What is the Infinitely Differentiability Theorem for Functions?

[Nedeljko]

Homework Statement

If $$f:R\longrightarrow R$$ is a infinitely differentiable function then the function $$g:R\longrightarrow R$$ defined as

$$g(x)=\left\{ \begin{array}{ll} \frac{f(x)-\sum_{k=0}^n\frac{f^{(k)}(0)}{k!}x^k}{x^{n+1}}, & x\neq 0, \vspace{0.5em}\\ \frac{f^{(n+1)}(0)}{(n+1)!}, & x=0, \end{array} \right.$$

is also infinitely differentiable. Prove it.

Homework Equations

No.

The Attempt at a Solution

It is easy to prove that the function $$g$$ is continuous. By computing derivatives I can prove that the function is three times differentiable for example, but I can not make inductive step. Is this theorem known under any name?

 

[jbunniii]

Try simplifying the $x \neq 0$ case by expanding $f(x)$ as a Taylor series.

 

[Nedeljko]

What if $$f$$ is not analytic near 0?

 

[jbunniii]

What if $$f$$ is not analytic near 0?

Good point.

Well, the numerator is simply the remainder term $R_n(x)$, which is well-defined provided $f$ has enough derivatives, whether or not it is analytic (Taylor's theorem). For $x \neq 0$, there exists $\xi$ such that $0 < \xi < x$ such that

$$R_n(x) = \frac{f^{n+1}(\xi)}{(n+1)!}x^{n+1}$$

Unfortunately, $\xi$ depends on $x$ so it's not a slam dunk from here.

 

[Nedeljko]

By this method I can prove that the function $$g$$ is infinitely Peano differentiable. It means that there are constants $$a_i$$ such that for any $$n$$ holds

$$\lim_{x\rightarrow 0}\frac{g(x)-\sum_{k=0}^n\frac{a_k}{k!}x^k}{x^{n+1}}=0$$.

But, this is necessary and insufficient condition for the infinite differentiability in the ordinary sense.

 

[Nedeljko]

Somebody has a idea?

 

[Nedeljko]

I solved the problem.