]> 30.7 Expressions for Coefficients of a Power Series

## 30.7 Expressions for Coefficients of a Power Series

We have for the most part so far discussed what to do when confronted with a series. You can test its convergence, estimate its limit, and try to find the function it represents, if it is a power series.

Another important question is: how can you find the coefficients in a power series expansion of a given function about some expansion point?

We know from our study of Taylor series in Section 10.2 that the coefficient of the j-th term will be the j-th derivative of the function at the expansion point, divided by $j$ factorial.

This is a useful fact, but not always useful enough, in part because it can be cumbersome to calculate or compute the higher derivatives of a complicated function.

Fortunately our standard functions can be defined in the complex plane, and in it we can give an integral representation of the coefficients of a power series, by using the residue theorem.

Suppose we have a function $f ( z )$ and wish to expand it in a series about the point $z '$ . We know that the integral of any function around a simple closed path in the complex plane that surrounds an isolated singular point $z '$ (and no other singular point) of $f$ is $2 π i$ times its residue at $z '$ , and the residue at $z '$ is the coefficient of $z − 1$ in the power series expansion of $f$ at the point $z '$ .

We can therefore deduce that the coefficient $a n$ of $z n$ in the power series expansion of $f ( z )$ about $z '$ , which is the residue of $f ( z ) ( z − z ' ) n + 1$ at $z = z '$ , is $( 2 π i ) − 1$ times the integral of $f ( z ) ( z − z ' ) n + 1$ on any simple closed path around $z '$ that does not include any singular point of $f$

$a n = 1 n ! f ( z ' ) ( n ) = 1 n ! d n f ( z ' ) d z ' n = 1 2 π i ∮ f ( z ) ( z − z ' ) n + 1 d z$

Integrals of this kind can be evaluated numerically for any $n$ without great difficulty.