Wikipedia:Reference desk/Archives/Mathematics/2023 September 16

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September 16

"Strict equidissection" of regular triangle (in Euclidean geometry, or EG) and all possible values of N

In this context, the equidissection is defined as the partition of a given planar figure into pieces (not need to be triangles) of the same area without overlapping. The term "strict" means that all of these pieces must be congruent (similar in both the area and shape). For example, one can strictly equidissects a disk into any number X of pieces (X is positive integer) by dividing the circle into equal non-overlapped X arcs, seperated by X points. Then connect these X points to the center using X continuous lines. The term "higher primes" stands for any prime congruent to either 1 or -1 modulo 6 (i.e, other than 2 and 3). I found that the regular triangle (in EG) can be strictly equidissected (by above definition) into N continuous pieces where most of the number N are of this form 2^x • 3^y • Δ! (1), x and y are both non-negative integers, Δ! is either 1 or the product of higher primes, each with positive even powers (they are the odd perfect squares not divisible by 3). The case N=1 (where x=y=0, Δ!=1) is OK but excluded here because it is truly trivial. The sequence for all number Ns of the (1) form: 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 25, 27, 32, 36, 48, 49, 50, 54, 64, 72, 75, 81, 96, 98, 100, 108, 121, 128, 144, 147, 150, 162, 169, 192, 196, 200, 216, 225, 242, 243, 256, 288, 289, 294, 300, 324, 338, 361, 363, 384, 392, 400, 432, 441, 450, 484, 486, 507, 512,... (Note that most of the number Ns are 3-smooth numbers). My question: Can the regular triangle in EG be strictly equidissected into N continuous pieces, where N is not of the (1) form? In other words, is the above sequence of number Ns complete or incomplete, are there any N-numbers missing from it - for example, can N=5? I have tried to found the results to this problem by myself since I read the WP article on "Monsky's theorem"- a related interesting problem, but i found nowhere on the Internet the article that discuss my problem or similar ones. 2402:800:63BC:DB8D:51EA:26FE:9925:F07B (talk) 08:47, 16 September 2023 (UTC)

It is possible that this precise problem has not been studied. Related problems can be found in Alexander Soifer's How does one cut a triangle? (Springer link; table of contents). Chapter 2 can be found online: Chapter 2. Most studies in cutting triangles, such as this one restrict the problem to cutting up into smaller triangles only. If this precise problem has been studied, publications will almost certainly reference the book.  --Lambiam 10:05, 16 September 2023 (UTC)

PS. Here is a proof that N=7 is impossible: "No triangle can be decomposed into seven congruent triangles".  --Lambiam 10:10, 16 September 2023 (UTC)

But my first sentence said that the pieces are not necessarily triangles (they can be any shapes, even ones whose border consists of only "curves", and no straight segments). So that the mathematics in the cited article is correct but obviously not sufficient for the case N=7. 2402:800:63BC:DB8D:51EA:26FE:9925:F07B (talk) 10:15, 16 September 2023 (UTC)

Based on the shapes in [1], it's possible to strictly equidissect the triangle into $180=2^{2}\cdot 3^{2}\cdot 5$ pieces and $405=3^{4}\cdot 5$ pieces. GalacticShoe (talk) 14:23, 16 September 2023 (UTC)

About trigonometric functions and hyperbolic functions and transcendental numbers

1. If x (in radians) is an algebraic number other than 0, must sin(x), cos(x), tan(x), cot(x), sec(x), csc(x) be transcendental numbers?

2. If x (in degrees) is an irrational algebraic number, must sin(x), cos(x), tan(x), cot(x), sec(x), csc(x) be transcendental numbers?

3. If x is an algebraic number other than 0, must sinh(x), cosh(x), tanh(x), coth(x), sech(x), csch(x) be transcendental numbers?

—— 36.234.121.52 (talk) 17:38, 16 September 2023 (UTC)

The answer is generally thought to be that this is indeed necessarily the case. But it is known for a fact for only a few isolated cases. For example, if $\cosh 2=\alpha ,$ it follows that

$e={\sqrt {\alpha +{\sqrt {\alpha ^{2}-1}}}}.$

So algebraicity of $\cosh 2$ would imply the algebraicity of $e,$ which is known to be transcendental.  --Lambiam 21:28, 16 September 2023 (UTC)

Lindemann–Weierstrass theorem#Transcendence of e and π says sin cos and tan and their hyperboic forms are transcendental for all non-zero algebraic numbers but hasn't got a citation. NadVolum (talk) 23:09, 16 September 2023 (UTC)

If we assume for sake of contradiction that $cos(\alpha )$ is algebraic for some algebraic $\alpha \neq 0$, then since ${\frac {1}{2}}$ and $-\cos(\alpha )$ would be algebraic and $\alpha i,-\alpha i,0$ would be algebraic and distinct, ${\frac {1}{2}}e^{\alpha i}+{\frac {1}{2}}e^{-\alpha i}-\cos(\alpha )e^{0}=0$ would contradict Baker's reformulation of the Lindemann–Weierstrass theorem as listed in the article. Similarly, ${\frac {1}{2i}}e^{\alpha i}-{\frac {1}{2i}}e^{-\alpha i}-\sin(\alpha )e^{0}=0$ would also create a contradiction if both $\alpha$ and $\sin(\alpha )$ were algebraic. Naturally, since reciprocals of algebraic numbers are also algebraic, it results that $\alpha \neq 0$ being algebraic implies that $\sec(\alpha ),\csc(\alpha )$ are not algebraic. Results for $\tan(x),\cot(x)$ can utilize the fact that $1+\tan ^{2}(x)=\sec ^{2}(x)$. GalacticShoe (talk) 00:41, 17 September 2023 (UTC)

I'm pretty sure that using analogous proofs extends this to hyperbolic functions too, which just leaves the second question open. GalacticShoe (talk) 01:10, 17 September 2023 (UTC)

Twin prime conjecture

The sum of the (-1)th power of the twin primes converges, but does the sum of the (-1/2)th power of the twin primes diverge? Of course if diverge, then there must be infinitely many twin primes. 36.234.121.52 (talk) 18:00, 16 September 2023 (UTC)

A conjecture by Hardy and Littlewood gives the number of twin primes up to $x$ as growing asymptotically like

$\pi _{2}(x)\sim 2C_{2}\int _{2}^{x}{\mathrm {d} t \over (\ln t)^{2}},$

in which $C_{2}$ is called the "twin prime constant". If the conjecture is correct, the sum of the $-(1{+}\varepsilon )$th powers converges for all $\varepsilon >0.$  --Lambiam 21:13, 16 September 2023 (UTC)

I mean (-1/2)th = (-0.5)th, not -(1+$\epsilon$)th with an $\epsilon$ > 0 36.234.122.29 (talk) 16:59, 18 September 2023 (UTC)

Let us consider the sum of the $-(1{-}\delta )$th powers, $\delta >0,$ of which $\delta ={\tfrac {1}{2}}$ is a special case. Assuming the conjecture, the question is then whether $\int _{2}^{\infty }{\frac {t^{-(1{-}\delta )}}{(\ln t)^{2}}}dt$ diverges. Substituting $t:=\exp z,$ we get the integral $\int _{\ln 2}^{\infty }{\frac {e^{\delta z}}{z^{2}}}dz,$ which clearly diverges.  --Lambiam 15:34, 19 September 2023 (UTC)

It diverges? So there must be infinitely many twin prime pairs. 220.132.230.56 (talk) 18:38, 20 September 2023 (UTC)

There are infinitely many twin prime pairs, if the sum diverges, if the first Hardy–Littlewood conjecture, which itself is based on the idea that there are infinitely many twin prime pairs, holds. In other words, using the conjecture to prove that there are infinitely many twin prime pairs would be circular reasoning. Obtaining a proof that the sum diverges independent of the conjecture would require some elaborate machinery, the discovery of which would probably be tantamount to solving the twin prime conjecture itself. GalacticShoe (talk) 18:54, 20 September 2023 (UTC)

Why the sum of the (-1)th power of the twin primes can be calculate, but the sum of the (-1/2)th power of the twin primes cannot? 220.132.230.56 (talk) 04:44, 21 September 2023 (UTC)