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August 22

Correlation for unordered sets

When I read the definition of correlation, it boils down to "as one attribute rises, the other rises. As one attribute falls, the other falls." That makes sense for two ordered sets of values. What if you have unordered sets. For example, I have 5 people and I want to know if weight is correlated to nostril radius. So, I have five weights like [ 150, 127, 210, 108, 250 ] and five nostril radii [ 2.4, 2.2, 3.0, 1.9, 2.7 ]. I can arrange those attribute pairs however I like. So, it isn't that one is rising or falling. They are independent values. If I say they are "correlated", what do I really mean? Is it called something other than "correlated" because they are not ordered? Is the basic calculation, which I often see named "Person's Correlation" the same? I hope that it is clear that I am asking for a layman's term for correlation to use when data is not rising or falling, assuming that correlation is a proper measure to use for this data. 75.136.148.8 (talk) 12:22, 22 August 2024 (UTC)

For correlation, you need to know which values in each set belong together - ie there are five individuals with the weights and nostril radii given, so each weight is linked to a specific nostril radius. Then you can put the weights in rising order, and if the corresponding nostril radii in that order also rise, then you have a correlation. If the radii in that order actually fall, then you have an inverse correlation. If the radii show no pattern, there is no correlation. -- Verbarson  ^talk_edits 12:54, 22 August 2024 (UTC)

Alternatively, plot weight as x-value and radius as y-value on a graph, giving five points. If they look like a diagonal line, there is a correlation of some sort. (Vertical or horizontal lines would only occur if all the values in one set were roughly identical, so no correlation.) -- Verbarson  ^talk_edits 12:58, 22 August 2024 (UTC)

So, what you are saying, is that it doesn't matter what the order of the values in the sets are, the correlation will be the same, correct? If that is true, I am having trouble with the definition "as one rises, the other rises" because I can rearrange them however I like to disrupt rising and falling data points. 75.136.148.8 (talk) 10:30, 23 August 2024 (UTC)

You have five people. Each person has a weight and a nostril radius. If you want, you can order the people from smallest to largest weight (which induces a particular order on the nostril radii). If you want, you can order the people from largest to smallest nostril radius (which induces an order on the weights). But no matter what, each weight always corresponds to the same nostril radius (unless you're doing some pretty serious surgical intervention), so what you really have is a list of five ordered pairs (weight, nostril radius). Correlation does not care about which of these points you label first, second, third, fourth, fifth; but it would care a lot if you changed the correspondence between the two coordinates of the pair. 100.36.106.199 (talk) 10:37, 23 August 2024 (UTC)

August 23

Perpendicular projection of Fourier rotating epicircles

Approximation of square wave

The animation shows how a Fourier series approximates a square wave.

I'm curious if instead of projecting the point horizontally, we project it vertically (or any other angle), what waveform do we get?

(I know all the terms will have cosine instead of sine, but is their sum meaningful?)

Thanks, cmɢʟee⎆τaʟκ 23:52, 23 August 2024 (UTC)

So basically, factoring out the constant, instead of

$\sum _{\overset {k=1}{k\ {\rm {odd}}}}^{\infty }{\frac {\sin kt}{k}}$

you want

$\sum _{\overset {k=1}{k\ {\rm {odd}}}}^{\infty }{\frac {\cos kt}{k}}.$

Notice that the series diverges at multiples of π. The graph zig-zags between positive and negative infinity, with asymptotes toward infinity at the even multiples of π, and asymptotes toward negative infinity at the odd multiples. I think it's possible to work out a closed formula, but I think I'll leave that as an "available upon request" kind of thing, since it will take time to work out and the result may not mean much to you. (Unless someone else wants to work it out.) See this Alpha result for a quick and dirty sketch. (Does anyone know how to get Alpha to find the closed formula for the infinite sum? I tried a few times but kept getting "could not determine the general term.") --RDBury (talk) 04:24, 24 August 2024 (UTC)

I did

$\sum _{k=0}^{\infty }{\frac {\cos((2k+1)t)}{2k+1}}$

But Alpha only gave me a partial sum formula with complicated Lerch transcendent terms. Under the assumption that those terms tend to $0$, it would seem that the function is explicitly equal to

${\frac {\tanh ^{-1}(e^{-ix})+\tanh ^{-1}(e^{ix})}{2}}$

GalacticShoe (talk) 07:58, 24 August 2024 (UTC)

Many thanks, @GalacticShoe and @RDBury. Good use of WolframAlpha. Guess the result isn't very interesting. Cheers, cmɢʟee⎆τaʟκ 09:44, 24 August 2024 (UTC)

I think that what Wolfram Alpha is telling you is trivial: Let $f_{N}(x)=\sum _{k=0}^{N}{\frac {\cos((2k+1)x)}{2k+1}}=\Re \left(\sum _{k=0}^{N}{\frac {e^{(2k+1)xi}}{2k+1}}\right).$ Then $f_{N}'(x)=\Re \left(i\cdot \sum _{k=0}^{N}e^{(2k+1)xi}\right)=-\Im \left({\frac {e^{xi}-e^{(2N+3)xi}}{1-e^{2xi}}}\right).$ Then do a bit of algebra and take the antiderivative. Basically all of the work is the final line (showing that the sum actually converges). 100.36.106.199 (talk) 15:08, 24 August 2024 (UTC)

Worth noting that this further simplifies down to

${\frac {1}{4}}\ln({\frac {1+\cos(x)}{1-\cos(x)}})$

Or

${\frac {1}{2}}\tanh ^{-1}(\cos(x))$

GalacticShoe (talk) 15:40, 24 August 2024 (UTC)

Nice closed formula. It doesn't seem to have much interesting application, unlike just the first term giving cosine. Thanks, cmɢʟee⎆τaʟκ 22:01, 24 August 2024 (UTC)

August 30

How does solving the finite’s fields discrete logarithm is easier on an extension field than with a prime degree ?

Hi,

simple question, I’m seeing discrete logarithms records are higher when the finite’s field degree is composite and that such degrees are expressed as the degree of prime and the composite part being the extension of the field.
But how does that makes solving the discrete logarithm easier ? Is it only something that apply to index calculus methods like ꜰꜰꜱ or xɴꜰꜱ ? 2A01:E0A:401:A7C0:6861:5696:FAEB:61D1 (talk) 19:14, 27 August 2024 (UTC)

I believe the function field sieve has much better asymptotic complexity for large powers of primes than other methods. Not sure about compositeness of degrees. Tito Omburo (talk) 20:36, 27 August 2024 (UTC)

I’m also seeing it applies to variant of the ɴꜰꜱ. The paper about 2⁸⁰⁹ discrete logarithm record told the fact 809 was a prime power was a key difficulty. And indeed, all the larger records happened on extension fields (with a lower base prime exponent than 809)

The problem is I don’t understand how it’s achieved to make it little easier. 2A01:E0A:401:A7C0:6861:5696:FAEB:61D1 (talk) 05:02, 28 August 2024 (UTC)

Projections of hypercubes

Looking at Hypercube#Graphs, it looks like the projection of the n-cube into the B_n Coxeter plane has a no central vertex exactly when n can be written as 2^m for some positive integer m. The pictures confirm this is true for 1 ≤ n ≤ 15.

So:

1. Is this true in general?
2. What's the general term of the sequence (a_n), where a_n is the number of vertices projected to the centre (i.e. 0, 0, 2, 0, 2, 4, 2, 0, ...) ?

Double sharp (talk) 18:21, 30 August 2024 (UTC)

Did you intend to include a "not" in the question? I get no central vertex for n=2, 4, 8. For n=9 I get 8 points projected to the center. --RDBury (talk) 23:02, 30 August 2024 (UTC)

Um, yes. Oops. T_T Double sharp (talk) 03:54, 31 August 2024 (UTC)

For those who want to play along at home, I'm pretty sure the projection in question, translated to R², is given by the matrix with columns $(1,\cos {\tfrac {\pi }{n}},\cos {\tfrac {2\pi }{n}},\cdots \cos {\tfrac {(n-1\pi }{n}})$ and $(0,\sin {\tfrac {\pi }{n}},\sin {\tfrac {2\pi }{n}},\dots \sin {\tfrac {(n-1\pi }{n}}).$ There may be a scaling factor involved if you're picky about distance being preserved, but this is irrelevant for the question. It's not too hard to show that these vectors are orthogonal and have the same length. So the question becomes, given n, how many combinations of $\pm (1,0)\pm (\cos {\tfrac {\pi }{n}},\sin {\tfrac {\pi }{n}})\pm (\cos {\tfrac {2\pi }{n}},\sin {\tfrac {2\pi }{n}})\dots \pm (\cos {\tfrac {(n-1)\pi }{n}},\sin {\tfrac {(n-1)\pi }{n}})$ add to $(0,0)?$ These vectors form half of the points on a regular 2n-gon. It's not hard to see that there are at least two ways of doing it if n is odd; just alternate signs. A similar sign alternating idea shows that the number must be at least 2^mp if n=mp where p is odd. So if n has an odd factor then there are points which project to 0. Proving the converse seems tricky though. --RDBury (talk) 00:04, 31 August 2024 (UTC)

PS. I think I have an argument for the converse. The n points on the circle are all of the from ρ^k where ρ = e^πi/n. We need to find a combination of these powers of ρ, which amounts to a polynomial in p of degree n-1, where all coefficients are ±1. If n is odd then ρ satisfies (ρⁿ+1)/(ρ+1) = 1 - ρ + ρ² - ... + ρ^n-1 = 0, and this polynomial has the desired properties. If n has an odd factor, say n=pq with p odd, then p satisfies (ρⁿ+1)/(ρ^q+1) = 1 - ρ^q + ρ^2q - ... + ρ^n-q = 0. Multiply by any polynomial of the form 1 ± ρ ± ρ² - ... + ρ^q-1 to get a polynomial with the desired properties. But if n is a power of 2 then the minimum polynomial for ρ is ρⁿ+1=0. The degree n is greater than n-1, so no integer combination of the powers of ρ from 1 to n-1 can add to 0 except when all the coefficients are 0. In other words, the condition that the coefficients are all ±1 isn't needed; we only need that they are not all 0, FWIW, it appears that the number of vertices projecting to the center is given by OEIS: A182256. It's a lower bound in any case. --RDBury (talk) 00:44, 31 August 2024 (UTC)

That's nice indeed! So it was really a question about roots of unity, after all. :) Double sharp (talk) 04:00, 31 August 2024 (UTC)

August 31

Term for prime to a power?

For the prime factorization of n: $n=p_{1}^{n_{1}}p_{2}^{n_{2}}\cdots p_{k}^{n_{k}}$

is there a term for an individual $p_{i}^{n_{i}}$? Bubba73 ^{You talkin' to me?} 04:17, 31 August 2024 (UTC)

Prime power. AndrewWTaylor (talk) 16:30, 31 August 2024 (UTC)

I thought of that but it isn't specific enough. What I'm looking for is for the largest power of the prime that divides the number. Bubba73 ^{You talkin' to me?} 20:26, 31 August 2024 (UTC)

It would be nice if there was a settled answer to this question. "Primary factor" would be appropriate in commutative ring theory. (Primary ideal) But this usage is not standard in this situation. Tito Omburo (talk) 21:44, 31 August 2024 (UTC)

The exponent of p in the prime factorization is called the p-adic valuation of n. 100.36.106.199 (talk) 01:49, 1 September 2024 (UTC)

Thanks. Bubba73 ^{You talkin' to me?} 02:28, 2 September 2024 (UTC)

A bit of a mouthful: "maximal prime-power factor".^[1][2][3]  --Lambiam 21:56, 1 September 2024 (UTC)

Thanks. That is a bit of a mouthful (i.e. too long to use repeatedly). In my mind, and in notes to myself, for years I have called it the "prime component". Soon I expect to be writing up something, so I wondering if there is a recognized term. Bubba73 ^{You talkin' to me?} 02:28, 2 September 2024 (UTC)