4
$\begingroup$

Construct a function with zero at $z=0$ and zeros at $z=-n$ with multiplicities $n$.

My answer is $f(z) = z\prod_{n=1}^{\infty}\left[E_n\left(-\frac zn\right)\right]^n,$ where $E_n(z)=(1-z)\exp\left(\sum_{k=1}^n\frac{z^k}k\right)$.

Is that right? And does the product converge?

  • 0
    @bgins: OP's is the form in the [Weierstrass factorization](http://en.wikipedia.org/wiki/Weierstrass_factorization_theorem), which ensures that the terms in the product tend to $1$ in the limit (notice the polynomial in the exponential is a truncation of the Maclaurin expansion of $\log(1-z)^{-1}$). The form you cite won't work because $(1+\frac{z}{n})^n\to e^z$, so the product won't converge unless $z$ is a nonpositive integer. | *Edit*: Even with $a_n$ coefficients it won't work because $e^z$ varies with $z$.2012-03-16

1 Answers 1

2

Your answer looks right.

It follows from the following theorem (Functions of One Complex Variable, John B Conway, Indian Edition, page number 169).

Theorem 5.12: Let $\{a_n\}$ be a sequence in $\mathbb{C}$ such that $\lim |a_n| = \infty$ and $a_n \neq 0$ for all $n \ge 1$. If $\{p_n\}$ is any sequence of integers such that

$ \sum_{n=1}^{\infty} \left(\frac{r}{|a_n|}\right)^{p_n+1} \lt \infty$

for all $r \gt 0$, then

$ f(z) = \prod_{n=1}^{\infty} E_{p_n}\left(\frac{z}{a_n}\right)$ converges and $f$ is an entire function with zeroes only at the points $a_n$. If $z_0$ occurs in $\{a_n\}$ exactly $m$ times, then $f$ has a zero $z_0$ of multiplicity $m$.

You have chosen $p_n = n$ which works.

This theorem is used to prove the Weirstrass Factorization theorem.

  • 0
    @John0417: Yes, all you need to do is get a sequence $\{a_n\}$ and show that the corresponding partial product (coming from the theorem) upto some number of terms can be recast into a form you have shown...2012-03-17