Power series

  • Thread starter Karlisbad
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  • #1
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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.
 

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  • #2
quasar987
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Do you have a question?

Btw - if the radius of convergence of

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

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

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

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

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

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

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

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

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

*(which in this case expresses the equality [itex]D^{n}g(0)=D^{n}f(0)[/itex])

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 [itex]n\rightarrow \infty[/itex].
 
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  • #3
HallsofIvy
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Karlisbad said:
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.
The power series expansion, about [itex]x= x_0[/itex] of a [itex]C^\infty[/itex] 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.
 

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