# Infinite product

#### Random Variable

##### Well-known member
MHB Math Helper
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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#### dwsmith

##### Well-known member
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?

#### Random Variable

##### Well-known member
MHB Math Helper
Re: infinite product

@dwmsmith

I don't know who that is.

#### Random Variable

##### Well-known member
MHB Math Helper
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.

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
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.

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

#### dwsmith

##### Well-known member
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.

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

#### Random Variable

##### Well-known member
MHB Math Helper
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.

#### Random Variable

##### Well-known member
MHB Math Helper
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:

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