Calculating Power Series Coefficients with Differentiability

[Karlisbad]

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]

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.

 

[HallsofIvy]

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.

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.