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[SOLVED] exhibit a function f

dwsmith

Well-known member
Feb 1, 2012
1,673
Use the Weierstrass Product Theorem to exhibit a function $f$ such that each positive integer $n$, $f$ has a pole of order $n$, and $f$ is analytic and nonzero at every other complex number.

So the solution goes as
Let $z_n$ be the nth term in the sequence $1,2,2,3,3,3,\ldots$.

Note that:
$$
\underbrace{\sum_{n=1}^{\infty}\frac{1}{|z_n|^3}}_{\text{Why to the 3rd power?}} = \underbrace{\sum_{n=1}^{\infty}\frac{n}{n^3}}_{ \text {Why is this equality true?}}
$$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Why is $k_n$ 3 here?

The Weierstrass Product is $\prod\limits_{n = 1}^{\infty}\left(1-\frac{z}{z_n}\right)^{-1}e^{-P\left(z/z_n\right)}$.

So that leaves us with $\displaystyle\prod\limits_{n = 1}^{\infty}\left(1-\frac{z}{z_n}\right)^{-1}e^{-P\left[\frac{z}{z_n}+\left(\frac{z}{z_n}\right)^2\right]}$

I was told that the above product can be simplified. Like this?

By expanding the product, we have
$$
\left(1 - z\right)^{-1}e^{\left[-z + \frac{z^2}{2}\right]}\left(1 - \frac{z}{2}\right)^{-2}e^{\left[-\frac{z}{2} + \frac{z^2}{4}\right]}\left(1 - \frac{z}{3}\right)^{-3}e^{\left[-\frac{z}{3} + \frac{z^2}{6}\right]}\cdots\left(1 - \frac{z}{n}\right)^{-n}e^{\left[-\frac{z}{n} + \frac{z^2}{2^n}\right]}\cdots
$$
So we have that
$$
\prod_{n = 1}^{\infty}\left(1 - \frac{z}{z_n}\right)^{-1}e^{-\left[\frac{z}{z_n} + \left(\frac{z}{z_n}\right)^2/2\right]} =
\prod_{n = 1}^{\infty}\left(1 - \frac{z}{n}\right)^{-n}\exp\left[{-\frac{z}{n} + \frac{z^2}{2n}}\right].
$$
 
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