the same trace on $Q$. – Joel David Hamkins Nov 17 2010 at 1:15
## 4 Answers
It seems to me that Sergei Ivanov's proof can be generalized to show that Martin's Axiom for countable posets implies that $\kappa$ is the continuum. By the characterization of $cov(\mathcal{M})$ (the smallest number of meager sets required to cover the real line) as the smallest cardinal for which MA(countable) fails, it follows that $cov(\mathcal{M})\leq\kappa$.
Given closed $F\subseteq Q$ both of size less than continuum we define a countable poset $\mathbb{P}$ as follows. First fix a countable collection $\mathcal{U}$ of open balls so that for any rational $q\in\mathbb{Q}^2$ and any rational $\epsilon_1<\epsilon_2$ there is $U\in\mathcal{U}$ centered at $q$ with some radius $\epsilon$ such that $\epsilon_1<\epsilon<\epsilon_2$ and the boundary of $U$ is disjoint from $Q$. (We can do this because $Q$ has size less than continuum and there are continuum many choices for $\epsilon$).
Now let $\mathbb{P}$ be the collection of finite disjoint unions of members of $\mathcal{U}$ which are disjoint from $F$. Then $\mathbb{P}$ is countable. For each $x\in X$, the set $D_x$ of $p$ with $x$ in $p$ is dense; we prove this as follows. Let $p\in\mathbb{P}$ and $x\in Q\setminus F$ be given. Take $\delta$ so that the $\delta$-ball $O$ around $x$ is disjoint from $F$ and $p$ (possible because $x$ doesn't lie on the boundary of any of the balls comprising $p$). Pick a rational $q$ within $\delta/3$ of $x$. Then there is a member $U$ of $\mathcal{U}$ insides $O$ and containing $x$. So $q=p\cup U$ belongs to $D_x$.
Now if $G$ is a filter intersecting each $D_x$, then $\cup G$ is a collection of pairwise disjoint balls disjoint from $F$ and containing every member of $Q\setminus F$. Let $C$ be the complement of the union. Then $C$ is as desired; it is path connected by exactly Sergei Ivanov's argument (some people seemed concerned about the radii of the balls but it doesn't appear to matter).
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Fantastic! I was playing around with much larger partial orders having the same goal, but you seem to have found the right way to do it. Since all nontrivial countable forcing is isomorphic to adding a Cohen real, this is a very weak forcing axiom. So this shows that it is consistent with ZFC that the continuum is large and Sergei's fact still holds for all sets of size less than the continuum. Now we need the converse result, where $\kappa$ is small and the continuum is large... – Joel David Hamkins Nov 17 2010 at 10:20
1
Congratulations! I was trying to do this with balls centered at the points of $Q-F$, with radii in appropriate countable sets. But not only isn't that poset countable, it doesn't even seem to be ccc, so not even the full MA would complete the proof. – Andreas Blass Nov 17 2010 at 14:41
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Andreas, I was doing the same thing with the same result---it is definitely not ccc if Q-F is large. I think Justin's answer is really great, and deserves many up-votes. – Joel David Hamkins Nov 17 2010 at 15:05
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
I suspect that Sergei Ivanov's proof can be extended to the case when $\mathbb R^2 \backslash Q$ is connected. I also suspect that the only case when $\mathbb R^2 \backslash Q$ is disconnected is precisely when $Q$ is a continuum.
(Consider this "answer" a comment, but with the details filled in, it would imply that $\kappa = 2^\omega$.)
Edit: I'm not so sure about the second point as the first: What is the cardinality of the smallest set $Q$ such that $\mathbb R^2 \backslash Q$ is disconnected?
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It seems that Sergei's argument used the countablility of $Q$, so I'm not clear on your proposal. (For your question in the edit, any set of size less than continuum has path-connected complement, since there is a continuous foliation of disjoint paths from $a$ to $b$, and one of them must be OK.) – Joel David Hamkins Nov 16 2010 at 22:48
1
Your second point is correct; removing fewer than continuum many points will leave the rest pathwise connected (by polygonal paths consisting of just two line segments). I'm not at all convinced by your first point, because Sergei Ivanov's proof depended on being able to choose disks (around the omitted points) that stay away from previously chosen disks, of which there were only finitely many. It will be much harder when there are infinitely many. Note also that, when there are uncountably many of these disks, uncountably many of them will have radii bounded away from zero. – Andreas Blass Nov 16 2010 at 22:51
The way I see the proof is that the countability and the covering of $Q \backslash F$ with disks was merely a means to avoid the "bad" points in $Q$ when path-connecting the points in $F$. Perhaps this works: For any $Q$ that has cardinality less than a continuum $\mathbb R^2 \backslash (Q \backslash F)$ is path connected. Let $G$ be a path in this set that passes through every point in $F$. Conveniently, paths are closed, so $F$ is closed. – trutheality Nov 16 2010 at 23:12
* I meant $G$ is closed. – trutheality Nov 16 2010 at 23:13
@trutheality: If $Q=F=\mathbb{Q}^2$ then how do you construct a closed path which passes through every point of $\mathbb{Q}^2$? – Guillaume Brunerie Nov 16 2010 at 23:49
show 4 more comments
I think you do not use the fact, that there is no point on the boundary of any of the balls. You only use the fact that $\kappa$ is less than continuum in the definition of MA. (actually the generic covering will cover everything from $G$ \ $F$ and will be disjoint from $F$)
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The no-points-on-the-boundary property is used in showing that the complement is path-connected, since you draw the straight line, and then whenever it is obstructed by a circle, you follow the boundary of the circle around the obstruction; and it is also used to know that the circles do not overlap, since otherwise you'd want to place another circle covering the boundary point. – Joel David Hamkins Nov 17 2010 at 15:08
Yes, partly you are right, I still think it is not a problem if there is an $F$ point on the boundary, since not $G$ \ $Q$ is connected, just $G$. (or can we make $G$ \ $Q$ connected? probably.) But it is a problem if there are some $Q$ \ $F$ points on the boundary.
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I'm studying soliton (solitary waves). They are many theory which explain the phenomenon, like sine-Gordon model. But sine-Gordon model has limitations when it applies to 4 dimension because it is ...
2answers
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### Derrick’s theorem
Consider a theory in D spatial dimensions involving one or more scalar fields $\phi_a$, with a Lagrangian density of the form L= \frac{1}{2} G_{ab}(\phi) \partial_\mu \phi_a \partial^\mu \phi_b- ...
1answer
34 views
### Creating a state of definite momentum and position(within uncertainty limit)
claim: $a^{\dagger}$= $\int d^{3}kf_{1}(k)a^{\dagger}(k)$ Creates a state with Localized momentum $k_{1}$and localized position near origin; where $f_{1}(k)$ \$\propto ...
1answer
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### sine-Gordon equation
I have derived a solition equation (2 dimensions) from scalar field theory $$\varphi(x) = v\tanh\Bigl(\tfrac{1}{2}m(x - x_0)\Bigr),\tag{1}$$ and also I have got sine-Gordon equation for solition ...
1answer
206 views
### Crazy Dirac Deltas
I'm not expecting any rigor in the following and the answers...since we're dealing with Dirac deltas in the context of QFT. Consider the integral \int d^4q\ \Theta(q_0)\Theta(p_{3,0}+q_0)\ ...
1answer
424 views
### Emergent symmetries
As we know, spontaneous symmetry breaking(SSB) is a very important concept in physics. Loosely speaking, zero temprature SSB says that the Hamiltonian of a quantum system has some symmetry, but the ...
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http://mathhelpforum.com/advanced-algebra/93297-change-basis-matrix.html
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# Thread:
1. ## Change of Basis Matrix
Hey guys, I have a question on change of basis.
Find the change of basis matrix from the standard basis {(1,0,0),(0,1,0),(0,0,1)} to {(-1,1,1),(1,-1,1),(1,1,-1)}.
I'm having some trouble in figuring out what to do. Thanks in advance!
2. Let $[a,b,c]=p[-1,1,1]+q[1,-1,1]+r[1,1,-1]$
giving
$a=-p+q+r$
$b= p-q+r$
$c= p+q-r$
which solves to
$p=\frac{b+c}{2}$
$q=\frac{a+c}{2}$
$r=\frac{b+a}{2}$
3. How is this a matrix though? I'm guessing there should be real numbers.
4. The "change of basis" matrix must map the coefficients of a given vector, as written in one basis, into the coefficients of the same vector, written in the other basis.
For example, the vector [1, 0, 0], in the first basis would be written, of course, as 1[1, 0, 0]+ 0[0, 1, 0]+ 0[0, 0, 1] so its coefficents are [1, 0, 0] (which is why it is the 'standard' basis). In terms of the other basis, [1, 0, 0]= a[-1, 1, 1]+ b[1, -1, 1]+ c[1, 1, -1]= [-a+b+c, a-b+c, a+b-c]. So we must have -a+ b+ c= 1, a- b+ c= 0, a+ b- c= 0. Adding the first two equatins, 2c= 1 so c= 1/2. Adding the last two equations, 2a= 0 so a= 0, Adding the first and last equations, 2b= 1 so b= 1/2. The coefficients are [0, 1/2, 1/2].
But it should be easy to see that if we multiply any matrix by [1, 0, 0] we get $\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}= \begin{bmatrix}a \\ d\\ f\end{bmatrix}$, the first column of the matrix. So the first column of the "change of basis" matrix must be $\begin{bmatrix}0 \\\frac{1}{2}\\ \frac{1}{2}\end{bmatrix}$. Do the same with the second and third vectors in each basis to get the second and third columns.
5. Originally Posted by HallsofIvy
So the first column of the "change of basis" matrix must be $\begin{bmatrix}0 \\\frac{1}{2}\\ \frac{1}{2}\end{bmatrix}$.
I don't understand how you got $\begin{bmatrix}0 \\\frac{1}{2}\\ \frac{1}{2}\end{bmatrix}$. How do we know what d and g are in the "change of basis" matrix?
Originally Posted by HallsofIvy
Do the same with the second and third vectors in each basis to get the second and third columns.
Shouldn't the first vector that you multiply by be (-1,1,1) and not (1,0,0)? And then the second and third vectors would be (1,-1,1) and (1,1,-1) respectively? Just a bit confused but thanks for taking the time to help.
6. Originally Posted by Fel
I don't understand how you got $\begin{bmatrix}0 \\\frac{1}{2}\\ \frac{1}{2}\end{bmatrix}$
I'm like you here.
Also, if I understood well, Halls made a little mistake when he wrote $\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}= \begin{bmatrix}a \\ d\\ f\end{bmatrix}$, I think it should be $\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}= \begin{bmatrix}a \\ d\\ g\end{bmatrix}$.
7. I am not sure that I fully understand what you need.
The matric $C = \left[ {\begin{array}{rrr}<br /> { - 1} & 1 & 1 \\<br /> 1 & { - 1} & 1 \\<br /> 1 & 1 & { - 1} \\<br /> <br /> \end{array} } \right]^{ - 1} = \frac{1}<br /> {2}\left[ {\begin{array}{ccc}<br /> 0 & 1 & 1 \\<br /> 1 & 0 & 1 \\<br /> 1 & 1 & 0 \\<br /> <br /> \end{array} } \right]$ when applied to any point will the “coordinates” in the new base.
Example: $<br /> C \cdot \left[ {\begin{array}{r}<br /> 2 \\<br /> { - 4} \\<br /> 6 \\<br /> <br /> \end{array} } \right] = \left[ {\begin{array}{r}<br /> 1 \\<br /> { - 4} \\<br /> 1 \\ \end{array} } \right]$ so $\left\langle {2, - 4,6} \right\rangle = \left\langle { - 1,1,1} \right\rangle + 4\left\langle {1, - 1,1} \right\rangle - \left\langle {1,1, - 1} \right\rangle$
8. Originally Posted by Plato
The matric $C = \left[ {\begin{array}{rrr}<br /> { - 1} & 1 & 1 \\<br /> 1 & { - 1} & 1 \\<br /> 1 & 1 & { - 1} \\<br /> <br /> \end{array} } \right]^{ - 1} = \frac{1}<br /> {2}\left[ {\begin{array}{ccc}<br /> 0 & 1 & 1 \\<br /> 1 & 0 & 1 \\<br /> 1 & 1 & 0 \\<br /> <br /> \end{array} } \right]$
How did you get that inverse? I keep getting something else.
9. Originally Posted by Fel
How did you get that inverse? I keep getting something else.
I used a computer algebra system.
Attached Thumbnails
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http://math.stackexchange.com/questions/239096/probability-for-winning?answertab=oldest
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# probability for winning
it is a well known game played in india.is there any theory for winning this game.it is possible where it falls How it works:in a shuffled deck the dealer offers a cut. You then bet on that card will be dealt on the inside or the outside. He then deals the cards one a time alternating between two spots; one spot called inside, the other spot called outside. He keeps dealing until the card that was originally cut appears. If it appears on the outside those who bet that spot win and the others lose; same is true if it appear on the inside (inside bettors win / outside bettors lose).
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If I understand the description correctly, the probability of winning is $\frac12$ for either bet: it’s equivalent to betting on the toss of a fair coin. – Brian M. Scott Nov 17 '12 at 8:20
do you explain elobarately.if i loss then where should i stand on same side or simply walkout – mohanasundaram Nov 17 '12 at 8:25
It makes no difference: each play is independent of the previous plays, just as each coin toss is independent of the previous tosses. Your probability of winning is always $\frac12$ (unless the dealer is cheating somehow). – Brian M. Scott Nov 17 '12 at 8:26
pls explain in mathematical analysis – mohanasundaram Nov 17 '12 at 8:34
There really isn’t much to explain, but give me a few minutes, and I’ll write up an answer. – Brian M. Scott Nov 17 '12 at 8:44
## 1 Answer
This game is essentially equivalent to betting on the toss of a fair coin: each of the possible outcomes, inside and outside, occurs with probability $\frac12$ unless the dealer is cheating somehow. To see this, suppose that the dealer deals first to the inside pile and then to the outside pile. Then the cut card will end up on the inside pile if its position in the deck is an odd number, and it will end up on the outside pile if its position in the deck is an even number. (Cards $1,3,5,\dots$ go on the inside pile; cards $2,4,6,\dots$ go on the outside pile.) Half of the cards are in odd-numbered positions and half are in even-numbered positions, and all positions in the deck are (in theory) equally likely to be cut, so the probability of cutting a card in an odd-numbered position is $\frac12$; the same goes for cards in even-numbered positions.
Each cut is independent of any earlier cuts: the deck has no ‘memory’ of the earlier cuts, just as a tossed coin has no ‘memory’ of earlier tosses. Thus, each game is a fresh start, just as if it were the first game. Consequently there is no strategy: no matter which way you bet, your chance of winning is $\frac12$.
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http://mathhelpforum.com/differential-geometry/189131-how-modify-hahn-banach-seperation-theorem.html
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# Thread:
1. ## how to modify Hahn-Banach Seperation Theorem?
Suppose X is a real normed space, B is a non-empty open convex subset of X, E is a linear subspace of X, B don't intersect with E.
Then there exists a linear functional f, such that f(x)=0 when x belong to E, and f(x)>0 when x belong to B.
i don't know how to use Hahn-Banach Seperation Theorem to get this.Who can help me?Thx a lot.
2. ## Re: how to modify Hahn-Banach Seperation Theorem?
We can find a linear functional $g$ such that $g(x)<g(y)$ for all $x\in B$ and $y\in E$, and since $E$ is a linear space we have for all $\alpha\in\mathbb R, g(x)<\alpha g(y)$, and considering $\alpha$ positive/ negative, we get $g(y)=0$ for all $y\in E$, and $g(x)<0$ for all $x \in B$. Now consider $f:=-g$.
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http://math.stackexchange.com/questions/3861/what-is-your-favorite-proof-that-eix-has-a-period-of-2-pi/3865
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# What is your favorite proof that $e^{ix}$ has a period of $2\pi$?
as a function of a real variable, apparently. Part of the freedom in choosing a proof is that you get to choose what definition of $e^{ix}$ to start from -- do you use a differential equation? a power series? a definition in terms of trig functions? Another bit of freedom is that you get to choose what definition of $\pi$ to start from.
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Even if x is complex, $\exp(ix)$ is still periodic. – J. M. Sep 2 '10 at 9:05
No takers for using the ODE?! – Mariano Suárez-Alvarez♦ Sep 2 '10 at 12:48
1
Maybe I didn't really frame the question clearly enough, but these are all kind of boring proofs. Perhaps I'd get some more interesting responses on MathOvervlow? – user1678 Sep 3 '10 at 6:55
I suppose the recognition that e^ix = cosx + isinx from Taylor series is really surprising when you first see it. But I find it inelegant because you're reasoning about an infinite number of terms (granted, they're really simple terms) in order to understand something about a finite number of functions. I kind of assumed there'd be another proof out there that doesn't rely so heavily on Taylor series... – user1678 Sep 3 '10 at 7:02
## 3 Answers
$$e^{ix} = \cos x + i \sin x \ .$$
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That pushes the question back: how do you define cosine and sine, and then--based on that definition--how do prove that their common period is 2 pi? One would like a definition that makes it relatively easy to prove the important properties of exp, especially that it is an entire analytic function of the complex plane. – whuber Sep 2 '10 at 16:07
You're right, but assuming the knowledge of cosine, sine and Euler identity it's an streamlined proof, isn't it? :-) – Agustí Roig Sep 2 '10 at 17:32
Well, yes, but then it's a tautology. Now you have to prove that the lcm of the periods of sine and cosine equals $2 \pi$! – whuber Sep 2 '10 at 22:10
1
Define (cos x, sin x) to be the rotation image of (1, 0) by x about the origin. They are thus obviously periodic with period 2π. The Maclaurin series for sine and cosine follow from this definition. Define exp(z) by its power series and it follows that $e^{ix}=\cos x+i\sin x$. – Isaac Sep 2 '10 at 22:14
My favorite has always been Walter Rudin's proof in the prologue to his "Real and Complex Analysis" (2nd Ed.). Here's a sketch:
• Define $\exp$ in terms of the power series.
• By manipulating the series, deduce that $\exp$ is a homomorphism from the additive group to the group of complex units.
• Show it satisfies the usual first order ODE.
• Define $\cos$ and $\sin$ as the real and imaginary parts of $\exp$, respectively.
• Define $\pi$ as twice the smallest positive real root of $\cos$.
• Deduce that $\exp( i \pi / 2) = i$.
• By multiplying, conclude that $2 \pi i$ is a period of $\exp$.
• Show, by means of the preceding properties, that no smaller period exists.
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$$e^{ix} = e^{i(x+T)} = e^{ix}e^{iT}$$
We have to find $T$ for which $e^{iT} = 1$
$$\rightarrow cos(T) + isin(T) = 1$$
$$\rightarrow sin(T) = 0$$ for all $$T = 2n\pi , n = 0,1,2,3...$$
So, period is $2\pi$.
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Can you explain the last line? You go from $T = 2n\pi$ to $2\pi$ – Tyler Hilton Sep 10 '10 at 18:51
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http://math.stackexchange.com/questions/67/what-is-an-elliptic-curve-and-how-are-they-used-in-cryptography?answertab=oldest
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# What is an elliptic curve, and how are they used in cryptography?
I hear a lot about Elliptic Curve Cryptography these days, but I'm still not quite sure what they are or how they relate to crypto...
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## 3 Answers
Here is a super nice powerpoint on the subject!
http://www.math.brown.edu/~jhs/Presentations/WyomingEllipticCurve.pdf
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10
Looks like a pdf to me. – Robin Chapman Sep 5 '10 at 10:38
1
@Robin A PDF of slides from a presentation actually – yydl Jun 22 '11 at 2:06
But obviously not a pdf created with Powerpoint. PP does not use Computer Modern. – Henning Makholm Sep 9 '11 at 16:28
The technical definition is a nonsingular projective curve of genus 1, which is an abelian variety under the group law: basially, this means that you draw the line through two points on the curve -- which can be embedded in the projective plane -- and find where that line intersects the curve again (and call that the negative of the sum).
We can always put elliptic curves in the (projectivization of the) form $y^2 = x^3 - Ax + B$.
So, the meaning of "abelian variety" is that you can add points on the elliptic curve, which is really useful; there isn't a way to do this for most objects in algebraic geometry. Then one can study things like the torsion points on an elliptic curve, with respect to this abelian group structure: it's a theorem that there are $m^2$ torsion points of order $m$ for instance, if you 're working in an algebraically closed field.
In fact, one way to think of this is that an elliptic curve is really--algebraically and topologically--a torus if you are working over the complex numbers, and the torsion points in the torus are easily determined. (Namely, a torus is algebraically the product of two copies of the unit circle.)
This also yields the theorem about torsion points for algebraically closed fields of characteristic zero via the "Lefschetz principle." (For characteristic p, you need a different argument.)
Other things one can consider include the group of points with coordinates in, say, the rational numbers (assuming the curve is defined by rational coefficients). One of the central theorems is that this group is finitely generated. The point is that the geometry of the elliptic curve leads to a rich algebraic structure.
That's a bit about elliptic curves; I know nothing about cryptography and can't comment on that.
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This is a nice answer and I feel sheepish that I took the lazy route. Kudos. – BBischof Jul 21 '10 at 4:39
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http://physics.stackexchange.com/questions/18062/how-do-electromagnetic-waves-carry-quantised-energy/18068
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# How do electromagnetic waves carry quantised energy?
If an electron oscillates about a mean position, it will create a time varying electric filed which in turn will create a time varying magnetic field and so on to create an electromagnetic wave. How does this wave carry energy , in which amount and how can this be quantised?
I know that energy is quantised when we see it through the particle nature of EM waves but how can it be defined in terms of time varying electric and magnetic fields?
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## 3 Answers
It is a difficult thing to visualize and connected with the wave-particle duality of photons. I think what you are interested in is the Second Quantization. This is where an electromagnetic wave is decomposed into its Fourier modes and each Fourier mode can be interpreted as simple harmonic oscillator. The energy levels of such oscillators corresponds to $E = nh\nu$, where each electromagnetic mode with that energy is a state that has $n$ photons with energy $h\nu$.
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Unless I'm making a mistake (somebody correct me if I'm wrong), that electromagnetic wave is identical to the wave of probability amplitude. (The word "amplitude" is the key. Think of it as a complex number spinning around in a circle.)
The way you get a particle out of it is by being uncertain of it's frequency.
If its frequency is totally certain, then its position is totally uncertain, and it's just infinitely "spread out".
If its frequency has a distribution, like a gaussian distribution about a mean, that's equivalent to an infinite sum of probability amplitude waves of different frequencies and powers. When they are added together, the only place they are "in sync" is one place, and either side of that, the power tapers off, because the waves all cancel out. The location of where they are in sync moves at what's called "group velocity".
That's the wave packet, whose power represents the probability of a photon appearing there.
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1
There has been lots of discussion on "wavefunction of a photon" over the years. The way I understand creation of a localized 1-photon state is to form $\int{f({\bf{k}})|1_{\bf{k}}\rangle d^{3}\bf{k}}$ where $|1_{\bf{k}}\rangle$ are the plane wave 1-photon states and $f(\bf{k})$ is a shaping function in momentum space. – twistor59 Dec 10 '11 at 13:02
As Einstein said and proved, with his photo-electric effect, light is quantized, photon's, 'packets' of light.
The 'sum' of one, two, three or more, simple harmonic oscillations, in space-time.
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http://mathoverflow.net/questions/21691/proofs-by-induction/21701
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## Proofs by induction [closed]
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Background
I'm interested in the issue of "explanatory" mathematical proofs and would like to try to find out what intuitions mathematicians have about induction, because there seems to be some disagreement as to whether such proofs can ever be explanatory.
Question
Does anyone have any proofs by induction that they think are particularly explanatory, and does anyone have any proofs by induction which they think fail to be explanatory at all? I'd be really grateful if you could post any examples, and also if you could say why you find it/don't find it explanatory.
Comment
I've just been reading through the comments and answers to my question and find them all very helpful. Thank you for taking the time to answer this question and post some interesting examples. Of course, more answers are still welcome!
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Unless you define what you mean by «explanatory» it is impossible to know what you have in mind... – Mariano Suárez-Alvarez Apr 17 2010 at 21:22
2
Also, in my specialty (evolutionary partial differential equations), almost every single existence/stability result ever put to paper is a proof by induction. Whether a proof is explanatory has relatively little to do with the technique employed (induction, contradiction, etc.) but more about the presentation by the author. I think this question is ill-posed. – Willie Wong Apr 17 2010 at 21:30
I'm sorry for the vague us of "explanatory", Mariano. I was just intending to try to appeal to people's intuitions about what is meant by "explanatory", if people have any, as I don't want to presuppose a particular theory of mathematical explanation. Willie, I'd be interested in knowing some more about what you mean by "presentation of the author". Thanks for your help. – Lea M Apr 17 2010 at 21:39
I agree that the question needs more -- um -- explanation. (In fact, the downvote is mine.) However, I think there is a question lurking in here, and I disagree with MSA's comment above in that I think I may know what the OP has in mind. I plan to post an answer in the near future. It would be nice if the question would stay open... – Pete L. Clark Apr 17 2010 at 21:41
Answers evince the fact that 'explanatory' is easily confused with 'constructive', if in fact they are different concepts... – Mariano Suárez-Alvarez Apr 17 2010 at 22:30
## 6 Answers
Let me see if I can address what I think is the underlying question. Probably most of the proofs by induction you've seen have been of the form "show that this identity $P(n)$ holds," where you are given the identity and must verify $P(1)$ and that $P(n)$ implies $P(n+1)$. It is hard to see how this "explains" identity $P(n)$, since you were given the identity and do not really know "why" it is true.
However, in practice this never happens. A mathematical situation usually gives various indications that it would be amenable to an inductive proof, and then a whole creative process - which students often don't get told about until much after they've seen their first inductive proof - goes into figuring out what the correct statement $P(n)$ should be. This is the part of the proof which is really explanatory, and it's a shame that often it never gets written down.
There are also situations, such as the one jamieweigandt mentions, where a mathematical situation has a natural recursive structure and the proof by induction clarifies the precise nature of this structure. This is somewhat different from the boring inductive proofs people are taught in school and while I don't have a good example at the moment I can assure you that such examples are ubiquitous.
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Your second paragraph here neatly answers the question Lea addressed to me. A proof that just lumps technical step one after another can certainly be logically sound, but it is the extra touch of the mathematician who writes about the intuition or the big picture of the proof that makes it understandable. To take an extreme example: Christodoulou's recent proof on black hole formation in relativity has its bulk (~300pps) in one big induction argument. Chapters 3 - 15, where the induction step is shown, is not explanatory at all. It is the three short chapters 1, 2, 16 that explains. – Willie Wong Apr 17 2010 at 22:38
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First of all, the phrase "explanatory" in the question should be replaced with "insightful", as I think you're asking for examples of inductive proofs which give the reader some real understanding of what is going on.
A lot of inductive proofs give insight. Here are a few elementary examples.
1. If $a$ and $b$ are relatively prime positive integers, we can write $ax + by = 1$ for some integers $x$ and $y$. The proof uses induction on $\max(a,b)$ and the inductive step is a replacement of $a$ and $b$ by $a$ and $b-a$ (if $a < b$). The maximum has decreased and this idea in fact leads to Euclid's algorithm, which is the efficient way to practically find $x$ and $y$.
2. For every positive integer $n$, $\cos(nx)$ is a polynomial in $\cos x$ (e.g., $\cos(2x) = 2\cos(x)^2 - 1$). The proof uses the addition formula for $\cos x$ and that idea even shows you how to build the polynomial expression if you want to find it explicitly, and that may be comforting to some people.
3. A polynomial of degree $n$ over a field has at most $n$ roots. The proof shows you how the existence of one root controls the cardinality of the number of possible other roots, using some algebra.
4. Many theorems about polynomials in several variables proceed by induction on the number of variables, and often watching how you can bootstrap a result in $n$ variables to a result in $n+1$ variables gives some kind of nice understanding (if only the understanding that the key case is one variable, suitably formulated). For example, a polynomial in several variables over an infinite field which is identically 0 as a function is the zero polynomial (all coefficients are 0). This is proved by induction and the base case is the previous example, which is really the only involved step.
Here are some theorems where the inductive proof does not give insight.
1. The "exchange lemma" from linear algebra, which is used to justify why any two bases of a finite-dimensional vector space have the same size, has never seemed particularly enlightening to me. It is proved (in part) using induction.
2. Different complex-valued characters of a finite abelian group are linearly independent functions. The proof goes by induction on the number of characters, but I never thought the proof itself really explains the linear independence in an "aha" kind of way. It verifies the truth and then you move on to use it.
3. Cauchy's forwards-backwards inductive proof of the arithmetic-geometric mean inequality is pretty remarkable (since so few theorems are amenable to a forwards-backwards inductive proof), but all the same it doesn't feel like it offers any useful understanding behind the inequality.
4. The sum of the first $n$ positive integers is $n(n+1)/2$. You can check the identity by induction but the mystery remains of how such a formula is found.
5. The sum of the squares of the first $n$ positive integers is $n(n+1)(2n+1)/6$. You can check the identity by induction but again there is no real understanding generated of where the identity comes from.
6. The sum of the cubes of the first $n$ positive integers is $(n(n+1)/2)^2$. You can check the identity by induction but not only does that approach not really explain the formula, it also doesn't explain the surprise that the formula is the square of the formula for the sum of the first $n$ positive integers.
I think I could give an argument by induction to generate infinitely many similar noninsightful proofs by induction starting with item 4, but such an argument wouldn't really give any insight itself, so I will stop here.
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#6 has a visual inductive proof which I find somewhat insight-generating, constructed by arranging i x i x i blocks for 0 <= i <= n diagonally on a (1 + 2 + ... + n)^2 board. – pelotom Jul 10 2011 at 9:43
A nice simple problem with natural inductive structure is the tower of Hanoi problem. It is not obvious that any solution exists, and it is hard to come up with one explicitly, until you assume that it is possible to move the top $n-1$ disks, after which it is obvious how to move $n$ disks.
There is in fact a "noninductive" solution, but since it takes $2^n-1$ steps, it is preferable not to think about it.
[Added later] Of course, there is really no such thing as a "noninductive" theorem about natural numbers because of the inductive structure of natural numbers themselves. However, the "right" induction in this case exponentially compresses the solution, by suppressing unnecessary details.
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Perhaps it is reasonable to regard Goodstein's Theorem and the Paris-Harrington Theorem as examples of "noninductive" theorems about natural numbers? After all, they cannot be proved by induction over the natural numbers ... – Simon Thomas Apr 17 2010 at 23:13
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Simon, I would regard Goodstein's theorem as more inductive than usual, since it has the natural structure of induction up to epsilon zero. :) – John Stillwell Apr 17 2010 at 23:21
Fair enough ... – Simon Thomas Apr 17 2010 at 23:36
The first example of an explanatory proof by induction that comes to mind is the solution to the following problem which originates (to my best knowledge) from Art Benjamin.
By a triomino I will mean an "L-shaped" union of 3 unit squares.
Claim: A $2^n \times 2^n$ grid of unit squares with one square removed can always be covered by triominos.
Proof:
Base case: if $n=1$ then we need to cover a $2 \times 2$ grid with one square removed by triominos. THat is, we need to cover a triomino by triominos. So we're good here.
Induction step: Assume that we can cover any $2^n$ by $2^n$ grid of squares with one square removed by triominos and suppose that we are presented with a $2^{n+1} \times 2^{n+1}$ grid of unit squares.
Separate this into a $2 \times 2$ grid of $2^n \times 2^n$ grids of squares one of which has one square removed. At the place where the four corners of these grids meet, place a triomino in such a way that it covers the corner square of each of the three $2^n \times 2^n$ grids without a square already having been removed.
Now what remains to be covered is the union of four $2^n \times 2^n$ grids of squares each with one square removed, so by the induction hypothesis we can cover it with triominos.
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So what is the difference between 'explanatory' and 'constructive'? – Mariano Suárez-Alvarez Apr 17 2010 at 21:56
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You have a point. This is definitely a constructive proof. The question made me think of a talk I saw Art Benjamin based on his book Proofs that Really Count. In the book, he presents combinatorial proofs of a number of claims that are often proven by induction. The combinatorial proof has the benefit that you understand why the result is true as you prove it instead of just pushing around symbols. This is an example of a proof by induction that I feel doesn't have that kind of issue. – Jamie Weigandt Apr 17 2010 at 22:09
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I would have let the base case be $n=1$, i.e., covering a $1\times1$ grid with one square removed with no triominos. But perhaps that is slightly pathological. – Harald Hanche-Olsen Apr 17 2010 at 22:56
In the last year, I have twice taught a course on mathematical reasoning for future undergraduate math majors. The first time I was surprised by how conceptually difficult induction was for many of the students. I think that part of it is related to the potentially non-explanatory nature of inductive proofs. Let me elaborate and say a little about what I did to try to be more explanatory.
The canonical first induction proof is that for all positive integers $n$, $1 + \ldots + n = \frac{n(n+1)}{2}$. After proving this by induction though, it is irresistible to muddy the waters by mentioning that Gauss, at age $10$, knew a better way. I think it is clear that little Gauss' proof [I assume you know it!] is more "explanatory" than the induction proof. For starters, it allows you to discover what the closed form expression is, and the induction proof does not.
Then we did more complicated such summation formulas, i.e., other power sums and variants of those. It took me a while to realize that, no matter how I had phrased the problem, the students were seriously concerned with "how I knew what to put on the other side". This is notwithstanding the fact that the statement of the problem had the closed form expression on the right hand side. After thinking about this for a while, I realized they had a point: in some sense we are just being asked to check an answer that someone has already found and this "checking" process is not explanatory. (And by the way, "How do you know what to put on the other side?" is much more interesting and natural than the question we are actually asking. We should be encouraging students to ask the natural and interesting questions, rather than discouraging them by asking questions which have less to them than meets the eye.)
I tried to reinforce this by stating in general terms what is done in this sort of induction proof. Suppose that you have two functions $f,g: \mathbb{Z}^+ \rightarrow \mathbb{R}$ and you want to show that for all $n$, $\sum_{k = 1}^n f(k) = g(n)$. Given the principle of mathematical induction, this is equivalent to showing that for all $n$, $g(n+1)-g(n) = f(n)$. Now it is clear that if $g$ is something reasonable like a polynomial, there is no "idea" or "explanation" involved in checking this identity: you just do the algebra.
After thinking yet more, I realized the analogy with differential and integral calculus. It's completely analogous to answering the question
"Why is it the case that $\int \sec^3 x \ dx = \frac{1}{2} \sec x \tan x + \frac{1}{2} \ln | \sec x + \tan x| + C$?"
by differentiating the right hand side, simplifying, and getting $\sec^3 x$.
I wrote up some thoughts on this in:
http://www.math.uga.edu/~pete/3200induction.pdf
and
http://www.math.uga.edu/~pete/finitecalc.pdf
Not all induction proofs have this aspect to them, of course, but it certainly is the case that it is often possible to bang out a brute force proof of something by induction and then feel that one has evaded the job of truly understanding why the statement is true. As one extreme example, the first proof of the law of quadratic reciprocity was given by Gauss as a teenager: he proved it by induction! There are now more than one hundred reasonably distinct proofs of QR; the reason for this unusually high multiplicity is, I think, that many people (including myself) have trouble with some of the standard proofs: they do not find them sufficiently explanatory.
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I was going to put Gauss' first proof of QR on the list of not-very-insightful inductive proofs in my answer, but it seemed out of proportion to the other examples so I didn't bother. Glad you mentioned it! – KConrad Apr 17 2010 at 23:39
Shouldn't it be $f(n) = g(n) - g(n - 1)$? – pelotom Jul 13 2011 at 23:39
It has been pointed out (see KConrad's answer) that the usual proof that the sum of the first $n$ squares is $n (n+1) (2n+1)/6$ does not give insight. However, I would like to show that a different inductive proof of this fact does give insight. Instead of performing induction on $n$, we can perform induction on the exponent. This approach is very instructive because it shows that the sum of the first $n$ $k$th powers is a polynomial in $n$. It also shows how we can derive the formula for any exponent given that we know the formulas for all the smaller exponents. Thus, in principle the only formula we need to know is that $1 + 2 + \dots n = n(n+1)/2$, which has a beautiful proof by picture here.
I will illustrate this idea by computing the sum of the first $n$ squares.
Observe that
$(n+1)^3-1=\sum_{i=1}^n (i+1)^3 -i^3 = \sum_{i=1}^n 3i^2 + 3i +1$.
Thus,
$3\sum_{i=1}^n i^2 = (n+1)^3-1 - n - 3 \sum_{i=1}^n i$.
However, we have already inductively determined that $\sum_{i=1}^n i = n(n+1)/2$. Thus, substituting and solving yields
$\sum_{i=1}^n i^2 = n (n+1)(2n+1)/6$,
as required.
We can now compute the sum of the first $n$ cubes using the same technique, given that we now know the sum of the first $n$ integers and the sum of the first $n$ squares. This gives a systematic way to compute the sum of the first $n$ $k$th powers, and why it is a polynomial in $n$ of degree $k+1$.
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Right -- see Theorem 4 in the first handout I linked to in my answer above. – Pete L. Clark Apr 18 2010 at 6:04
this is not induction on n. – Martin Brandenburg Apr 18 2010 at 9:23
I never said it was induction on $n$, although I can rename the variables to make it induction on $n$. The theorem that I am proving is my last sentence. The sum of the first $n$ $k$th powers is a polynomial in $n$ of degree $k+1$. This can be proven by induction on $k$. I illustrated the inductive step for $k=2$, but it is clear that it works for any $k$ by (strong) induction. Moreover, this inductive proof shows how the polynomial can be computed. – Tony Huynh Apr 18 2010 at 13:47
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http://math.stackexchange.com/questions/130354/surface-measure-for-lipschitz-domain?answertab=votes
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# Surface measure for Lipschitz domain
Let $D\subset R^d$ be a bounded Lipschitz domain. Must there exist a bounded function $\Phi$ on $\partial D$ and collections of subsets $(\partial D )^{\epsilon} \subset \partial D$ (indexed by $\epsilon$) such that for all bounded continuous function $g$ on $\partial D$ we have $$\lim_{\epsilon\to 0} \sum_{(\partial D )^{\epsilon}}g\,\Phi\,\epsilon^{d-1}= \int_{\partial D }g\,d\sigma$$
We may suppose $\partial D$ is just the graph of a Lipschitz function over a ball in $R^{d-1}$.
This is a discrete approximation to the surface measure $\sigma$. The result is true if we add the condition that "the unit outward normal vector field on $\partial D$ is continuous $\sigma-$almost everywhere on $\partial D$." But I wonder if we can get rid of it.
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## 1 Answer
Let $\sigma$ be the restriction of the $(d-1)$-dimensional Hausdorff measure to $\partial D$. Since $D$ is bounded with Lipschitz boundary, $\sigma$ is a finite measure. The Krein-Milman theorem implies that $\sigma$ is a weak*-limit of sums of point masses: that is, there exists a sequence $\nu_k$ of finite combinations of point masses (with nonnegative coefficients and total mass equal to $\sigma(\mathbb R^d)$ ) such that $\int g\,d\nu_k\to \int g\,d\sigma$ for every function $g\in C(\mathbb R^d)$. This yields the desired convergence.
Putting the measures $\nu_k$ into the desired form, even with $\Phi\equiv 1$, should be straightforward: take the mass of each point to be a rational number, hide their common denominator in $\epsilon^{d-1}$, and split integer masses into tightly bunched unit masses.
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http://mathoverflow.net/questions/47952/proving-interesting-theorems-about-s-n-using-its-character-table/48003
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## Proving interesting theorems about S_n using its character table.
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Hi, i wonder if there are interesting proofs about $S_n$ (group theoretic or not) using its character table. Using the Murnaghan-Nakayama rule you can for example prove that for $n>4$ $A_n$ is the only normal subgroup of $S_n$ because there are no nonlinear characters $x$ and $g$(not 1) in $S_n$ with $x(g)=x(1)$, since $x(1)>x(g)$ . Do you know any other nontrivial theorems about $S_n$ with a proof using its charactertable ?
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## 4 Answers
In an answer to an earlier question, I showed how to prove that the square root counting function `$r_2: S_n\rightarrow \mathbb{N},\;g\mapsto \#\{h\in S_n|h^2=g\}$` assumes its maximum at the identity, using the representation theory of $S_n$. Admittedly, you need to know slightly more than the character table. You need to be able to compute the Frobenius-Schur indicators of the characters, so you need to know how the conjugacy classes multiply. Alternatively, you just need to know that all representations are defined over $\mathbb{R}$, which you prove along to way to computing the character table anyway. In a comment to my answer, Richard Stanley remarks that, also using the representation theory of $S_n$, you can generalise this to the $k$-th root counting function for any positive integer $k$. In an answer to the same question, Alon Amit remarks on possible generalisations to solving other polynomial equations in the elements of $S_n$.
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
I'm not sure if you will consider this nontrivial, but from the character table you can very quickly show that the number of conjugacy classes of even permutations is always greater than or equal to the number of conjugacy classes of odd permutations.
One just applies the general fact that the sum of the entries in any row of a character table (not weighted by the size of the conjugacy class) is a nonnegative integer (because the character `$\chi ( \pi ) = \frac{\# S_n}{\# C_\pi}$`, where $C_\pi$ is the set of conjugates of the permutation $\pi$, corresponds to an actual representation, namely, take a vector space with basis $\{ e_\pi \mid \pi \in S_n\}$ and define $g(e_\pi) = e_{g \pi g^{-1}}$ for $g \in S_n$) to the sign character.
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The displayed equation isn't showing properly, so I fixed it. Why change it back? Or is there some strange behavior here? – Charles Rezk Dec 2 2010 at 4:26
I mean, strange MathJax TeX behavior ... – Charles Rezk Dec 2 2010 at 4:27
Oh sorry...I missed your fix before I changed it back...I was struggling to get it to display properly...if you have a chance, please refix it! Thanks so much! – Ken Fan Dec 2 2010 at 4:33
The following is not strictly speaking something that can be read off from the character table. However, it is an elementary combinatorial identity about partitions which one can deduce from understanding the character theory of symmetric groups well enough, and looking at the character table does play a central role:
For $\lambda \vdash n$ a partition of $n$ (i.e., $n = 1 \lambda_1 + 2 \lambda_2 + \cdots + n \lambda_n$) define $$A(\lambda) = \prod_{i=1}^{n} n^{\lambda_n}, \qquad B(\lambda) = \prod_{i=1}^{n} (\lambda_n)!$$ Claim: $$\prod_{\lambda \vdash n} A(\lambda) = \prod_{\lambda \vdash n} B(\lambda)$$
The character-theoretic proof proceeds as follows:
1. For an element in the conjugacy class of $S_n$ indexed by the partition $\lambda$, it's centralizer has cardinality $A(\lambda) B(\lambda)$, i.e., the number of elements in the conjugacy class is $$\frac{n!}{A(\lambda) B(\lambda)}$$
2. Take the character matrix $M$. The orthogonality relations tells us that a suitable rescaling of the character matrix is orthogonal, so has $\det = \pm 1$. From this, together with 1 to find the scaling factors for the columns, we obtain $$(\det M)^2 = \prod_{\lambda} A(\lambda) B(\lambda)$$
3. $M$ relates two bases for the spaces of class-functions: The characters of irreps (indexed in the Schur ordering by partitions), and the delta functions on conjugacy classes (indexed obviously by partitions). For symmetric groups, there is a third nice basis: For $\lambda \vdash n$, let $$S_\lambda = \prod_{i=1}^{n} S_i^{\lambda_i}$$ and consider the characters of the induced reps $Ind_{S_\lambda}^{S_n} \mathbb{C}$.
4. Consider the change of basis matrix relating characters of induced reps and the delta functions on conjugacy classes: Easy character theory shows that it is triangular with diagonal entries equal to $B(\lambda)$.
5. Consider the change of basis matrix relating characters of induces reps and characters of irreps: Knowing how character theory for symmetric groups works over $\mathbb{Z}$ (i.e., that both span integrally), we can show that it has determinant $\pm 1$. More precisely, knowing the character theory well enough we can show that the change of basis matrix between them is upper triangular with ones on the diagonal.
Putting together 2, 4, 5 we obtain $$\det(M)^2 = \prod_{\lambda \vdash n} A(\lambda) B(\lambda) = \left(\prod_{\lambda \vdash n} B(\lambda)\right)^2$$ and so the claimed identity.
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Your notation for partitions is not the standard one: you denote by $\lambda_i$ the number of all $i$'s in the partition, while usually it means the $i$-th part. Just noting this down for future readers... – darij grinberg Dec 2 2010 at 10:54
Because all the entries in the character table are integers and not just algebraic integers, you get that a proof that every permutation $\sigma$ of order $n$ is conjugate to all $\sigma^j$ for $j$ coprime to $n$. (Of course, one usually uses this in the opposite direction, to deduce that all entries are integers!)
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thank you!does someone know if you need to use that result to prove Murnaghan-Nakayama,which also gives you that the character table entries are integers?I think so. – trew Dec 1 2010 at 22:22
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http://math.stackexchange.com/questions/46871/how-to-calculate-dual-frames-under-constraints
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# How to calculate dual frames under constraints?
Denote orthonormal basis in $\mathbb{R}^2$: $(\epsilon_1,\epsilon_2)=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ and $A=\begin{pmatrix}1&0&\frac{\sqrt{2}}{2}&\frac{\sqrt{2}}{2}\\0&1&\frac{\sqrt{2}}{2}&-\frac{\sqrt{2}}{2}\end{pmatrix}$.
Now let frames $(e_1,e_2,e_3,e_4)=(\epsilon_1,\epsilon_2)A$ and its dual $(\tilde{e}_1,\tilde{e}_2,\tilde{e}_3,\tilde{e}_4)=(\epsilon_1,\epsilon_2)B$. For $x=(1,1)^T$, one can decompose and reconstruct $x$ through $\{e_j\}$ and $\{\tilde{e}_j\}$: $$x=\sum_{j}\langle x,\tilde{e}_j\rangle e_j=AB^Tx=\sum_j\alpha_j e_j$$ The problem is to determine $B$ such that: 1) $\|\alpha\|_2$ is minimized; 2) $\alpha$ has the least non-zero entries.
The first one can be solved by Moore–Penrose pseudoinverse, but I have no idea about P2. Intuitively it appears to be some analogues of Moore–Penrose pseudoinverse but minimizing $L^1$ norm. Can anyone give hints? Thank you.
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## 1 Answer
OK I think I've got it.
By SVD, $$A=U\Sigma V^T= \begin{pmatrix}0&1\\1&0\end{pmatrix} \cdot \begin{pmatrix}\sqrt{2}&0&0&0\\0&\sqrt{2}&0&0\end{pmatrix} \cdot \begin{pmatrix}0 & \frac{1}{\sqrt{2}} & \frac{1}{2} & -\frac{1}{2} \\ \frac{1}{\sqrt{2}} & 0 & \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} & 0 & \frac{1}{\sqrt{2}} \\ -\frac{1}{2} & -\frac{1}{2} & \frac{1}{\sqrt{2}} & 0\end{pmatrix}$$
Let $\beta=V^T\alpha$, then $$\begin{pmatrix}\sqrt{2}&0&0&0\\0&\sqrt{2}&0&0\end{pmatrix} \cdot\begin{pmatrix}\beta_1\\\beta_2\\\beta_3\\\beta_4\end{pmatrix} =\begin{pmatrix}1\\1\end{pmatrix}\qquad\cdots\cdots(*)$$
Therefore $\beta_1=\beta_2=\frac{1}{\sqrt{2}}$. Further, $$\alpha=V\beta=\left(\frac{1}{2}-\frac{\beta _3}{2}-\frac{\beta _4}{2},\frac{1}{2}+\frac{\beta _3}{2}-\frac{\beta _4}{2},\frac{1}{\sqrt{2}}+\frac{\beta _4}{\sqrt{2}},\frac{\beta _3}{\sqrt{2}}\right)^T$$
To achieve sparsity, it must hold that $\beta_3=0$ and $\beta_4=1$, yielding $\alpha=(0,0,\frac{1}{\sqrt{2}},0)^T$
Now from $(*)$ and $AB^T=I$, we have $$\begin{pmatrix}\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\\0\\1\end{pmatrix} =\begin{pmatrix}\frac{1}{\sqrt{2}}&0\\0&\frac{1}{\sqrt{2}}\\0&0\\0&1\end{pmatrix} \cdot \begin{pmatrix}1\\1\end{pmatrix}=\tilde{B}^T\begin{pmatrix}1\\1\end{pmatrix}$$
The entries in lower part of $\tilde{B}^T$ are chosen out of simplicity.
Finally, $$B=\left(V\tilde{B}^TU^T\right)^T=\begin{pmatrix} 0 & -\frac{1}{2} & \frac{3}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} \\ 0 & \frac{1}{2} & \frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} \end{pmatrix}$$
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http://mathoverflow.net/revisions/58828/list
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5 added 534 characters in body; added 6 characters in body
Actually, one can show the following stronger result:
Proposition
Assume that a finite abelian group $A$ admits a non-degenarate, bilinear alternating form $\psi$. Then $A$ has a lagrangian decomposition, i.e. there exists a subgroup $G$, isotropic for $\psi$, such that
$A \cong G \times \widehat{G}$,
where $\widehat{G}$ denotes as usual the group of characters of $G$. In particular, $|A|=|G|^2$.
Therefore, the elements of $A$ can be written as $(x, \chi)$, with $x \in G$ and $\chi \in \widehat{G}$. Moreover, in such a presentation the form $\psi$ take the following shape:
$\psi((x, \chi), (y, \eta))=\chi(y)\eta(x)^{-1}$.
An easy proof, by induction on the order of the group, can be found in Lemma 5.2 of A. Davydov, Twisted automorphisms of group algebras, arXiv:0708.2758
Remark. It is interesting to notice the analogy with symplectic vector spaces. In fact, any symplectic vector space $(V, \omega)$ can be written as $V = W \oplus W^{*}$, where $W$ is a lagrangian (=isotropic of maximal dimension) subspace for $\omega$. In particular, $\dim V = 2 \dim W$. Moreover, with respect to this decomposition, $\omega$ has the following shape:
$\omega(x \oplus \chi, y \oplus \eta) = \chi(y) - \eta(x)$.
In the case of finite abelian groups the "dual role" is played by the group of characters, as usual.
4 added 54 characters in body
Actually, one can show the following stronger result:
Proposition
Assume that a finite abelian group $A$ admits a non-degenarate, bilinear alternating form $\psi$. Then $A$ has a lagrangian decomposition, i.e. there exists a subgroup $G$, isotropic for $\psi$, such that
$A \cong G \times \widehat{G}$,
where $\widehat{G}$ denotes as usual the group of characters of $G$. In particular, $|A|=|G|^2$.
Therefore, the elements of $A$ can be written as $(x, \chi)$, with $x \in G$ and $\chi \in \widehat{G}$. Moreover, in such a presentation the form $\psi$ take the following shape:
$\psi((x, \chi), (y, \eta))=\chi(y)\eta(x)^{-1}$.
For a
An easy proof, see by induction on the order of the group, can be found in Lemma 5.2 of A. Davydov, Twisted automorphisms of group algebras, arXiv:0708.2758
3 added 110 characters in body; deleted 1 characters in body
Actually, one can show the following stronger result:
Proposition
Assume that a finite abelian group $A$ admits a non-degenarate, bilinear alternating form $\psi$. Then $G$ A$has a lagrangian decomposition, i.e. there exists a subgroup$G$, isotropic for$\psi\$, such that
$A \cong G \times \widehat{G}$,
where $\widehat{G}$ denotes as usual the group of characters of $G$. In particular, $|A|=|G|^2$.
Therefore, the elements of $A$ can be written as $(x, \chi)$, with $x \in G$ and $\chi \in \widehat{G}$. Moreover, in such a presentation the form $\psi$ take the following shape:
$\psi((x, \chi), (y, \eta))=\chi(y)\eta(x)^{-1}$.
For a proof, see Lemma 5.2 of A. Davydov, Twisted automorphisms of group algebras, arXiv:0708.2758
2 added 160 characters in body; deleted 12 characters in body
Actually, one can show the following stronger result:
Proposition
Assume that a finite abelian group $A$ admits a non-degenarate, bilinear alternating form . $\psi$. Then $G$ has a lagrangian decomposition, i.e. there exists a finite abelian group subgroup $G$ G$, isotropic for$\psi\$, such that
$A \cong G \times \widehat{G}$,
where $\widehat{G}$ denotes as usual the group of characters of $G$. In particular, $|A|=|G|^2$. Moreover, in such a presentation the form $\psi$ take the following shape:
$\psi((x, \chi), (y, \eta))=\chi(y)\eta(x)^{-1}$.
For a proof, see Lemma 5.2 of A. Davydov, Twisted automorphisms of group algebras, arXiv:0708.2758
1
Actually, one can show the following stronger result:
Proposition
Assume that a finite abelian group $A$ admits a non-degenarate, bilinear alternating form. Then $G$ has a lagrangian decomposition, i.e. there exists a finite abelian group $G$ such that
$A \cong G \times \widehat{G}$,
where $\widehat{G}$ denotes as usual the group of characters of $G$. In particular, $|A|=|G|^2$.
For a proof, see Lemma 5.2 of A. Davydov, Twisted automorphisms of group algebras, arXiv:0708.2758
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http://en.wikisource.org/wiki/Page:Zur_Thermodynamik_bewegter_Systeme.djvu/7
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# Page:Zur Thermodynamik bewegter Systeme.djvu/7
From Wikisource
and isochorically brought to velocity zero. The arising function of $\beta$ plays here the role of a constant, so it is irrelevant. Of course, it must have the same value for all bodies.
Since we see $\beta$ as constant in this case, it is
$dQ=\frac{\partial H}{\partial U_{0}}(dU_{0}+p_{0}dv)=dH+pdv.$
In a system whose velocity isn't changing, $H$ plays the role of the inner energy for a co-moving observer; between the quantities $H, v, p, t$ the same relations exist, which follows from the thermodynamic main-theorems for $U_{0}, v, p_{0}, T_{0}$.
Now, let a body pass through Carnot's circular process, in which the two reservoirs have different velocities; namely, $T_{1}, \beta_{1}c$ are the temperature and velocity of one reservoir; $T_{2}, \beta_{2}c$ are the relevant quantities for the other one. If the theorem of the impossibility of a thermal perpetual motion machine is also valid when it assumes different velocities in its different stages, then the ratio of the heat quantities given off to the reservoirs cannot depend on the nature of the circular process. Then it must be
$\frac{dQ_{1}}{dQ_{2}}=\frac{\left[\frac{\partial H}{\partial U_{0}}(dU_{0}+p_{0}dv)\right]_{1}}{\left[\frac{\partial H}{\partial U_{0}}(dU_{0}+p_{0}dv)\right]_{2}}=\varphi(T_{1},T_{2},\beta_{1},\beta_{2})$
One can easily see, that this function must have the form
$\frac{\phi(T_{1},\beta_{1})}{\phi(T_{2},\beta_{2})}$
Furthermore
$\frac{\phi(T_{1},\beta)}{\phi(T_{2},\beta)}=\frac{T_{1}}{T_{2}}$
must be the case as well,
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http://mathhelpforum.com/geometry/46491-easy-question.html
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# Thread:
1. ## Easy Question
When we want to find the area of a two dimensional figure and there are more than one type of shapes that can be used to find the whole figure, we find the areas of the shapes comprising the figure one step at a time. Then finally adding the areas of the shapes together to find area of the whole figure. But do we round twice or just at the end of the calculation? And does it always have to be rounded to the least number of significant figures. What does this achieve exactly? Please show an example and show your reasoning.
-Thanks
2. Originally Posted by Joker37
When we want to find the area of a two dimensional figure and there are more than one type of shapes that can be used to find the whole figure, we find the areas of the shapes comprising the figure one step at a time. Then finally adding the areas of the shapes together to find area of the whole figure.
Yes, this is the correct method of finding area of complex shapes. Such as a 'L' 2D shape where you would consider two rectangles.
Originally Posted by Joker37
But do we round twice or just at the end of the calculation?
We dont always round. It depends on the question.
Originally Posted by Joker37
And does it always have to be rounded to the least number of significant figures. What does this achieve exactly? Please show an example and show your reasoning.
-Thanks
^ For this shape, the area would be:
$(3\times4) +(5\times2) = 22\mathrm{cm}^2$
$22\mathrm{cm}^2$ is the area. We did not need to round here.
^ For this shape, the area would be:
$(4.2\times3.2)+(5.5\times 2.7) = 28.29\mathrm{cm}^2$
$28.29\mathrm{cm}^2$ is the answer and it can be kept as that (without rounding). But, if you wish to round, then you could round to 1dp as all the values were given to 1dp so the rounded area would be $28.3\mathrm{cm}^2$.
It's all a matter of judgement. Personally, I wouldn't round it unless stated or if the value of the area has too many decimal places (e.g. $15.89374987439732740323\mathrm{cm}^2$).
(All diagrams are not to scale)
3. So, just to clarify, under test conditions it wouldn't matter to which number we round it to unless it is stated in the question?
4. Originally Posted by Joker37
So, just to clarify, under test conditions it wouldn't matter to which number we round it to unless it is stated in the question?
In UK, under test conditions, we are asked to give the answer to 3sf unless told otherwise.
5. Originally Posted by Air
In UK, under test conditions, we are asked to give the answer to 3sf unless told otherwise.
6. Well, does anyone know? I don't want to get a zero just because I put it in the wrong the number of significant figures.
7. Originally Posted by Joker37
Well, does anyone know? I don't want to get a zero just because I put it in the wrong the number of significant figures.
If the writer of the exam fails to specify the accuracy to which an answer is to be given then the best you can do is to use your common sense.
In my experience this is usually only a problem with internal assessment at secondary school level. It will then depend on your teacher, not your country. Despite my personal opinion reagrding the competency of the external assessment authorities, I would be astonished if this was an issue in external examinations.
If an accuracy is prescribed then you should always use greater accuracy during the calculation than the final answer requires. Otherwise you get accumulation of rounding error and it's likely that your final answer, after rounding, won't be correct to the required accuracy. This can be a particular problem when non-trivial statistics and probability questions require answers correct to, say, 4 decimal places.
By the way: Unless the question is only worth 1 mark you will not get zero for correct working but incorrect answer.
8. Originally Posted by Joker37
When we want to find the area of a two dimensional figure and there are more than one type of shapes that can be used to find the whole figure, we find the areas of the shapes comprising the figure one step at a time. Then finally adding the areas of the shapes together to find area of the whole figure. But do we round twice or just at the end of the calculation? And does it always have to be rounded to the least number of significant figures. What does this achieve exactly? Please show an example and show your reasoning.
-Thanks
Very quick answer ...
In general, only round at the end. Keep as many sig.fig. as "seem sensible" in the initial workings.
Beware when subtracting one number from another that is of a similar size!
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http://math.stackexchange.com/questions/209019/linearly-separate-a-cube/209210
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# Linearly separate a cube
I am facing the problem of the linear separability of a three dimensional cube.
Let's take the opposite vertexes of the cube as $(0, 0, 0), (1, 1, 1)$. It is possible to split it with a plane in two tetrahedron-like parts, and so define two sets, each containing lying points lying on a specific side of the plane. Let's take one of the set $t=\{(0, 1, 1), (1, 1, 1), (1, 0, 1), (1, 1, 0)\}$ and the other the obvious complement.The question is: what is the simplest form of the boolean function $P$ such that $\forall x \in t: P(x)=1$?
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$(x \wedge y) \vee (x \wedge z) \vee (y \wedge z)$ is nice and symmetric. – mjqxxxx Oct 8 '12 at 1:37
@mjqxxxx that sounds good for my ultimate aim, which is to predict how many ways there are to linearly separate the cube. Starting from the 0-d, 1-d and 2-d case, I was including functions in which at most 0, 1, 2 or 3 arguments (between x, y, z) appeared, but I was missing exactly 8 cases, which I reckon to be a variation of your suggestion (x -> !x, etc.). Now I should figure out why exactly at space dimension 3 there is the need of this additional function. – Lorenzo Pistone Oct 8 '12 at 1:47
I'm not sure I understand your question. Are you looking for a function like $P(x,y,z)=1$ if $x+y+z\geq 3/2$ and $0$ otherwise? This effectively cuts your cube into two congruent pieces - not tetrahedra, of course, but it meets the other criteria of your question the way you've stated it. – user22805 Oct 8 '12 at 8:12
@DavidWallace Unfortunately I can express this $P$ only in terms of the boolean operators (and, or, not). – Lorenzo Pistone Oct 8 '12 at 9:16
So you're not working in ${\Bbb R}^3$ at all then? Your question really didn't make this clear. Your talk of points, planes, cubes and tetrahedra makes the question sound like a geometry problem. If x,y,z can only be true or false, then the expression that you want would be something like ( x and (y or z )) or (y and z ). Is that what you had in mind? – user22805 Oct 8 '12 at 17:38
## 1 Answer
Any subset of the collection of triples (a,b,c) with a,b,c all in {0,1} can easily be described using and, or, not: Make the corresponding truth table, and for each row write that row as a conjunction of "literals", i.e. v or (not v) for a variable v.
For example if the vars are x,y,z in that order, one row of a truth table would have x true, y false, and z true. Then this row corresponds to the conjunction
x and (not y) and z.
Now just put the rows that appear each in this format, put parentheses around each row result, and put "or" between them in case there are more than one row giving true. This is a standard method to get a boolean from the initial collection of triples.
I think you're asking whether one can realize any such collection of triples by means of a plane which cuts the cube without going through any vertices, and using one side of that plane for the "true" triples.
But this can't give all possible subsets, since the set {(0,0,0), (1,1,1)} would be two diagonally opposite points on the cube, and any plane cutting the cube with these two on one side will definitely have more triples on that same side. The boolean for this is of course (not x and not y and not z) or (x and y and z).
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"But this can't give all possible subsets". Can you explain why this start happening with 3 dimensions? For example, if you redo the calculation on a 2-d space, you don't have to consider more than 2 coordinates in the $P$ function (counting also the number of times that a variable appear) to get all the possible "splits". I was led to think that the necessary number of coordinates in $P$ was equal to the dimension of the space, but indeed this way I miss exactly 8 cases in the 3 dimensional case (and the question shows one of them, the others correspond to the other vertexes of the cube). – Lorenzo Pistone Oct 8 '12 at 16:09
For 2 dimensions, there are 4 rows in a truth table using two variables, for a total of 2^4=16 subsets of the set of 4 vertices of the square. It's easy to get the singletons e.g the set {(0,1)} by cutting the square with a line which "cuts of that one corner" (0,1). And using complements we get the three element subsets. We can get the empty subset or the complete set of all eight using a line that doesn't cut the square at all. But I can't see how to get a "diagonal doubleton" like {(0,0),(1,1)} by cutting the square with one line. So I think you miss some combinations even in the 2d case. – coffeemath Oct 8 '12 at 22:57
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http://mathoverflow.net/questions/41830?sort=votes
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## isogenies between abelian varieties that induce isomorphisms?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $\varphi : A \to B$ be an isogeny between 2 abelian varieties of dimension $g$. Are there known conditions for the $\ker\varphi$ so that this induces an isomorphism between $A$ and $B$? For example, if $\ker\varphi \cong (\mathbb{Z}/n\mathbb{Z})^{2g}$, then $A \cong B$, because $\varphi$ factors through the multiplication map $A \xrightarrow{\times n} A$ followed by an isomoprhism $A \to B$. I wonder if there are other cases that induce isomorphisms.
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5
If I understand your question correctly, then there are many other cases. Take for example an elliptic curve $E$ with complex multiplication, and take any non-zero map $\phi:E\to E$. The kernel can be cyclic but $E$ is still isomorphic to $E$ mod $\ker(\phi)$. In general this has nothing much to do with abelian varieties, it's just group theory really. You're just asking, in some slightly cryptic manner, of examples of isogenies from an abelian variety to itself. You know about multiplication by $n$ but there can be far more than that; the situation is well-understood buttoolongtofitintothisbo – Kevin Buzzard Oct 11 2010 at 20:24
1
x.$\mathfrak{}$ – S. Carnahan♦ Oct 12 2010 at 2:51
Already the simplest criterion given here, seems to have non trivial applications. Here is a paper disproving the 30 year old Donagi conjecture on the Torelli problem for P=rym varieties, by Izadi and Lange. math.uga.edu/~izadi/papers/prymtornew3.pdf – roy smith Nov 13 2010 at 20:29
## 1 Answer
Kevin's comment is right on the money, but here it is in more detail: I will give a general criterion for an isogeny $\varphi: A \rightarrow B$ of abelian varieties to induce an isomorphism upon passage to the kernel.
Let me work over an unnamed algebraically closed field. Suppose that $A = B$ and $\eta \in \operatorname{End}(A)$ is a surjective endomorphism of $A$. (N.B.: If $A$ is simple -- i.e., contains no proper nontrivial subvariety -- then any nonzero endomorphism is surjective. In particular this holds for all elliptic curves.) Then $\eta$ is also an isogeny: i.e., its kernel is a finite subgroup scheme, say $K$ and -- essentially, by the first isomorphism theorem for groups, as Kevin says -- it follows that there is an induced isomorphism
`$A/K \stackrel{\sim}{\rightarrow} A$`.
This condition is also necessary: if $\varphi: A \rightarrow B$ is an isogeny such that $B \cong A$, then composing with this isomorphism gives a surjective endomorphism of $A$ and the resulting map factors through an isomorphism `$A/(\operatorname{ker}(\varphi)) \rightarrow B$`. Thus all examples arise from a surjective endomorphism of $A$ as above, well-defined up to isomorphisms on the source and target.
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http://www.sacredduty.net/2013/01/18/revenge-combinatorics/
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A Protection Paladin Blog
# Revenge Combinatorics
Posted on January 18, 2013 by
Since January is apparently “Warrior Month” here at Sacred Duty, I figured I’d share a little side project I’ve been working on. Omega (of DBM fame) and I were having a conversation about warriors on Lei Shi, and what effect the lack of melee attacks did to your rage generation through Revenge. He mentioned that he’d love to see the calculation, and I said those famous last words, “Oh, that should be easy…”
Several hours later….
Several days later….
OK, to be fair, it didn’t actually take days. It was maybe 8 or 9 hours spread over 3 or 4 days, in bits and pieces. And the vast majority of that time was spent trying to get two different expressions to agree with one another. I worked up a haphazard solution for one limit (boss swing timer longer than the GCD) and then a separate but similar solution for short swing timers, and then found out they didn’t agree with one another. So most of that time was spent hunting down the error causing the inconsistency. Eventually I decided that it would be easier to put together a completely general derivation, and about an hour later had the correct solution. But since I wasted about 8 hours on this, we’ll pretend it took 9 hours just to make me feel better. It can be our little secret.
What follows is the full derivation of the average cooldown.
Derivation
The diagram below illustrates the situation. We consider a string of n GCDs, each having a duration G seconds. During that time, we have a boss attacking us with a swing timer of S seconds. It will be useful to define the maximum number of swings we’ll take during those nG seconds, so we’ll call that N.
Figure 1: Mathematical model for Revenge cooldown calculations. We’re interested in calculating the probability of avoiding a boss attack during the nth GCD (blue) given a boss swing timer S.
It’s fairly easy to find a mathematical expression for N. We note that the elapsed time from t=0 to the time at which attack N lands is $(N-1)S$, and since this must be less time than nG (otherwise attack N would happen after the nth GCD), we can write:
$\large (N-1)S\leq nG$
$\large N-1 \leq nG/S$
$\large N \leq 1 + nG/S$
Since we want to stipulate that N is the very last attack that falls within the window, we apply a floor() function and turn the $\leq$ into an equals sign:
$\large N = 1 + {\rm floor} \left ( \frac{nG}{S} \right)$
To try and put that another way, N is 1 plus the largest integer number of times you can fit S into nG. The 1 is due to the first attack at t=0, which we include in the tally.
Of course, the boss isn’t always going to attack at the very start of the first GCD. It could happen half a second into the GCD, or a full second into the GCD, or anything up to S seconds into the GCD (but not larger than S, because at that point there’ll be a new swing happening at t=0). To see what this looks like, all we need to do is shift the green boxes to the right by some amount $\tau$:
Figure 2: Shifting the time of the first attack causes the total number of attacks occurring between t=0 and t=nG to change.
Note that the amount of shifting we do has a significant effect on the math. In the figure above, we’ve shifted the green boxes over enough that the Nth attack crosses over t=nG and no longer falls within the nth GCD. Now we’re no longer taking N attacks in n GCDs, we’re taking N-1 attacks in that period of time. That changes the statistics slightly, giving us a different probability for refreshing Revenge early due to an avoid.
In the diagram above, I’m using a swing timer that’s exactly 1/3 of a GCD. But if G isn’t evenly divisible by S, we get another complication, which is that the number of attacks that occur during the nth GCD changes as $\tau$ changes. If S were just a bit bigger (see Figure 3) we would push the Nth attack “off” of the blue section before attack N-m-1 gets pushed “on” to the blue section. In that case, we’d only get two attacks during the blue GCD instead of three, as in the lower part of Figure 3.
We have to account for this variation somehow. In practice, the time that elapses between t=0 and the first incoming boss attack is essentially random, can be modeled as a uniform probability distribution. That’s good, because it makes the math pretty easy. We want to perform a weighted average of the three possible situations we can run into, which will end up being an integral. A uniform probability distribution makes that integral easier to evaluate.
Let’s set up that expression now. We’ll consider the nth GCD, which spans the period of time from t=(n-1)G to t=nG. The period of time between the start of the nth GCD and the first boss attack that occurs during that GCD will be important, so we’ll call that x. To be even more specific, x is a function of $\tau$, so let’s define $x_0$ as the value of x when $\tau=0$. In other words, for $x(\tau)$, $x_0=x(\tau=0)$.
In performing our integral, we’re going to need to consider one full period of the swing timer. That means we’ll need to integrate from $\tau=0$ to $\tau=S$, or equivalently from $x=x_0$ to $x=x_0+S$. However, due to the periodicity, we’re not bound to those exact values. We’ll get exactly the same answer if we integrate from $x=0$ to $x=S$, for example – in other words, starting in the condition where a boss attack occurs immediately upon the start of the nth GCD and ending when that boss attack has been shifted by S seconds. It turns out that it’s easier to perform this integration from $x=0$ to $x=S$, so we’ll do that instead. The situation is depicted graphically in Figure 3, where I’ve also increased the swing timer by 20% so that G is no longer evenly divisible by S:
Figure 3: Model for the development of the analytical expressions we need to calculate the average Revenge cooldown. The top panel shows the system near the lower limit of integration (x=0) and the lower panel shows the system near the upper limit of integration (x=S).
The top half of Figure 3 shows the case where x is near zero. We get three attacks within the nth GCD in this case. The lower half of the figure shows the system when x is almost at the upper limit of integration, x=S. There are only two attacks occurring during G in that case. However, we want a more general expression. It will be useful to define a value m to represent the minimum number of attacks we will take during G seconds – in other words, the number of attacks we’re guaranteed to have. We can define this mathematically in a very similar fashion to the way we defined N, giving us
$\large m={\rm floor}\left ( \frac{G}{S} \right )$
For our case, it’s easy to see that m=2. It should also be clear that we’ll only ever have m or m+1 attacks during G as a result of our definition.
Now let’s look at what happens as we increase x from 0 (i.e. slide the green boxes to the right). At a certain value of x we push attack N “off” of the blue box, meaning it no longer occurs during the nth GCD. When this happens, the sum $x+S+S$ must be equal to G (if you’re having trouble seeing that, try and draw it out on paper, it’s somewhere between the top and bottom halves of Figure 3). In the general case, the sum is actually $x+mS$ because there are $m$ full green boxes that fit inside the blue box. Setting those two equal to one another, we have an expression for this value of $x$:
$\large x = G-mS$.
Between $x=0$ and $x=G-ms$, we always have $m+1$ attacks occurring during the nth GCD. Then, from $x=G-mS$ to $x=S$, we have $m$ attacks occurring during the nth GCD. Why is this important, you ask? Well, our chance to reset the cooldown of Revenge during the nth GCD depends on how many attacks occur during that GCD. In fact, let’s figure out what that probability is now.
Assume that we have decimal avoidance $a$, such that 20% avoidance gives us $a=0.20$. It will be more convenient to use the chance that we don’t avoid, so we’ll define that as $q = 1-a$. If we take $y$ swings during a particular GCD, the probability $\rho$ that we avoid at least one of them is:
$\large \rho \left ( y \right) =1-q^y$
This should look familiar if you’ve got a bit of background in statistics, but basically it’s just saying that the probability we avoid at least one attack is 100% minus the probability that we don’t avoid any of them. For a proof, feel free to consult any introductory statistics textbook.
So far, we’ve figured out everything we need to know about what’s happening during the nth GCD. There’s one more piece to the puzzle that we haven’t dealt with yet, and it’s the one that throws the largest wrench into the works. See, we only care about the chance that we avoid an attack during the nth GCD if we didn’t already avoid one in all of the previous GCDs – if we avoided an attack in the 2nd GCD since our last Revenge cast, then we stop counting and the whole thing starts over at 0 again. We only ever get to the 3rd GCD if we didn’t avoid anything during GCDs 1 and 2, and similarly we only ever get to the nth GCD if we don’t avoid anything during the previous (n-1) GCDs.
So we need to know how many attacks occurred during the previous (n-1) GCDs. You might look at Figure 3 and guess that this is just $N-m-1$. It’s not though. This is trickier than it sounds because of the limits of integration we chose. We decided that we’d shift the entire attack stream to the left by $x_0$ so that we could consider the range $x=0$ to $x=S$. But in doing so, we pushed attack #1 to the left as well, so that it no longer occurs during the first GCD. So in fact, we have $N-m-2$ attacks occurring during the first $(n-1)$ GCDs.
However, at some value of $x$ that first attack is going to cross $t=0$ and rejoin the group, which will change the number of attacks occurring between $t=0$ and $t=(n-1)G$. That will bring us back up to $N-m-1$ attacks, and change all of our formulas. Oops.
Luckily, we know what value of $x$ this happens at – it’s easy to see from Figure 1 that it happens at $x=x_0$. And, also using Figure 1, we can write out a pretty simple definition of $x_0$ with a simple geometric argument. $N-m-1$ green blocks is the same length as the orange block, which is $(n-1)G$, plus $x_0$. Rearranging we get:
$\large x_0 = (N-m-1)S-(n-1)G$
This bifurcates our first situation (m+1 attacks during the nth GCD) into two different sections, as illustrated in Figure 4. If attack #1 crosses $t=0$ before we push attack #N out of the nth GCD, then we have to split the term where we have m+1 attacks during the nth GCD into two parts: one with N-m-1 attacks preceding the nth GCD and one with N-m-2 attacks preceding the nth GCD.
Figure 4: Diagram illustrating the four possible probability combinations for GCD #n.
We don’t know the value of $x_0$ until we chose values for G and S, of course, but we do know it occurs between $x=0$ and $x=G-mS$. It can’t happen later than $x=G-mS$, though the reason may not be obvious at first. Looking back at Figure 1, remember that $x_0$ is the value of $x$ if we line up the very first attack that occurs with $t=0$. But $G-mS$ is the value of $x$ where the Nth attack gets pushed out of the nth GCD. For $x_0$ to be greater than $G-mS$, it would mean that the Nth attack occurred after $t=nG$, which can’t happen – remember, we defined attack #N as occurring before $t=nG$ when $x=x_0$.
At this point, we have enough information to write out the expression for the probability that we reset the Revenge cooldown during the nth GCD. To see how this works, let’s write the first integral out and analyze it. From $x=0$ to $x=x_0$, we have $m+1$ attacks during the nth GCD and $N-m-2$ attacks preceding the nth GCD. We can write an integral representing the probability contribution of this first component as follows:
$\Large \displaystyle P_1(n)=\left ( \int_0^{x_0}\frac{dx}{S}\right ) \left ( 1-q^{m+1} \right ) q^{N-m-2}$
Let’s go through that piece by piece. The first part is an integral of $dx/S$, which gives us the portion of our sweep from 0 to $S$ during which the above conditions are true. If we hadn’t used a uniform probability distribution, the arguments of the integrals would have been $p(x)dx$, where $p(x)$ was the probability distribution for the incoming attack time. However, the assumption of uniformity vastly simplifies things, because $p(x)=1/S$ which is independent of $x$.
The second factor is the probability that we avoid at least one of the $m+1$ attacks that occur during the nth GCD. And the third is the probability that we didn’t avoid any of the $N-m-2$ attacks that occurred during the previous $n-1$ GCDs. So this term in its entirety gives us the probability of resetting the cooldown of Revenge between $x=0$ and $x=x_0$.
We can write similar expressions for the other two situations. If we do, we get an expression that looks like this:
$\large \displaystyle P(n)=\left (\int_0^{x_0} \frac{dx}{S}\right )\left (1-q^{m+1}\right )q^{N-m-2} + \left(\int_{x_0}^{G-mS} \frac{dx}{S}\right )\left (1-q^{m+1}\right )q^{N-m-1} \\+ \left(\int_{G-mS}^S\frac{dx}{S}\right )\left(1-q^m\right )q^{N-m-1}$
$P(n)$ is the complete probability that we reset Revenge during GCD #n. I’ve written this as if $P$ was a function of $n$ even though $n$ doesn’t’ show up on the right hand side explicitly because $N$ is a function of $n$. In general, we have all three terms. When $x_0$ is equal to zero or $G-mS$, then one of the integrals has the same upper and lower limits, which makes it identically zero, eliminating one of the first two terms.
We can simplify this a little further by evaluating the integrals. Doing so gives us our final expression for $P(n)$:
$\large \displaystyle P(n) = \left ( \frac{x_0}{S} \right ) \left (1-q^{m+1}\right )q^{N-m-2} + \left (\frac{G-mS-x_0}{S}\right )\left ( 1-q^{m+1}\right )q^{N-m-1} \\+\left (\frac{(m+1)S-G}{S}\right )\left (1-q^m\right )q^{N-m-1}$
Now that we have the probability that we’ll reset Revenge’s cooldown during the nth GCD, there’s just one more thing we need to do to finish our calculation. We want to know the average number of GCDs it takes to reset the cooldown. The average of a discrete random variable is defined to be $\bar{X}=\sum X p(X)$, where $p(X)$ is the normalized probability density function for $X$. In our case, we want $\bar{n}$, which is
$\large \displaystyle \bar{n}=\left (\sum_{n=1}^{\infty} nP(n)\right )$.
However, there’s one other detail we need to fix. The sum in the above equation should be carried out over all values of $n$ from 1 to $\infty$. However, that’s now how our system works. Revenge’s cooldown isn’t infinite, it’s 6 GCDs. So we need to truncate our sum at $n=6$, and somehow account for all of the probability we’re throwing away when we do that.
Luckily this last part is easy. We just need to express the probability that we didn’t avoid any attacks during all 6 GCDs. Looking back to Figures 1 and 2, we see that at some value of $\tau$, the Nth attack lines up with the end of the nth GCD (i.e. attack #N occurs at $t=nG$). For smaller values of $\tau$, we have $N$ attacks in n GCDs; for higher values, we have $N-1$ attacks. Geometrically we see that this value of $\tau$ is $nG-(N-1)S$. So for $\tau$ seconds out of our $S$-second period, we have $N$ attacks that weren’t avoided, which happens with a probability of $q^N$. For the other $S-\tau$ seconds, we have $N-1$ attacks that weren’t avoided, with a probability of $q^{N-1}$. If we let $n=6$, these two probabilities are
$\large Q(6,N)=\left (\frac{6G-(N-1)S}{S}\right )q^N$
$\large Q(6,N-1)=\left (\frac{NS-6G}{S}\right )q^{N-1}$
I’ve used $Q(6,x)$ to represent this, because it’s the probability that we have not reset Revenge after 6 GCDs given $x$ attacks occurring during that time. The contribution to $\bar{n}$ from both of these terms is going to be $6\cdot Q(6,x)$, because when these events occur the cooldown is the full 6 GCDs. Thus, our final expression for $\bar{n}$ is
$\large \displaystyle \bar{n}=\left (\sum_{n=1}^{6} nP(n)\right )+6\left ( \frac{6G-(N-1)S}{S} \right )q^N + 6\left (\frac{NS-6G}{S}\right ) q^{N-1}$.
And of course, the average revenge cooldown is just $\bar{n}G$.
Special Case
In testing this expression to make sure it was correct for all values of $S$ and $G$, I ran into a curious special case. Let’s let $S=2$ and $G=1.5$ and consider the first GCD $(n=1)$. Then our equations tell us that $N=1$ and $m=0$. And the expression for $x_0$ gives us $x_0=(1-0-1)S-(1-1)G=0$, which means that $x_1=0$ and $x_2=G$ (if you didn’t follow that, go back up to the expressions and plug in $N=1$ and $m=0$). Then the first term in our expression for $P(1)$ is
$\large P_1(1) = \frac{x_1}{S}\left (1-q^{m+1}\right )q^{N-m-2} = \frac{0}{S}(1-q)\frac{1}{q}$
OK, so no problem, the first term is zero, right? For most cases, that’s true. But let’s consider the special case of 100% avoidance. While it’s not realistic in-game, it’s a good extreme limit to test to make sure our equation is working properly. 100% avoidance is equivalent to $a=1$, which means $q=1-a=0$. Which is a problem – if we plug this into a computer, it sees that we’re trying to divide by zero and tells us that the result is invalid (MATLAB spits out “NaN” for “not a number” any time you divide by zero).
Before we fix this, let’s think about what the math means. We’re considering the very first GCD, with a swing timer $S$ that’s longer than the GCD. That $q^{N-m-2}$ term is accounting for the $N-m-2$ attacks that we didn’t avoid during the previous GCDs. Except wait a minute… we’re in the very first GCD! There were no earlier GCDs! So that should be identically $q^0$ whenever $n=1$.
This isn’t an issue for the other two terms because $N-m-1$ is identically 0 when $n=1$. But it leads to a weird result for this first term. There’s probably an alternative way we could write $N-m-2$ in terms of $n$, $G$, and $S$ that would fix this problem. However, it would almost certainly not be pretty. Instead of trying to spend a lot of time finding that expression, I’ve implemented a slightly easier fix: I’ve constrained the exponent $N-m-2$ to be greater or equal to zero with a max() function. Thus, if we ever get into that $N=1$, $m=0$ situation, $q^{N-m-2}$ becomes $q^0$ and no longer causes problems. This will be strictly correct for our situation as well, because we know that we can’t have a negative number of attacks preceding a GCD.
Just to check, let’s finish that example with 100% avoidance. The first term now goes to zero because $x_0=0$. The second term is simply
$\large P_2(1)=\left ( \frac{G}{S}\right )$
or the probability that the attack falls within the first GCD. The last term is also zero:
$\large P_3(1)=\left ( \frac{S-G}{S}\right ) \left (1-q^0\right )q^0 =\left (\frac{S-G}{S}\right ) (1-1)\cdot 1 = 0$
Thus, the probability of resetting Revenge’s cooldown within one GCD is just the probability that an attack happens in that GCD. It’s straightforward to show that for $n=2$, $N=2$ and $x_0=S-G$, giving just
$\large P(2)=P_1(2) = \left ( \frac{S-G}{S}\right )\left (1-q\right )$.
Or, in other words, just $1-P(1)$. Thus, as long as we have 100% avoidance, we’re guaranteed to reset the cooldown of Revenge within two GCDs if we’re getting attacked every two seconds, just as one would expect. Every other term becomes zero since $q=0$. So the result is exactly what we expect for this simple special case. If we plug in $S=2$ and $G=1.5$, we find that the result is
$\bar{n}G = 1.5 \left [1\cdot \left ( \frac{1.5}{2} \right ) + 2 \cdot \left ( \frac{2-1.5}{2} \right ) \right ]= 1.5\cdot 0.75 + 3 \cdot 0.25 = 1.875$
or 1.875 seconds.
Numerical Solutions
Now that we have the proper expressions, we can start cranking out some more general results. For all of these cases we’ll choose a GCD length of $G=1.5$, as warriors have no mechanism for haste scaling. The first thing we might want to know is how the average Revenge cooldown scales with avoidance. We’ll assume a swing timer of $S=2$ seconds and let $a$ vary from 0 to 1. Doing so gives us this plot:
Average Revenge cooldown as a function of avoidance for a swing timer of S=2 seconds.
As expected, this function seems continuous and slowly decreases to the minimum of just under 2 seconds. At around 20% avoidance, which seems like a reasonable guess for a well-geared warrior, the average cooldown is 6.2 seconds, or about a 31% reduction. The reduction very sharp between 0% and 50% and becomes shallower above 50%. This steepness at low avoidance is part of the reason dodge and parry perform as well as they do for warriors as compared to paladins. You get a pretty significant resource increase at low levels of avoidance.
This also lets us answer the original question that sent us down this path in the first place. For an average tank with ~20% avoidance, Revenge is available every 6.2 seconds. If we’re generous and say that they get lucky, we could say that they cast it every ~6 seconds instead since that’s a GCD multiple. Thus, they’d be gaining 15 rage from Revenge every 6 seconds, for a net rage generation rate of 2.5 RPS. With 0% avoidance (equivalent to a boss that doesn’t melee, like Lei Shi), you get that 15 rage every 9 seconds, for 1.67 RPS. So the net loss is about 0.83 RPS, which is pretty large (over 10% of your rage generation).
However, that’s assuming Revenge is your highest priority. In practice, it isn’t, so it will get pushed back sometimes for Shield Slam. As such it might be a better estimate to round up in GCDs, turning that 6.2 seconds into 7.5 seconds. And 15 rage every 7.5 seconds is an even 2 RPS. In that case, the loss in rage generation by not being meleed drops to 0.33 RPS. That’s only about 4% of your total rage generation. Though we should note that as we saw in the last round of Monte-Carlo simulations, even that small amount makes a significant difference in damage patterns, especially in queues that involve Shield Barrier (which Lei Shi obviously would).
We could also ask how the average cooldown varies with boss swing timer. Keeping avoidance fixed at 20% and letting $S$ vary from 1.5 to 2.5 seconds gives us the following plot:
Average Revenge cooldown as a function of boss swing timer for 20% avoidance.
This plot has some interesting features. The cooldown isn’t a continuous function of boss swing timer, which seems strange. What we’re seeing here is similar to a Moiré pattern, caused by the interference of two periodic effects at different frequencies. Whenever we hit a point at which an integral number of swing timers is divisible by an integer number of GCDs (i.e. $bS=cG$, for arbitrary integers $b$ and $c$), we get a discontinuity. So, for example, at $S=2$ (3S=2G), or $S=1.8$ (5S=6G). These variations aren’t very large, and there’s really nothing we can do about them anyway, so it’s not worth worrying about. It’s just an interesting quirk of the combinatorics.
And that quirk gets even more noticeable for swing timers shorter than 1.5 seconds. Let’s extend that plot all the way down to $S=0$:
Average Revenge cooldown vs. swing timer for 20% avoidance, now with illogically small swing timers included!
As you can see from the plot, there are lots of discontinuities occurring below $S=1.5$ because we’re rapidly scanning through different integer multiple conditions. Since few bosses (if any, nowadays) have swing timers as short as even one second, these features aren’t likely to manifest themselves in game. In cases where you have multiple mobs attacking you, such that the average swing timer is less than one second, there’s an inherent randomness that breaks the periodicity and averages out these discontinuities. If we ran a simulation instead of an analytical model, we could perform that averaging and see that it just smooths this curve out a bit. Since the variations aren’t that great to begin with, it’s not worth doing that, because this plot is good enough to give us good ballpark estimates of what we expect in those situations.
Since it’s not always easy to read off of the graph that accurately, in closing I want to provide a table you can use to look up the average Revenge cooldown given a certain amount of avoidance and swing timer. The table below contains calculated values for avoidance ranging from 0% to 30% and swing timers from 0.5s to 2.5s in steps of 0.25s.
``` -----------------------------------S-----------------------------------
| A | 0.50s | 0.75s | 1.00s | 1.25s | 1.50s | 1.75s | 2.00s | 2.25s | 2.50s |
| 0% | 9.000 | 9.000 | 9.000 | 9.000 | 9.000 | 9.000 | 9.000 | 9.000 | 9.000 |
| 1% | 8.358 | 8.564 | 8.670 | 8.734 | 8.778 | 8.809 | 8.833 | 8.851 | 8.866 |
| 2% | 7.776 | 8.155 | 8.355 | 8.476 | 8.562 | 8.623 | 8.669 | 8.705 | 8.733 |
| 3% | 7.249 | 7.770 | 8.053 | 8.225 | 8.351 | 8.440 | 8.507 | 8.560 | 8.602 |
| 4% | 6.772 | 7.410 | 7.766 | 7.982 | 8.147 | 8.262 | 8.349 | 8.418 | 8.472 |
| 5% | 6.340 | 7.071 | 7.491 | 7.747 | 7.947 | 8.088 | 8.194 | 8.278 | 8.344 |
| 6% | 5.947 | 6.754 | 7.228 | 7.518 | 7.753 | 7.917 | 8.041 | 8.141 | 8.218 |
| 7% | 5.591 | 6.455 | 6.977 | 7.296 | 7.564 | 7.751 | 7.892 | 8.005 | 8.092 |
| 8% | 5.267 | 6.175 | 6.737 | 7.080 | 7.381 | 7.588 | 7.745 | 7.872 | 7.969 |
| 9% | 4.972 | 5.912 | 6.508 | 6.871 | 7.202 | 7.429 | 7.601 | 7.740 | 7.847 |
| 10% | 4.704 | 5.665 | 6.290 | 6.669 | 7.028 | 7.273 | 7.460 | 7.611 | 7.726 |
| 11% | 4.460 | 5.433 | 6.081 | 6.472 | 6.859 | 7.121 | 7.321 | 7.484 | 7.607 |
| 12% | 4.237 | 5.215 | 5.881 | 6.281 | 6.695 | 6.973 | 7.185 | 7.359 | 7.489 |
| 13% | 4.034 | 5.010 | 5.691 | 6.096 | 6.535 | 6.828 | 7.052 | 7.236 | 7.372 |
| 14% | 3.849 | 4.818 | 5.509 | 5.916 | 6.380 | 6.686 | 6.921 | 7.115 | 7.257 |
| 15% | 3.679 | 4.637 | 5.335 | 5.742 | 6.229 | 6.548 | 6.792 | 6.996 | 7.143 |
| 16% | 3.523 | 4.466 | 5.169 | 5.573 | 6.082 | 6.413 | 6.667 | 6.878 | 7.031 |
| 17% | 3.381 | 4.306 | 5.011 | 5.409 | 5.939 | 6.281 | 6.543 | 6.763 | 6.920 |
| 18% | 3.250 | 4.156 | 4.860 | 5.250 | 5.800 | 6.152 | 6.422 | 6.650 | 6.810 |
| 19% | 3.129 | 4.014 | 4.715 | 5.096 | 5.665 | 6.026 | 6.304 | 6.538 | 6.702 |
| 20% | 3.018 | 3.880 | 4.577 | 4.947 | 5.534 | 5.903 | 6.188 | 6.429 | 6.595 |
| 21% | 2.916 | 3.755 | 4.445 | 4.802 | 5.407 | 5.783 | 6.074 | 6.321 | 6.490 |
| 22% | 2.822 | 3.636 | 4.320 | 4.661 | 5.283 | 5.666 | 5.962 | 6.215 | 6.385 |
| 23% | 2.735 | 3.525 | 4.199 | 4.525 | 5.162 | 5.552 | 5.853 | 6.111 | 6.282 |
| 24% | 2.655 | 3.419 | 4.084 | 4.392 | 5.046 | 5.440 | 5.746 | 6.009 | 6.180 |
| 25% | 2.580 | 3.320 | 3.975 | 4.264 | 4.932 | 5.332 | 5.641 | 5.908 | 6.080 |
| 26% | 2.511 | 3.226 | 3.870 | 4.140 | 4.822 | 5.225 | 5.538 | 5.809 | 5.980 |
| 27% | 2.447 | 3.138 | 3.770 | 4.019 | 4.715 | 5.122 | 5.438 | 5.712 | 5.882 |
| 28% | 2.387 | 3.054 | 3.674 | 3.903 | 4.611 | 5.021 | 5.339 | 5.617 | 5.785 |
| 29% | 2.331 | 2.975 | 3.582 | 3.789 | 4.510 | 4.922 | 5.243 | 5.523 | 5.690 |
| 30% | 2.279 | 2.900 | 3.495 | 3.680 | 4.412 | 4.826 | 5.148 | 5.431 | 5.595 |```
As a final note, I want to apologize for the formatting in this post. I’ve used LaTeX fairly inconsistently because the WordPress plugin for latex is ugly as hell for inline equations (you might note that it “sinks” the letters with respect to the baseline of text). It was much prettier to just use italicized letters (i.e. N, S, G, instead of $N$, $S$, $G$), but for some of the equations and symbols it was just easier to use TeX. Sorry for the ugly, if anyone knows of a good/better LaTeX plugin for WordPress, I’d love to know about it.
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### 7 Responses to Revenge Combinatorics
1. bryjoered says:
Honestly, I don’t think regular people will have any idea what most of these equations actually mean. The amount of math that you do for this game is astounding, what would we do without you? I mean I get the gist, I think, you are trying to develop (or already have) a reliable way to measure revenge procs.
2. bryjoered says:
Also, I’m not positive on this, but isn’t revenge a priority when AOE tanking? Maybe it doesn’t surpass shield slam in pure rage generation, but it sure as hell helps you keep aggro better. I normally hit thunderclap, shockwave, revenge when available, then continue my normal rotation, of course always keeping shield block up which is clearly better unless they are caster mobs.
• Sly says:
I think Shield Slam almost always hits harder than Revenge (at least since they nerfed it to do reduced damage to the second and third targets). So you’re better off keeping Shield Slam as your highest priority and tab targeting rather than moving Revenge up. This also results in better rage generation.
Furthermore, there’s really only one situation I can think of where AoE aggro is really a concern: two-tanking Windlord and trying to keep a somewhat even distribution of adds (note I haven’t completed all Heroic bosses).
My usual AoE queue is: Shield Slam > Revenge > Shout > Shockwave/Dragon Roar > Thunderclap > Devastate. Thunderclap doesn’t even beat Devastate until you get to three or four targets, either.
3. Pantherj says:
As a math major in Applied Stats 2 and Probability, this was amazing to look at. I can’t wait to open more posts to see what other satisfying concepts to observe in (mostly) real life.
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• ### Theck’s Twitter
• @Slootbag I could maybe see an efficiency argument though. Large DPS gain for small relative HPG impact due to proc 1 day ago
• @Slootbag it is a sizable damage increase though, but so would just prioritizing AS with or w/o proc 1 day ago
• @Slootbag it's actually an HPG loss though. The probability of wasting a proc is very small, so the loss incurred by CS pushback is greater 1 day ago
• @Slootbag @ConverttoRaid TLDR rarely worth separating AS from ASGC, and none of the edge cases boost one over CS/J but not the other 1 day ago
• @Slootbag @ConverttoRaid or you want max DPS, in which case AS> all regardless of proc status. 1 day ago
• ### Ana’s Twitter
• Third time I've heard Clarity in the last hour. Good thing I still love this track! 5 hours ago
• Hey @riftmaker I made you assist again- I might be late and miss the start! Getting in the car soon, 2.5 hrs away 7 hours ago
• My grandfathers old friend, taking a photo of me, my two sisters and my mom, calls out: "OK ladies, tits up!" ......... What. 22 hours ago
• Turned my radio back on at the awesome \$25/6mo deal. Live EDC NYC playing!! Oh my xm, I missed you! 1 day ago
• Only 30 min into this car trip and already I desperately miss my satellite radio. I wonder how fast it would be active if I called to resub. 1 day ago
• ### Mel’s Twitter
• @Esoth @KihraOfTemerity @Kerriodos @methodtreckie @Anafielle Say what now? That sounds almost exactly like my philosophy. 4 days ago
• @SerrinneWoW Co-incidence, or the start of a pattern? :P 1 month ago
• @Rhidach @_vidyala @SerrinneWoW Clearly Sara isn't a foodie. Anyone who knows me knows how I go out of my way to avoid a good meal. 1 month ago
• @SerrinneWoW Are you stalking me? 1 month ago
• @_vidyala @SerrinneWoW @Jasyla_ @Kaleri_ @Chronis_ Liekwise! Sorry I had to miss dinner yesterday. 1 month ago
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http://nrich.maths.org/1882/note
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### Exploring Wild & Wonderful Number Patterns
EWWNP means Exploring Wild and Wonderful Number Patterns Created by Yourself! Investigate what happens if we create number patterns using some simple rules.
### I'm Eight
Find a great variety of ways of asking questions which make 8.
### Dice and Spinner Numbers
If you had any number of ordinary dice, what are the possible ways of making their totals 6? What would the product of the dice be each time?
# A Conversation Piece
### Why do this
problem?
This problem offers a lot of practice of division as well as revision of ideas about prime numbers in a situation in which children want to get to the answer - much more fun than a page of calculations but just as much practice. It requires a calculator unless someone really relishes long division!
### Key questions
Have you made a list of the first ten prime numbers?
Remember that $1$ is not a prime number.
Which number are you going to divide by first?
How will you know whether you have divided by that number already?
### Possible extension
Learners who find it easy could be encouraged to invent their own similar problems which will give them practice with multiplication as well as division.
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.
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http://math.stackexchange.com/questions/152167/laplace-transform-of-simple-function/268487
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# Laplace transform of simple function
I have the function below:
$$3-2t+3t^2$$
I tried to get the transfer function, but my solution seems to be wrong, can someone please tell me what I am doing wrong:
$$\frac{1}{s}+\frac{1}{s^3}$$
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3
Are we just finding $\int_0^\infty (3-2t+3t^2)e^{-st}\,dt$? If so (using integration by parts for the last two terms) I think we get $\frac{3}{s}-\frac{2}{s^2}+\frac{6}{s^3}$. As to what you did wrong, that is hard to know if one does not see what you did. – André Nicolas May 31 '12 at 19:33
You can use the following properties: $\mathcal{L} \{1\} =\frac{1}{s}$ , $\mathcal{L} \{t^n \} =\frac{n!}{s^{n+1}}$ – passenger May 31 '12 at 19:37
It would be better if you added what was done in a clear way. Indeed, Andre got what you are looking for. – Babak S. May 31 '12 at 19:40
Is the function of $t$ meant to be the impulse response? – copper.hat May 31 '12 at 19:44
Thanks guys, this was so simple I was ashamed to continue the question :( got it right now. – Sean87 May 31 '12 at 20:51
## 2 Answers
Just looking at a table and seeing how $$\mathcal L\{1\}=\frac{1}{s},\quad \mathcal L\{t\}=\frac{1}{s^2},\quad \mathcal L\{t^2\}=\frac{2}{s^3}$$... add them up, combine the constants, hope to get something like $$F(s)=\frac{3}{s}-\frac{2}{s^2}+\frac{6}{s^3}$$
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$$\begin{align*}f(t)&=3-2t+3t^2\\ \\ F(s)&=\mathcal L\{3-2t+3t^2\}\\ \\ &=\mathcal L\{3\}-\mathcal L\{2t\}+\mathcal L\{3t^2\}\\ \\ &=3\mathcal L\{t^0\}-2\mathcal L\{t^1\}+3\mathcal L\{t^2\}\\\\&=\end{align*}$$
use $$\boxed{\begin{align*}\mathcal L\{t^n\}&=\frac{\Gamma(n+1)}{s^{n+1}},\qquad n>-1\\&=\frac{n!}{s^{n+1}},\qquad \qquad n\in\mathbb Z> 0\qquad\end{align*}}$$
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http://math.stackexchange.com/questions/243043/decompose-matrix
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# Decompose matrix
The following $m$ x $n$ matrix, decompose the first standard basis vector $e_1 = w + z \in \mathbb{R^n}$, where $w \in$ rowspace(A) and $z\in kerA$. Verify your answer by expressing $w$ as a linear combination of the rows of $A$.
a.) $\pmatrix{1&-2&1\\2&-3&2}$
b.) $\pmatrix{1&-1&0&3\\2&1&3&3\\1&2&3&0}$
For a, I got the $kerA = \pmatrix{-1\\0\\1}$ and the $rowspaceA = \pmatrix{1\\-2\\1},\pmatrix{2\\-3\\2}$.
The answer is:
$z = \pmatrix{\frac{1}{2}\\0\\-\frac{1}{2}}, w = \pmatrix{\frac{1}{2}\\0\\\frac{1}{2}} = - \frac{3}{2} \pmatrix{1\\-2\\1} + \pmatrix{2\\-3\\2}$
How did they get that?
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## 1 Answer
Try:
$e_1=\pmatrix{1\\0\\0}=z+w=c_1\pmatrix{-1\\0\\1}+c_2\pmatrix{1\\-2\\1}+c_3\pmatrix{2\\-3\\2}$ and solve the equations.
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http://mathoverflow.net/questions/28644/the-isomorphism-inference-rule/28785
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## The isomorphism inference rule
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Suppose we are writing very detailed proofs, absolutely without any gaps (for example, for checking proofs by computer).
In such formal proofs every step (even a trivial one) must be justified.
For example, when we have proved that A = B and A is a prime number, we can infer that B is also a prime number.
We can justify this step by Leibniz Law (which may be represented by an axiom of Predicate Calculus with Equality).
When a group A is isomorphic to a group B and the group A is simple, then we can infer that the group B is also simple.
We can say , that "the isomorphism inference rule" was used in that case.
To justify such inferences, Bourbaki developed a general theory of isomorphism (see their book "Theory of Sets").
Because the Bourbaki theory is rather complicated, my questions are:
1) Are there other approaches for justification of "the isomorphism inference rule"?
2) Or maybe there are more simple expositions of Bourbaki approach?
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You wrote "When a group A is isomorphic to a group B and the group G is simple, then we can infer that the group B is also simple." Don't you mean A instead of G? – Pierre-Yves Gaillard Jun 19 2010 at 4:08
Yes, of course, I am sorry for the misprint. – Victor Makarov Jun 19 2010 at 10:22
Why don't you correct it? – Pierre-Yves Gaillard Jun 19 2010 at 10:36
I just did it and added a question about possible simplification of Bourbaki theory if we allow only one principal base set. Will be still there examples of untrasportable formulas? – Victor Makarov Jun 19 2010 at 11:11
## 7 Answers
Well, if you prove a predicate P respects any equivalence relation R, you can then use a similar inference from one member of an equivalence class to another. So you can deduce P(y) from P(x) under those circumstances from x R y. Now are you objecting to the complexity of the isomorphism concept (R), or the fact that certain concepts P are in some a priori way known to depend only on the isomorphism class? I suspect the latter. Bourbaki likes "transport of structure" as a concept; I suppose the point may be that "structure" in this sense is by no means confined to first-order logic.
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### 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.
First of all, I suspect that whenever a formal system has such an "isomorphism inference rule," all proofs using that rule can be converted to proofs not using it. (I don't know the details of Bourbaki's set theory, though.)
So what could an "isomorphism rule" look like? First of all, you need a metamathematical characterization of what constitutes an "isomorphism" between arbitrary "structures." I guess category theory can answer this question, but only if you encode your specific class of structures as a category. So that does not really define isomorphism on a metamathematical level. (Unless, apparently, one works within fully categorial foundations: http://cs.nyu.edu/pipermail/fom/2003-July/007064.html)
However, there is indeed a comparatively simple way to characterize isomorphisms. This is actually part of a formal system I have developed for a proof assistant, but you can apply the same principle in naive set theory if you don't mind a little vagueness. (Don't try to apply it to first-order axiomatic set theory, though.)
There is one rule of this system that matters here: Roughly speaking, for given x and S, you may not ask whether x is a member of S unless x was introduced as a member of some superset of S. The introduction of a variable as a member of a given set is considered primitive, so the rule is purely syntactical. An example:
We want to say: "Let S be the intersection of the set of primes with the set of even integers, and n be a member of S. Then n is in N." Surely, the set of primes is defined as some subset of N, which in turn is a subset of Z, and the set of even integers is also defined as a subset of Z. So n is syntactically known to be a member of Z, and we can ask whether it is also in N. However, we cannot ask whether it is e.g. in the set of finite graphs. Or whatever else. Formally, this amounts to a type system.
Another aspect of the formal system is that when you want to define what a "group" is, you need to specify when two "groups" are considered equal. So suppose you are given two sets S and T, as well as group operations on each. There is no syntactically "known" superset, so you cannot ask whether S=T because that would involve taking an element of S and asking whether it is in T. Now convince yourself that the most you can do is ask whether the two groups are isomorphic. That is, no other formula you can come up with will ever distinguish two isomorphic groups. (It is required to be reflexive, symmetrical, and transitive, of course.)
To conclude, it is possible to construct a system in which you cannot even talk about structures except up to isomorphism, for arbitrary structures. (That does not mean you cannot take their concrete sets into account in special cases. For example, if you have a group with two isomorphic subgroups, these will be considered equal as groups, but the sets can be different. Note how it makes sense to ask whether they are equal or different because we know a common superset.)
Now here is my equivalent of your "isomorphism inference rule": Since two isomorphic groups are in fact equal in this system, any property you can specify about one of them will be considered true of the other.
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1
I tend to think considering isomorphic groups to be "equal" is the wrong way to go, because it seems to lose information about the particular isomorphism used. When talking about "properties", the particular isomorphism doesn't matter, but when transporting things other than properties, it tends to matter which isomorphism you choose to transport them along. – Mike Shulman Jun 20 2010 at 4:52
I'm not saying one should necessarily identify isomorphic groups in mathematical practice, but in this particular system, it just comes out that way, by virtue of having to specify when two groups are supposed to be equal. So what I called "group" is strictly speaking an "isomorphism class of groups." Properties like "simple" can be defined on such classes, and they are automatically guaranteed to be well-defined. In the case you mention, however, you would need to refer to the actual sets, not just the isomorphism classes. This is analogous to what I wrote about subgroups. – Sebastian Reichelt Jun 20 2010 at 10:32
Dealing with this question is one of the goals behind Voevodsky's equivalence axiom for type theory.
In ordinary Martin-Löf dependent type theory, an equality A=B is represented by a term in an identity type, $p:Id_T(A,B)$ where T is the type of A and B. The term p is regarded as a "proof" that A=B, under the "propositions as types" paradigm. The fact that any property of A then transfers to a property of B is implemented by the elimination rule for identity types. For example, if P(X) is the property "X is prime" and we have a proof of P(A), i.e. a term $q:P(A)$ (under propositions-as-types), then from p and q we can construct a new term $J(q;A,B,p): P(B)$ which "proves" that B is also prime. This is the type-theoretic incarnation of "Leibniz' law," but since it applies to terms of arbitrary types (not just proofs of propositions) it is more powerful.
The equivalence axiom says, roughly, that If A and B are types and $Type$ is a type-theoretic universe containing them, then terms of the identity type $Id_{Type}(A,B)$ are the same as isomorphisms between A and B. In particular, this means that given an isomorphism between A and B, we can apply the equivalence axiom to obtain a term $p:Id_{Type}(A,B)$, and then use the elimination rule for identity types, as above, to show that any property of A is also possessed by B.
The equivalence axiom greatly restricts the class of models of type theory. Notably it excludes all "extensional" approaches which model types by sets and identity types by equality. Models of the equivalence axiom tend to be categorial or homotopical, in which the models of types are higher categories or homotopy types containing internal structure with which to record isomorphisms. The ur-example models types by homotopy-types, a.k.a. ∞-groupoids, and the universe $Type$ by the ∞-groupoid of all small ∞-groupoids.
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This approach is related to my other answer, but simpler.
Note that if we define groups and so on in terms of classical membership-based set theory, then it is not true that all properties transfer along isomorphisms. For instance, the property "the identity element of this group is the empty set" is not preserved under passage to an isomorphic group. So if we want an "isomorphism inference rule," we need to somehow restrict the allowable statements. A very natural way to do this is by using a category-theoretic foundation of mathematics. If we write the axioms of a category in dependent type theory in a way which excludes the notion of "equality between objects" (which reflects how we usually reason about objects of a category anyway), then any fact about objects or morphisms in a category which can be expressed in this language will automatically be invariant under isomorphisms of objects, and under equivalences of categories. This was first shown by Peter Freyd in the paper "Properties invariant within equivalence types of categories" and by Georges Blanc in the paper "Équivalence naturelle et formules logiques en théorie des catégories," and more recently it has been explored in much more depth and generality by Michael Makkai in the theory of FOLDS.
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There is more than one kind of isomorphism, because isomorphism preserves structure, and there is more than one kind of structure.
• Set isomorphisms preserve cardinality.
• Group isomorphisms preserve the way the group may be generated.
and so on. They are meant to show that two instances are merely 're-presentations' of a common structure. An 'isomorphism inference rule' would simply be the observation that any property P which is phrased directly in terms of the properties preserved by isomorphisms, will also be preserved by isomorphisms.
Whether an integer number is prime does not depend on whether it is expressed in decimal or binary, but on e.g. the Peano Axioms used to define the integers and the particular integer in question; whether a group is simple does not depend on anything but the group axioms and the way in which the group is generated. If you prove isomorphisms of the two instances, the preservation of properties defined in terms of those structures follow ipso facto.
Leibnitz's law is the same principle as this, except applied to an equivalence relation which preserves "all" structures. (Leibnitz's law is tantamount to the declaration that such an equivalence relation exists.)
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It requires that there is an equivalence relation that preserves the structures which happen to be definable by the theory in question. This is why we can define an equality relation satisfying Leibniz's rule in ZFC: because anything we can define using only ∈ will respect extensional equality. But it's perfectly possible for Leibniz's rule to hold in a structure for which = is not interpreted as true equality, in which case it does not capture everything. However, by convention, we always mod out by = so that the = symbol does represent actual equality. – Carl Mummert Jun 19 2010 at 11:59
I don't see why a special rule should be introduced to make inferences about properties preserved by isomorphisms: if a property P is preserved by isomorphisms then it should hopefully be possible to prove this preservation using only the axioms and the standard logic inference rules.
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The "isomorphism inference rule" can be understood in terms of the concept of elementary equivalence in model theory: http://en.wikipedia.org/wiki/Elementary_equivalence
The wikipedia article above discusses elementary equivalence for first order logic, but it may be generalised to higher order logic. Given a language $\mathcal{L}$ with semantics $\models$, two structures $\mathbb{M}$ and $\mathbb{N}$ are elementarily equivalent whenever for all $\phi \in \mathcal{L}$, $\mathbb{M} \models \phi$ holds if and only if $\mathbb{N} \models \phi$.
One can prove for first-order, second-order and higher order logic the following theorem:
if two structures are isomorphic, then they are elementarily equivalent
The proof almost always employs structural induction on the composition of formulae, making use of the fact that the semantics for various logics are usually defined compositionally using recursion. The converse of the above statement is not true in general.
Here is how we can apply the concept of elementary equivalence to the example provided. We can express in second order logic a formula $\phi$ such that $\mathbb{M} \models \phi$ if and only if $\mathbb{M}$ is a simple group. We then know that if $\mathbb{M} \cong \mathbb{N}$, then $\mathbb{M} \models \phi$ if and only if $\mathbb{N} \models \phi$. This means that if $\mathbb{M}$ and $\mathbb{N}$ are isomorphic, $\mathbb{M}$ is a simple group if and only if $\mathbb{N}$ is a simple group.
TL;DR: The "isomorphism inference rule" can be understood model theoretically in terms of elementary equivalence, and is a consequence of the compositionality of semantics.
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http://unapologetic.wordpress.com/2011/08/02/integration-on-the-standard-cube/
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# The Unapologetic Mathematician
## Integration on the Standard Cube
Sorry for the delay, I’ve had a packed weekend.
Anyway, we’re ready to start getting into integration on manifolds. And we start with a simple case that everything else will be built on top of.
We let $[0,1]^k\subseteq\mathbb{R}^k$ be the “standard $k$-cube”. We know that the space $\Omega^k([0,1]^k)$ of “top forms” — top because $k$ is the highest degree possible for a differential form on a differential form — has rank $1$ over the algebra $\mathcal{O}([0,1]^k)$ of smooth functions. That is, if $\omega$ is a top form then we can always write
$\displaystyle\omega=f\,du^1\wedge\dots\wedge du^k$
for some smooth function $f$ on the standard cube. Then we write
$\displaystyle\int\limits_{[0,1]^k}\omega=\int\limits_{[0,1]^k}f\,du^1\wedge\dots\wedge du^k=\int\limits_{[0,1]^k}f\,d(u^1\dots du^k)$
here we sorta pull a fast one, notationally speaking. On the left we’re defining the integral of a $k$-form $\omega$. In the middle we rewrite the form as above, in terms of a function and the canonical basis $k$-form $du^1\wedge\dots\wedge du^k$ made from wedging together the basic $1$-forms in order. And then on the right we suddenly switch to a $k$-dimensional Riemann integral over the standard $k$-cube. The canonical basis $k$-form $du^1\wedge\dots\wedge du^k$ corresponds to the volume element $d(u^1,\dots,u^k)$, and top forms are often also called “volume forms” because of this correspondence. In fact, it’s not hard to see that they’re related to signed volumes. This is the starting point from which all integration on manifolds emerges, and everything will ultimately come back to this definition.
### Like this:
Posted by John Armstrong | Differential Topology, Topology
## 7 Comments »
Comment by | August 2, 2011 | Reply
2. [...] next step after standard cubes in our integration project is to define integration on “singular [...]
Pingback by | August 3, 2011 | Reply
3. [...] can integrate on the standard cube, and on singular cubes in arbitrary manifolds. What if it’s not very easy to describe a [...]
Pingback by | August 5, 2011 | Reply
4. [...] we get by setting that component to . Explicitly, we’ll define the following faces of the standard cube [...]
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5. [...] start by considering the case where is the standard cube . Whipping out the definition of the boundary operator , the integral on the right proceeds as [...]
Pingback by | August 18, 2011 | Reply
6. [...] that we’ve proven Stokes’ theorem in the case of the standard cube, we can now tackle the general [...]
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7. [...] Armstrong: Integrals are Independent of Parametrization, Integration on Singular Cubes, Integration on the Standard Cube, Stokes’ Theorem [...]
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## 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.
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http://catalog.flatworldknowledge.com/bookhub/reader/128?e=fwk-redden-ch06_s07
|
# Elementary Algebra, v. 1.0
by John Redden
Study Aids:
Click the Study Aids tab at the bottom of the book to access your Study Aids (usually practice quizzes and flash cards).
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Study Pass is our latest digital product that lets you take notes, highlight important sections of the text using different colors, create "tags" or labels to filter your notes and highlights, and print so you can study offline. Study Pass also includes interactive study aids, such as flash cards and quizzes.
Highlighting and Taking Notes:
If you've purchased the All Access Pass or Study Pass, in the online reader, click and drag your mouse to highlight text. When you do a small button appears – simply click on it! From there, you can select a highlight color, add notes, add tags, or any combination.
Printing:
If you've purchased the All Access Pass, you can print each chapter by clicking on the Downloads tab. If you have Study Pass, click on the print icon within Study View to print out your notes and highlighted sections.
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## 6.7 Applications Involving Quadratic Equations
### Learning Objectives
1. Set up and solve applications involving relationships between real numbers.
2. Set up and solve applications involving geometric relationships involving area and the Pythagorean theorem.
3. Set up and solve applications involving the height of projectiles.
## Number Problems
The algebraic setups of the word problems that we have previously encountered led to linear equations. When we translate the applications to algebraic setups in this section, the setups lead to quadratic equations. Just as before, we want to avoid relying on the “guess and check” method for solving applications. Using algebra to solve problems simplifies the process and is more reliable.
Example 1: One integer is 4 less than twice another integer, and their product is 96. Set up an algebraic equation and solve it to find the two integers.
Solution: First, identify the variables. Avoid two variables by using the relationship between the two unknowns.
The key phrase, “their product is 96,” indicates that we should multiply and set the product equal to 96.
Once we have the problem translated to a mathematical equation, we then solve. In this case, we can solve by factoring. The first step is to write the equation in standard form:
Next, factor completely and set each variable factor equal to zero.
The problem calls for two integers whose product is +96. The product of two positive numbers is positive and the product of two negative numbers is positive. Hence we can have two sets of solutions. Use $2n−4$ to determine the other integers.
Answer: Two sets of integers solve this problem: {8, 12} and {−6, −16}. Notice that (8)(12) = 96 and (−6)(−16) = 96; our solutions check out.
With quadratic equations, we often obtain two solutions for the identified unknown. Although it may be the case that both are solutions to the equation, they may not be solutions to the problem. If a solution does not solve the original application, then we disregard it.
Recall that consecutive odd and even integers both are separated by two units.
Example 2: The product of two consecutive positive odd integers is 99. Find the integers.
Solution:
The key phase, “product…is 99,” indicates that we should multiply and set the product equal to 99.
Rewrite the quadratic equation in standard form and solve by factoring.
Because the problem asks for positive integers, $n=9$ is the only solution. Back substitute to determine the next odd integer.
Answer: The consecutive positive odd integers are 9 and 11.
Example 3: Given two consecutive positive odd integers, the product of the larger and twice the smaller is equal to 70. Find the integers.
Solution:
The key phrase “twice the smaller” can be translated to $2n$. The phrase “product…is 70” indicates that we should multiply this by the larger odd integer and set the product equal to 70.
Solve by factoring.
Because the problem asks for positive integers, $n=5$ is the only solution. Back substitute into n + 2 to determine the next odd integer.
Answer: The positive odd integers are 5 and 7.
Try this! The product of two consecutive positive even integers is 168. Find the integers.
Answer: The positive even integers are 12 and 14.
## Geometry Problems
When working with geometry problems, it is helpful to draw a picture. Below are some area formulas that you are expected to know. (Recall that $π≈3.14$.)
Area of a rectangle$A=lw$, where l represents the length and w represents the width.: $A=l⋅w$
Area of a square$A=s2$, where s represents the length of each side.: $A=s2$
Area of a triangle$A=12bh$, where b represents the length of the base and h represents the height.: $A=12bh$
Area of a circle$A=πr2$, where r represents the radius and the constant $π≈3.14$.: $A=πr2$
Example 4: The floor of a rectangular room has a length that is 4 feet more than twice its width. If the total area of the floor is 240 square feet, then find the dimensions of the floor.
Solution:
Use the formula $A=l⋅w$ and the fact that the area is 240 square feet to set up an algebraic equation.
Solve by factoring.
At this point we have two possibilities for the width of the rectangle. However, since a negative width is not defined, choose the positive solution, $w=10$. Back substitute to find the length.
Answer: The width is 10 feet and the length is 24 feet.
It is important to include the correct units in the final presentation of the answer. In the previous example, it would not make much sense to say the width is 10. Make sure to indicate that the width is 10 feet.
Example 5: The height of a triangle is 3 inches less than twice the length of its base. If the total area of the triangle is 7 square inches, then find the lengths of the base and height.
Solution:
Use the formula $A=12bh$ and the fact that the area is 7 square inches to set up an algebraic equation.
To avoid fractional coefficients, multiply both sides by 2 and then rewrite the quadratic equation in standard form.
Factor and then set each factor equal to zero.
In this case, disregard the negative answer; the length of the base is 7/2 inches long. Use $2b−3$ to determine the height of the triangle.
Answer: The base measures 7/2 = 3½ inches and the height is 4 inches.
Try this! The base of a triangle is 5 units less than twice the height. If the area is 75 square units, then what is the length of the base and height?
Answer: The height is 10 units and the base is 15 units.
### Video Solution
Recall that a right triangle is a triangle where one of the angles measures 90°. The side opposite of the right angle is the longest side of the triangle and is called the hypotenuse. The Pythagorean theoremGiven any right triangle with legs measuring a and b units and hypotenuse measuring c units, then $a2+b2=c2$. gives us a relationship between the legs and hypotenuse of any right triangle, where a and b are the lengths of the legs and c is the length of the hypotenuse:
Given certain relationships, we use this theorem when determining the lengths of sides of right triangles.
Example 6: The hypotenuse of a right triangle is 10 inches. If the short leg is 2 inches less than the long leg, then find the lengths of the legs.
Solution:
Given that the hypotenuse measures 10 inches, substitute its value into the Pythagorean theorem and obtain a quadratic equation in terms of x.
Multiply and rewrite the equation in standard form.
Once it is in standard form, factor and set each variable factor equal to zero.
Because lengths cannot be negative, disregard the negative answer. In this case, the long leg measures 8 inches. Use $x−2$ to determine the length of the short leg.
Answer: The short leg measures 6 inches and the long leg measures 8 inches.
Example 7: One leg of a right triangle measures 3 centimeters. The hypotenuse of the right triangle measures 3 centimeters less than twice the length of the unknown leg. Find the measure of all the sides of the triangle.
Solution:
To set up an algebraic equation, we use the Pythagorean theorem.
Solve by factoring.
Disregard 0. The length of the unknown leg is 4 centimeters. Use $2x−3$ to determine the length of the hypotenuse.
Answer: The sides of the triangle measure 3 centimeters, 4 centimeters, and 5 centimeters.
Try this! The hypotenuse of a right triangle measures 13 units. If one leg is 2 units more than twice that of the other, then find the length of each leg.
Answer: The two legs measure 5 units and 12 units.
## Projectile Problems
The height of an object launched upward, ignoring the effects of air resistance, can be modeled with the following formula:
Using function notation, which is more appropriate, we have
With this formula, the height can be calculated at any given time t after the object is launched. The coefficients represent the following:
$−12g$ The letter g represents the acceleration due to gravity.
$v0$ “v-naught” represents the initial velocity of the object.
$s0$ “s-naught” represents the initial height from which the object is launched.
We consider only problems where the acceleration due to gravity can be expressed as . Therefore, in this section time will be measured in seconds and the height in feet. Certainly though, the formula is valid using units other than these.
Example 8: The height of a projectile launched upward at a speed of 32 feet/second from a height of 128 feet is given by the function $h(t)=−16t2+32t+128$. How long does it take to hit the ground?
Solution: An inefficient method for finding the time to hit the ground is to simply start guessing at times and evaluating. To do this, construct a chart.
Use the table to sketch the height of the projectile over time.
We see that at 4 seconds, the projectile hits the ground. Note that when this occurs, the height is equal to 0. Now we need to solve this problem algebraically. To find the solution algebraically, use the fact that the height is 0 when the object hits the ground. We need to find the time, t, when $h(t)=0$.
Solve the equation by factoring.
Now set each variable factor to zero.
As expected, the projectile hits the ground at $t=4$ seconds. Disregard −2 as a solution because negative time is not defined.
Answer: The projectile hits the ground 4 seconds after it is launched.
Example 9: The height of a certain book dropped from the top of a 144-foot building is given by $h(t)=−16t2+144$. How long does it take to hit the ground?
Solution: Find the time t when the height $h(t)=0$.
Answer: The book takes 3 seconds to hit the ground when dropped from the top of a 144-foot building.
Try this! The height of a projectile, shot straight up into the air from the ground, is given by $h(t)=−16t2+80t$. How long does it take to come back down to the ground?
Answer: It will take 5 seconds to come back down to the ground.
### Key Takeaways
• It is best to translate a word problem to a mathematical setup and then solve using algebra. Avoid using the “guess and check” method of solving applications in this section.
• When solving applications, check that your solutions make sense in the context of the question. For example, if you wish to find the length of the base of a triangle, then you would disregard any negative solutions.
• It is important to identify each variable and state in a sentence what each variable represents. It is often helpful to draw a picture.
### Topic Exercises
Part A: Number Problems
Set up an algebraic equation and then solve.
1. One integer is five times another. If the product of the two integers is 80, then find the integers.
2. One integer is four times another. If the product of the two integers is 36, then find the integers.
3. An integer is one more than four times another. If the product of the two integers is 39, then find the integers.
4. An integer is 3 more than another. If the product of the two integers is 130, then find the integers.
5. An integer is 2 less than twice another. If the product of the two integers is 220, then find the integers.
6. An integer is 3 more than twice another. If the product of the two integers is 90, then find the integers.
7. One integer is 2 units more than another. If the product of the two integers is equal to five times the larger, then find the two integers.
8. A positive integer is 1 less than twice another. If the product of the two integers is equal to fifteen times the smaller, then find the two integers.
9. A positive integer is 3 more than twice a smaller positive integer. If the product of the two integers is equal to six times the larger, then find the integers.
10. One positive integer is 3 more than another. If the product of the two integers is equal to twelve times the smaller, then find the integers.
11. An integer is 3 more than another. If the product of the two integers is equal to 2 more than four times their sum, then find the integers.
12. An integer is 5 more than another. If the product of the two integers is equal to 2 more than twice their sum, then find the integers.
13. The product of two consecutive positive even integers is 120. Find the integers.
14. The product of two consecutive positive odd integers is 99. Find the integers.
15. The product of two consecutive positive integers is 110. Find the integers.
16. The product of two consecutive positive integers is 42. Find the integers.
17. The product of two consecutive positive odd integers is equal to 1 less than seven times the sum of the integers. Find the integers.
18. The product of two consecutive positive even integers is equal to 22 more than eleven times the sum of the integers. Find the integers.
19. The sum of the squares of two consecutive positive odd integers is 74. Find the integers.
20. The sum of the squares of two consecutive positive even integers is 100. Find the integers.
21. The sum of the squares of two consecutive positive integers is 265. Find the integers.
22. The sum of the squares of two consecutive positive integers is 181. Find the integers.
23. For two consecutive positive odd integers, the product of twice the smaller and the larger is 126. Find the integers.
24. For two consecutive positive even integers, the product of the smaller and twice the larger is 160. Find the integers.
Part B: Geometry Problems
Set up an algebraic equation and then solve.
25. The width of a rectangle is 7 feet less than its length. If the area of the rectangle is 170 square feet, then find the length and width.
26. The length of a rectangle is 2 feet more than its width. If the area of the rectangle is 48 square feet, then find the length and width.
27. The width of a rectangle is 3 units less than the length. If the area is 70 square units, then find the dimensions of the rectangle.
28. The width of a rectangle measures one half of the length. If the area is 72 square feet, then find the dimensions of the rectangle.
29. The length of a rectangle is twice that of its width. If the area of the rectangle is 72 square inches, then find the length and width.
30. The length of a rectangle is three times that of its width. If the area of the rectangle is 75 square centimeters, then find the length and width.
31. The length of a rectangle is 2 inches more than its width. The area of the rectangle is equal to 12 inches more than three times the perimeter. Find the length and width of the rectangle.
32. The length of a rectangle is 3 meters more than twice the width. The area of the rectangle is equal to 10 meters less than three times the perimeter. Find the length and width of the rectangle.
33. A uniform border is to be placed around an 8-inch-by-10-inch picture. If the total area including the border must be 224 square inches, then how wide should the border be?
34. A 2-foot brick border is constructed around a square cement slab. If the total area, including the border, is 121 square feet, then what are the dimensions of the slab?
35. The area of a picture frame including a 2-inch wide border is 99 square inches. If the width of the inner area is 2 inches more than its length, then find the dimensions of the inner area.
36. A box can be made by cutting out the corners and folding up the edges of a square sheet of cardboard. A template for a cardboard box with a height of 2 inches is given. What is the length of each side of the cardboard sheet if the volume of the box is to be 50 cubic inches?
37. The height of a triangle is 3 inches more than the length of its base. If the area of the triangle is 44 square inches, then find the length of its base and height.
38. The height of a triangle is 4 units less than the length of the base. If the area of the triangle is 48 square units, then find the length of its base and height.
39. The base of a triangle is twice that of its height. If the area is 36 square centimeters, then find the length of its base and height.
40. The height of a triangle is three times the length of its base. If the area is 73½ square feet, then find the length of the base and height.
41. The height of a triangle is 1 unit more than the length of its base. If the area is 5 units more than four times the height, then find the length of the base and height of the triangle.
42. The base of a triangle is 4 times that of its height. If the area is 3 units more than five times the height, then find the length of the base and height of the triangle.
43. The diagonal of a rectangle measures 5 inches. If the length is 1 inch more than its width, then find the dimensions of the rectangle.
44. The diagonal of a rectangle measures 10 inches. If the width is 2 inches less than the length, then find the area of the rectangle.
45. If the sides of a right triangle are consecutive even integers, then what are their measures?
46. The hypotenuse of a right triangle is 13 units. If the length of one leg is 2 more than twice the other, then what are their lengths?
47. The shortest leg of a right triangle measures 9 centimeters and the hypotenuse measures 3 centimeters more than the longer leg. Find the length of the hypotenuse.
48. The long leg of a right triangle measures 24 centimeters and the hypotenuse measures 4 centimeters more three times the short leg. Find the length of the hypotenuse.
Part C: Projectile Problems
Set up an algebraic equation and then solve.
49. The height of a projectile launched upward at a speed of 32 feet/second from a height of 48 feet is given by the function $h(t)=−16t2+32t+48$. How long will it take the projectile to hit the ground?
50. The height of a projectile launched upward at a speed of 16 feet/second from a height of 192 feet is given by the function $h(t)=−16t2+16t+192$. How long will it take to hit the ground?
51. An object launched upward at a speed of 64 feet/second from a height of 80 feet. How long will it take the projectile to hit the ground?
52. An object launched upward at a speed of 128 feet/second from a height of 144 feet. How long will it take the projectile to hit the ground?
53. The height of an object dropped from the top of a 64-foot building is given by $h(t)=−16t2+64$. How long will it take the object to hit the ground?
54. The height of an object dropped from an airplane at 1,600 feet is given by $h(t)=−16t2+1,600$. How long will it take the object to hit the ground?
55. An object is dropped from a ladder at a height of 16 feet. How long will it take to hit the ground?
56. An object is dropped from a 144-foot building. How long will it take to hit the ground?
57. The height of a projectile, shot straight up into the air from the ground at 128 feet/second, is given by $h(t)=−16t2+128t$. How long does it take to come back down to the ground?
58. A baseball, tossed up into the air from the ground at 32 feet/second, is given by $h(t)=−16t2+32t$. How long does it take to come back down to the ground?
59. How long will it take a baseball thrown into the air at 48 feet/second to come back down to the ground?
60. A football is kicked up into the air at 80 feet/second. Calculate how long will it hang in the air.
Part D: Discussion Board Topics
61. Research and discuss the life of Pythagoras.
62. If the sides of a square are doubled, then by what factor is the area increased? Why?
63. Design your own geometry problem involving the area of a rectangle or triangle. Post the question and a complete solution on the discussion board.
64. Write down your strategy for setting up and solving word problems. Share your strategy on the discussion board.
### Answers
1: {4, 20} or {−4, −20}
3: 3, 13
5: {11, 20} or {−22, −10}
7: {5, 7} or {−2, 0}
9: 6, 15
11: {7, 10} or {−2, 1}
13: 10, 12
15: 10, 11
17: 13, 15
19: 5, 7
21: 11, 12
23: 7, 9
25: Length: 17 feet; width: 10 feet
27: Length: 10 units; width: 7 units
29: Length: 12 inches; width: 6 inches
31: Length: 14 inches; width: 12 inches
33: 3 inches
35: 5 inches by 7 inches
37: Base: 8 inches; height: 11 inches
39: Base: 12 centimeters; height: 6 centimeters
41: Base: 9 units; height: 10 units
43: 3 inches by 4 inches
45: 6 units, 8 units, and 10 units
47: 15 centimeters
49: 3 seconds
51: 5 seconds
53: 2 seconds
55: 1 second
57: 8 seconds
59: 3 seconds
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http://www.sagemath.org/doc/reference/rings_standard/sage/rings/finite_rings/integer_mod_ring.html
|
Ring $$\ZZ/n\ZZ$$ of integers modulo $$n$$¶
EXAMPLES:
```sage: R = Integers(97)
sage: a = R(5)
sage: a**100000000000000000000000000000000000000000000000000000000000000
61
```
This example illustrates the relation between $$\ZZ/p\ZZ$$ and $$\GF{p}$$. In particular, there is a canonical map to $$\GF{p}$$, but not in the other direction.
```sage: r = Integers(7)
sage: s = GF(7)
sage: r.has_coerce_map_from(s)
False
sage: s.has_coerce_map_from(r)
True
sage: s(1) + r(1)
2
sage: parent(s(1) + r(1))
Finite Field of size 7
sage: parent(r(1) + s(1))
Finite Field of size 7
```
We list the elements of $$\ZZ/3\ZZ$$
```sage: R = Integers(3)
sage: list(R)
[0, 1, 2]
```
AUTHORS:
• William Stein (initial code)
• David Joyner (2005-12-22): most examples
• Robert Bradshaw (2006-08-24): convert to SageX (Cython)
• William Stein (2007-04-29): square_roots_of_one
• Simon King (2011-04-21): allow to prescribe a category
class sage.rings.finite_rings.integer_mod_ring.IntegerModFactory¶
Bases: sage.structure.factory.UniqueFactory
Return the quotient ring $$\ZZ / n\ZZ$$.
INPUT:
• order - integer (default: 0), positive or negative
EXAMPLES:
```sage: IntegerModRing(15)
Ring of integers modulo 15
sage: IntegerModRing(7)
Ring of integers modulo 7
sage: IntegerModRing(-100)
Ring of integers modulo 100
```
Note that you can also use Integers, which is a synonym for IntegerModRing.
```sage: Integers(18)
Ring of integers modulo 18
sage: Integers() is Integers(0) is ZZ
True
```
create_key(order=0, category=None)¶
An integer mod ring is specified uniquely by its order.
EXAMPLES:
```sage: Zmod.create_key(7)
7
sage: Zmod.create_key(7, Fields())
(7, Category of fields)
```
create_object(version, order)¶
EXAMPLES:
```sage: R = Integers(10)
sage: TestSuite(R).run() # indirect doctest
```
class sage.rings.finite_rings.integer_mod_ring.IntegerModRing_generic(order, cache=None, category=None)¶
Bases: sage.rings.quotient_ring.QuotientRing_generic
The ring of integers modulo N, with N composite.
EXAMPLES:
```sage: R = IntegerModRing(97)
sage: a = R(5)
sage: a**(10^62)
61
```
cardinality()¶
Returns the cardinality of this ring.
EXAMPLES:
```sage: Zmod(87).cardinality()
87
```
characteristic()¶
EXAMPLES:
```sage: R = IntegerModRing(18)
sage: FF = IntegerModRing(17)
sage: FF.characteristic()
17
sage: R.characteristic()
18
```
degree()¶
Return 1.
EXAMPLE:
```sage: R = Integers(12345678900)
sage: R.degree()
1
```
factored_order()¶
EXAMPLES:
```sage: R = IntegerModRing(18)
sage: FF = IntegerModRing(17)
sage: R.factored_order()
2 * 3^2
sage: FF.factored_order()
17
```
factored_unit_order()¶
Returns a list of Factorization objects, each the factorization of the order of the units in a $$\ZZ / p^n \ZZ$$ component of this group (using the Chinese Remainder Theorem).
EXAMPLES:
```sage: R = Integers(8*9*25*17*29)
sage: R.factored_unit_order()
[2^2, 2 * 3, 2^2 * 5, 2^4, 2^2 * 7]
```
field()¶
If this ring is a field, return the corresponding field as a finite field, which may have extra functionality and structure. Otherwise, raise a ValueError.
EXAMPLES:
```sage: R = Integers(7); R
Ring of integers modulo 7
sage: R.field()
Finite Field of size 7
sage: R = Integers(9)
sage: R.field()
Traceback (most recent call last):
...
ValueError: self must be a field
```
is_field(proof=True)¶
Return True precisely if the order is prime.
EXAMPLES:
```sage: R = IntegerModRing(18)
sage: R.is_field()
False
sage: FF = IntegerModRing(17)
sage: FF.is_field()
True
```
is_finite()¶
Return True since Z/NZ is finite for all positive N.
EXAMPLES:
```sage: R = IntegerModRing(18)
sage: R.is_finite()
True
```
is_integral_domain(proof=True)¶
Return True if and only if the order of self is prime.
EXAMPLES:
```sage: Integers(389).is_integral_domain()
True
sage: Integers(389^2).is_integral_domain()
False
```
is_noetherian()¶
EXAMPLES:
```sage: Integers(8).is_noetherian()
True
```
is_prime_field()¶
Return True if the order is prime.
EXAMPLES:
```sage: Zmod(7).is_prime_field()
True
sage: Zmod(8).is_prime_field()
False
```
krull_dimension()¶
EXAMPLES:
```sage: Integers(18).krull_dimension()
0
```
list_of_elements_of_multiplicative_group()¶
Returns a list of all invertible elements, as python ints.
EXAMPLES:
```sage: R = Zmod(12)
sage: L = R.list_of_elements_of_multiplicative_group(); L
[1, 5, 7, 11]
sage: type(L[0])
<type 'int'>
```
modulus()¶
Return the polynomial $$x - 1$$ over this ring.
Note
This function exists for consistency with the finite-field modulus function.
EXAMPLES:
```sage: R = IntegerModRing(18)
sage: R.modulus()
x + 17
sage: R = IntegerModRing(17)
sage: R.modulus()
x + 16
```
multiplicative_generator()¶
Return a generator for the multiplicative group of this ring, assuming the multiplicative group is cyclic.
Use the unit_gens function to obtain generators even in the non-cyclic case.
EXAMPLES:
```sage: R = Integers(7); R
Ring of integers modulo 7
sage: R.multiplicative_generator()
3
sage: R = Integers(9)
sage: R.multiplicative_generator()
2
sage: Integers(8).multiplicative_generator()
Traceback (most recent call last):
...
ValueError: multiplicative group of this ring is not cyclic
sage: Integers(4).multiplicative_generator()
3
sage: Integers(25*3).multiplicative_generator()
Traceback (most recent call last):
...
ValueError: multiplicative group of this ring is not cyclic
sage: Integers(25*3).unit_gens()
(26, 52)
sage: Integers(162).unit_gens()
(83,)
```
multiplicative_group_is_cyclic()¶
Return True if the multiplicative group of this field is cyclic. This is the case exactly when the order is less than 8, a power of an odd prime, or twice a power of an odd prime.
EXAMPLES:
```sage: R = Integers(7); R
Ring of integers modulo 7
sage: R.multiplicative_group_is_cyclic()
True
sage: R = Integers(9)
sage: R.multiplicative_group_is_cyclic()
True
sage: Integers(8).multiplicative_group_is_cyclic()
False
sage: Integers(4).multiplicative_group_is_cyclic()
True
sage: Integers(25*3).multiplicative_group_is_cyclic()
False
```
We test that #5250 is fixed:
```sage: Integers(162).multiplicative_group_is_cyclic()
True
```
multiplicative_subgroups()¶
Return generators for each subgroup of $$(\ZZ/N\ZZ)^*$$.
EXAMPLES:
```sage: Integers(5).multiplicative_subgroups()
((2,), (4,), ())
sage: Integers(15).multiplicative_subgroups()
((11, 7), (4, 11), (8,), (11,), (14,), (7,), (4,), ())
sage: Integers(2).multiplicative_subgroups()
((),)
sage: len(Integers(341).multiplicative_subgroups())
80
```
TESTS:
```sage: IntegerModRing(1).multiplicative_subgroups()
((0,),)
sage: IntegerModRing(2).multiplicative_subgroups()
((),)
sage: IntegerModRing(3).multiplicative_subgroups()
((2,), ())
```
order()¶
Returns the order of this ring.
EXAMPLES:
```sage: Zmod(87).order()
87
```
quadratic_nonresidue()¶
Return a quadratic non-residue in self.
EXAMPLES:
```sage: R = Integers(17)
sage: R.quadratic_nonresidue()
3
sage: R(3).is_square()
False
```
random_element(bound=None)¶
Return a random element of this ring.
If bound is not None, return the coercion of an integer in the interval [-bound, bound] into this ring.
EXAMPLES:
```sage: R = IntegerModRing(18)
sage: R.random_element()
2
```
square_roots_of_one()¶
Return all square roots of 1 in self, i.e., all solutions to $$x^2 - 1 = 0$$.
OUTPUT:
• tuple - the square roots of 1 in self.
EXAMPLES:
```sage: R = Integers(2^10)
sage: [x for x in R if x^2 == 1]
[1, 511, 513, 1023]
sage: R.square_roots_of_one()
(1, 511, 513, 1023)
```
```sage: v = Integers(9*5).square_roots_of_one(); v
(1, 19, 26, 44)
sage: [x^2 for x in v]
[1, 1, 1, 1]
sage: v = Integers(9*5*8).square_roots_of_one(); v
(1, 19, 71, 89, 91, 109, 161, 179, 181, 199, 251, 269, 271, 289, 341, 359)
sage: [x^2 for x in v]
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
```
unit_gens()¶
Returns generators for the unit group $$(\ZZ/N\ZZ)^*$$.
We compute the list of generators using a deterministic algorithm, so the generators list will always be the same. For each odd prime divisor of N there will be exactly one corresponding generator; if N is even there will be 0, 1 or 2 generators according to whether 2 divides N to order 1, 2 or $$\ge 3$$.
OUTPUT:
A tuple containing the units of self.
EXAMPLES:
```sage: R = IntegerModRing(18)
sage: R.unit_gens()
(11,)
sage: R = IntegerModRing(17)
sage: R.unit_gens()
(3,)
sage: IntegerModRing(next_prime(10^30)).unit_gens()
(5,)
```
TESTS:
```sage: IntegerModRing(2).unit_gens()
()
sage: IntegerModRing(4).unit_gens()
(3,)
sage: IntegerModRing(8).unit_gens()
(7, 5)
```
unit_group_exponent()¶
EXAMPLES:
```sage: R = IntegerModRing(17)
sage: R.unit_group_exponent()
16
sage: R = IntegerModRing(18)
sage: R.unit_group_exponent()
6
```
unit_group_order()¶
Return the order of the unit group of this residue class ring.
EXAMPLES;
```sage: R = Integers(500)
sage: R.unit_group_order()
200
```
sage.rings.finite_rings.integer_mod_ring.crt(v)¶
INPUT: v - (list) a lift of elements of rings.IntegerMod(n), for various coprime moduli n.
EXAMPLES:
```sage: from sage.rings.finite_rings.integer_mod_ring import crt
sage: crt([mod(3, 8),mod(1,19),mod(7, 15)])
1027
```
sage.rings.finite_rings.integer_mod_ring.is_IntegerModRing(x)¶
Return True if x is an integer modulo ring.
EXAMPLES:
```sage: from sage.rings.finite_rings.integer_mod_ring import is_IntegerModRing
sage: R = IntegerModRing(17)
sage: is_IntegerModRing(R)
True
sage: is_IntegerModRing(GF(13))
True
sage: is_IntegerModRing(GF(4, 'a'))
False
sage: is_IntegerModRing(10)
False
sage: is_IntegerModRing(ZZ)
False
```
Previous topic
Elements of the ring $$\ZZ$$ of integers
Next topic
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|
http://mathoverflow.net/questions/44086/is-there-a-ruler-and-compass-construction-of-the-common-perpendicular-of-two-geod
|
## Is there a ruler and compass construction of the common perpendicular of two geodesics in H^3?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Assume we have two geodesics in the Poincaré ball model of $\mathbb{H}^3$, viewed as arcs intersecting the boundary of and contained in the Euclidean unit sphere in $\mathbb{R}^3$. Is there a ruler and compass construction in $\mathbb{R}^3$ to construct their common perpendicular geodesic?
It seems that there is in 2 dimensions, see here. However the obvious generalisation to 3 dimensions would give the common perpendicular to geodesic planes containing the original two geodesics, rather than between the geodesics themselves.
-
I can't tell what your axioms might be. If you gave me two skew lines in $R^3$ and a compass and straightedge to wave about in midair, but no plane on which to draw, I'm not sure what I could accomplish. So, as I think you are probably envisioning a computer model of some kind, perhaps you could solve the $R^3$ problem and describe what was involved. Given enough detail on that I can probably solve your problem in $H^3$ or show it cannot always be done. – Will Jagy Oct 29 2010 at 22:22
@Will: Yes, this is a practical problem with drawing a diagram in a 3d computer model. I don't know if an exhaustive list of operations would be useful, but I can do things like draw lines between two points, circles or planes through three non-colinear points, spheres through 4 points, find intersections and I can do Euclidean isometries to objects (by distances and angles that I already know). – Henry Segerman Oct 29 2010 at 23:30
A solution to the $R^3$ version is not immediately apparent to me (although I think it should be possible), I'll have a think. For the Poincaré ball model embedded in $R^3$ there is more to work with: you have the endpoints of the $H^3$ geodesics to start from. – Henry Segerman Oct 29 2010 at 23:35
1
Now that I see what you are prepared to use, it is easy in $R^3.$ Take the two direction vectors for your skew lines and take their cross product, call that $N.$ Construct the plane normal to $N$ that contains one of the lines. Take two distinct points on the other line and drop their perpendiculars to the plane, thses being parallel to $N.$ Draw the line between the two image points, mark where it connects with the first line. Draw another line parallel to $N$ through that point. – Will Jagy Oct 30 2010 at 4:37
Ah yes, that works for $R^3$. – Henry Segerman Oct 30 2010 at 13:05
show 2 more comments
## 2 Answers
If you really want to use just straight-edge and compass, don't go into $3$-space at all! Coxeter points out the equivalence between the inversive plane and hyperbolic space in the paper The inversive plane and hyperbolic space. In this particular situation one can think of lines as pairs of points in the inversive plane. Two pairs of points correspond to perpendicular lines of hyperbolic space when the two pairs are harmonic. So the problem is:
Given two pairs of points $(a,b)$ and $(c,d),$ to construct using straightedge and compass the (unique) pair $(x,y)$ harmonic to both.
Fenchel asserts the existence of such a thing, and relates them to square roots, but without mentioning the geometric argument, so here goes.
We first start off easy, and assume that $a = \infty.$ We define the (two) geometric means of $(c,d)$ with respect to b to be given as follows. (Basically they are the complex square roots of $d$ where $b = 0$ and $c = 1$.) Bisect the angle $\angle c b d$ by a line $\ell.$ Let $C_c, C_d$ be the circles centered at $b$ through $c,d$ respectively. Intersect these with $\ell$ to get four points $p_c, p_c', p_d, p_d',$ so that $(p_c,p_d)$ separate $(p_c',p_d').$ Let $q$ be the midpoint of $p_c$ and $p_d.$ Draw the circle $D$ centered at $q$ through $p_c$ and $p_d.$ Draw the perpendicular $\ell_\perp$ to $\ell$ at $b,$ and intersect it with $D$ at a point $r$. Then draw the circle $E$ centered at $b$ through $r$, and intersect it with $\ell$ at the points $s_{(c,d)},t_{(c,d)}$. These are the geometric means of $(c,d)$ with respect to $b$.
Now, it should be that $(s_{(c,d)}, t_{(c,d)})$ is harmonic to both $(\infty, b)$ and $(c,d).$ (I haven't worked this out yet.) So that's what we were looking for. In the case that $a \neq \infty,$ just let $C$ be the circle centered at $a$ through $b,$ and denote by $I_C$ the inversion in $C.$ Then the points we're looking for are $I_C(s_{I_C(c,d)}, t_{I_C(c,d)}) = (\sigma_{(c,d)}, \tau_{(c,d)}).$ These give the endpoints for the common perpendicular to the hyperbolic lines with endpoints $(a,b)$ and $(c,d).$
It's probably all in Fenchel, anyway....
-
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
I remembered that the isometries for the upper half space model of $\mathbf H^3$ are given by the action of $PSL_2\mathbf C$ on the points at infinity (the $xy$-plane as the complex numbers). So, given four geodesic endpoints, move one to $0$ with a pure translation. We now have three complex numbers $A,B,C,$ where $A$ is the other endpoint of the geodesic with an endpoint at $0.$ Define a complex number $\gamma$ that solves $$ABC \gamma^2 + 2 BC \gamma + (B+C-A) =0,$$ where you might as well pick $\gamma = 0$ if the constant term $B+C-A=0.$
Next, apply the linear fractional transformation $$h(z) = \frac{z}{\gamma z + 1}$$ to the plane. The result is $$h(A) = h(B) + h(C)$$
That is, the midpoint of $0$ and $h(A)$ is the same as the midpoint of $h(B)$ and $h(C).$ So the common orthogonal geodesic is the vertical ray through $h(A)/2$ and allowing the third coordinate to vary. Then map everything back to your originals.
Note that there is no answer if two of your original geodesic endpoints coincide. In that case, they lie in a common flat, and lines asymptotic at infinity do not share a perpendicular.
See if I can do the link the right way this time, Milnor's survey on the first 150 years of hyperbolic geometry can be downloaded.
-
Yes, if we allow linear fractional transformations then this works. However, I don't know if they would really fit into the spirit of "ruler and compass construction", even in my somewhat loose interpretation: my natively Euclidean computer modelling program wouldn't do them for instance. However, using this construction it would be possible to write a script to do the calculations. – Henry Segerman Jan 3 2011 at 0:42
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|
http://math.stackexchange.com/questions/35399/is-this-true-mathrmcovx-y-sqrt-mathrmvarx-sqrt-mathrmvary/35404
|
# Is this true: $\mathrm{Cov}(X,Y) = \sqrt{\mathrm{Var(X)}}\sqrt{\mathrm{Var}(Y)}$?
I just want to confirm whether this is correct or not:
The covariance of $X$ and $Y$ is equal to the standard deviation of $X$ times the standard deviation of $Y$.
or, in mathematical notation,
$$\mathrm{Cov}(X,Y) = \sqrt{\mathrm{Var(X)}}\sqrt{\mathrm{Var}(Y)}$$
-
i see. Thanks for help – user10128 Apr 27 '11 at 12:07
@cardinal: ... and perhaps also posted the answer as an answer instead of as a comment? As it is now, the question is answered even though it appears as unanswered on the front page. – Hans Lundmark Apr 27 '11 at 12:24
@Hans, fair enough. I've copied my comment into an answer and deleted the comments. – cardinal Apr 27 '11 at 12:30
@Hans, also, if you don't mind, please review my edit of the question and approve it, if appropriate. I think that will also make my answer match up better with the question. Cheers. – cardinal Apr 27 '11 at 12:32
@cardinal: Looks fine to me. :) – Hans Lundmark Apr 27 '11 at 12:49
## 2 Answers
This is false in general. In fact, it is true if and only if $X=aY+b$ (almost surely) for some fixed constants $a \geq 0$ and $b \in \mathbb{R}$. That is $X$ and $Y$ must be positively linearly related for this to hold. Wikipedia also has a decent page on this.
For a counterexample to your statement, consider any two independent random variables $X$ and $Y$ each with strictly positive variance. Then, $\mathrm{Cov}(X,Y) = 0 \>,$ but, $\sqrt{\mathrm{Var}(X)} \sqrt{\mathrm{Var}(Y)} > 0 \>.$
A quick proof (and a slick one, I think; it's not my own) of the if and only if assertion uses the Cauchy–Schwarz inequality and goes as follows. Let $U$ and $V$ be random variables such that $\newcommand{\e}{\mathbb{E}}\e U^2 < \infty$ and $\e V^2 < \infty$. Then, $|\e UV|^2 < \e U^2 \e V^2$ if and only if $\e (t U - V)^2 > 0$ for all $t$. But, if $\e(t_0 U - V)^2 = 0$ for some $t_0$, then $t_0 U = V$ almost surely. Now, set $U = X - \e X$ and $V = Y - \e Y$ to get the desired conclusion.
-
1
+1 though with a pedantic point: suppose $Z$ is a uniform random variable from $[0,1)$, $Y=1$ if $Z$ is rational and $Y=0$ if $Z$ is irrational, and $X=1-Y$. Then $\mathrm{Cov}(X,Y) = \sqrt{\mathrm{Var(X)}}\sqrt{\mathrm{Var}(Y)}$ even though $a \lt 0$. – Henry May 14 '12 at 22:35
@Henry: (+1) Thank you; it's not pedantic at all. I'm not sure how you came across such an old answer, but I appreciate very much your comment. I'll try to think of a way to clean up the statement. This should only be possible if one of the random variables is almost surely constant. – cardinal May 15 '12 at 2:46
While cardinal has answered the question, I wanted to add a little something extra.
The Cauchy-Schwarz inequality shows that $\left|\operatorname{Cov}(X,Y)\right|\leq \sqrt{\operatorname{Var}(X)\operatorname{Var}(Y)}$, with equality if and only if $X$ and $Y$ are linearly dependent. In general, you can take the ratio $\rho=\frac{\operatorname{Cov}(X,Y)}{ \sqrt{\operatorname{Var}(X)\operatorname{Var}(Y)}}$ which will be between $-1$ and $1$. This gives the correlation between $X$ and $Y$. It is $1$ when the variables are perfectly correlated (one goes up exactly when the other does), $-1$ when they are perfectly anti-correlated (one goes up exactly when the other goes down), $0$ when they are independent (knowing that one goes up tells you nothing about the behavior of the other), and in general gives a measure of the behavior when things are somewhere in between.
One of the first places where this comes up is least squares approximation. . While you can always do a least squares approximation to get a line of best fit for your data points, the correlation coefficient tells you whether your line of best fit is actually a good fit. It will be near $0$ for data that is essentially random, and small in magnitude for data which is non-linear.
-
(+1) Your last statement depends strongly on the type of nonlinearity present. There are many cases where even in the presence of considerable nonlinearity, the correlation coefficient will not be small in magnitude. – cardinal Apr 27 '11 at 13:52
@cardinal Yes. I guess that by "non-linear" I actually mean "not well approximated by anything linear" or "very non-linear". On the other hand, this could be used as a definition of what it means for noisy data to be linear/non-linear. – Aaron Apr 27 '11 at 14:27
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|
http://pediaview.com/openpedia/Associative_algebra
|
# Associative algebra
In mathematics, an associative algebra A is a (not necessarily unital) associative ring that has a compatible structure of a vector space over a certain field K or, more generally, of a module over a commutative ring R. Thus A is endowed with binary operations of addition and multiplication satisfying a number of axioms, including associativity of multiplication and distributivity, as well as compatible multiplication by the elements of the field K or the ring R.
For example, a ring of square matrices over a field K is an associative K algebra. More generally, given a ring S with center C, S is an associative C algebra.
In some areas of mathematics, associative algebras are typically assumed to have a multiplicative unit, denoted 1. To make this extra assumption clear, these associative algebras are called unital algebras. Additionally, some authors demand that all rings be unital; in this article, the word "ring" is intended to refer to potentially non-unital rings as well.
Algebraic structures
Group-like
Ring-like
Lattice-like
Module-like
Algebra-like
• Algebra
• Associative
• Non-associative
## Formal definition
Let R be a fixed commutative ring. An associative R-algebra is an additive abelian group A which has the structure of both a ring and an R-module in such a way that ring multiplication is R-bilinear:
$r\cdot(xy) = (r\cdot x)y = x(r\cdot y)$
for all r ∈ R and x, y ∈ A. We say A is unital if it contains an element 1 such that
$1 x = x = x 1$
for all x ∈ A. Note that such an element 1 must be unique if it exists at all.
If A itself is commutative (as a ring) then it is called a commutative R-algebra.
### From R-modules
Starting with an R-module A, we get an associative R-algebra by equipping A with an R-bilinear mapping A × A → A such that
$x(yz) = (xy)z\,$
for all x, y, and z in A. This R-bilinear mapping then gives A the structure of a ring and an associative R-algebra. Every associative R-algebra arises this way.
Moreover, the algebra A built this way will be unital if and only if there exists an element 1 of A such that every element x of A satisfies 1x = x1 = x. This definition is equivalent to the statement that a unital associative R-algebra is a monoid in R-Mod (the monoidal category of R-modules).
### From rings
Starting with a ring A, we get a unital associative R-algebra by providing a ring homomorphism $\eta\colon R \to A$ whose image lies in the center of A. The algebra A can then be thought of as an R-module by defining
$r\cdot x = \eta(r)x$
for all r ∈ R and x ∈ A.
If A is commutative then the center of A is equal to A, so that a commutative unital R-algebra can be defined simply as a homomorphism $\eta\colon R \to A$ of commutative rings.
## Algebra homomorphisms
A homomorphism between two associative R-algebras is an R-linear ring homomorphism. Explicitly, $\phi : A_1 \to A_2$ is an associative algebra homomorphism if
$\phi(r\cdot x) = r\cdot \phi(x)$
$\phi(x+y) = \phi(x)+\phi(y)\,$
$\phi(xy) = \phi(x)\phi(y)\,$
For a homomorphism of unital associative R-algebras, we also demand that
$\phi(1) = 1\,$
The class of all unital associative R-algebras together with algebra homomorphisms between them form a category, sometimes denoted R-Alg.
The subcategory of commutative R-algebras can be characterized as the coslice category R/CRing where CRing is the category of commutative rings.
## Examples
The most basic example is a ring itself; it is algebra over its center or any subring lying in the center. In particular, any commutative ring is an algebra over any of its subring. Other examples abound both from algebra and other fields of mathematics.
Algebra
• Any (unital) ring A can be considered as a unital Z-algebra in a unique way. The unique ring homomorphism from Z to A is determined by the fact that it must send 1 to the identity in A. Therefore rings and unital Z-algebras are equivalent concepts, in the same way that abelian groups and Z-modules are equivalent.
• Any ring of characteristic n is a (Z/nZ)-algebra in the same way.
• Given an R-module M, the endomorphism ring of M, denoted EndR(M) is an R-algebra by defining (r·φ)(x) = r·φ(x).
• Any ring of matrices with coefficients in a commutative ring R forms an R-algebra under matrix addition and multiplication. This coincides with the previous example when M is a finitely-generated, free R-module.
• The square n-by-n matrices with entries from the field K form a unital associative algebra over K. In particular, the 2 × 2 real matrices form an associative algebra useful in plane mapping.
• The complex numbers form a 2-dimensional unital associative algebra over the real numbers.
• The quaternions form a 4-dimensional unital associative algebra over the reals (but not an algebra over the complex numbers, since if complex numbers are treated as a subset of the quaternions, complex numbers and quaternions do not commute).
• The polynomials with real coefficients form a unital associative algebra over the reals.
• Every polynomial ring R[x1, ..., xn] is a commutative R-algebra. In fact, this is the free commutative R-algebra on the set {x1, ..., xn}.
• The free R-algebra on a set E is an algebra of polynomials with coefficients in R and noncommuting indeterminates taken from the set E.
• The tensor algebra of an R-module is naturally an R-algebra. The same is true for quotients such as the exterior and symmetric algebras. Categorically speaking, the functor which maps an R-module to its tensor algebra is left adjoint to the functor which sends an R-algebra to its underlying R-module (forgetting the ring structure).
• Given a commutative ring R and any ring A the tensor product R⊗ZA can be given the structure of an R-algebra by defining r·(s⊗a) = (rs⊗a). The functor which sends A to R⊗ZA is left adjoint to the functor which sends an R-algebra to its underlying ring (forgetting the module structure).
Analysis
• Given any Banach space X, the continuous linear operators A : X → X form a unital associative algebra (using composition of operators as multiplication); this is a Banach algebra.
• Given any topological space X, the continuous real- or complex-valued functions on X form a real or complex unital associative algebra; here the functions are added and multiplied pointwise.
• An example of a non-unital associative algebra is given by the set of all functions f: R → R whose limit as x nears infinity is zero.
Geometry and combinatorics
• The Clifford algebras, which are useful in geometry and physics.
• Incidence algebras of locally finite partially ordered sets are unital· associative algebras considered in combinatorics.
## Constructions
Subalgebras
A subalgebra of an R-algebra A is a subset of A which is both a subring and a submodule of A. That is, it must be closed under addition, ring multiplication, scalar multiplication, and it must contain the identity element of A.
Quotient algebras
Let A be an R-algebra. Any ring-theoretic ideal I in A is automatically an R-module since r·x = (r1A)x. This gives the quotient ring A/I the structure of an R-module and, in fact, an R-algebra. It follows that any ring homomorphic image of A is also an R-algebra.
Direct products
The direct product of a family of R-algebras is the ring-theoretic direct product. This becomes an R-algebra with the obvious scalar multiplication.
One can form a free product of R-algebras in a manner similar to the free product of groups. The free product is the coproduct in the category of R-algebras.
Tensor products
The tensor product of two R-algebras is also an R-algebra in a natural way. See tensor product of algebras for more details.
## Associativity and the multiplication mapping
Associativity was defined above quantifying over all elements of A. It is possible to define associativity in a way that does not explicitly refer to elements. An algebra is defined as a vector space A with a bilinear map
$M: A \times A \rightarrow A$
(the multiplication map). An associative algebra is an algebra where the map M has the property
$M \circ (\mbox {Id} \times M) = M \circ (M \times \mbox {Id})$
Here, the symbol $\circ$ refers to function composition, and Id : A → A is the identity map on A.
To see the equivalence of the definitions, we need only understand that each side of the above equation is a function that takes three arguments. For example, the left-hand side acts as
$( M \circ (\mbox {Id} \times M)) (x,y,z) = M (x, M(y,z))$
Similarly, a unital associative algebra can be defined as a vector space A endowed with a map M as above and, additionally, a linear map
$\eta: K \rightarrow A$
(the unit map) which has the properties
$M \circ (\mbox {Id} \times \eta ) = s;\ M \circ (\eta \times \mbox {Id}) = t$
Here, the unit map η takes an element k in K to the element k1 in A, where 1 is the unit element of A. The map t is just plain-old scalar multiplication: $t:K\times A \rightarrow A, \ \left(k,a\right)\mapsto ka$; the map s is similar: $s:A\times K \rightarrow A, \ \left(a,k\right)\mapsto ka$.
## Coalgebras
Main article: Coalgebra
An associative unital algebra over K is given by a K-vector space A endowed with a bilinear map A×A→A having 2 inputs (multiplicator and multiplicand) and one output (product), as well as a morphism K→A identifying the scalar multiples of the multiplicative identity. If the bilinear map A×A→A is reinterpreted as a linear map (i. e., morphism in the category of K-vector spaces) A⊗A→A (by the universal property of the tensor product), then we can view an associative unital algebra over K as a K-vector space A endowed with two morphisms (one of the form A⊗A→A and one of the form K→A) satisfying certain conditions which boil down to the algebra axioms. These two morphisms can be dualized using categorial duality by reversing all arrows in the commutative diagrams which describe the algebra axioms; this defines the structure of a coalgebra.
There is also an abstract notion of F-coalgebra. This is vaguely related to the notion of coalgebra discussed above.
## Representations
Main article: Algebra representation
A representation of a unital algebra A is a unital algebra homomorphism ρ: A → End(V) from A to the endomorphism algebra of some vector space (or module) V. The property of ρ being a unital algebra homomorphism means that ρ preserves the multiplicative operation (that is, ρ(xy)=ρ(x)ρ(y) for all x and y in A), and that ρ sends the unity of A to the unity of End(V) (that is, to the identity endomorphism of V).
If A and B are two algebras, and ρ: A → End(V) and τ: B → End(W) are two representations, then it is easy to define a (canonical) representation A ⊗ B → End(V ⊗ W) of the tensor product algebra A ⊗ B on the vector space V ⊗ W. Note, however, that there is no natural way of defining a tensor product of two representations of a single associative algebra in such a way that the result is still a representation of that same algebra (not of its tensor product with itself), without somehow imposing additional conditions. Here, by tensor product of representations, the usual meaning is intended: the result should be a linear representation of the same algebra on the product vector space. Imposing such additional structure typically leads to the idea of a Hopf algebra or a Lie algebra, as demonstrated below.
### Motivation for a Hopf algebra
Consider, for example, two representations $\sigma:A\rightarrow \mathrm{End}(V)$ and $\tau:A\rightarrow \mathrm{End}(W)$. One might try to form a tensor product representation $\rho: x \mapsto \sigma(x) \otimes \tau(x)$ according to how it acts on the product vector space, so that
$\rho(x)(v \otimes w) = (\sigma(x)(v)) \otimes (\tau(x)(w)).$
However, such a map would not be linear, since one would have
$\rho(kx) = \sigma(kx) \otimes \tau(kx) = k\sigma(x) \otimes k\tau(x) = k^2 (\sigma(x) \otimes \tau(x)) = k^2 \rho(x)$
for k ∈ K. One can rescue this attempt and restore linearity by imposing additional structure, by defining an algebra homomorphism Δ: A → A ⊗ A, and defining the tensor product representation as
$\rho = (\sigma\otimes \tau) \circ \Delta.$
Such a homomorphism Δ is called a comultiplication if it satisfies certain axioms. The resulting structure is called a bialgebra. To be consistent with the definitions of the associative algebra, the coalgebra must be co-associative, and, if the algebra is unital, then the co-algebra must be unital as well. A Hopf algebra is a bialgebra with an additional piece of structure (the so-called antipode), which allows not only to define the tensor product of two representations, but also the Hom module of two representations (again, similarly to how it is done in the representation theory of groups).
### Motivation for a Lie algebra
See also: Lie algebra representation
One can try to be more clever in defining a tensor product. Consider, for example,
$x \mapsto \rho (x) = \sigma(x) \otimes \mbox{Id}_W + \mbox{Id}_V \otimes \tau(x)$
so that the action on the tensor product space is given by
$\rho(x) (v \otimes w) = (\sigma(x) v)\otimes w + v \otimes (\tau(x) w)$.
This map is clearly linear in x, and so it does not have the problem of the earlier definition. However, it fails to preserve multiplication:
$\rho(xy) = \sigma(x) \sigma(y) \otimes \mbox{Id}_W + \mbox{Id}_V \otimes \tau(x) \tau(y)$.
But, in general, this does not equal
$\rho(x)\rho(y) = \sigma(x) \sigma(y) \otimes \mbox{Id}_W + \sigma(x) \otimes \tau(y) + \sigma(y) \otimes \tau(x) + \mbox{Id}_V \otimes \tau(x) \tau(y)$.
This shows that this definition of a tensor product is too naive. It can be used, however, to define the tensor product of two representations of a Lie algebra (rather than of an associative algebra).
## References
• Bourbaki, N. (1989). Algebra I. Springer. ISBN 3-540-64243-9 [Amazon-US | Amazon-UK].
• Ross Street, Quantum Groups: an entrée to modern algebra (1998). (Provides a good overview of index-free notation)
## Source
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http://en.wikipedia.org/wiki/Quadratic_function
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# Quadratic function
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$x^2 - x - 2\!$
A quadratic function, in mathematics, is a polynomial function of the form
$f(x)=ax^2+bx+c,\quad a \ne 0.$[1]
The graph of a quadratic function is a parabola whose axis of symmetry is parallel to the y-axis.
The expression $ax^2+bx+c$ in the definition of a quadratic function is a polynomial of degree 2 or second order, or a 2nd degree polynomial, because the highest exponent of x is 2.
If the quadratic function is set equal to zero, then the result is a quadratic equation. The solutions to the equation are called the roots of the equation.
## Origin of word
The adjective quadratic comes from the Latin word quadrātum (“square”). A term like x2 is called a square in algebra because it is the area of a square with side x.
In general, a prefix quadr(i)- indicates the number 4. Examples are quadrilateral and quadrant. Quadratum is the Latin word for square because a square has four sides.
## Roots
Further information: Quadratic equation
The roots (zeros) of the quadratic function
$f(x) = ax^2+bx+c\,$
are the values of x for which f(x) = 0.
When the coefficients a, b, and c, are real or complex, the roots are
$x=\frac{-b \pm \sqrt{\Delta}}{2 a},$
where the discriminant is defined as
$\Delta = b^2 - 4 a c \, .$
## Forms of a quadratic function
A quadratic function can be expressed in three formats:[2]
• $f(x) = a x^2 + b x + c \,\!$ is called the standard form,
• $f(x) = a(x - x_1)(x - x_2)\,\!$ is called the factored form, where $x_1$ and $x_2$ are the roots of the quadratic equation, it is used in logistic map
• $f(x) = a(x - h)^2 + k \,\!$ is called the vertex form, where h and k are the x and y coordinates of the vertex, respectively.
To convert the standard form to factored form, one needs only the quadratic formula to determine the two roots $x_1$ and $x_2$. To convert the standard form to vertex form, one needs a process called completing the square. To convert the factored form (or vertex form) to standard form, one needs to multiply, expand and/or distribute the factors.
## Graph
$f(x) = ax^2 |_{a=\{0.1,0.3,1,3\}} \!$
$f(x) = x^2 + bx |_{b=\{1,2,3,4\}} \!$
$f(x) = x^2 + bx |_{b=\{-1,-2,-3,-4\}} \!$
Regardless of the format, the graph of a quadratic function is a parabola (as shown above).
• If $a > 0 \,\!$ (or is a positive number), the parabola opens upward.
• If $a < 0 \,\!$ (or is a negative number), the parabola opens downward.
The coefficient a controls the speed of increase (or decrease) of the quadratic function from the vertex, bigger positive a makes the function increase faster and the graph appear more closed.
The coefficients b and a together control the axis of symmetry of the parabola (also the x-coordinate of the vertex) which is at $x = -\frac{b}{2a}$.
The coefficient b alone is the declivity of the parabola as y-axis intercepts.
The coefficient c controls the height of the parabola, more specifically, it is the point where the parabola intercept the y-axis.
### Vertex
The vertex of a parabola is the place where it turns, hence, it's also called the turning point. If the quadratic function is in vertex form, the vertex is $(h, k)\,\!$. By the method of completing the square, one can turn the general form
$f(x) = a x^2 + b x + c \,\!$
into
$f(x) = a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2-4ac}{4 a} ,$
so the vertex of the parabola in the vertex form is
$\left(-\frac{b}{2a}, -\frac{\Delta}{4 a}\right).$
If the quadratic function is in factored form
$f(x) = a(x - r_1)(x - r_2) \,\!$
the average of the two roots, i.e.,
$\frac{r_1 + r_2}{2} \,\!$
is the x-coordinate of the vertex, and hence the vertex is
$\left(\frac{r_1 + r_2}{2}, f\left(\frac{r_1 + r_2}{2}\right)\right).\!$
The vertex is also the maximum point if $a < 0 \,\!$ or the minimum point if $a > 0 \,\!$.
The vertical line
$x=h=-\frac{b}{2a}$
that passes through the vertex is also the axis of symmetry of the parabola.
#### Maximum and minimum points
Using calculus, the vertex point, being a maximum or minimum of the function, can be obtained by finding the roots of the derivative:
$f(x)=ax^2+bx+c \quad \Rightarrow \quad f'(x)=2ax+b \,\!,$
giving
$x=-\frac{b}{2a}$
with the corresponding function value
$f(x) = a \left (-\frac{b}{2a} \right)^2+b \left (-\frac{b}{2a} \right)+c = -\frac{(b^2-4ac)}{4a} = -\frac{\Delta}{4a} \,\!,$
so again the vertex point coordinates can be expressed as
$\left (-\frac {b}{2a}, -\frac {\Delta}{4a} \right).$
## The square root of a quadratic function
The square root of a quadratic function gives rise to one of the four conic sections, almost always either to an ellipse or to a hyperbola. If $a>0\,\!$ then the equation $y = \pm \sqrt{a x^2 + b x + c}$ describes a hyperbola. The axis of the hyperbola is determined by the ordinate of the minimum point of the corresponding parabola $y_p = a x^2 + b x + c \,\!$.
If the ordinate is negative, then the hyperbola's axis is horizontal. If the ordinate is positive, then the hyperbola's axis is vertical.
If $a<0\,\!$ then the equation $y = \pm \sqrt{a x^2 + b x + c}$ describes either an ellipse or nothing at all. If the ordinate of the maximum point of the corresponding parabola $y_p = a x^2 + b x + c \,\!$ is positive, then its square root describes an ellipse, but if the ordinate is negative then it describes an empty locus of points.
## Iteration
Given an $f(x)=ax^2+bx+c$, one cannot always deduce the analytic form of $f^{(n)}(x)$, which means the nth iteration of $f(x)$. (The superscript can be extended to negative number referring to the iteration of the inverse of $f(x)$ if the inverse exists.) But there is one easier case, in which $f(x)=a(x-x_0)^2+x_0$.
In such case, one has
$f(x)=a(x-x_0)^2+x_0=h^{(-1)}(g(h(x)))\,\!$,
where
$g(x)=ax^2\,\!$ and $h(x)=x-x_0\,\!$.
So by induction,
$f^{(n)}(x)=h^{(-1)}(g^{(n)}(h(x)))\,\!$
can be obtained, where $g^{(n)}(x)$ can be easily computed as
$g^{(n)}(x)=a^{2^{n}-1}x^{2^{n}}\,\!$.
Finally, we have
$f^{(n)}(x)=a^{2^n-1}(x-x_0)^{2^n}+x_0\,\!$,
in the case of $f(x)=a(x-x_0)^2+x_0$.
See Topological conjugacy for more detail about such relationship between f and g. And see Complex quadratic polynomial for the chaotic behavior in the general iteration.
## Bivariate (two variable) quadratic function
Further information: Quadric and Quadratic form
A bivariate quadratic function is a second-degree polynomial of the form
$f(x,y) = A x^2 + B y^2 + C x + D y + E x y + F \,\!$
Such a function describes a quadratic surface. Setting $f(x,y)\,\!$ equal to zero describes the intersection of the surface with the plane $z=0\,\!$, which is a locus of points equivalent to a conic section.
### Minimum/maximum
If $4AB-E^2 <0 \,$ the function has no maximum or minimum, its graph forms an hyperbolic paraboloid.
If $4AB-E^2 >0 \,$ the function has a minimum if A>0, and a maximum if A<0, its graph forms an elliptic paraboloid.
The minimum or maximum of a bivariate quadratic function is obtained at $(x_m, y_m) \,$ where:
$x_m = -\frac{2BC-DE}{4AB-E^2}$
$y_m = -\frac{2AD-CE}{4AB-E^2}$
If $4AB- E^2 =0 \,$ and $DE-2CB=2AD-CE \ne 0 \,$ the function has no maximum or minimum, its graph forms a parabolic cylinder.
If $4AB- E^2 =0 \,$ and $DE-2CB=2AD-CE =0 \,$ the function achieves the maximum/minimum at a line. Similarly, a minimum if A>0 and a maximum if A<0, its graph forms a parabolic cylinder.
## References
1. "Quadratic Equation -- from Wolfram MathWorld". Retrieved January 06, 2013.
2. Hughes-Hallett, Deborah; Connally, Eric; McCallum, William G. (2007), College Algebra, John Wiley & Sons Inc, p. 205, ISBN 0-471-27175-6, 9780471271758 Check `|isbn=` value (help) , Search result
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http://physics.stackexchange.com/questions/1586/what-if-physical-constants-were-increased-or-decreased
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What if physical constants were increased or decreased?
(Probably related to this one, and probably should be CW.)
A very long time ago, I had the good fortune to read George Gamow's excellent series of Mr. Tompkins books. That introduced me to the idea of a world where the usual physical constants (e.g. the speed of light and Planck's constant) were changed such that "paradoxical" effects became apparent in the macroworld.
My memory is hazy now, but I do recall the concepts of relativity (e.g. dilation) becoming more pronounced when the speed of light is reduced to "human-sized" speeds.
In this vein, I ask this: assuming all other physical constants being fixed, what exactly can be expected to happen if (physical constant of your choice) is increased/or decreased?
One physical constant per answer, please.
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I suppose that you realize that books such Mr. Tompkins are not physical at all. In them it is supposed that speed of light is decreased but that people are still the same old people which is inconsistent because people are made of elementary particles that would also need to obey new speed limit and so life (and probably also any composite matter) would cease to be possible. So are you asking for real physical effects (which usually imply that life becomes impossible) or unphysical magnified effects for better illustration of the phenomena (as in Mr. Tompkins)? – Marek Dec 3 '10 at 14:08
@Marek: I meant the question in a speculative spirit; that is to say, what would we be observing if we increase or decrease a certain physical constant? – user172 Dec 3 '10 at 14:11
I cannot answer your question since I am not a physicist, but João Magueijo has worked on the theory of variable speed of light (VSL) in a vacuum. – Jaime Soto Dec 3 '10 at 15:00
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M.: it all depends on how precise you want to be. If you want to be very precise you can't really say anything because we don't know anything about complex structures in any other universe than our own. If you don't want to be precise, you are really just asking for personification of some natural effect (like Doppler shifts, length contraction, time dilation, etc.). So please answer which question are you asking: complete implication for real physics, or just personified explanation of some individual effects? – Marek Dec 3 '10 at 16:07
As a side note, I think a much more interesting/entertaining question is "what if the physical constants were allowed to change very slowly over space-time"? – Sklivvz♦ Dec 4 '10 at 13:52
show 1 more comment
3 Answers
I think you can find one of the answers here.
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Increasing the value of $\hbar$ significantly would be pretty interesting, as matter has a wavelength proportional to it. Walking through a doorway might become a new experience as you get diffracted into a wall.
However, this is all backwards thinking. The physical constants calibrate our mathematical models of the universe, like $\lambda = h/p$, not the other way around. So it is better to say that if you got diffracted every time you walked through a doorway, you would need a large $\hbar$ to model that.
Additionally, since constants like $\hbar$ have units, their value is somewhat arbitrary anyway. Which brings us to dimensionless constants... here's a good one:
$F_{\rm{grav}} / F_{\rm{em}} \propto \frac{\frac{G m m}{r^2}}{\frac{K q q }{ r^2}} \propto \frac{G}{K}$
the ratio of the gravitational to electromagnetic force for a unit charge and mass at a unit distance.
Currently the electromagnetic force kicks gravity's butt, but I wonder what life would be like if that weren't the case.
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Value of dimensionful constants is not really arbitrary. This is because there exist various scales in the nature. That is, speed of light is big with respect to normal velocities we observe around us (and SI system reflects this in the concept of second and meter). It's true that these are arbitrary scales of human beings but it's precisely they what allows Gamow to talk about small speed of light. Problem is, once you redefine some constant the matter will not able to form compounds and you'll lose all scales needed to talk about the change in (dimensionful) constants anyway. – Marek Dec 3 '10 at 16:02
The existence of scales does not negate the fact that you can make dimensionful constants whatever you want. "speed of light is big with respect to normal velocities we observe around us". That's a dimensionless constant $c/v_{\rm{normal}}$. Can't I find units where c = hbar = e = G = K = ... = 1? – Pete Dec 3 '10 at 16:32
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@Pete: oh, if you meant your statements in this way then I completely agree with you. Scales are what enables us to define dimensionless constants and then it's indeed true that only those are really physical. – Marek Dec 3 '10 at 20:35
I wonder, can I really define units where all the physical constants are 1? Or do I run out of degrees of freedom at some point? – Pete Dec 3 '10 at 20:51
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@Pete, you run out way before that. It's one of the most important problems in modern physics. E.g. the standard model has a big bunch of finely tuned constants whose measured value is unexplained at the moment. – Sklivvz♦ Dec 3 '10 at 21:06
show 2 more comments
See Smolin's book "The Life of the Cosmos", here and here, wherein he suggests common descent and Darwinian selection amongst multiverses for peturbed physical constants maximizing black hole formation in that universe. His most recent paper on arxiv on this topic is http://arxiv.org/abs/hep-th/0612185, where he argues that this is still a live hypothesis, which has already survived several experimental tests.
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http://mathoverflow.net/revisions/64577/list
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## Return to Question
3 deleted 72 characters in body
Let $X$ be a compact complex manifold. By definition, $Pic(X)={\rm H^1}(X,\mathcal{O}^\times)$. We know a lot about this group. What is known about the groups ${\rm H^n}(X,\mathcal{O}^\times)$ for $n\ge 2$? Another question: when $X$ is an Abelian variety, are there some explicit constructions similar to theta bundles?
A bit more specialized question. It is well known that for a nonsingular projective complex variety $X$ the natural map $${\rm H^1}(X,\mathcal{O}^\times)\to{\rm H^1}(X,\mathcal{M}^\times)$$ is trivial. What is known about the kernel of the same map for $n=2$ or $n=3$? (Here $\mathcal{M}^\times$ is the sheaf of nonzero meromorphic functions)functions, and the topology is the strong one).
2 added 347 characters in body
Let $X$ be a compact complex manifold. By definition, $Pic(X)={\rm H^1}(X,\mathcal{O}^\times)$. We know a lot about this group. What is known about the groups ${\rm H^n}(X,\mathcal{O}^\times)$ for $n\ge 2$? Another question: when $X$ is an Abelian variety, are there some explicit constructions similar to theta bundles?
A bit more specialized question. It is well known that for a nonsingular projective variety $X$ the natural map $${\rm H^1}(X,\mathcal{O}^\times)\to{\rm H^1}(X,\mathcal{M}^\times)$$ is trivial. What is known about the kernel of the same map for $n=2$ or $n=3$? (Here $\mathcal{M}^\times$ is the sheaf of nonzero meromorphic functions).
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# References for some analogs of the Picard group.
Let $X$ be a compact complex manifold. By definition, $Pic(X)={\rm H^1}(X,\mathcal{O}^\times)$. We know a lot about this group. What is known about the groups ${\rm H^n}(X,\mathcal{O}^\times)$ for $n\ge 2$? Another question: when $X$ is an Abelian variety, are there some explicit constructions similar to theta bundles?
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http://physics.stackexchange.com/questions/tagged/gauss-law?sort=faq&pagesize=30
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# Tagged Questions
The gauss-law tag has no wiki summary.
5answers
2k views
### Does Coulomb's Law, with Gauss's Law, imply the existence of only three spatial dimensions?
Coulomb's Law states that the fall-off of the strength of the electrostatic force is inversely proportional to the distance squared of the charges. Gauss's law implies that a the total flux through a ...
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### Why are so many forces explainable using inverse squares when space is three dimensional?
It seems paradoxical that the strength of so many phenomena (Newtonian gravity, Coulomb force) are calculable by the inverse square of distance. However, since volume is determined by three ...
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### Paradox with Gauss' law when space is uniformly charged everywhere
Consider that space is uniformly charged everywhere, i.e., filled with a uniform charge distribution, $\rho$, everywhere. By symmetry, the electric field is zero everywhere. (If I take any point in ...
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### Intuitive explanation of the inverse square power $\frac{1}{r^2}$ in Newton's law of gravity
Is there an intuitive explanation why it is plausible that the gravitational force which acts between two point masses is proportional to the inverse square of the distance $r$ between the masses (and ...
1answer
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### Why doesn't a gaussian surface pass through discrete charges?
I have read that Gaussian surface cannot pass through discrete charges. Why is it so? I have even seen in application of Gauss' Law when we imagine a Gaussian Surface passing through a charge ...
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### Infinitely charged wire and Differential form of Gauss' Law
I have tried calculating the potential of a charged wire the direct way. If lambda is the charge density of the wire, then I get \phi(r) = \frac{\lambda}{4 \pi \epsilon_0 r} \int_{-\infty}^\infty ...
1answer
542 views
### How is Gauss' Law (integral form) arrived at from Coulomb's Law, and how is the differential form arrived at from that?
On a similar note: when using Gauss' Law, do you even begin with Coulomb's law, or does one take it as given that flux is the surface integral of the Electric field in the direction of the normal to ...
2answers
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### In which cases is it better to use Gauss' law?
I could, for example calculate the electric field near a charged rod of infinite length using the classic definition of the electric field, and integrating the: \overrightarrow{dE} = \frac{dq}{4 ...
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http://mathoverflow.net/questions/67576?sort=votes
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## basic measure theory question - measure on the natural numbers [closed]
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I am looking for a succinct way to describe a subset of the natural numbers which has measure zero" in the following sense: Let X_1 \subset X_2 \subset .... be any strictly nesting sequence of finite subsets of natural numbers, I want to describe a subset S \subset \N which has the property that the limit of the usual counting measures applied to S restricted to X_i is zero.
I don't know that much about measure theory and am having trouble finding any measures on countable sets such as the natural numbers. There should be a natural measure on them such that the sets I describe above have zero measure with respect to it. Is there a name for such a natural measure on the natural numbers?
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7
The word "measure" in this context is not the appropriate one. You are probably looking for subsets of the naturals having zero density, even if your definition is not the correct definition of density. In fact, you should divide the counting measure by the size of $X_i$ before taking the $\liminf$ (or the $\limsup$). See en.wikipedia.org/wiki/Natural_density for more details. – Francesco Polizzi Jun 12 2011 at 12:33
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I hate to say it, but I feel obliged to: this site is for mathematics approximately at a research level. Please read the faq for appropriate places to ask this type of question. – Todd Trimble Jun 12 2011 at 13:47
While I agree that the question is not really suited to MO, I would at least offer some encouragement: this kind of question is both natural and good when one is trying to get to grips with notions of `size' for infinite subsets of the natural numbers. Hopefully the wikipedia link in Francesco Polizzi's comment may be of use. – Yemon Choi Jun 13 2011 at 2:52
## 1 Answer
A measure, here, isn't going to work. You can modify your idea to get near the idea of density, as Francesco pointed out. But, since you're new with measure theory, remember that a measure is countably additive, and (assuming you do want to divide by the size of the $X_i$ in your explanation, as was suggested) you're going to get the measure of any single point to be zero. Then by countable additivity, you're going to get the measure of any set to be zero.
This is why you're having trouble finding measures on countable sets; if you want small sets to be measure zero, you're going to get the zero measure, which is not what you want.
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Yes this seems to be precisely why finding such a measure will fail. If one wishes to weight all natural numbers equally, then only the trivial `zero measure' can assign a finite measure to an infinite set. – Spencer Jun 12 2011 at 23:52
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http://mathematica.stackexchange.com/questions/tagged/expression-manipulation?sort=faq&pagesize=15
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# Tagged Questions
How to manipulate expressions structurally, not necessarily complying with the rules of algebra.
2answers
696 views
### Head and everything except Head?
I have been working on picking expressions apart using Head and Part and encountered a little mystery. Consider the canonical ...
3answers
395 views
### How can I completely ban usage of some functions in output and mandate use of others?
For example, I hate that Mathematica uses Pochhammer symbol in outputs and prefer all the expressions in Gamma function. How can ...
1answer
175 views
### Transform an expression into a graph that can be plotted with TreeGraph (not TreeForm)
I would like to plot an expression (like TreeForm does), but using the new TreeGraph functionality. ...
1answer
253 views
### How do I expand a sum?
I have a problem with Mathematica's symbolic manipulations. As an example, consider the following expression: $$\sum _{i=1}^n -2 x_i \left(-a x_i-b+y_i\right)=0$$ How do I get Mathematica to expand ...
5answers
484 views
### How do I get my equation to have the form $(x-a)^2 + (y-b)^2 + (z-c)^2-d = 0$?
I want Mathematica to express the equation $$-11 - 2 x + x^2 - 4 y + y^2 - 6 z + z^2=0$$ in the form $$(x - 1)^2 + (y - 2)^2 + (z - 3)^2 - 25=0$$ How do I tell Mathematica to do that?
1answer
190 views
### How can I see which transformations Simplify attempts?
The documentation for Simplify[expr] says that it performs a sequence of algebraic and other transformations on expr, and returns the simplest form it finds. How can I see which transformations it ...
3answers
195 views
### Comparing Mathematica expressions like diff
I am looking for a way to compare (or "diff") two Mathematica expressions, similarly to how to diff utility can compare two text files and report the differences. Has anyone already written such a ...
3answers
302 views
### Pattern matching a pattern with patterns
Confusing title, I know. But the question is, if we have two patterns which have the same general structure but different names used in the patterns and different names: ...
2answers
355 views
### Displaying a series obtained by evaluating a Taylor series
Description of problem I would like to use Mathematica to display the series obtained by substituting a value for $x$ in a Taylor series expansion. The terms of the series will be rational numbers, ...
1answer
160 views
### List manipulation to build a functional expression
Since Mathematica deals with Head[a, b] as an expression in this way The first Part of Head[a, b] is ...
2answers
442 views
### Google Code Jam: Mountain View code review
After participating in Google Code Jam 2012 Round 2, it occurred to me that the Mountain View problem boiled down to solving a set of linear inequalities, which could be done in Mathematica. I've only ...
2answers
668 views
### How to expand tan(x+y) as normal form?
TrigExpand@Tan[x + y] gives $\frac{\sin (x) \cos (y)}{\cos (x) \cos (y)-\sin (x) \sin (y)}+\frac{\cos (x) \sin(y)}{\cos (x) \cos (y)-\sin (x) \sin (y)}$ but I ...
4answers
129 views
### Extracting variables from an expression
I'm looking for a way to extract a list of variables from an expression, for example with an input like: ...
4answers
292 views
### Extracting equations from Piecewise expressions
Say I have a PDF: PDF[LogNormalDistribution[1.75, 0.65], x] Calculating it, Mathematica gives me an expression that looks like this: I want to extract the ...
1answer
151 views
### Behavior of expression evaluation in Plot
I'm confused with Mathematica's way of parsing expressions. I've been struggling with this for a while and never found an exhaustive answer, sometimes things don't parse the way I think they would ...
2answers
95 views
### How to change the Integer to a special form in an expression?
I have an expression. I want to change all the Integers to a new form. The rule is, x_Integer->x._f. But we should consider some special cases. For example, ...
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http://math.stackexchange.com/questions/134229/derive-density-of-log-of-theta-when-theta-follows-a-uniform-distribution
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# Derive density of log of theta when theta follows a uniform distribution
I'm learning bayesian and I need to do some proofs that I'm not quite sure I can do. Any help will be very welcome!
If $\theta \text{~Unif}(0,1)$
What's the density of $y = - \log\theta$?
And what's density of $\theta \ (1-\theta)$?
Finally, what's density of $\log (\theta \ (1 - \theta))$?
Thanks
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## 1 Answer
All three are similar:
1) $-\log\theta$ is monotonically decreasing on $(0;1]$ from $+\infty$ to $0$, therefore, for $y\ge 0$: $F_{-\log\theta}(y)=Pr\{-\log\theta<y\}=Pr\{\theta>e^{-y}\}=1-e^{-y}$, and density is $e^{-y}$.
2) $\theta(1-\theta)$ is increasing on $[0;1/2]$ from $0$ to $1/4$, and decreasing on $[1/2;1]$ from $1/4$ to $0$, therefore, for $y\in[0;1/4]$: $F_{\theta(1-\theta)}(y)=Pr\{\theta(1-\theta)<y\}=1-Pr\{\theta(1-\theta)>y\}=1-Pr\{\theta^2-\theta+y<0\}=1-\sqrt{1-4y}$, and density is just the derivative.
3) Similarly. I am pretty sure you can now get it done by yourself.
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http://math.stackexchange.com/questions/126732/minimal-subset-of-x-1-x-2-x-100-that-xors-to-y
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# Minimal subset of $x_1, x_2, …, x_{100}$ that XORs to $y$
Given a 64-bit positive integer $y$ and a set of $100$ $64$-bit positive integers: $X = \{ x_1, x_2, \dots, x_{100} \}$
I want to find a smallest possible $Z = \{z_1, z_2, \dots, z_n\} \subset X$, such that:
$$\begin{align} y = z_1 ⊕ z_2 ⊕ \dots ⊕ z_n \end{align}$$
if such a $Z$ exists.
Clearly there are $2^{100}$ possible subsets of $X$, so iterating through them would take too long.
Does anyone know of, or can anyone think of, some sort of dynamic programming solution or other feasible algorithm?
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– user2468 Apr 1 '12 at 1:00
@J.D.: No, I think because it's XOR and not addition that it is not NP-Complete. I'm sure there is a solution that takes advantage of the independence of the bit positions in XOR, I just can't see it. – user1131467 Apr 1 '12 at 1:19
## 1 Answer
If you introduce binary variables $c_i$ that are $1$ when $x_i\in Z$ and $0$ otherwise, you can write
$$y=\bigoplus_ic_ix_i\;.$$
This is a $64\times100$ system of linear equations over the field $\mathbb F_2$ for the $c_i$ that you can easily solve using Gaussian elimination; the number of steps is of the order of a million. If the $x_i$ have full rank $64$, which is almost certain to be the case if they're randomly chosen with uniform distribution, the solution space will have $36$ dimensions, so you just need to enumerate $2^{36}$ different solutions to find the optimal one; this is doable in reasonable time on a present-day computer.
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http://mathoverflow.net/questions/72009/truel-extended-to-n-persons/72097
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## Truel extended to n persons
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
n players numbered 1~n play a shooting game. Their accuracy rates p1~pn are strictly between 0 and 1, and strictly increases from p1 to pn. This is common knowledge.
Before the game starts, the referee arranges the n players in some order. When game starts, players take turns to fire at one another according to that order. (For example, if n=4, and the referee arranges the players in order (3,4,1,2), when game starts, 3 fires first, then 4 fires, then 1, then 2, then 3 again, so on and so forth, as long as they are alive) The last person left is the winner of the game.
To be more specific, define $S_{i}$={1,2,3,...,i-1,i+1,i+2,...,n}. Let $S_{i}^{k}$ be any subset of $S_{i}$ that contains k elements(n>k>0). Then the strategy of player i is a function that maps $S_{i}^{k}$ to one of its element, for any $S_{i}^{k}$. What this definition means is that, given any k players (excluding i himself) left, player i's strategy tells him whom to shoot first. Notice that we rule out the possibility that any player can hold fire in his turn: he must choose someone to shoot.
After the referee arranges the firing order, all players must announce their strategies simultaneously. A player's payoff is his winning probability.
Question 1: Is there always an Nash equilibrium in this game, for any firing order?
Question 2: Suppose firing order is (1,2,3,...,n). Which players have fixed optimal strategies with respect to changes in (p1,p2,...,pn)? (When n=3, all have fixed optimal strategies; when n=4, player 1,2 and 4 have fixed optimal strategies) Furthermore, these fixed optimal strategies are intuitive and simple in the sense that they always instruct the player to fire at the most accurate person alive. I guess there could be regularities as n gets larger? At least can we say for player 1 this strategy is always optimal?
EDIT: "dominant strategy" in Question 2 is changed to "fixed optimal strategy with respect to changes in probabilities", which is more appropriate.
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I had to edit the problem several times because I have to avoid using "<" after "0", or there will be errors with the display. Sorry for that. I don't know what's wrong. Please take a look at the problem again. – unknown (google) Aug 3 2011 at 17:30
2
< is interpreted as a start of an HTML element by the system. There are two ways how to avoid problems with it: (1) Leave a blank space after each < . This tells markdown it's not a tag, and the extra space is ignored inside TeX math. (2) Put backticks around the whole math expression (including the dollars) which contains the < . – Emil Jeřábek Aug 3 2011 at 17:41
2
@Emil and unknown: alternatively, you should be able to use "\lt" instead of <. – Thierry Zell Aug 3 2011 at 19:44
@Thierry: That will just make it look messier. A cleaner markdown usage makes for easier edits later on. – Rodrigo A. Pérez Dec 15 at 5:37
## 2 Answers
Consider a scenario with firing order $(1,2,\ldots,n)$ where players 1 to $n-3$ have hit probabilities so close to 0 that their only real purpose is to allow the top 3 players to waste shots, while players n-2 to n have hit probabilities very close to 1. Clearly player $n$ will be the target whenever player $n-2$ or $n-1$ makes a non-wasted shot, so his best chance at survival will always be to shoot player $n-1$ and hope that player $n-2$ misses. Therefore player $n-1$ will always shoot player $n$ and hope that player $n-2$ misses. On the other hand, whenever players $n-1$ and $n$ and some of $1$ to $n-3$ are still alive, player $n-2$ would be foolish to shoot player $n$ or $n-1$ (becoming the target for the next shot): instead he shoots a low-ranking player (presumably number $n-3$ if available), and next will get to shoot at the survivor of players $n-1$ and $n$.
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1
I tried the case $n=10$ with hitting probabilities $1/20, 3/20, \ldots, 19/20$. If my programming is correct, there are very many cases where the optimal strategy does not say to shoot at the most accurate person alive. For example, when all are still alive players 1 to 10 should target players 8, 7, 5, 2, 9, 8, 3, 5, 5, 9 respectively. – Robert Israel Aug 4 2011 at 21:49
@Robert: If I computed it correctly, with the same probabilities, if players are allowed to hold fire in his turn, then 1 to 10 should target: 4,5,5,2,3,H,1,1,7,8, where H means hold fire. Interestingly, anyone could be a first target, even if he's most inaccurate. – unknown (google) Aug 5 2011 at 3:27
### 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.
This is a non-cooperative game of perfect information. In the absence of degeneracy, there is always a unique optimal pure strategy for each player. Note that your probability of hitting your target doesn't depend on who the target is, and the situation if you miss also doesn't depend on who the target is. Thus your only concern is what happens if you hit your target. You shoot at the target whose death would give you the highest probability of being the survivor. These probabilities for all participants can be computed by "dynamic programming", starting with the case of only one surviving participant and working backwards. The only complication is how ties might be broken.
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@Robert: Thanks! I agree there's always a unique pure strategy for a player. And indeed with probabilities given, we can recursively determine everyone's optimal strategy. But this doesn't answer Quesiton 2 (which i've modified): Will the naive strategy always remain most players' optimal strategy as n gets larger? Or can we predict which players will always be able to adopt the naive strategy as optimal, as n gets larger? – unknown (google) Aug 4 2011 at 2:16
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http://math.stackexchange.com/questions/180307/a-geometry-problem-measure-of-an-angle?answertab=active
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# a geometry problem - measure of an angle
We have a square and the following information:
1) $E \in [AB]$, $E$ an arbitrary point
2) $[AC] \cap [DE]= \{P\}$ and
3)$FP \perp ED$, where $F \in BC$ .
We have to prove that the measure of the angle $\angle EDF = 45^{\circ}$.
Thanks a lot !
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## 2 Answers
It's easy to see that $PFCD$ is a cyclic quadrilateral, $\angle DPF$ + $\angle FCD = 180^{\circ}$. Therefore, we have $$\angle EDF = \angle PDF = \angle PCF=\angle ACB = 45^{\circ}$$
Q.E.D. (a very simple problem)
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This follows becuse $DP$ and $FP$ have the same length, being segments hitting the sides from the diagonal at equal angles.
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it is not clearly. and how can I prove that the segments $DP$ and $FP$ are hitting the sides from the diagonal at equal angles? Thank ! – Iuli Aug 8 '12 at 14:57
How do you mean "it is not clearly"? Do you mean the answer isn't clear, or that it's clearly wrong? The segments hit the sides at equal angles because both the segments and the sides are at right angles to each other. – joriki Aug 8 '12 at 14:59
@Iuli: There was a comment from you here earlier, now deleted, asking me to explain this in mathematical notation: $DP\perp FP\land CD\perp CB\rightarrow \angle PDC=\angle PFB$. – joriki Aug 9 '12 at 6:16
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http://mathoverflow.net/questions/28118?sort=newest
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## Does the norm of a normed linear space determine the form of its dual spaces elements?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Hello everybody,
As an introductory example, suppose $U \subset R^n$ is open and bounded, let $p = 2$. Then there is a constant $c>0$ s.t. `$\forall u \in W^{1,p}_0 : \Vert u \Vert _ {W^{1,p}_0} \le c \Vert \vert \nabla u \vert\Vert_{L^p(U)}$`. This implies that the latter expression defines an equivalent norm on ${W^{1,p}_0}$.
Let $f \in L^2(U), g \in C^1(\bar{U})$. Then there exists an unique solution $u \in W^{1,p}$ to the system $\triangle u = f$ over $U$, $u = g$ over $\partial U$ - or equivalently, there exists an unique solution $u \in W^{1,p}_0$ to the system $\triangle u = f - \triangle g$ over $U$, (in the distributional sense).
Proof: $W^{1,2}_0$ is a Hilbert space, hence self-dual. The rhs $f - \triangle g$, defines an element of $D'$, which by density can extended to $W^{1,2}_0$. On the other side, the equivalent norm as introduced above is defined the inner product $(u,v) = \int \nabla u \cdot \nabla v$, by riesz' representation theorem, there is an $u \in W^{1,2}_0$ s.t. the induced form $(u, \cdot)$ coincides with the form defined by the rhs. But then this $u$ is a weak solution to $\triangle u = f - \triangle g$.
So far, so good. I would like to ask some questions on this.
i) Can this be extended to other dual exponents $p$, $q$ ?
ii) The equivalent norm that regards first derivatives only is not only an equivalent norm for $p=2$, but also for $1 \leq p < \infty$. In the above case, it seems the norm imposes a form the dual vectors are subject to. I wonder whether in general - not only in the case of $L^p$ and its friends - there is some way how the form of linear functionals on some normed space $X$ are determined by the norm attached to the vector space $X$.
I hope this questions ain't too vacuous and there are interesting answers. In either case, thanks.
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I've cleaned up some of the TeX, but I don't know which of several symbols you intend by "del". Please make use of the preview feature when writing a question. – Mark Meckes Jun 14 2010 at 13:47
Thanks. Unfortunately, I didn't see the preview feature, maybe due to js being disabled. – Martin Jun 14 2010 at 14:38
## 1 Answer
The argument you have presented is an adaptation of the Lax-Milgram theorem which is essentially equivalent to the Riesz representation theorem (and generally speaking both of these results hold only in the Hilbert space framework). The Lax-Milgram theorem fails for the Laplace equation in $L^p$-spaces with $p\neq2$. Instead, some analogous results based on the ideas of coercivity, duality and monotonicity can be obtained in any reflexive Banach space.
The Dirichlet problem for the $p$-Laplace operator
$$-\nabla(|\nabla u|^{p-2}\nabla u)=f,\quad x\in\Omega,\qquad (*)$$ $$u=0,\qquad x\in\partial\Omega,$$ might be a "correct" $L^p$-analogue of the problem described in the question.
The right hand side of $(*)$ gives rise to the mapping $A: W_{0}^{1,p}\to(W_{0}^{1,p})^{*}$ defined by the identity $$\langle Au,v\rangle=\int_{\Omega}|\nabla u|^{p-2}\nabla u\cdot\nabla v\ dx\quad \mbox{for all } v\in W_{0}^{1,p}.$$ A straightforward check shows that $A$ satisfies the conditions of the following theorem (which might be viewed as an $L^p$-analogue of the Lax-Milgram theorem).
Theorem. Let $A$ be a strictly monotone, coercive operator from a reflexive Banach space $E$ to its dual $E^{* }$. If $A$ is continuous on finite-dimensional subspaces of $E$ then for every $f\in E^{*}$ there exists a unique solution to the problem $$Au=f.$$
Have a look at the textbook by Chipot or the free monograph by Showalter where the approach is explained in detail.
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http://psychology.wikia.com/wiki/Root_mean_square?oldid=26564
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# Root mean square
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In mathematics, root mean square (abbreviated RMS or rms), also known as the quadratic mean, is a statistical measure of the magnitude of a varying quantity. It can be calculated for a series of discrete values or for a continuously varying function. The name comes from the fact that it is the square root of the mean of the squares of the values. It is a power mean with the power $t=2$.
## Calculating the root mean square
The rms for a collection of $N$ values $\{x_1,x_2,\dots,x_N\}$ is:
$x_{\mathrm{rms}} = \sqrt {{1 \over N} \sum_{i=1}^{N} x_i^2} = \sqrt {{x_1^2 + x_2^2 + \cdots + x_N^2} \over N}$
and the corresponding formula for a continuous function $f(t)$ defined over the interval $T_1 \le t \le T_2$ (for a periodic function the interval should be a whole number of complete cycles) is:
$f_{\mathrm{rms}} = \sqrt {{1 \over {T_2-T_1}} {\int_{T_1}^{T_2} {[f(t)]}^2\, dt}}$
## Uses
The RMS value of a function is often used in physics and electronics. For example, we may wish to calculate the power $P$ dissipated by an electrical conductor of resistance $R$. It is easy to do the calculation when a constant current $I$ flows through the conductor. It is simply,
$P = I^2 R$
But what if the current is a varying function $I(t)$? This is where the rms value comes in. It may be trivially shown that the rms value of $I(t)$ can be substituted for the constant current $I$ in the above equation to give the average power dissipation:
$P_\mathrm{avg}\,\!$ $= \mathrm{E}(I^2R)\,\!$ (where $\mathrm{E}(\cdot)$ denotes the arithmetic mean)
$= R\mathrm{E}(I^2)\,\!$ (R is constant so we can take it outside the average)
$= I_\mathrm{rms}^2R\,\!$ (by definition of RMS)
We can also show by the same method
$P_\mathrm{avg} = {V_\mathrm{rms}^2\over R}\,\!$
By square rooting both these equations and multiplying them together we get the equation
$P_\mathrm{avg} = V_\mathrm{rms}I_\mathrm{rms}\,\!$
However it is important to stress that this is based on the assumption that voltage and current are proportional (that is the load is resistive) and is not true in the general case (see AC power for more information).
In the common case of alternating current, when $I(t)$ is a sinusoidal current, as is approximately true for mains power, the rms value is easy to calculate from equation (2) above. The result is:
$I_{\mathrm{rms}} = {I_\mathrm{p} \over {\sqrt 2}}$
where $I_\mathrm{p}$ is the peak amplitude.
It should be noted that the peak amplitude is half of the peak-to-peak amplitude. When the peak-to-peak amplitude is known, the same formula is applied by using half of the p-p value.
The RMS value can be calculated using equation (2) for any waveform, for example an audio or radio signal. This allows us to calculate the mean power delivered into a specified load. For this reason, listed voltages for power outlets (e.g. 110 V or 240 V) are almost always quoted in RMS values, and not peak values.
From the formula given above, we can calculate also the peak-to-peak value of the mains voltage which is approx. 310 (U.S.A) and 677 (Europe) volts respectively.
In the field of audio, mean power is often (misleadingly) referred to as RMS power. This is probably because it can be derived from the RMS voltage or RMS current. Furthermore, because RMS implies some form of averaging, expressions such as "peak RMS power", sometimes used in advertisements for audio amplifiers, are meaningless.
In chemistry, the root mean square velocity is defined as the square root of the average velocity-squared of the molecules in a gas. The RMS velocity of an ideal gas is calculated using the following equation:
${u_\mathrm{rms}} = {\sqrt{3RT \over {M}}}$
where $R$ represents the ideal gas constant (in this case, 8.314 J/(mol⋅K)), $T$ is the temperature of the gas in kelvins, and $M$ is the molar mass of the compound in kilograms per mole.
## Relationship to the arithmetic mean and the standard deviation
If $\bar{x}$ is the arithmetic mean and $\sigma_{x}$ is the standard deviation of a population then
$x_{\mathrm{rms}}^2 = \bar{x}^2 + \sigma_{x}^2$
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http://crypto.stackexchange.com/questions/869/sim-security-for-two-messages
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# SIM security for two messages
Here SIM means the simulation based security
Consider a two message encryption scheme:$$Enc:K \times M \times M \rightarrow C \times C$$
and $Enc(K, m, m')=(K \oplus m, K \oplus m')$, In the real world experiment, the adversary $Eve$ gets $(K \oplus m, K \oplus m')$ as ciphertext, and thus, the information leaked to $Eve$ is $(K\oplus m)\oplus(K \oplus m')=m \oplus m'$, now how should I modify the original ideal world experiment so that in the ideal world, the adversary $Eve$ also gets $m \oplus m'$?
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2
Looking for a clarification here. You say "In the real world experiment..." and "how should I modify the original ideal world experiment...". I'm not seeing the difference between the "real world experiment" and the "ideal world experiment". What is the difference? – mikeazo♦ Oct 2 '11 at 11:28
## 3 Answers
You did not state what the original ideal world experiment is, so we cannot say how it should be modified. In most ideal functionalities for encryption, when Alice encrypts a message for Bob, the ciphertext is not written to the adversary's (simulator's) tape however the length of the message is (and maybe a record of Alice sending something to Bob). In this case, you would modify it so that the simulator would also receive $m \oplus m'$ at the same time.
PS. If you are going to completely change your question (after answers have been posted), you should leave the original question intact and append your new question to the end.
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Actually, the ideal world is defined as following: Ideal world experiment: Repeatedly do: The adversary Eve arbitrarily interacts with the environment. The environment sends a message m to Alice. Alice sends the message m to Bob through a secure channel. – huyichen Oct 3 '11 at 17:19
Eve receives notification that a message was sent through the channel (but does not receive a ciphertext). (Here we assume that the messages come from a finite message space. Otherwise, Eve is notified the length of the message as well.) Eve continues to arbitrarily interact with the environment. The environment outputs a bit (whether Eve caused a particular observable effect on the environment). The adversary is said to "succeed" in the experiment if the environment outputs 1. – huyichen Oct 3 '11 at 17:22
So when Eve receives notification, the notification will include $m \oplus m'$. – PulpSpy Oct 3 '11 at 17:35
I would highlight Thomas's last point; not only is leaking $m \oplus m'$ failing to provide perfect secrecy, it is often practically insecure.
Exactly how bad this is depends on what those messages actually are (and how they are encoded), but it can be bad indeed. As one example, in my experience, recovering two ASCII English texts from their xor is about as difficult as solving a newspaper cryptogram.
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For a given pair of cipher text $(c, c')$, you can compute $x = c \oplus c'$ which is the XOR of the two ciphertexts and is also equal, as you note, to the XOR of the two corresponding plaintexts. Then, for any pair of potential messages $(m, m')$ such that $m \oplus m' = x$, you can compute a corresponding key $K = m \oplus c$; you can verify that $Enc(K, m, m') = (c, c')$. Therefore, the information you have (the ciphertexts $c$ and $c'$) gives you the XOR of the two plaintexts, but since any pair of messages which matches that information is still a possibility, you have no more information than that.
An other way to see it: given a One-Time Pad encryption of a message $m$ with a key $K$ (you know $c = m \oplus K$, but not $m$ or $K$), you can create a random $x$ and compute $c' = c \oplus x$. You then know that $Enc(K, m, m') = (c, c')$ and $m \oplus m' = x$. If you can extract out of that more information than the XOR of $m$ and $m'$ (which the value $x$ you chose), then, congratulations, you have broken One-Time Pad. Given the proven impossibility thereof, one must conclude that out of $(c, c')$ you learn only $m \oplus m'$ and no more.
Mind you, for messages $m$ and $m'$ which are not completely random (e.g. plaintext messages which make sense), the XOR of $m$ and $m'$ is a lot of information; this is the infamous "Two-Times Pad" which has lead to actual decryption in some historical cases (the Wikipedia page on OTP gives a few example, with the Soviet Union in the role of the loser).
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http://mathhelpforum.com/advanced-algebra/64779-find-splitting-field.html
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# Thread:
1. ## Find the splitting field
Find the splitting field for f(x)=((x^2)+x+2)((x^2)+2x+2) over Z_3[x]
Write f(x) as a product of linear factors
2. Originally Posted by mandy123
Find the splitting field for f(x)=((x^2)+x+2)((x^2)+2x+2) over Z_3[x]
Write f(x) as a product of linear factors
Let $f(x)$ be a non-constant polynomial over a field $F$. Given a field extension $K$ over $F$ we say that $f(x)$ splits over $K$ iff $f(x) = a(x-\alpha_1)(x-\alpha_2)...(x-\alpha_r)$ where $\alpha_i \in K$ i.e. into linear factors. Furthermore, we say that $K$ is a splitting field of $f(x)$ over $F$ iff $K = F(\alpha_1, ...,\alpha_r)$ i.e. $K$ is the smallest such field extension for which the polynomial splits over.
Returning back to your problem with $f(x) = (x^2 + x + 2)(x^2 + 2x + 2)$ in the field $\mathbb{F}_3$. Look at the first factor, $x^2+x+2$, it has no zeros in $\mathbb{F}_3$ by simply checking. Therefore this polynomial is irreducible. Form the factor ring $\mathbb{F}_3[t]/(t^2+t+2)$ which turns out being a field since $t^2+t+2$ is an irreducible polynomial, this is also a finite field with $3^2 = 9$ elements, so we will refer to it as $\mathbb{F}_9$. The mapping $a \mapsto a + (t^2+t+2)$ embeds $\mathbb{F}_3$ in $\mathbb{F}_9$ and therefore we can think of $\mathbb{F}_3$ being contained in $\mathbb{F}_9$. Let $\alpha = t + (t^2+t+2) \in \mathbb{F}_9$ and so $\alpha^2 + \alpha + 2 = 0$, this means that $\alpha$ is root of the polynomial $x^2 + x + 2$. It is not hard to see that $-\alpha - 1$ is another root of $x^2 + x + 2$. Therefore, $f(x) = (x - \alpha)(x + \alpha + 1)(x^2+2x+2)$. Now the question is whether the second factor, $x^2+2x+2$, splits over $x^2+2x+2$. A little guessing shows that $2\alpha$ is a root of this polynomial because $(2\alpha)^2 + 2(2\alpha) + 2 = \alpha^2 + \alpha + 2 = 0$. And so the other root is easy to find which is $\alpha + 1$ which means $f(x) = (x-\alpha)(x+\alpha + 1)(x - 2\alpha)(x - \alpha - 1)$ and $\mathbb{F}_9 = \mathbb{F}_3(\alpha)=\mathbb{F}_3 (\alpha,\alpha+1,-2\alpha,-\alpha - 1)$. Therefore, $\mathbb{F}_9$ is a splitting field.
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http://mathhelpforum.com/calculus/31649-setting-g-x-zero.html
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# Thread:
1. ## Setting g'(x) to zero
(I know this is the pre-cal forum, but the calculus isn't my problem here ... it's the basics that are dogging me.)
In order to make a wiggle graph, I need to find where my slopes are zero. The formula of my derivative is:
$g'(x) = 6x^2 - 7x - 3$
So my question is, when g'(x) = 0, x = ?
I don't think I'm just supposed to plug random numbers in and hope to hit 0, but simplifying has only gotten me a little closer. I think I can simplify to:
$x = 1/(2x) + 7/6$
It would be a little easier to just start plugging numbers in, but I think I'm supposed to actually solve this one before moving on. I'm stuck - how do I know what x is when g'(x) = 0?
2. Originally Posted by Boris B
(I know this is the pre-cal forum, but the calculus isn't my problem here ... it's the basics that are dogging me.)
In order to make a wiggle graph, I need to find where my slopes are zero. The formula of my derivative is:
$g'(x) = 6x^2 - 7x - 3$
So my question is, when g'(x) = 0, x = ?
I don't think I'm just supposed to plug random numbers in and hope to hit 0, but simplifying has only gotten me a little closer. I think I can simplify to:
$x = 1/(2x) + 7/6$
It would be a little easier to just start plugging numbers in, but I think I'm supposed to actually solve this one before moving on. I'm stuck - how do I know what x is when g'(x) = 0?
You need to solve the quadratic equation
$0 = 6x^2 - 7x - 3$ .....
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http://math.stackexchange.com/questions/82030/size-of-a-union-of-two-sets
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# Size of a union of two sets
We were ask to prove that $|A \cup B| = |A| + |B| - |A \cap B|$. It was easy to prove it using a Venn diagram, but I think we might be expected to do if more formally. Is there a formal way?
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What does $|A|$ mean? Is $|\cdot|$ a measure? Or maybe the sets are finite? – Jeff Nov 14 '11 at 16:44
2
just observe that if you do $|A|+|B|$ you count twice the elements in $A\cap B$. – Valerio Capraro Nov 14 '11 at 16:48
Suppose that you want to give a dollar to everyone in $A\cup B$. If you give a dollar to everyone in $A$, and then give a dollar to everyone in $B, \dots$. – André Nicolas Nov 14 '11 at 17:21
I am fairly certain that this question appeared here perhaps infinitely many times before. I cannot find it though... – Asaf Karagila Nov 14 '11 at 18:22
## 1 Answer
$A\cup B = (A\setminus B) \cup (B\setminus A) \cup (A\cap B)$. These three sets are disjoint, so $$|A\cup B| = |A\setminus B| + |B\setminus A| + |(A\cap B)|$$ But $A\setminus B = A\setminus(A\cap B)$, so $|A\setminus B|=|A|-|A\cap B|$. A similar equality holds for $|B \setminus A|$. Substitution of these into the displayed equation above yields your result.
Of course, one might need to formally show that $|A\setminus (A\cap B)| = |A|-|A\cap B|$. I can't decide if this is any less obvious than the original proposition...
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2
I have seen $A \setminus B$, $A \smallsetminus B$ and $A - B$ used to denote set difference, but not $A / B$. Do you normally use that notation? =) – Srivatsan Nov 14 '11 at 16:59
1
Is using a forward slash for set difference standard notation? – Henning Makholm Nov 14 '11 at 17:00
3
bleech... I'm left/right backslash challenged...I got points taken off for doing that on a topology test once. – David Mitra Nov 14 '11 at 17:00
@David Mitra: The forward slashes were bothersome, so I was changing things to setminus, and in the middle they changed to backslash. Not thinking, I continued. Can turn them to backslash if you wish. – André Nicolas Nov 14 '11 at 17:40
Thanks. \setminus is better of course. I'm not sure if \backslash is a mathop... – David Mitra Nov 14 '11 at 17:44
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http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.ijm/1258138306
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### The spectrum of differential operators in $H\sp p$ spaces
Dashan Fan, Liangpan Li, Xiaohua Yao, and Quan Zheng
Source: Illinois J. Math. Volume 49, Number 1 (2005), 45-62.
#### Abstract
This paper is concerned with linear partial differential operators with constant coefficients in $H^p(\mathbf{R} ^n)$. In the case $0<p\le1$, we establish some basic properties and the spectral mapping property, and determine completely the essential spectrum, point spectrum, approximate point spectrum, continuous spectrum, and residual spectrum of such differential operators. In the case $p>2$, we show that the point spectrum of such differential operators in $L^p(\mathbf{R} ^n)$ is the empty set for $p\in(2,{2n\over n-1})$, but not for $p>{2n\over n-1}$ in general. Moreover, we make some remarks on the case $p>1$ and give several examples.
First Page:
Primary Subjects: 35P05
Secondary Subjects: 42B15, 42B30, 46E15, 47F05
Full-text: Open access
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http://mathoverflow.net/questions/27089/necessary-and-sufficient-criteria-for-non-trivial-derivations-to-exist/27093
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## Necessary and sufficient criteria for non-trivial derivations to exist?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Off hand, does anyone know of some useful conditions for checking if a ring (or more generally a semiring) has non-trivial derivations? (By non-trivial, I mean they do not squish everything down to the additive identity.) Part of the motivation for this is that I was thinking about it the other day, and had trouble finding any good example of a semiring with an interesting derivation.
For example, the multiplicative Banach algebra of positive functions is an algebra of the semifield of nonnegative reals. However, the usual definition for derivative breaks down due to the fact that you can have positive functions with negative slope. So, this leads me to wonder if there are any semirings with derivations at all?
As a related question, is there a known classification of all the derivations for an algebra? It feels like this should be a pretty standard thing, but I don't think I've ever encountered it in one of my courses and my initial googling around was not too successful at finding references.
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I assume you mean for all rings (and semirings) to be commutative in this case? – Harry Gindi Jun 4 2010 at 19:50
Yes, that is correct. – Mikola Jun 4 2010 at 20:02
## 3 Answers
There is a notion of a universal derivation for an algebra. I'll assume everything is commutative for simiplcitity. If $A$ is a $k$-algebra ($k$ a commutative ring) then there is an $A$-module $\Omega_{A/k}$, the module of Kahler differentials of $A$ over $k$ and a $k$-derivation $d:A\to\Omega_{A/k}$ which is universal for derviations of $A$. That is, if $\delta:A\to M$ is a $k$-derivation from $A$ to an $A$ module, then $\delta=f\circ d$ for a unique $A$-module homomorphism $f$.
There is an explicit desciption of $\Omega_{A/k}$ as $I/I^2$ where $I$ is the ideal in the ring $B=A\otimes_k A$ which is the kernel of the map $\mu:B\to A$ with $\mu(x\otimes y)=xy$. Then $d$ maps $x\in A$ to $1\otimes x-x\otimes 1$. So to find a derivation from $A$ with values in your favourite $A$-module $M$ all one has to do is to find an $A$-homomorphism from $\Omega_{A/k}$ to $M$. Of course $\Omega_{A/k}$ may be hard to determine concretely, and even if that is possible, perhaps it may not be easy to find a homomorphism from that into $M$. Indeed using this method may be no easier than finding a derivation directly :-)
For details see the commutative algebra texts by Eisenbud or Matsumura.
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Thanks! That exactly answers the second question, and combined with the below paper it pretty much settles what I was asking. – Mikola Jun 4 2010 at 20:08
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
The Banach algebra of bounded functions on a finite set turns out to be semisimple, and therefore carries no nonzero derivations by the results of Johnson, B. E. "Continuity of derivations on commutative algebras". Amer. J. Math. 91, 1 (1969).
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That said, if the set is naturally realized as a "poised" set in $\mathbb{R}^n$, one can use Lagrange interpolation to define "effective" derivations. This is particularly doable in the case of the set of nonnegative multiindices of fixed weight. I would like one day to try to apply this fact to renormalization of lattice gauge theories. – Steve Huntsman Jun 4 2010 at 20:05
For your question in the second paragraph, any semiring can be embedded in one with non-trivial derivation.
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http://mathhelpforum.com/trigonometry/168162-never-learned-how-change-cosine-into-frational-form.html
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# Thread:
1. ## never learned how to change a cosine into frational form
I was wondering if someone could help me out with how to change a cosine into fractional from: example cos 45 = sqrt2/2 and cos 150 = sqrt3/2
I know that if I take the inverse of the cosine decimal form I can take the answer do square it, make a fraction out of it with the sqrt of the square dived by the squared number. that gets really messy as well though.
2. Are you saying...
When solving $x = \cos 45^o$ , how can you get a fractional answer rather than some nasty decimal?
3. yeah, I can do it the way I explained but with most angles, that will still achieve a nasty decimal
4. Only particular angles will give you something nice.
In general these are 0,30,45,60,90 and multiples of these. You can achieve a rational answer for others, but start by working on these ones first.
Have you seen the special triangles?
5. I haven't seen the special triangles.
6. They are my favourite triangles. Commit them to memory and you will be a trig-wiz!
Special Triangles
7. Essentially your construct a triangle with angles 60-30-90 and 45-45-90
Special right triangles - Wikipedia, the free encyclopedia
8. awesome, thanks a bunch
9. You can also google the unit circle.
10. A side note: Most of those decimal answers you are getting are not exact answers. They are only approximations (although good approximations) to the actual answers. The actual answers are irrational and therefore cannot be written as finite decimals.
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http://mathoverflow.net/questions/16003/what-is-rich-structure-actually/16040
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## What is “rich structure”, actually? [closed]
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
An ubiquitous claim in mathematics is that such-and-such mathematical entity has a rich structure or more structure than another one. Most oftenly the entity is a structure - a set explicitly equipped with a structure, i.e. a graph, a group, etc. - but often enough it's just a set not explicitly equipped with a structure.
Examples:
1. Ash/Gross in Fearless Symmetry claim that the solution set [sic!] of a polynomial equation has "more structure" than just its cardinality (p.66).
2. They state on the other hand that the absolute Galois group $G_{\mathbb Q}$ "has a rich structure - much of it still unknown" (p. 87).
Two questions arise (for the beginner):
Question #1: If one considers the structure-richness of a genuine structure - as a set equipped with a structure (as in Example #2) -, what is a possible quantitative measure for richness?
The only answer that comes to my mind is something like the diversity of non-isomorphic (induced?) sub-structures. Is this formalizable?
Question #2: If only a set is mentioned - not explicitly equipped with a structure (as in Example #1) -, what sense does it make to talk of the richness of its structure?
I guess I have to presume a structure imposed on the set, and go back to Question #1. But often enough the structure to impose isn't obvious from the context (like in Example #1). Is this only because this is a book for the general audience? My impression is, that it's fairly often "left to the reader" to literally guess, which structure has to be imposed (to be a rich one).
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8
I think this is way too philosophical and discussion-y to work on MO. I don't think it's a question that "has an answer" (i.e. it's really bait for a discussion). I'd vote to close it as "not a real question", but I don't want to close unilaterally. If it acquires two or three other votes to close, then I'll vote to close too. – Anton Geraschenko♦ Feb 22 2010 at 0:35
6
Why is there a rush to label questions as ones which might lead to a discussion? If someone gives a concrete example of what is meant when someone says "rich structure," is that so horrible? – Douglas Zare Feb 22 2010 at 0:56
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It's not my question, but I'd like to see the answers, not to have this closed out of a bizarre fear that someone might try to discuss something instead of answering the question. I'm aware of the limitations of MO. – Douglas Zare Feb 22 2010 at 1:04
4
@Douglas: while what you're saying sounds reasonable, I disagree with it. If you want to discuss it further, let's do it on meta, where this topic has already received a lot of attention. Here is another one of my posts that elaborates on why we should close questions at all: meta.mathoverflow.net/discussion/130/2/…. Please also have a look at this thread: meta.mathoverflow.net/discussion/155/…. – Anton Geraschenko♦ Feb 22 2010 at 1:19
8
It seems to me that the most honest answer is that "richness of structure" is used only informally by mathematicians: like a deep theorem or an elegant proof, it involves an aesthetic judgment on the part of the speaker. You could try to create some quantitative measure of richness of structures, in which e.g., richness is non-increasing under forgetful functors, but this is not what people mean when they use the term. Thus I agree with Anton G. that the question is not a good one for MO: it is not possible to decide whether a given answer is correct. – Pete L. Clark Feb 22 2010 at 4:21
show 11 more comments
## 8 Answers
If one's intent is to understand the comments in Fearless Symmetry, then an investigation of the actual subject at hand (Galois theory and, more particularly, algebraic number theory) will be more revealing than an enquiry into generalities about the concepts of structure and richness.
In this direction:
Regarding Example 1/Question 2: Literally, the structure on the set of solutions is that it is a set equipped with an action of the Galois group (as was already noted by several people).
Regarding Example 2/Question 1: There has already been an MO question about what number theorists mean when they speak of understanding the group $G_{\mathbb Q}$, which is relevant to this question too. In any event, in the quoted example, as in many similar statements in the literature, "structure of $G_{\mathbb Q}$" is a short-hand for a number of problems related to the study of $G_{\mathbb Q}$, first and foremost being the study of the representations of $G_{\mathbb Q}$ and their relationships to (i) automorphic forms and (ii) motives.
Note that $G_{\mathbb Q}$ itself has to be thought of not just as a group, but as a group equipped with conjugacy classes of embeddings $G_K \hookrightarrow G_{\mathbb Q}$ for each completion $K$ of $\mathbb Q$ (so $K$ is $\mathbb R$ or $\mathbb Q_p$ for some prime $p$), whose images topologically generated $G_{\mathbb Q}$ (by Cebotarev density) in an extremely overdetermined way.
The theory of representations of $G_{\mathbb Q_p}$ itself has a deep theory, involving $p$-adic Hodge theory among other tools, and this should be thought of as being included in the "rich structure" being alluded to.
In summary: I think that such assertions are typically short-hand allusions to a deep and important (in the eyes of those making the assertion, at least) set of problems, techniques, and theorems related to the object at hand; that is the certainly the case in this instance. At least in the case of Example 2, they do not admit a superficial description in terms of some formal notion of "rich structure".
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
A good case to look at would be Michiel Hazewinkel's 'star example' of a rich structure Symm, the ring of symmetric functions in a countably infinite number of indeterminates.
Symm, the Hopf algebra of the symmetric functions is a truly amazing and rich object. It turns up everywhere and carries more extra structure than one would believe possible. For instance it turns up as the homology of the classifying space BU and also as the cohomology of that space, illustrating its self-duality. It turns up as the direct sum of the representation spaces of the symmetric group and as the ring of rational representations of the infinite general linear group. This time it is Schur duality that is involved. It is the free $\lambda$-ring on one generator. It has a nondegenerate inner product which makes it self-dual and the associated orthonormal basis of the Schur symmetric functions is such that coproduct and product are positive with respect to these basis functions...Symm is also the representing ring of the functor of the big Witt vectors and the covariant bialgebra of the formal group of the big Witt vectors (another manifestation of its auto-duality)...
As the free $\lambda$-ring on one generator it of course carries a $\lambda$-ring structure. In addition it carries ring endomorphisms which define a functorial $\lambda$-ring structure on the rings $W(A) = CRing(Symm, A)$ for all unital commutative rings $A$. A sort of higher $\lambda$-ring structure. Being self dual there are also co-$\lambda$-ring structures and higher co-$\lambda$-ring structures (whatever those may be).
Of course, Symm carries still more structure: it has a second multiplication and a second comultiplication (dual to each other) that make it a coring object in the category of algebras and, dually, (almost) a ring object in the category of coalgebras.
The functor represented by Symm, i.e. the big Witt vector functor, has a comonad structure and the associated coalgebras are precisely the $\lambda$-rings.
All this by no means exhausts the manifestations of and structures carried by Symm. It seems unlikely that there is any object in mathematics richer and/or more beautiful than this one, and many more uniqueness theorems are needed. (Witt vectors. Part 1: 7)
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Sounds promising, I will try to get into it. I also found very interesting your post on Mathematical robustness at the n-Category Café: golem.ph.utexas.edu/category/2008/11/… – Hans Stricker Feb 22 2010 at 10:02
I think maybe arxiv.org/abs/0810.5691 might be a more interesting reference. I would really like to thank you though for pointing out the Witt Vector article, I have been trying to get into them. – Sean Tilson Mar 3 2010 at 20:09
I'm answering with a stub that I hope others will fill out. Please feel free to edit this.
The rich structure on "the solution set of a polynomial" can mean the nice properties of the function which assigns to a polynomial its multiset of roots rather than a value of that function. It can mean that we have a permutation representation of the absolute Galois group on the roots.
The rich structure on the absolute Galois group may describe the actions on many objects, the lattice of (normal) subgroups, and that it carries a topology as a profinite group. Perhaps to say an object has a rich structure in this sense, there should be natural morphisms to many interesting objects, and we can recognize it as an image of a forgetful functor.
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I would interpret things like "has richer structure than just X" to mean that there's more there that you get for free, in some sense. For instance, in your example, the set of solutions to a system of polynomial equations at first seems to just have a cardinality, which is the only thing you can say automatically about a set. However, in this case, it's not just any set, by virtue of being solutions to some polynomial equations over a ring, it has a structure of a scheme (variety if the ring is a field and the equations are nice) where, without making any choices at all, we can make it into something more than just a set. In some cases (elliptic curves) the structure is yet richer, and is in fact also a group in a natural way.
At the least, things like this are how I generally hear the phrase used. But generally it's used informally to say that something is more interesting than a random object that looks like it at first glance, or that an object is very interesting (richness of structure of the absolute Galois group) and in nontrivial ways.
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Regarding question 1:
Formalizing the idea of richness of structure is very difficult. This is not because the question is inherently impossible to answer, but because many of the relevant questions are open problems in logic which do not yet have satisfactory explanations. As just one example, we don't have entirely satisfactory explanations of what the complexity of a proof is, which seems necessary to make the idea of rich structure a fruitful one.
Concretely, let's restrict our attention to set equipped with structure, and then further restrict our attention to purely algebraic structures on a set -- that is, we posit that our set has a collection of operations whose equations are universally quantified equalities on the operations. Now, notice that for any such algebraic structure, the one-element set provides a trivial model.
Of course, it's utterly absurd to say that the one-element set has rich structure because it's a model of all algebraic theories. The reason it's absurd to say such a thing, is because it's a model of all algebraic theories for trivial reasons -- in other words, there's a trivial, generic, proof that the one-element set satisfies all those equalities (namely, all functions into a one-element set are equal, by the extensionality of equality of functions).
In order to rule out these boring counterexamples, we need a concept of the complexity of proofs, so that we have some formal way of saying that an equation is satisfied for an interesting reason. For example, we may wish to say that the natural numbers have nontrivial commutative monoid structure $(\mathbb{N}, 0, +)$, because the proofs of the associativity and commutativity of addition are not trivial (though of course they are very easy -- which is itself another idea that calls for formalization!).
However, even the equality theory of proofs is very difficult, let alone being able to measure their relative complexities. The basic technology here is Gentzen's sequent calculus, and the basic question here is how to deal with lemmas (i.e., the "cuts" of his cut-elimination theorem). Namely, if we take a proof, and factor it into a series of lemmas, is it still the same proof? If we answer the question with a "yes", then we have a theory of proof which says we should compare the normal forms of proofs -- but it's easy to show that even in the propositional case, the presence of lemmas can make proofs doubly-exponentially smaller. It seems extremely weird to say that such a dramatic change makes no difference! If we answer the question with a "no", then some trivial rearrangements can cause us to declare two proofs not the same, which is also strange.
One way of squaring this circle is to suggest that perhaps we should look at the complexity of proof-normalization, and equate two proofs when it only takes a small amount of work to go to a common normal form. However, this suggestion has the problem that we don't know how to do this. Perhaps the closest we can presently come to this is with Jean-Yves Girard's Geometry of Interaction, which gives a fixed point semantics to proof normalization, which can be used to get numerical estimates of the complexity of normalizing individual proofs. However, this is very much current research, and many basic questions about GoI remain open.
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Very interesting note. As regards to complexity of proofs: I check that in Mizar proof assistant, the longest proofs are usually assigned to not-very-interesting or most-known theorems. Of course it s not so simple to check just size of certain proof to say something about general complexity, but in other way it is interesting to see it on certain example. – kakaz Feb 22 2010 at 11:51
At least in the examples you mentioned, richness can be interpreted as the structure of the group of symmetries associated to the object (sub-object graph and permutation group in the examples).
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Then the question goes for the richness-of-structure of the group of symmetries. – Hans Stricker Feb 22 2010 at 0:27
No, I think the claim is "more structure = smaller automorphism group" (which seems pretty reasonable, at least in these cases). – Reid Barton Feb 22 2010 at 1:14
So it's just a matter of the size of the automorphism group? But cannot two groups of the same size have different complexities? – Hans Stricker Feb 22 2010 at 8:40
Clearly it's the partial ordering by inclusion, not the size, that's relevant here. – Scott Morrison♦ Feb 23 2010 at 6:09
It sounds like richness shares with beauty: the beholder's eye determines what is rich. While I will not attempt to give a quantitative answer, I think the following points can be considered.
1. The object may have many properties that are desired for study or application. An example of this is a set system {{1,2,3},{1,4,5},{1,6,7},{2,4,6},{2,5,7},{3,4,7},{3,5,6}} on seven points. It has a lot of symmetry but the structure as a system is not apparent until you give it an interpretation: finite projective plane, complement of a D-optimal design, arrangement of statistical tests. Richness of application may be meant even though richness of structure was stated.
2. The richness may come from using the structure to build more things. The two-element lattice generates the variety of all distributive lattices, and its equational theory is the same as the equational theory of any larger distributive lattice in the same language. So in a proper context the richness may stem from how the structure will be used in creating other structures or in determining properties of related structures.
Granted these are not literal interpretations, but I think they are in mind when someone makes a statement about richness of structure.
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In my opinion the richness of the structure is related to endomorphisms of such set. It i s note necessary the group of symmetry but definitely it has something in common to possible relations within elements of such set. Even if You cannot have symmetry within set, it may have rich structure, for example even if there is no inverse elements or identity, there may be many interesting properties.
Remarks:
If You agree with above, it probably we may go further and try to define the measure for this "richness". For example You may try to construct some functors from given class of endomorphisms over structure to some predefined and known categories for example. Maybe there is possible to find something analogous to class of homotopy etc. in such construction? Probably it would be very interesting thing to define "richness invariants" which would be some constructions staying at some level whilst between similarly rich structures.
Note that it is not that the number ( cardinality) of such endomorphism decide about richness, but rather its "strangeness". For example class of endomorphisms of general function between two complex planes is much bigger that class of conformal mapping of complex planes, but the last class is definitely more interesting than first one. The last one is too big so not very interesting...
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http://mathematica.stackexchange.com/questions/8296/plotting-a-path-in-a-vector-field
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# Plotting a path in a vector field
I am trying to make a visual representation of the curve $y=x^{2}$, with $x\in\left[0,1\right]$ in the vector field defined by $\vec{F}=\left[-y,x\right]$.
However, as far as I can tell there does not seem to be any facility which allows me to do this in Mathematica (I've looked through as much of the documentation as I could find which seemed pertinent).
I know how to do each of these separately, e.g.
````Plot[x^2,{x,0,1}]
VectorPlot[{-y,x},{x,-3,3},{y,-3,3}]
````
However, I am unsure how I can combine these two plots (or plot them both at the same time).
Thanks in advance!
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## 2 Answers
You should make use of `Show` and appropriate options e.g. we added `PlotRange`, `AspectRatio` etc. :
````Show[{ Plot[x^2, {x, -3, 3}, PlotRange -> 3, PlotStyle -> {Thick, Darker @ Green}],
VectorPlot[{-y, x}, {x, -3, 3}, {y, -3, 3}]}, AspectRatio -> 1]
````
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Thank you, this seems to be what I was looking for! +1 – Shaktal Jul 13 '12 at 23:21
Instead of using `Show` you might prefer the paradigm of David Park's Presentations package:
````<< Presentations`
Draw2D[{VectorDraw[{-y, x}, {x, -3, 3}, {y, -3, 3}],
Darker@Green, Draw[x^2, {x, -3, 3}]},
PlotRange->3, Axes->True]
````
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lang-mma
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http://mathoverflow.net/questions/44523/riemann-roch-as-an-index-theorem
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## Riemann-Roch as an index theorem [closed]
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I am sorry to make this a new question. I would have liked to leave a comment, but I suppose I don't have enough rep for that....
So, in the accepted answer to this question I don't understand why in the second term it is $\dim H^0(X,L\otimes \Lambda^{0,1})$ and not $L^{-1}$ with the same.thing. I suspect this is some Hodge theory thing that I am not recognizing, and it is probably completely trivial, but I would appreciate if someone could enlighten me.
Thank you.
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I think you are right. It should be $ind (\partial_L)=dim H^0 (X;L) - dim H^0 (X;L^{-1} \otimes \Lambda^{0,1})$, the second summand is dual to the cokernel of $\partial_L$ by Serre duality. – Johannes Ebert Nov 2 2010 at 8:16
4
I both upvoted, and voted to close. So you would have enough rep to repost this good question as a comment. – Daniel Moskovich Nov 2 2010 at 11:30
Thank you for that. I made the comment where it belonged. Since I am new to MO, I am not sure what I should do now. Should I delete this, or "vote to close"? Or just wait for the moderators to take action? Thanks. – Alfonz Nov 3 2010 at 0:08
I made a comment on the accepted answer to the other question (I also agree with you). I want to give people (especially the answerer) a little time to respond, and then I or someone else can simply edit the answer. @Alfonz: perhaps it is best for you to wait until the answer is corrected (or the situation is otherwise resolved) and then delete this question. – Pete L. Clark Nov 3 2010 at 15:24
@Pete: OK, thanks. – Alfonz Nov 3 2010 at 15:57
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http://math.stackexchange.com/questions/5088/number-of-continuous-0-1-to-0-1-functions-for-given-arc-length
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# Number of continuous $[0; 1] \to [0; 1]$ functions for given arc length
Just out of pure curiosity ...
Suppose I want to connect the two points $(0|0)$ and $(1|1)$ with the graph of some continuous and differentiable function
$$f : [0; 1] \to [0; 1]$$
and let $s$ be the arc length of that function in $[0; 1]$.
Of course, the function with minimum $s$ that satisfies the above conditions is $f(x) = x$ with $s = \sqrt 2$. So for $s = \sqrt 2$, exactly one matching function can be found.
But what happens to the number of these functions if $s$ increases?
Surely, more functions can be found to match the given arc length - uncountably many more I suppose due to the nature of the real numbers.
But intuitively, I'd think that the number of such functions grows even more the greater $s$ gets, since there is more "space" the graph can use.
So, despite continuum cardinality, are there any means of measuring the number of such functions against $s$ or is it all the same once that minimal way of $f(x) = x$ as been taken?
And would this change if we limited the ways of constructing such functions to e.g. some elementary ones?
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1
I think the question would seem less naive if you replaced "number" with "size" or "measure." – Qiaochu Yuan Sep 20 '10 at 20:19
## 2 Answers
There is probably a sophisticated answer to this; let me just mention an unsophisticated one. Several possible discrete analogues of the question you ask exhibit the desired behavior. Consider, for example, the problem of finding a path from $(0, 0)$ to $(n, n)$ in the lattice $\mathbb{Z}^2$ where the only allowable steps are between diagonally adjacent lattice points. Let the length of such a path be the total number of steps. Then, of course, there is only one path of length $n$, which is the obvious diagonal one. However, there are ${n+2 \choose 2} + n$ paths of length $n+2$, already a big leap.
The general answer can be computed as follows. In a path of length $n+2k$ we need to choose $2j$ of the steps to be up-left or down-right (for some $j$ between $0$ and $k$) and we need to choose $n+2k-2j$ of the steps to be up-right or down-left. Of the first set of steps, half need to be of one type and half need to be the other, and of the second set of steps, $n+k-j$ need to be of one type and $k-j$ need to be the other. This gives the total number of paths of length $n+2k$ as
$$\sum_{j=0}^{k} {n+2k \choose 2j} {2j \choose j} {n+2k-2j \choose k-j}.$$
The dominant term when $n$ is large compared to $k$ is the term associated to $j = k$, which is ${n+2k \choose 2k}$ and hence grows asymptotically like $\frac{n^{2k}}{(2k)!}$. So, again when $n$ is large compared to $k$, it is indeed true that the number of allowable paths grows with the length of the path.
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Here's my take on an intuitive answer:
Since there are $2^{\aleph_0}$ continuous functions from the unit interval to itself, it is clear that there can be at most that many functions.
Now, given $s > \sqrt{2}$ (which as you well observed the minimal distance), we define $t = s - \sqrt{2}$ and for any $x\in (0,1)$ we can construct a function that has the wanted arc length, simply by "pulling" the value of $f(x)$ smoothly in a way that extends the length of the arc by $t$.
(The construction is very clear when you don't request $f$ to be differentiable, as you can just stick a triangle in the proper length, however the idea should be clear)
So you have at least $|(0,1)| = 2^{\aleph_0}$ functions as such, which means that you have exactly that many.
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http://mathoverflow.net/questions/123159/when-is-a-substack-closed/123165
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## When is a substack closed?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
For this question we will consider the Zariski site of affine schemes and a stack $\mathcal{M}$ over it. I don't know what a substack is, but I have a guess. The stack $\mathcal{M}$ has an underlying path-component sheaf $\pi_0\mathcal{M}$, and we have a map $\mathcal{M}\rightarrow \pi_0\mathcal{M}$. A stack $\mathcal{N}\rightarrow \mathcal{M}$ is substack if the induced map of sheaves $\pi_0\mathcal{N}\rightarrow\pi_0\mathcal{M}$ is an inclusion, and the resulting square is $2$-categorical pullback. I am not exactly sure this definition is right, but at least the notion is "homotopic". So my first question is:
$1$. Is this a definition of a substack?
I also have a guess of what a closed substack is supposed to be (of course, assuming that the definition of a substack is more or less what I wrote above). We can first define it for sheaves. A map of sheaves $F\rightarrow G$ is a closed immersion of sheaves, if any pullback along a map $\mathrm{Spec}(R)\rightarrow G$ is a map of form $\mathrm{Spec}(R/I)\rightarrow \mathrm{Spec}(R)$. For the case of stacks, we say $\mathcal{N}\rightarrow\mathcal{M}$ is a closed immersion if it is a substack, such that the induced map $\pi_0\mathcal{N}\rightarrow\pi_0\mathcal{M}$ is a closed immersion. My second question:
$2$. Is this the correct definition of a closed substack?
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1
A substack $Y \subset X$ is a stack $Y$ equipped with a map to $X$ such that for every affine $A$, $Y(A) \to X(A)$ is the inclusion of a full, replete subcategory (replete: every object isomorphic to one in the image is in the image). At least for topological stacks, the property of a map being a closed embedding is local on the target and invariant under base change, so we can define a representable map of stacks to be closed if it is so locally. – David Roberts Feb 27 at 23:25
## 2 Answers
No need to guess, just look it up. http://ens.math.univ-montp2.fr/~toen/cours8.pdf Definition 1.1.
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Here is a definition in terms of presentation of the stack: Definition 1.5 in Angelo Vistoli's paper "Intersection theory on algebraic stacks and on their moduli spaces".
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http://math.stackexchange.com/questions/229912/roc-analysis-detection-probability-when-false-alarm-is-maximized
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# ROC analysis: detection probability when false alarm is maximized.
I am wondering if detection probability always goes to 1 as false alarm probability goes to 1.
Let's assume binary hypothesis problem:
$\mathcal{H}_0: x(t) =n(t)$
$\mathcal{H}_1: x(t) = s(t) + n(t)$ where $s(t)$ and $n(t)$ are the desired signal and arbitrary noise respectively.
And we have the two probability density: $p(x|\mathcal{H}_0)$ and $p(x|\mathcal{H}_1)$ with a threshold $\gamma$ where the threshold constrains the false alarm probability.
If the two densities are identical, the ROC curve shows strain line with slop=1.
If $p(x|\mathcal{H}_1)$ has larger variance than $p(x|\mathcal{H}_0)$ and has smaller mean than $p(x|\mathcal{H}_0)$, I thought the detection probability could be less than 1 even though the false alarm probability is 1.
Am I wrong...?
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## 1 Answer
You cannot have a false alarm (false positive) without a detection (positive). So the answer to your initial query (detection probability to one as FA probability to one) is yes.
(edited because the only "question" you wrote was "Am I wrong..." and I didn't want to seem callous).
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http://mathoverflow.net/questions/74359/algorithm-to-compute-certain-poset-from-a-given-poset/74476
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## Algorithm to compute certain poset from a given poset.
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Hi. Associated with a finite poset $P$, one can consider the poset $S(P)$, whose elements are the intervals of $P$, ordered by inclusion. (See http://mathoverflow.net/questions/73640 for some motivation why to look at this).
Does anyone know if there is an algorithm around which, given $P$, computes $S(P)$? I think it is not very difficult to come up with one, but I just want to know if it has already been implemented, say, in some computer algebra system, or studied in the literature.
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Short answer: I don't know. Long answer: There is literature on the poset (lattice?) of ideals and filters in lattices and likely also arbitrary posets. Intervals being an intersection of a filter and an ideal, they may have been considered alongside such structures (possible keyword: convex). You might consider looking at the literature on the lattice of order ideals for lattices or other specific posets. You might be able to extend whatever algorithms they have to your situation. Gerhard "Ask Me About System Design" Paseman, 2011.09.02 – Gerhard Paseman Sep 2 2011 at 16:47
I am not sure whether intervals can always be seen as the intersection you say (according to the definition of the cited post, i.e. something of the form [x,y]). Consider for instance the case where the filter and/or ideal are unions of principal ideals/filters. Then you have the union of two (disjoint) intervals, which is not something of the form [x,y] and is, in particular, not "convex". – Camilo Sarmiento Sep 5 2011 at 12:37
Indeed, your notion of interval may not include the general case of an intersection of a filter and ideal. However, the literature may use convex, or it may use interval, in describing such intersections. If you are cautious in reading such literature, you may find it useful for answering your question. Gerhard "Ask Me About System Design" Paseman, 2011.09.07 – Gerhard Paseman Sep 7 2011 at 19:14
Also, I cannot think of a poset where the intersection of an upset and a downset could be both nonempty and not convex. Your example of two disjoint (using your notion) intervals is convex in the context of a general poset. Of course, I am using what I understand for a notion of convexity; I have not seen your notion. Gerhard "Ask Me About System Design" Paseman, 2011.09.07 – Gerhard Paseman Sep 7 2011 at 19:21
so, what is your definition of "convexity" in this case? – Camilo Sarmiento Sep 9 2011 at 8:43
show 1 more comment
## 2 Answers
Assuming that by "interval" you mean something of the form [x,y] (as in the earlier question you linked to), you could just represent intervals by ordered pairs $(x,y)\in P\times P$ with $x\leq y$. The inclusion relation on intervals is easily expressible in terms of the order relation of $P$ because $[x,y]\subseteq[u,v]$ if and only if $u\leq x\leq y\leq v$.
If, on the other hand, you mean by "interval" an arbitrary order-convex subset of $P$, as Gerhard Paseman's comment seems to suppose, then I agree with his comment, and I would caution you that $S(P)$ could, under this interpretation, be exponentially bigger than $P$.
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I cannot tell if this will help, but mention it just in case. The area of formal concept analysis deals with algorithms for constructing lattices from sets of objects and attributes. A concept is a tuple $C = (O, A, R)$, where $O$ and $A$ are sets and $R$ is a binary relation from $O$ to $A$. The relation gives rise to a standard function $f$ from the powerset of $O$ to the powerset of $A$.
$f$ maps $X \subseteq O$ to $\{ y \in A \mid \text{for all }x \in X, (x,y) \in R \}$
A function $g$ from the powerset of $A$ to powerset of $O$ is similarly defined such that $f$ and $g$ form a Galois connection. The concept lattice consists of the Galois stable subsets of $A$. By choosing the relation $R$, one can generate lattices with various properties. Algorithms for lattice construction are surveyed in:
Algorithms for the Construction of Concept Lattices and Their Diagram Graphs, Kuznetsov, Sergei O.; Obiedkov, Sergei A. (2001)
The Formal Concept Analysis site contains links to relevant material and software.
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http://math.stackexchange.com/questions/tagged/galois-representations
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The galois-representations tag has no wiki summary.
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http://medlibrary.org/medwiki/Continuous_group
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# Continuous group
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Group theory
• Integers (Z)
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In mathematics, a topological group is a group G together with a topology on G such that the group's binary operation and the group's inverse function are continuous functions with respect to the topology.[1] A topological group is a mathematical object with both an algebraic structure and a topological structure. Thus, one may perform algebraic operations, because of the group structure, and one may talk about continuous functions, because of the topology.
Topological groups, along with continuous group actions, are used to study continuous symmetries, which have many applications, for example in physics.
## Formal definition
A topological group G is a topological space and group such that the group operations of product:
$G\times G \to G : (x,y)\mapsto xy$
and taking inverses:
$G\to G : x \mapsto x^{-1}$
are continuous functions. Here, G × G is viewed as a topological space by using the product topology.
Although not part of this definition, many authors[2] require that the topology on G be Hausdorff; this corresponds to the identity map $* \to G$ being a closed inclusion (hence also a cofibration). The reasons, and some equivalent conditions, are discussed below. In the end, this is not a serious restriction—any topological group can be made Hausdorff in a canonical fashion .
In the language of category theory, topological groups can be defined concisely as group objects in the category of topological spaces, in the same way that ordinary groups are group objects in the category of sets. Note that the axioms are given in terms of the maps (binary product, unary inverse, and nullary identity), hence are categorical definitions. Adding the further requirement of Hausdorff (and cofibration) corresponds to refining to a model category.
### Homomorphisms
A homomorphism between two topological groups G and H is just a continuous group homomorphism G $\to$ H. An isomorphism of topological groups is a group isomorphism which is also a homeomorphism of the underlying topological spaces. This is stronger than simply requiring a continuous group isomorphism—the inverse must also be continuous. There are examples of topological groups which are isomorphic as ordinary groups but not as topological groups. Indeed, any nondiscrete topological group is also a topological group when considered with the discrete topology. The underlying groups are the same, but as topological groups there is not an isomorphism.
Topological groups, together with their homomorphisms, form a category.
## Examples
Every group can be trivially made into a topological group by considering it with the discrete topology; such groups are called discrete groups. In this sense, the theory of topological groups subsumes that of ordinary groups.
The real numbers R, together with addition as operation and its usual topology, form a topological group. More generally, Euclidean n-space Rn with addition and standard topology is a topological group. More generally yet, the additive groups of all topological vector spaces, such as Banach spaces or Hilbert spaces, are topological groups.
The above examples are all abelian. Examples of non-abelian topological groups are given by the classical groups. For instance, the general linear group GL(n,R) of all invertible n-by-n matrices with real entries can be viewed as a topological group with the topology defined by viewing GL(n,R) as a subset of Euclidean space Rn×n.
An example of a topological group which is not a Lie group is given by the rational numbers Q with the topology inherited from R. This is a countable space and it does not have the discrete topology. For a nonabelian example, consider the subgroup of rotations of R3 generated by two rotations by irrational multiples of 2π about different axes.
In every Banach algebra with multiplicative identity, the set of invertible elements forms a topological group under multiplication.
## Properties
The algebraic and topological structures of a topological group interact in non-trivial ways. For example, in any topological group the identity component (i.e. the connected component containing the identity element) is a closed normal subgroup. This is because if C is the identity component, a*C is the component of G (the group) containing a. In fact, the collection of all left cosets (or right cosets) of C in G is equal to the collection of all components of G. Therefore, the quotient topology induced by the quotient map from G to G/C is totally disconnected.[3]
The inversion operation on a topological group G is a homeomorphism from G to itself. Likewise, if a is any element of G, then left or right multiplication by a yields a homeomorphism G → G.
Every topological group can be viewed as a uniform space in two ways; the left uniformity turns all left multiplications into uniformly continuous maps while the right uniformity turns all right multiplications into uniformly continuous maps. If G is not abelian, then these two need not coincide. The uniform structures allow one to talk about notions such as completeness, uniform continuity and uniform convergence on topological groups.
As a uniform space, every topological group is completely regular. It follows that if a topological group is T0 (Kolmogorov) then it is already T2 (Hausdorff), even T3½ (Tychonoff).
Every subgroup of a topological group is itself a topological group when given the subspace topology. If H is a subgroup of G, the set of left or right cosets G/H is a topological space when given the quotient topology (the finest topology on G/H which makes the natural projection q : G → G/H continuous). One can show that the quotient map q : G → G/H is always open.
Every open subgroup H is also closed, since the complement of H is the open set given by the union of open sets gH for g in G \ H.
If H is a normal subgroup of G, then the factor group, G/H becomes a topological group when given the quotient topology. However, if H is not closed in the topology of G, then G/H will not be T0 even if G is. It is therefore natural to restrict oneself to the category of T0 topological groups, and restrict the definition of normal to normal and closed.
The isomorphism theorems known from ordinary group theory are not always true in the topological setting. This is because a bijective homomorphism need not be an isomorphism of topological groups. The theorems are valid if one places certain restrictions on the maps involved. For example, the first isomorphism theorem states that if f : G → H is a homomorphism then G/ker(f) is isomorphic to im(f) if and only if the map f is open onto its image.
If H is a subgroup of G then the closure of H is also a subgroup. Likewise, if H is a normal subgroup, the closure of H is normal.
A topological group G is Hausdorff if and only if the trivial one-element subgroup is closed in G. If G is not Hausdorff then one can obtain a Hausdorff group by passing to the quotient space G/K where K is the closure of the identity. This is equivalent to taking the Kolmogorov quotient of G.
The fundamental group of a topological group is always abelian. This is a special case of the fact that the fundamental group of an H-space is abelian, since topological groups are H-spaces.
## Relationship to other areas of mathematics
Of particular importance in harmonic analysis are the locally compact groups, because they admit a natural notion of measure and integral, given by the Haar measure. The theory of group representations is almost identical for finite groups and for compact topological groups. In general, σ-compact Baire topological groups are locally compact.
## Generalizations
Various generalizations of topological groups can be obtained by weakening the continuity conditions:[4]
• A semitopological group is a group G with a topology such that for each c in G the two functions G → G defined by $x\mapsto xc$ and $x\mapsto cx$ are continuous.
• A quasitopological group is a semitopological group in which the function mapping elements to their inverses is also continuous.
• A paratopological group is a group with a topology such that the group operation is continuous.
## Notes
1. Waerden, Bartel Leendert et al (2003). "Topological algebra". Algebra. Vol. 2. Springer. p. 256. ISBN 978-0-387-40625-1 [Amazon-US | Amazon-UK].
2. Armstrong, p. 73; Bredon, p. 51; Willard, p. 91.
3. O.V. Mel'nikov (2001), "Topological group", in Hazewinkel, Michiel, , Springer, ISBN 978-1-55608-010-4 [Amazon-US | Amazon-UK]
4. Arhangel'skii & Tkachenko, p12
## References
• Arhangel'skii, Alexander; Tkachenko, Mikhail (2008). Topological Groups and Related Structures. Atlantis Press. ISBN 90-78677-06-6 [Amazon-US | Amazon-UK].
• Armstrong, M. A. (1997). Basic Topology (1st ed.). Springer Verlag. ISBN 0-387-90839-0 [Amazon-US | Amazon-UK].
• Bredon, Glen E. (1997). Topology and Geometry. Graduate Texts in Mathematics (1 ed.). Springer. ISBN 0-387-97926-3 [Amazon-US | Amazon-UK].
• Husain, Taqdir (1981). Introduction to Topological Groups. Philadelphia: R.E. Krieger Pub. Co. ISBN 0-89874-193-9 [Amazon-US | Amazon-UK].
• Pontryagin, Lev S. (1986). Topological Groups. trans. from Russian by Arlen Brown and P.S.V. Naidu (3rd ed.). New York: Gordon and Breach Science Publishers. ISBN 2-88124-133-6 [Amazon-US | Amazon-UK].
• Porteous, I.R. (1969). Topological Geometry. Van Nostrand Reinhold. pp. 336–352. ISBN 0-442-06606-6 [Amazon-US | Amazon-UK]. Zbl 0186.06304.
• Willard, Stephen (2004). General Topology. Dover Publications. ISBN 0-486-43479-6 [Amazon-US | Amazon-UK].
Content in this section is authored by an open community of volunteers and is not produced by, reviewed by, or in any way affiliated with MedLibrary.org. Licensed under the Creative Commons Attribution-ShareAlike 3.0 Unported License, using material from the Wikipedia article on "Continuous group", available in its original form here:
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