The Unapologetic Mathematician

Mathematics for the interested outsider

Derivatives of Power Series

The uniform convergence of a power series establishes that the function it represents must be continuous. Not only that, but it turns out that the limiting function must be differentiable.

A side note here: we define the derivative of a complex function by exactly the same limit of a difference quotient as before. There’s a lot to be said about derivatives of complex functions, but we’ll set the rest aside until later.

Now, to be specific: if the power series \sum\limits_{n=0}^\infty a_n(z-z_0)^n converges for |z-z_0|<r to a function f(z), then f has a derivative f', which itself has a power series expansion

\displaystyle f'(z)=\sum\limits_{n=1}^\infty na_n(z-z_0)^{n-1}

which converges within the same radius r.

Given a point z_1 within r of z_0, we can expand f as a power series about z_1:

\displaystyle f(z)=\sum\limits_{k=0}^\infty b_k(z-z_1)^k

convergent within some radius R of z_1. Then for z in this smaller disk of convergence we have

\displaystyle\frac{f(z)-f(z_1)}{z-z_1}=b_1+\sum\limits_{k=1}^\infty b_{k+1}(z-z_1)^k

by manipulations we know to work for series. Then the series on the right must converge to a continuous function, and continuity tells us that each term vanishes as z approaches z_1. Thus f'(z_1) exists and equals b_1. But our formula for b_1 tells us

\displaystyle f'(z_1)=b_1=\sum\limits_{n=1}^\infty\binom{n}{1}a_n(z_1-z_0)^{n-1}=\sum\limits_{n=1}^\infty na_n(z_1-z_0)^{n-1}

Finally, we can apply the root test again. The terms are now \sqrt[n]{n}\sqrt[n]{|a_n|}. Since the first radical expression goes to {1}, the limit superior is the same as in the original series for f: \frac{1}{r}. Thus the derived series has the same radius of convergence.

Notice now that we can apply the exact same reasoning to f'(z), and find that it has a derivative f''(z), which has a power series expansion

\displaystyle f'(z)=\sum\limits_{n=2}^\infty n(n-1)a_n(z-z_0)^{n-2}

which again converges within the same radius. And so on, we determine that the limiting function of the power series has derivatives of arbitrarily large orders.

September 17, 2008 Posted by | Analysis, Calculus, Power Series | 3 Comments

   

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