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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?

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$$E_n(z)$$ is $$E_n(z)=(1-z)e^{z+z^2/2+z^3/3+...+z^n/n}$$ –  John0417 Mar 16 '12 at 3:50
    
Why not something like $f(z)=z\prod_{n=1}^\infty a_n \left(1+\frac{z}n\right)^n$ for some suitable choice of constants $a_n$ (so that we get nice convergence)? As long as $\lim_{n\rightarrow\infty}a_n<\frac1e$ we should be okay. –  bgins Mar 16 '12 at 18:50
    
@bgins: OP's is the form in the Weierstrass factorization, 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 $z$. –  anon Mar 16 '12 at 18:52

1 Answer 1

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.

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hmm, here I used En to the power n,to satisfies the multiplicity condition. it still converges right? Followed by the radius of convergence of $$\sum(a_n z^n)=\sum (na_n z^n)$$ –  John0417 Mar 17 '12 at 0:04
    
@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... –  Aryabhata Mar 17 '12 at 0:07

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