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Suppose I have two complex valued functions $f(z) = e^z$ and $g(z) = \frac{1}{z^2 + 1}$ How can I derive a series expansion for $f(x)g(x) = \frac{e^z}{z^2 + 1}$ using multiplication of power series? Note that $e^z = \sum^{\infty}_{n=0}{\frac{z^n}{n!}}$ and we can write $\frac{1}{z^2 + 1} = \frac{-1}{1-(z^2 + 2)}$ so that $\frac{-1}{1-(z^2 + 2)} = -\sum^{\infty}_{n=0}{(z^2 + 2)^n}$

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Why would you write $\frac{1}{1+z^2} = \sum^{\infty}_{n=0}{(z^2 + 2)^n}$ rather than $\frac{1}{1+z^2} =\sum^{\infty}_{n=0}{(-z^2)^n}$? –  Dennis Gulko Mar 6 '13 at 15:07
    
That would be the same I guess so it doesn't matter what representation to use –  user61835 Mar 6 '13 at 15:13
    
In fact - it does: you can't use your form to compute the series around $0$ –  Dennis Gulko Mar 6 '13 at 15:15
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2 Answers

The power series of $e^z=\sum_{n\geq 0}b_nz^n$ has infinite radius of convergence.

Now, actually, you should write $$ \frac{1}{1+z^2}=\frac{1}{1-(-z^2)}=\sum_{n\geq 0}(-z^2)^n=\sum_{n\geq 0}(-1)^nz^{2n}=\sum_{n\geq 0}c_nz^n $$ which has radius of convergence $1$.

It follows that $\frac{e^z}{1+z^2}$ can be developed as a power series $$ \frac{e^z}{1+z^2}=\sum_{n\geq 0}a_nz^n $$ with radius of convergence $\geq 1$ (consequence of the fact that the Cauchy product of two absolutely convergent series converges absolutely, itself a consequence of Mertens' theorem).

The coefficients $a_n$ are computed by convolution $$ a_n=\sum_{k=0}^{n}b_{n-k}c_k=\sum_{0\leq 2k\leq n}\frac{(-1)^k}{(n-2k)!}. $$

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why $\ge1$ and not $\le1$? –  nbubis Mar 6 '13 at 15:17
    
@nbubis Pick $|z|< \min(R_1,R_2)$. Then both series converge absolutely, so the product converges absolutely. Hence $R\geq |z|$. Let $|z|$ tend to this min, you get $R\geq\min(R_1,R_2)$. And it is not equal in general (only generically). Since the product can converge (sometimes) beyond the min of the radii. –  1015 Mar 6 '13 at 15:20
    
Nicely done. It seems that the series converges very very slowly at $x=1$, but that in the end it does converge there as well. I'm not sure yet about points $x>1$ –  nbubis Mar 6 '13 at 15:42
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The general way to multiply power series is the generalization of the way we multiply polynomials: if $f(z)=\sum_{n=0}^\infty a_nz^n$ and $g(z)=\sum_{n=0}^\infty b_nz^n$ then: $$f(z)g(z)=\sum_{n=0}^\infty\left(\sum_{k=0}^na_kb_{n-k}\right)z^n$$ (try this for polynomials and you'll see that it works)
In you case, $e^z = \sum^{\infty}_{n=0}{\frac{z^n}{n!}}$ and $\frac{1}{1+z^2} =\sum^{\infty}_{n=0}(-1)^nz^{2n}$. Hence: $$\frac{e^z}{z^2+1}=\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac{1}{k!}b_{n-k}\right)z^n, \hspace{5pt} b_k=\begin{cases}(-1)^m&k=2m\\0&k=2m+1\end{cases}$$

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