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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|.

Progress

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| 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|

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

EDIT 1

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?2011-12-07
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    Recall *partial fractions*.2011-12-07
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    @AD: Not sure what you mean I'm afraid.2011-12-07
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    @AndréNicolas: How would I use that in this case?2011-12-07
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    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).2011-12-07
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    It is better to do $N=2$ for concreteness. But in general there exist (complex) numbers $c_1,\dots,c_N$ such that $\frac{1}{p(z)}=\frac{c_1}{z-a_1}+\cdots+\frac{c_N}{z-a_N}$. The power series for the $\frac{c_i}{z-a_i}$ are done like your case $N=1$.2011-12-07
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    Hint: $$\frac{1}{p(z)}=\sum_{j=1}^N\ \frac{1}{p'(a_j)\ (z-a_j)}\quad.$$2011-12-07
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    It seems I've made a number of 'absurd' errors here. I'm very new to the forum so I apologise. Thanks for the help though; hadn't thought about simply writing $\frac{1}{p(z)}$ as a linear combination of the $\frac{1}{z-a_k}$ fractions. @Pierre-YvesGaillard: How do we know that the coefficient for each fraction is $\frac{1}{p'(a_j)}$? Thanks again by the way.2011-12-07
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    Dear @TJO: Short answer: "clear denominators". If it's too short, tell me! (You're welcome.)2011-12-07
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    @Pierre-YvesGaillard: I fear that may be too short.2011-12-07
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    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 $a_1$.2011-12-07
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    @Pierre-YvesGaillard Magic! Once again, thanks!2011-12-07
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    @TJO: Again, you're welcome, and, again, I didn't invent this! (+1 for your question!)2011-12-07
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    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.2011-12-07
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    @AD.: Apologies for the confusion there; I've edited appropriately.2011-12-07
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    You might be interested in the [Mittag-Leffler theorem](http://en.wikipedia.org/wiki/Mittag-Leffler%27s_theorem) applied to the set $\{a_1,\ldots,a_N\}$.2011-12-07
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    Regarding the "Further Problem": No, it can't, since there's a pole at distance $M$ from the origin and the radius of convergence can't extend beyond a pole.2011-12-07

2 Answers 2

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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| 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 $\rho=M$.