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.

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September 17, 2008 - Posted by John Armstrong | Analysis, Calculus, Power Series

3 Comments »

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
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
[...] 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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