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Prove gamma function's trig identity?


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This is Euler's reflection formula, and a derivation can be found at ProofWiki: https://proofwiki.org/wiki/Euler's_Reflection_Formula

A more detailed derivation starts near the bottom of page 5 here: http://homepage.tudelft.nl/11r49/documents/wi4006/gammabeta.pdf

Both use concepts from complex analysis, of which I know very little. So perhaps someone more knowledgable can provide some insight. But anyway, there are two starting points if you'd like to explore further.

Edited by John
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I'm not sure what you mean by that, since in both the ProofWiki article and the PDF, the derivations are explained step-by-step. The author of either one could have made mistakes somewhere along the line, which I'm not qualified to judge since I'm not familiar with the requisite concepts and I'm not particularly inclined to spend that much time on learning them right now, but they certainly don't lack explanation.

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On wiki, it just plain doesn't explain anything. In the PDF, it makes a giant leap between two theorems without explaining the intermediate information. It just talks about the bell curve, then the Beta integral, then out of nowhere the paper's talking about (cos^5)(sin^7) without explaining how trig functions are at all related to anything in the entire paper.


The Beta functions is already just a quotient of gamma functions so it makes the logic seem rather circular.

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Alright, having looked into this a bit more (namely on Wikipedia here and here), here's a derivation.

 

First, we have that [math]\Gamma(1 - x) = -x \Gamma(-x)[/math].

 

Using Weierstrass factorization, we have

 

[math]\sin{\pi x} = \pi x \prod_{n=1}^{\infty} \left(1 - \frac{x^{2}}{n^{2}}\right)[/math]

 

and

 

[math]\frac{1}{\Gamma(x)} = xe^{\gamma x} \prod_{n=1}^{\infty} \left(1 + \frac{x}{n}\right)e^{-\frac{x}{n}}[/math].

This is enough to derive the formula, as follows:

[math]\begin{array}{rcl}\frac{1}{\Gamma(x) \Gamma(1 - x)} & = & \frac{1}{-x\Gamma(x) \Gamma(-x)} \\ & = & \left(\frac{1}{-x}\right) \left(xe^{\gamma x} \prod_{n=1}^{\infty} \left(1 + \frac{x}{n}\right)e^{-\frac{x}{n}}\right) \left(-xe^{-\gamma x} \prod_{n=1}^{\infty} \left(1 - \frac{x}{n}\right)e^{\frac{x}{n}}\right) \\ & = & \left(\frac{1}{-x}\right) \left(xe^{\gamma x}\right) \left(-xe^{-\gamma x}\right) \prod_{n=1}^{\infty} \left(1 + \frac{x}{n}\right)e^{-\frac{x}{n}}\left(1 - \frac{x}{n}\right)e^{\frac{x}{n}} \\ & = & x\prod_{n=1}^{\infty}\left(1 - \frac{x^{2}}{n^{2}}\right) \\ & = & \frac{\sin{\pi x}}{\pi}\end{array}[/math].

 

Since [math]\frac{1}{\Gamma(x) \Gamma(1 - x)} = \frac{\sin{\pi x}}{\pi}[/math], we have [math]\Gamma(x) \Gamma(1 - x) = \frac{\pi}{\sin{\pi x}}[/math].

Edited by John
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That proof looks legitimate when I look at the culmination of theorems, but there's still this huge gap. Where did the first derivation of sin(x) not being related to a circle but rather only factorials and multiples come from? How was that factorization theorem discovered? Lets say I never heard of a circle before. How would I discovery or derive sin(x) from scratch using factorials and multiplication?

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Nothing about this implies that sine is unrelated to the circle.

 

Here's a resource related to the factorization of sine: http://www.ams.org/bookstore/pspdf/gsm-97-prev.pdf, and here's some more general stuff: http://www.math.umn.edu/~garrett/m/complex/hadamard_products.pdf.

 

To get a real understanding of all this, you'll probably want to study some analysis. I know I've enjoyed learning a bit about it in trying to answer your questions, but while I know enough to point in some potentially fruitful directions, I still don't know near enough to provide the clarification you seem to want.

Edited by John
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