# The Unapologetic Mathematician

## Power Series Expansions

Up to this point we’ve been talking about power series like $\sum\limits_{n=0}^\infty c_nz^n$, where “power” refers to powers of $z$. This led to us to show that when we evaluate a power series, the result converges in a disk centered at ${0}$. But what’s so special about zero?

Indeed, we could just as well write a series like $\sum\limits_{n=0}^\infty c_n(z-z_0)^n$ for any point $z_0$. The result is just like picking up our original power series and carrying it over a bit. In particular, it still converges — and within the same radius — but now in a disk centered at $z_0$.

So when we have an equation like $f=\sum\limits_{n=0}^\infty c_n(z-z_0)^n$, where the given series converges within the radius $R$, we say that the series “represents” $f$ in the disk of convergence. Alternately, we call the series itself a “power series expansion” of $f$ about $z_0$.

For example, consider the series $\sum\limits_{n=0}^\infty\left(\frac{2}{3}\right)^{n+1}\left(z+\frac{1}{2}\right)^n$. A simple application of the root test tells us that this series converges in the disk $\left|z+\frac{1}{2}\right|<\frac{3}{2}$, of radius $\frac{3}{2}$ about the point $z_0=-\frac{1}{2}$. Some algebra shows us that if we multiply this series by $1-z=\frac{3}{2}-\left(z+\frac{1}{2}\right)$ we get ${1}$. Thus the series is a power series expansion of $\frac{1}{1-z}$ about $z_0=-\frac{1}{2}$.

This new power series expansion actually subsumes the old one, since every point within ${1}$ of ${0}$ is also within $\frac{3}{2}$ of $-\frac{1}{2}$. But sometimes disks overlap only partly. Then each expansion describes the behavior of the function at values of $z$ that the other one cannot. And of course no power series expansion can describe what happens at a discontinuity.

September 15, 2008 - Posted by | Analysis, Calculus, Power Series

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