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Problem Let $p(z)=(z-a_1)(z-a_2)...(z-a_N)$, where $a_1, a_2, ..., a_N$ are distinct complex numbers. Let $M=\min_{1\le{k}\le{N}}|a_k|$. Prove that it is possible to express $\frac{1}{p(z)}$ as a power series $\sum_{n=0}^{\infty}c_nz^n$, for $|z|<M$.


We look to prove this by induction on $N$. The case $N=1$ is rather simple. We see that $$\frac{1}{p(z)}=\frac{1}{z-a_1}=-\frac{1}{a_1}\sum_{n=0}^{\infty}(\frac{z}{a_1})^n$$ which converges for $|z|<M$ by comparison with $\sum_{n=0}^{\infty}z^n$.

We assume now that the proposition holds for arbitrary $N$ and consider the '$N+1$' case.

Now, $p(z)=(z-a_1)(z-a_2)\cdots(z-a_N)(z-a_{N+1})$. By the assumption of our inductive hypothesis, $\frac{1}{(z-a_1)(z-a_2)\cdots(z-a_N)}$ can be expressed in the form $\sum_{n=0}^{\infty}c_nz^n$ for some complex coefficients $c_n$.

As such, $$\frac{1}{p(z)}=-\frac{1}{a_{N+1}}\sum_{n=0}^{\infty}c_nz^n\sum_{n=0}^{\infty}\frac{1}{(a_{N+1})^n}z^n=-\frac{1}{a_{N+1}}\sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{c_n}{(a_{N+1})^{n-k}} z^n$$ which is of the correct form, but I'm not sure how to demonstrate that convergence holds for $|z|<M$

Any help would be very appreciated. I'm not sure if induction is the best approach to proving this.


I'll leave previous working here for reference. It seems simply writing $\frac{1}{p(z)}$ as a linear combination of the fractions $\frac{1}{z-a_k}$ for $1\le{k}\le{N}$ is a far less cumbersome approach. Thanks to all who have helped.

Further Problem: Could the radius of convergence exceed $M$?

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If you try, can you read this if I wrote the question? –  AD. Dec 7 '11 at 14:48
Hint: One can decompose $1/p(z)$ as a linear combination of the fractions $1/(z-a_k)$, and expand each $1/(z-a_k)$ as a power series in $z$. The result is $1/p(z)=\sum\limits_{n=0}^{+\infty}b_nz^n$ for some complex numbers $b_n$ (and not what you write in the question, which is absurd). –  Did Dec 7 '11 at 15:19
Hint: $$\frac{1}{p(z)}=\sum_{j=1}^N\ \frac{1}{p'(a_j)\ (z-a_j)}\quad.$$ –  Pierre-Yves Gaillard Dec 7 '11 at 15:26
I think the simplest way is this: Put $a:=a_1$. We have $$\frac{1}{p(z)}=\frac{b}{z-a}+f(z)$$ with $f$ defined at $a$. Multiplying by $z-a$ and letting $z$ tend to $a$, we get $$\frac{1}{p'(a)}=b.$$ And there is nothing special about $a_1$. –  Pierre-Yves Gaillard Dec 7 '11 at 16:03
What I meant is that there are plenty of misprints e.g. in $\sum p(z)z^n$ (which is actually equal to $p(z)/(1-z)$) do you really mean that or do you mean $\sum p_n(z)z^n$ for certain sequence $p_n$? I just want you to think about your problem carefully and then read the question carefully. I myself often makes the mistake to skip some necessary assumption etc. –  AD. Dec 7 '11 at 18:40

2 Answers 2

up vote 2 down vote accepted

Write $f = C \prod_i \frac{1}{1 - z/a_i}$, each has the regular power series expansion for $\frac{1}{1+z}$ for $|z| < |a_i|$, done!

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A standard theorem of complex analysis says: If $z\mapsto f(z)$ is analytic in the disk $D_M:=\{z\in{\mathbb C}\ |\ |z|<M\}$ then the Taylor series of $f$ at the origin has a convergence radius $\rho\geq M$. In your case the function $f(z):={1\over p(z)}$ satisfies the assumption of the theorem, and in addition it has a pole on $\partial D_M$. Therefore one definitely has $\rho=M$.

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