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picking k_n

dwsmith

Well-known member
Feb 1, 2012
1,673
I don't understand how one picks a $k_n$.

For example, lets look at $\sin\pi z = \prod\limits_{n\in\mathbb{Z}-\{0\}}\left[\left(1-\frac{z}{n}\right)e^{z/n}\right]$.

For all n, $k_n = 2$. With this $k_n$, the product is entire. What is $k_n$? I know for that product we can write it as $\prod\limits_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)$. So $k_n$ is the power of z and n. I don't know why this was chosen.

Then there is the corollary:
If $\sum\limits_{n =1}^{\infty}\frac{1}{\left|n\right|^2}$ converges, then $\prod\limits_{n=1}^{\infty}\left[\left(1-\frac{z}{n}\right)e^{z/n}\right]$ converges.

I guessing this all related to picking $k_n$ but I don't get it.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
I don't understand how one picks a $k_n$.

For example, lets look at $\sin\pi z = \prod\limits_{n\in\mathbb{Z}-\{0\}}\left[\left(1-\frac{z}{n}\right)e^{z/n}\right]$.

For all n, $k_n = 2$. With this $k_n$, the product is entire. What is $k_n$? I know for that product we can write it as $\prod\limits_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)$. So $k_n$ is the power of z and n. I don't know why this was chosen.

Then there is the corollary:
If $\sum\limits_{n =1}^{\infty}\frac{1}{\left|n\right|^2}$ converges, then $\prod\limits_{n=1}^{\infty}\left[\left(1-\frac{z}{n}\right)e^{z/n}\right]$ converges.

I guessing this all related to picking $k_n$ but I don't get it.
Hi dwsmith,

I don't understand how one picks a $k_n$.
What is \(k_n\)? Can you please give us more information.

For example, lets look at $\sin\pi z = \prod\limits_{n\in\mathbb{Z}-\{0\}}\left[\left(1-\frac{z}{n}\right)e^{z/n}\right]$.
How did you get this equation, and what are the limits of the infinite product? The Gamma function has a infinite product representation similar to this, but I have never seen a representation of this kind for \(\sin\pi z\).

Then there is the corollary:
If $\sum\limits_{n =1}^{\infty}\frac{1}{\left|n\right|^2}$ converges, then $\prod\limits_{n=1}^{\infty}\left[\left(1-\frac{z}{n}\right)e^{z/n}\right]$ converges.
This has no meaning since \(\sum\limits_{n =1}^{\infty}\frac{1}{\left|n\right|^2}=\sum\limits_{n =1}^{\infty}\frac{1}{n^2}\) is always convergent.

I know for that product we can write it as \(\prod\limits_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)\,.\)
\(\displaystyle\sin \pi z = \pi z \prod_{n=1}^{\infty} \left(1 - \frac{z^2}{n^2}\right)\) not \(\displaystyle\prod_{n=1}^{\infty} \left(1 - \frac{z^2}{n^2}\right)\).