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


  1. How did you obtain the equation of:

    \frac{f(z) - f(z_1)}{z - z_1}?

    Since f(z_1) = 0 about z_1, the above becomes

    \frac{b_0}{z-z_1} + b_1 + b_2(z-z_1) + \cdots,

    which is different from your result.

    Comment by Bernd | September 23, 2008 | Reply

  2. Where do you get that f(z_1)=0? I just gave the power series expansion about z_1, which clearly indicates that f(z_1)=b_0.

    Comment by John Armstrong | September 23, 2008 | Reply

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

    Pingback by Analytic Functions « The Unapologetic Mathematician | September 27, 2008 | Reply

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