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

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

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 &lt; \xi &lt; 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.