The Unapologetic Mathematician

Mathematics for the interested outsider

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.

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

6 Comments »

  1. […] Power Series So we know that we can have two power series expansions of the same function about different points. How are they related? An important step in this […]

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  2. […] Now, to be specific: if the power series converges for to a function , then has a derivative , which itself has a power series expansion […]

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  3. […] we have power series expansions of functions around various points, and within various radii of convergence. We even have formulas […]

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  4. […] Okay, we know that power series define functions, and that the functions so defined have derivatives, which have power series expansions. And thus […]

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  5. […] what functions might we try finding a power series expansion for? Polynomials would be boring, because they already are power series that cut off after a finite […]

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  6. […] to stop what I was working on about linear algebra. Instead, I set off on power series and how power series expansions can be used to express analytic functions. Then I showed how power series can be used to solve […]

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