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Infinite product

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
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)$.
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
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
Jan 31, 2012
253
Re: infinite product

@dwmsmith

I don't know who that is.
 

dwsmith

Well-known member
Feb 1, 2012
1,673

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
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)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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 .
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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.
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
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
Jan 31, 2012
253
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) $