Given two power series,




It is easy to form their product




But many of the series I come across only contain negative powers of $x$, that is


Is there any tricks or methods anyone knows of to find the series representation of the product $f(x)h(x)$ ??

  • 2
    $\begingroup$ Formally, I think this is just a matter of changing $x$ into $\frac{1}{x}$ in your formula. $\endgroup$
    – Joel Cohen
    Jun 11, 2011 at 22:27
  • $\begingroup$ @Joel Cohen But if it was a simple matter of making the substitution $x\mapsto\frac{1}{x}$ the resulting power series would be for the product $f(x)h(\frac{1}{x})$. I dont see an obvious way to transform the this series to obtain a series for $f(x)h(x)$ $\endgroup$
    – aukie
    Jun 12, 2011 at 0:09

1 Answer 1


If it exists, then we can say that the function is defined using the infinite Laurent series $$f(z)h(z) = \sum_{j=-\infty}^{\infty} c_j z^j$$ where $c_j$ is defined for nonnegative $j$ as $$c_j=\sum_{k=0}^{\infty} a_k d_{k+j}$$ and for nonpositive $j$ as

$$c_j=\sum_{k=0}^{\infty} a_{k-j} d_k$$

  • $\begingroup$ .. Thanks. I was aware of this result but I was hoping something more managable/practicle existed in the particular case the terms laurant series do not involve any non-negative powers of $x$ $\endgroup$
    – aukie
    Jun 12, 2011 at 0:13
  • $\begingroup$ I'm confused by that comment. Which series does not involve any non-negative powers? Suppose $f(z) = \sum_{n=0}^\infty a_n z^n$ (converging, say, for $|z| < R_1$) and $h(z) = \sum_{n=0}^\infty d_n z^{-n}$ (converging, say, for $|z| > R_2$ where $R_2 < R_1$). Thus $g(z) = f(z) h(z)$ is analytic in $R_2 < |z| < R_1$. If the Laurent series of this function has no nonnegative powers of $z$, it extends to an analytic function on $|z| > R_2$ (including $\infty$), and $f(z) = g(z)/h(z)$ is analytic on the whole Riemann sphere except for a finite number of poles, and thus is a rational function. $\endgroup$ Jun 12, 2011 at 20:08
  • $\begingroup$ @Robert Israel. I want to find the series representation of the product $fh$. The series representation of $h$ consists only of terms involving non-positive integer powers of $x$ and $f$ has a power series representation. If I treat them both as Laurent series I have little hope of determining the coefficients $c_n$. Thats the point i was trying to make. Perhaps i'm asking for something that does not exist but i was hoping i could compute the coefficients $c_n$ from $a_n$ and $d_n$. As opposed to dealing with infinite sums $\endgroup$
    – aukie
    Jun 13, 2011 at 23:25

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