- #1
Karlisbad
- 131
- 0
Let be the powe series:
[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.
[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.