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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⎆τたうaʟκかっぱ 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ぱい. The graph zig-zags between positive and negative infinity, with asymptotes toward infinity at the even multiples of πぱい, 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⎆τたうaʟκかっぱ 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⎆τたうaʟκかっぱ 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ろー^k where ρろー = e^πぱいi/n. We need to find a combination of these powers of ρろー, which amounts to a polynomial in p of degree n-1, where all coefficients are ±1. If n is odd then ρろー satisfies (ρろーⁿ+1)/(ρろー+1) = 1 - ρろー + ρろー² - ... + ρろー^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 (ρろーⁿ+1)/(ρろー^q+1) = 1 - ρろー^q + ρろー^2q - ... + ρろー^n-q = 0. Multiply by any polynomial of the form 1 ± ρろー ± ρろー² - ... + ρろー^q-1 to get a polynomial with the desired properties. But if n is a power of 2 then the minimum polynomial for ρろー is ρろーⁿ+1=0. The degree n is greater than n-1, so no integer combination of the powers of ρろー 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)

The prime component? $3240$ has three mpp factors: $2^{3},3^{4}$ and $5^{1}.$  --Lambiam 12:58, 2 September 2024 (UTC)

I'd call each of those a prime component. Bubba73 ^{You talkin' to me?} 21:09, 2 September 2024 (UTC)

We hereby authorise you to name it the "Bubba factor". -- Jack of Oz ^{[pleasantries]} 20:02, 2 September 2024 (UTC)

Ha ha. Bubba73 ^{You talkin' to me?} 21:09, 2 September 2024 (UTC)

September 2

Coin flip

This has to be a really dumb question but it seems slightly paradoxical. Say you are going to bet $1 on a coin flip. One way to look at this is you plunk down your $1, the coin is flipped, and if heads you get back $2 (your original bet plus $1 winnings), net result +1. If tails, you lose your $1, net result -1. So the expected value is 0.5(+1) + 0.5(-1) which is 0, not surprising.

Another way to see the same proposition is you start with nothing and the coin is flipped. If heads, you receive $2. If tails, you receive $(-1) (i.e. you now have to pay $1). So the expectation is 0.5*2 + 0.5*(-1)= 0.5.

What has happened? It's the same proposition both ways, I think. Is there a systematic way to tell which analysis is the right one? The second calculation has to be wrong, but it's not obvious how. Thanks. 2601:644:8581:75B0:0:0:0:C030 (talk) 22:12, 2 September 2024 (UTC)

The first way, you gain net $1 for a win; the second way, you gain $2. (The dollar in escrow does not change that.) They are not the same bet. —Tamfang (talk) 23:37, 2 September 2024 (UTC)

The calculations by themselves are both correct, but (as noted by Tamfang), they represent different betting propositions. Assume the coin comes up tails. In the first version your loss is the $1 paid in advance, in the second you pay $1 afterwards for losing the bet. So that amounts to the same loss. But now assume the coin comes up heads. In the first you pay $1 in advance and then receive $2. In the second version you just receive $2 without having to make an advance payment. That is clearly more advantageous. To make your second version equivalent to the first, replace "you receive $2" by "you receive $1".  --Lambiam 06:00, 3 September 2024 (UTC)

To supplement to the arithmetical explanations above: It's not a paradox, and since you know which answer is right, you know the basic analysis is to step through it slowly and carefully. This may not be possible in a real-world cash transaction, and this is how quick change scams work (and similar for some street gambling scams) -- in other words, it's not a dumb question, it's not obvious, and if you can think of a truly easy generalized way to work this stuff out for people in real-time social situations, you'll have done a huge service for humanity. (See video examples of the quick change scam from Noah Da Boa and The Real Hustle.) (Right now, most people online say just not to give change to strangers -- the best way to win is not to play.) SamuelRiv (talk) 18:23, 3 September 2024 (UTC)

Incidentally, one way to see that they are different is to determine the (maximum) amount you would be willing to pay to be in the second position instead of the first. (I get $0.50) Tito Omburo (talk) 20:10, 3 September 2024 (UTC)

Thanks everyone, I must not have been thinking clearly. Another way to see it is imagine playing twice, winning one and losing one. You end up with $1 instead of with $0. 2601:644:8581:75B0:0:0:0:C030 (talk) 22:34, 3 September 2024 (UTC)

Anomalous result

Solve for x:

• ${\sqrt {x+1}}$ = x - 1.

Here's my approach, step by step:

• Square both sides:

x + 1 = ${x^{2}}$ - 2x + 1

• Cancel 1's:

x = ${x^{2}}$ - 2x

• Collect x's:

${x^{2}}$ - 3x = 0

• Factorise:

x (x - 3) = 0

• Solution:

x = 0 or 3.

So far, so good. Or so it seems.

Plug 3 back into the original equation:

• ${\sqrt {3+1}}$ = 3 - 1
• ${\sqrt {4}}$ = 2 = 3 - 1. Correct

Plug 0 back into the original equation:

• ${\sqrt {0+1}}$ = 0 - 1
• ${\sqrt {1}}$ = 1 =/= 0 - 1. Incorrect.

I've gone over this a dozen or more times but cannot see what really basic error I must be making.

Any ideas? -- Jack of Oz ^{[pleasantries]} 22:06, 5 September 2024 (UTC)

You didn't do anything really wrong, but just discovered that 0 is an "Extraneous solution to the problem. As our article says, they "result from performing operations that are not invertible for some or all values of the variables involved, which prevents the chain of logical implications from being bidirectional."

The problem is that squaring is not a one-to-one function, so its inverse, square-rooting needs to be carefully defined. That is, -5 and 5 squared are both 25. So we must pick one of them as "the" square root if we want to define a function, something that spits out just one value "5" when fed "25". If we defined "square root" as Euler did and said either are square roots, then 0 is perfectly good solution. In modern terms it would amount to solving ±${\sqrt {x+1}}$ = x - 1John Z (talk) 23:19, 5 September 2024 (UTC)

You proved that ${\sqrt {x+1}}=x-1$ implies $(x=0$ or $x=3).$ Indeed, if $x=3$ (the only true solution), it is the case that $x=0$ or $x=3.$ You appear to assume that the converse implication also holds, but this assumption is unwarranted. The false solution is introduced by the squaring operation; it adds solutions of the equation $~{-}{\sqrt {x+1}}=x-1.$ A simpler puzzle based on the same issue is the following:

• Solve for $x$ the equation

  ${\sqrt {x}}=-1.$

• Square both sides:

  $x=1.$

• Plug $1$ for $x$ into the original equation:

  ${\sqrt {1}}=-1.$

• What gives?

What we found is the solution of the equation $~{-}{\sqrt {x}}=-1.$  --Lambiam 23:17, 5 September 2024 (UTC)