# Power series

Let be the powe series:

$$f(x)=\sum_{n=0}^{\infty}a(n)x^{n}$$

then if f(x) is infinitely many times differentiable then for every n we have:

$$n!a(n)=D^{n}f(0)$$ (1) of course we don't know if the series above is

of the Taylor type, but (1) works nice to get a(n) at least for finite n.

quasar987
Homework Helper
Gold Member
Do you have a question?

Btw - if the radius of convergence of

$$\sum_{n=0}^{\infty}a(n)x^{n}$$

is R, then for all -R<x<+R, define

$$f(x)=\sum_{n=0}^{\infty}a(n)x^{n}$$

Then automatically, f defined in this way is $C^{\infty}$ because that is a property of power series. and additionally, the coefficients are related to f by $n!a(n)=D^{n}f(0)$.

On the other hand, if you start with a function f(x), that you know is $C^{\infty}$ on some interval (-a,a), and you define the series

$$\sum_{n=0}^{\infty}\frac{D^{n}f(0)}{n!}x^{n}$$

then the function g(x), defined by the value of the series where it converges, i.e.

$$g(x)=\sum_{n=0}^{\infty}\frac{D^{n}f(0)}{n!}x^{n} \ \ \ x\in(-R,R)$$

is not necessarily f(x). Although by the above, we do have that $n!a(n)=D^{n}g(0)$*, as you pointed out in your thread. It's a little subtlety.

*(which in this case expresses the equality $D^{n}g(0)=D^{n}f(0)$)

There are theorems however that give conditions for the equality of g(x) to f(x). One of them requires that the Lagrange remainder from Taylor's formula vanishes as $n\rightarrow \infty$.

Last edited:
HallsofIvy
Homework Helper
$$f(x)=\sum_{n=0}^{\infty}a(n)x^{n}$$
$$n!a(n)=D^{n}f(0)$$ (1) of course we don't know if the series above is
The power series expansion, about $x= x_0$ of a $C^\infty$ function is unique. It doesn't matter how you get it, it will be identical to the Taylor's series for the function about that point.