Representing Functions by Power Series

Once we know that a power series has a radius of convergence, we can use it to define a function. Suppose the power series $\sum$a_nxⁿ has radius of convergence R > 0, and let I = (- R, R). We now define a function f on this open interval I as follows:

f (x) = $$\sum_{{n=1}}^{\infty}$$a_nxⁿ        for x $\in$ I.    

It turns out that this is the last, and best behaved of the classes of functions we study in this course.

In fact all of what we say below remains true when R = $\infty$, provided we interpret the open interval I as $\mathbb {R}$.

Theorem 7.9   Let $\sum$a_nxⁿ be a power series with radius of convergence R > 0. Let I be the open interval (- R, R), and define f (x) = $\sum$a_nxⁿ for x $\in$ I. Then $\sum$na_nx^n-1 has radius of convergence R, f is differentiable on I, and

f'(x) = $$\sum_{{n=1}}^{\infty}$$na_nx^n-1    for x $\in$ I.    

We summarise this result by saying that we can differentiate a power series term - by - term everywhere inside the circle of convergence. If R = $\infty$, then this can be done for all x.

Proof. Quite a lot harder than it looks; we need to be able to re-arrange power series, and then use the Mean Value Theorem to estimate differences, and show that even when we add an infinite number of errors, they don't add up to too much. It can be found e.g. in (Spivak, 1967). $\qedsymbol$

Corollary 7.10   Let f and I be defined as 7.9. Then f has an indefinite integral defined on I, given by

G(x) = $$\sum_{{n=0}}^{\infty}$$$${\frac{{a_n}}{{(n+1)}}}$$xⁿ⁺¹    for x $\in$ I.    

Proof. Apply 7.9 to G to see that G'(x) = f (x), which is the required result. $\qedsymbol$