Infinite Product: Showing & Evaluating

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    Infinite Product
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Discussion Overview

The discussion centers around evaluating infinite products, specifically focusing on the formula for products of the form $\prod_{k=1}^{\infty} \left( 1- \frac{z^{n}}{k^{n}} \right)$ and its applications. Participants explore derivations and evaluations related to these products, referencing mathematical literature and personal insights.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants present the formula $\prod_{k=1}^{\infty} \left( 1- \frac{z^{n}}{k^{n}} \right) = \prod_{k=0}^{n-1} \frac{1}{\Gamma\left[ 1-\exp (2 \pi i k/n) z\right]}$ for $n > 1$ as a basis for further exploration.
  • Others propose using the above formula to show that $\prod_{k=1}^{\infty} \left(1- \frac{z^{2}}{k^{2}} \right) = \frac{\sin \pi z}{\pi z}$, noting the derivation is not commonly seen.
  • One participant mentions the challenge of evaluating $\prod_{k=2}^{\infty} \left(1- \frac{1}{k^{3}} \right)$, indicating it requires a trick.
  • Several participants express uncertainty about the originality of their findings, with one acknowledging that the formula for $\sin$ is referenced in Wolfram MathWorld and was likely known prior to their discovery.
  • Another participant shares Euler's limit definition of the gamma function and its relevance to the discussion, providing a detailed mathematical derivation.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the originality of the formulas discussed, with some believing they have derived new insights while others reference existing literature. The evaluation of the infinite products remains unresolved, with differing opinions on the methods and tricks required.

Contextual Notes

Participants note that the derivations depend on specific mathematical definitions and properties of the gamma function, and there are unresolved steps in the evaluations presented.

polygamma
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1) Show that for $n >1$, $\displaystyle \prod_{k=1}^{\infty} \left( 1- \frac{z^{n}}{k^{n}} \right) = \prod_{k=0}^{n-1} \frac{1}{\Gamma\left[ 1-\exp (2 \pi i k/n) z\right]}$.2) Use the above formula to show that $ \displaystyle \prod_{k=1}^{\infty} \left(1- \frac{z^{2}}{k^{2}} \right) = \frac{\sin \pi z}{\pi z}$.

3) Evaluate $ \displaystyle \prod_{k=2}^{\infty} \left(1- \frac{1}{k^{3}} \right)$.
 
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Re: infinite product

Random Variable said:
1) Show that for $n >1$, $\displaystyle \prod_{k=1}^{\infty} \left( 1- \frac{z^{n}}{k^{n}} \right) = \prod_{k=0}^{n-1} \frac{1}{\Gamma\left[ 1-\exp (2 \pi i k/n) z\right]}$.2) Use the above formula to show that $ \displaystyle \prod_{k=1}^{\infty} \left(1- \frac{x^{2}}{k^{2}} \right) = \frac{\sin \pi z}{\pi z}$.

3) Evaluate $ \displaystyle \prod_{k=2}^{\infty} \left(1- \frac{1}{k^{3}} \right)$.

Are you reading Serge Lang's book?
 
Re: infinite product

@dwmsmith

I don't know who that is.
 
Re: infinite product

I thought I had put together three interesting questions since I've never seen $\displaystyle \prod_{k=1}^{\infty} \left(1- \frac{z^{2}}{k^{2}} \right) = \frac{\sin \pi z}{\pi z}$ derived from that formula.And to evaluate the second product using that formula requires a bit of a trick.(Sadface)
 
Re: infinite product

Random Variable said:
I thought I had put together three interesting questions since I've never seen $\displaystyle \prod_{k=1}^{\infty} \left(1- \frac{z^{2}}{k^{2}} \right) = \frac{\sin \pi z}{\pi z}$ derived from that formula.And to evaluate the second product using that formula requires a bit of a trick.(Sadface)

It happened a lot for me last days . Nevertheless , I had interesting time deriving ''what I thought" new formulas .
 
Re: infinite product

Random Variable said:
I thought I had put together three interesting questions since I've never seen $\displaystyle \prod_{k=1}^{\infty} \left(1- \frac{z^{2}}{k^{2}} \right) = \frac{\sin \pi z}{\pi z}$ derived from that formula.And to evaluate the second product using that formula requires a bit of a trick.(Sadface)

Part 3 isn't there or I can't find it.
 
Re: infinite product

ZaidAlyafey said:
It happened a lot for me last days . Nevertheless , I had interesting time deriving ''what I thought" new formulas .

I knew it wasn't a new formula. It's listed on Wolfram MathWorld with a reference to a book from 1986. And I'm sure it was known well before then. I had just never seen it used anywhere to evaluate the infinite product for $\sin$ or anything else.
 
Re: infinite product

dwsmith didn't post the solution he said he had for the first problem, so I'm going to at least post that.Euler's limit definition of the gamma function is $ \displaystyle \Gamma(z) = \lim_{m \to \infty} \frac{m! \ m^{z}}{z(z+1) \cdots (z+m)} $.

So $\displaystyle \Gamma (1+z) = z \Gamma (z) = \lim_{m \to \infty} \frac{m! \ m^{z}}{(z+1) \cdots (z+m)} = \lim_{m \to \infty} m^{z} \prod_{k=1}^{m} \Big(1+ \frac{z}{k} \Big)^{-1}$$\displaystyle \Gamma \big[ 1-\exp(2 \pi i l /n)z \big] = \lim_{m \to \infty} m^{-\exp(2 \pi i l /n)z} \prod_{k=1}^{m} \Big(1- \frac{\exp(2 \pi i l/n)z}{k} \Big)^{-1}$

$ \displaystyle \prod_{l=0}^{n-1} \Gamma \big[1-\exp(2 \pi i l/n)z \big] = \lim_{m \to \infty} m^{- \sum_{l=0}^{n-1} \exp(2 \pi i l/n)z} \prod_{l=0}^{n-1} \prod_{k=1}^{m} \Big(1- \frac{\exp(2 \pi i l/n)z}{k} \Big)^{-1}$

$ \displaystyle = \lim_{m \to \infty} \prod_{k=1}^{m} \prod_{l=0}^{n-1} \Big(1- \frac{\exp(2 \pi i l/n)z}{k} \Big)^{-1}$And since $\displaystyle x^n-z = \prod_{k=0}^{n-1} \Big( x- z^{\frac{1}{n}} \exp(2 \pi i k /n) \Big) \implies 1- z^{n} = \prod_{k=0}^{n-1} \Big( 1- \exp(2 \pi i k /n)z\Big)$,

$ \displaystyle \lim_{m \to \infty} \prod_{k=1}^{m} \prod_{l=0}^{n-1} \Big(1- \frac{\exp(2 \pi i l/n)z}{k} \Big)^{-1} = \lim_{m \to \infty} \prod_{k=1}^{m} \Big( 1- \frac{z^{n}}{k^{n}} \Big) ^{-1} = \prod_{k=1}^{\infty} \left(1 - \frac{z^{n}}{k^{n}} \right)^{-1}$
Then for the second question, $\displaystyle \prod_{k=1}^{\infty} \left( 1- \frac{z^{2}}{k^{2}} \right) = \frac{1}{\Gamma(1-z) \Gamma(1+z)} = \frac{1}{z \Gamma(1-z) \Gamma(z)} = \frac{\sin \pi z}{\pi z}$Hint for the third question:

$ \displaystyle \prod_{k=2}^{\infty} \left( 1- \frac{1}{k^{3}} \right) = \lim_{z \to 1} \frac{1}{1-z^{3}} \prod_{k=1}^{\infty} \left(1 - \frac{z^{3}}{k^{3}} \right) $
 

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