MHB Infinite Product: Showing & Evaluating

[polygamma]

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)$.

 

[Dustinsfl]

Re: infinite product

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?

 

[polygamma]

Re: infinite product

@dwmsmith

I don't know who that is.

 

[Dustinsfl]

Re: infinite product

@dwmsmith

I don't know who that is.

This is who is https://en.wikipedia.org/wiki/Serge_Lang

and those are worked questions and examples from his book

Complex Analysis (Graduate Texts in Mathematics): Serge Lang: 9781441931351: Amazon.com: Books

 

[polygamma]

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)

 

[alyafey22]

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)

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

 

[Dustinsfl]

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)

Part 3 isn't there or I can't find it.

 

[polygamma]

Re: infinite product

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.

 

[polygamma]

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 defintion 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:

Spoiler

$ \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) $