1. ## Euler's Reflection Formula...

What is the first formula in "Proof" and how can I prove it?

Euler's Reflection Formula - ProofWiki

Thank you...

2. ## Re: Euler's Reflection Formula...

Originally Posted by Also sprach Zarathustra
What is the first formula in "Proof" and how can I prove it?...
If as 'first formula' You mean that...

$\displaystyle \sin \pi z = \pi\ z\ \prod_{n \ne 0} (1-\frac{z}{n})\ e^{\frac{z}{n}$ (1)

... then that's a perfect rubbish and I suggest You don't waste time trying to 'demonstrate' it...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. ## Re: Euler's Reflection Formula...

Originally Posted by chisigma
If as 'first formula' You mean that...

$\displaystyle \sin \pi z = \pi\ z\ \prod_{n \ne 0} (1-\frac{z}{n})\ e^{\frac{z}{n}$ (1)

... then that's a perfect rubbish and I suggest You don't waste time trying to 'demonstrate' it...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

I came up with idea, tell me if that is wrong....

We need to show that:

$\displaystyle \sin \pi z = \pi\ z\ \prod_{n \ne 0} (1-\frac{z}{n})\ e^{\frac{z}{n}$

If we know that $\displaystyle \sin \pi z = \pi\ z\ \prod_{n \ne 0} (1-\frac{z^2}{n^2})$, then we can re-write it in the form:

$\displaystyle \pi\ z\ \prod_{n \ne 0} (1-\frac{z^2}{n^2})=\pi\ z\ \prod_{n \ne 0} (1-\frac{z}{n})(1+\frac{z}{n})$

And the question now is:

Is it true that,

$\displaystyle \prod_{n \ne 0} (1+\frac{z}{n})= \prod_{n \ne 0} e^{\frac{z}{n}}$

4. ## Re: Euler's Reflection Formula...

Originally Posted by Also sprach Zarathustra
What is the first formula in "Proof" and how can I prove it?

Euler's Reflection Formula - ProofWiki

Thank you...
Weierstrass factorization theorem - Wikipedia, the free encyclopedia

5. ## Re: Euler's Reflection Formula...

Originally Posted by Also sprach Zarathustra
I came up with idea, tell me if that is wrong....

We need to show that:

$\displaystyle \sin \pi z = \pi\ z\ \prod_{n \ne 0} (1-\frac{z}{n})\ e^{\frac{z}{n}$

If we know that $\displaystyle \sin \pi z = \pi\ z\ \prod_{n \ne 0} (1-\frac{z^2}{n^2})$, then we can re-write it in the form:

$\displaystyle \pi\ z\ \prod_{n \ne 0} (1-\frac{z^2}{n^2})=\pi\ z\ \prod_{n \ne 0} (1-\frac{z}{n})(1+\frac{z}{n})$

And the question now is:

Is it true that,

$\displaystyle \prod_{n \ne 0} (1+\frac{z}{n})= \prod_{n \ne 0} e^{\frac{z}{n}}$
I said the the formula was 'a perfect rubbish' because it is written in ambigous form... the usual and fully clear formula is...

$\displaystyle \sin \pi\ z = \pi\ z\ \prod_{n=1}^{\infty} (1-\frac{z^{2}}{n^{2}})$ (1)

The 'demonstration' of (1) is done using the Weierstrass factorisation theorem...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$