- #1

Karlisbad

- 131

- 0

[tex] f(x)=\sum_{n=0}^{\infty}a(n)x^{n} [/tex]

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

[tex] n!a(n)=D^{n}f(0) [/tex] (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.