# Power series

1. Nov 1, 2006

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.

2. Nov 1, 2006

### quasar987

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: Nov 1, 2006
3. Nov 1, 2006

### HallsofIvy

Staff Emeritus
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.