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February 13

If no question is asked then any answer written here is correct. Count Iblis (talk) 12:32, 18 February 2014 (UTC)

And you didn't even link to vacuous truth! —Quondum 15:26, 18 February 2014 (UTC)

Thanks, I didn't know that this is called that way (English is not my mother tongue) :). You also encounter this when taking the intersection of a collection of sets. If the collection of the sets is empty then the intersection is the whole space. In some statements in topology or measure theory you then don't have to explicitly address the exceptional cases when a collection of sets is empty. Count Iblis (talk) 15:59, 18 February 2014 (UTC)

February 23

Linking Geometry to Combinatorics

• Is there a branch of mathematics that (systematically) connects these two fields with one another ?
• Are there any books, papers, articles (both online or offline) that treat this connection systematically ?

For instance, we know that geometric shapes of the form $x^{n}+y^{m}=r^{n}$ are connected with binomial coefficients, inasmuch as the former's area is nothing else than the latter's reciprocal, or multiplicative inverse. (Wallis' integrals also come here to mind). Likewise, Vandermonde's identity $\sum _{0}^{n}{n \choose k}^{2}={2n \choose n}$ gives, for $n={\tfrac {1}{2}},$ a formula for π : $\sum _{0}^{\infty }\left[{\frac {(2k-3)!!}{(2k)!!}}\right]^{2}={\frac {4}{\pi }}.$

• Are there any other such similar results, connecting these two fields together ? — 79.113.198.225 (talk) 16:58, 23 February 2014 (UTC)

Your examples are more like connections between calculus and combinatorics, so it's not clear exactly what you're trying to get. There's a whole field of discrete geometry which explores combinatorial aspects of geometry/geometrical aspects of combinatorics. --RDBury (talk) 11:59, 24 February 2014 (UTC)

Do you also have any suggestions concerning possible connections with calculus ? I was thinking, maybe generating functions ? — 79.114.140.233 (talk) 00:13, 25 February 2014 (UTC)

In addition to discrete geometry, there is also geometric combinatorics. In analysis, people have looked at analytic properties of generating functions in combinatorics, .e.g, Analytic Combinatorics—A Calculus of Discrete Structures. Analytic number theory also has a number of connections to combinatorics. --Mark viking (talk) 01:22, 25 February 2014 (UTC)

Equations

Through what process do these two equations...

${\begin{aligned}N\cdot \!\,sin\alpha \ &=m{\frac {v^{2}}{r}}\ \\N\cdot \!\,cos\alpha \ &=m\cdot \!\,g\\\end{aligned}}$

...become this:

${\begin{aligned}{\frac {N\cdot \!\,sin\alpha }{N\cdot \!\,cos\alpha }}\ &={\frac {m{\frac {v^{2}}{r}}}{m\cdot \!\,g}}\ \\\end{aligned}}$

Th4n3r (talk) 20:49, 23 February 2014 (UTC)

Er, division? So long as $N\cos \alpha \neq 0$… --Tardis (talk) 21:55, 23 February 2014 (UTC)

To elaborate slightly, either (A) The two sides of eq. 1 are equal, so if each side is divided by the same quantity, then they remain equal - but the two sides of eq. 2 are equal (i.e. are the same quantity), so it's a valid move to divide side 1 of eq. 1 by side 1 of eq. 2 and side 2 of eq. 1 by side 2 of eq. 2 - or alternatively (B) Divide both sides of eq. 1 by N cos(α), then substitute for N cos(α) on the right hand side only using eq. 2. --catslash (talk) 23:01, 23 February 2014 (UTC)

Thank you. Th4n3r (talk) 23:23, 23 February 2014 (UTC)

I find it helps to rephrase the equations as A=B, C=D, and then A/C=B/D. HTH, 164.11.203.58 (talk) 08:31, 24 February 2014 (UTC)

You can do the same with all sorts of operations, for instance squaring and adding you get:

$\left(N\cdot \!\,sin\alpha \ \right)^{2}+\left(N\cdot \!\,cos\alpha \ \right)^{2}=\left(m{\frac {v^{2}}{r}}\right)^{2}+\left(m\cdot \!\,g\right)^{2}$

which simplifying and taking the square root becomes:

$N=m\cdot {\sqrt {\left({\frac {v^{2}}{r}}\right)^{2}+g^{2}}}$

which eliminates the dependence on $\alpha$. Dmcq (talk) 08:54, 24 February 2014 (UTC)

February 24

Question on what the name of a step is

Simply, I have forgotten the name of what the mathematical steps of solving a problem are; for example:

4(x+4) = 6(x+5)-x
4x+16 = 6x+30-x
4x+16 = 5x+30
16 = x + 30
-14 = x

What is the term used to describe the solving of "6(x+5)" to "6x+30", as well as the canceling on each side of the equation (i.e. "4x+16 = 5x+30", subtracting 4x from 5x and subtracting 30 from 16 to get "-14 = x")? Thanks :) -- 140.202.10.134 (talk) 19:50, 24 February 2014 (UTC)

"6(x+5)" to "6x+30" is the distribution of multiplication over addition (i.e. using the distributive property). RJFJR (talk) 20:52, 24 February 2014 (UTC)

Is this for an assignment, in which case you would want to refer to your book or teacher. If not, when you are subtracting 4x from 5x, you are really subtracting 4x from each side of the equality, you can do this because equal numbers are identical; so you could say this is justified by the identity of equals, or something to that effect - you could also say that you are using the cancellative property of addition, if you wanted to. Why are you looking for the names? If you are looking for some specific idea/rule that you have forgotten, and need, we may be able to help you given more context. :-) Phoenixia1177 (talk) 08:53, 25 February 2014 (UTC)

Animated GIFs of complex 4D rotations?

Our article tesseract has wonderful animated gifs of rotations of these simple 4D objects. Do we have any more complex gifs, with more or different objects, or are any available on the web? Thanks. μηδείς (talk) 21:08, 24 February 2014 (UTC)

If you look at Category:Featured animations, there are other nice animations at WP. --Mark viking (talk) 21:24, 24 February 2014 (UTC)

A quick google of tasseract gif or 4d gif shows a huge amount of gifs. I am sure that you have already done such a search, so I was hoping that you might clarify what more you are asking for. They are wonderful, aren't they? I especially like this 4D image. http://www.moillusions.com/wp-content/uploads/2009/11/greatestgifever.gif . Please delete my comment if it is irrelevant to what you are looking for. I will not mind, since I am not sure what you are after. DanielDemaret (talk) 21:32, 24 February 2014 (UTC)

Commons has commons:Category:Animations_of_polyhedra with a lot of similar models. This is a sub-sub category to commons:Category:Mathematical animations with many other animations. --Salix alba (talk): 22:43, 24 February 2014 (UTC)

If you look at the bottom of my home page, there's a few I created. (One is 3D, and requires red/blue 3D glasses.) StuRat (talk) 02:15, 25 February 2014 (UTC)

Just to confirm your intuition, an animation of the 4D sphere is just as boring as an animation of a rotating 3D sphere [1]. You can get animations for any of the Convex regular polychoron by searching google with their proper names (though I think we have animations for all of them.) SemanticMantis (talk) 21:31, 26 February 2014 (UTC)

• Thanks for the answers so far. I guess what I am really looking for is something like a Penrose tiling of the plane in a rotating 4D analog, with pentagonal faces being dodecagons, or the like. A sea of boiling hyper-dodecahedra. The 4D torus was excellent, hadn't seen that before. Even two rotating tesseracts that touch at a face would be excellent. μηδείς (talk) 21:34, 26 February 2014 (UTC)

February 25

Pearson's chi square for null values in the "expected" range

I wanted to use MS Excel to calculate p-values using Pearson's (new and improved!) Chi square formula. The problem I've run into is that some of the values in my "expected" range are zero. Excel therefore gives the error #DIV/0. How can I proceed? Do I need to find an alternative statistical test or could I just put very low values (e.g 0.001). In case that was incoherent, I made an example to see if the red-blue berry ratio was significantly different in my garden compared to my friend's garden; the relevant cells are highlighted. — Preceding unsigned comment added by 129.215.47.59 (talk) 12:00, 25 February 2014 (UTC)

Actually, nevermind. There are a load of promising results after searching Google for "pearson chi square zero". 129.215.47.59 (talk) 12:04, 25 February 2014 (UTC)

February 26

Fourier transform

What kind of restriction does the requirement $|{\hat {f}}(\omega )|=|{\hat {f}}(-\omega )|$ place on the time domain function $f(t)\in \mathbb {C}$.

Is there an easy interpretation?

I can see that both even and odd $f(t)$ satisfy this but I think there is a more general restriction. — Preceding unsigned comment added by 128.40.61.82 (talk) 11:17, 26 February 2014 (UTC)

I'm not sure very much can be got, even with a real function one could swap the sign at random every time it passed zero. Dmcq (talk) 12:59, 26 February 2014 (UTC)

Don't take my word for it, but I think that this translates into the autocorrelation being a real function. —Quondum 14:42, 26 February 2014 (UTC)

For complex functions the absolute values being equal means a value for −ω is the +ω one multiplied by any $e^{i\theta (\omega )}$. Dmcq (talk) 15:57, 26 February 2014 (UTC)

I'm not sure what point you're trying to make. Sum them for a given |ω|, which means that each frequency component is "plane polarized" with arbitrary complex argument (no circular component, graphed as a complex number against time). This has the effect that when multiplied by a time-shifted version of the complex complement, the result is real. Integrate over time. Repeat for every time offset, and you have the definition of autocorrelation of complex functions. Hence my conclusion. —Quondum 16:49, 26 February 2014 (UTC)

Sorry yes I believe you're right. That's quite interesting. Dmcq (talk) 17:42, 26 February 2014 (UTC)

Um, no the cross-products shouldn't be formed but you get them from the sums being multiplied together. Dmcq (talk) 08:24, 27 February 2014 (UTC)

I don't follow you. (Aside: I left out mention of how products of distinct frequencies interact, but the time integral of these cross-frequency products is zero and does not affect the result.) I'll check my conclusion as originally stated using an FFT on constrained random data and let you know. —Quondum 15:07, 27 February 2014 (UTC)

I don't think it works, e.g., with a fundamental solution of the Schrodinger equation. Sławomir Biały (talk) 15:21, 27 February 2014 (UTC)

I've verified my original conjecture with an FFT (generate random complex values in the frequency domain, then adjust only magnitude to match that of negative frequency, do inverse FFT, do autocorrelation). The autocorrelation has to be done in a cyclic fashion, but the imaginary part of the autocorrelation is zero to the precision of the internal representation. —Quondum 16:55, 27 February 2014 (UTC)

@Dmcq and Sławomir Biały: C'mon, guys, you can't express doubt and then evaporate. The Wiener–Khinchin theorem (though that article seems a mess, so see Autocorrelation#Properties) says essentially of the autocorrelation that

$R_{f}(\tau )=\int _{-\infty }^{+\infty }f(t)f^{*}(t-\tau )\,dt={\frac {1}{2\pi }}\int _{-\infty }^{+\infty }|{\hat {f}}(\omega )|^{2}e^{i\omega \tau }\,d\omega .$

Since it is given that $|{\hat {f}}(\omega )|$ is a symmetric function, the conclusion that $R_{f}(\tau )$ is real follows pretty directly. —Quondum 05:45, 28 February 2014 (UTC)

Sorry, you're quite right I believe. I had to do something else and it isn't something I'm altogether well informed about. It is still surprising I think. Dmcq (talk) 11:09, 28 February 2014 (UTC)

Neither am I (undergraduate days, too long ago), and my reaction is the same. And since I derived it afresh, I guess I felt I needed confirmation that I was correct, though as is often the case, this is merely a special case of a more general theorem. In hindsight, this looks like it could have been a homework question for someone who'd just covered the theorem in class. —Quondum 14:53, 28 February 2014 (UTC)

February 28

Conservative field in high dimension

Given a bounded, continuously differentiable vector field on a bounded simply connected domain in $\mathbb {R} ^{n}$ (where n>3); How can I prove that this field is conservative? Differentiating the field gives the Hessian matrix of the potential (if it exists). Clearly it is a necessary condition that the Hessian is symmetric, and for n ≤ 3 a symmetric Hessian is equivalent to zero curl, so in that case it is also a sufficient condition. However, curl is not defined for n>3, and I can't find any citeable reference that will tell me if a symmetric derivative matrix implies a conservative field. Any help will be appreciated. PeR (talk) 09:07, 28 February 2014 (UTC)

This can be formulated in the language of differential forms. Instead of your vector field $F=(F_{1},\dots ,F_{n})$, you look at the 1-form $v_{F}=F_{1}dx_{1}+F_{2}dx_{2}+\dots +F_{n}dx_{n}$. By Poincaré's lemma, if F is defined on all of $\mathbb {R} ^{n}$, $v_{F}$ is exact if and only if it is closed. Closed means $dv_{F}=0$, which is (look at exterior derivative) another way of writing $\partial _{x_{i}}F_{j}=\partial _{x_{j}}F_{i}$. Exact means there is a 0-form (which is the same as a function) U such that $v_{F}=dU$, which is the same as $F=\nabla U$.

So a "symmetric derivative matrix" implies that the field is conservative, provided the domain on which it is defined is contractible (although simply connected is enough).

I hope that was not too confusing. There are elementary treatments of the fact (and they are not difficult), but all citable references I have on my desk at the moment are in German. For example, H. Heuser, Lehrbuch der Analysis 2, 8th edition, Stuttgart: Teubner 1993, Satz 182.2. Basically, the proof is the same as for two or three dimensions. —Kusma (t·c) 09:53, 28 February 2014 (UTC)

Great! Thank you so much for the explanation. PeR (talk) 08:09, 1 March 2014 (UTC)

Sum of Reciprocal (prime) powers.

Let D be the set of integers > 1. Let E be the set of Prime Numbers (2,3,5,7,11...).

• W = (sum over s in D, sum over t in D, (1/(s^t))
• X = (sum over s in D, sum over t in E, (1/(s^t))
• Y = (sum over s in E, sum over t in D, (1/(s^t))
• Z = (sum over s in E, sum over t in E, (1/(s^t))

Any ideas on how to prove any of these being finite or infinite. Note that W>Y>Z and W>X>Z.Naraht (talk) 19:14, 28 February 2014 (UTC)

Using the sum of a geometric series,

$W=\sum _{n\geq 2}\sum _{k\geq 2}{\frac {1}{n^{k}}}=\sum _{n\geq 2}{\frac {1}{n^{2}(1-{\frac {1}{n}})}},$

and that sum converges. So all of your sums converge. —Kusma (t·c) 20:06, 28 February 2014 (UTC)