1. ## TAYLOR-function null

hello,

f is a function $C^{\infty}$ from [a,b] to $\mathbb{R}$.

We suppose :

$\exists x_0 \in ]a,b[ \ and \ k \ real \ >0 | \forall n\in \mathbb{N},f^{(n)}(x_0)=0 \ \ and \ \ sup_{[a,b]}\{|f^{(n)}(x)|\}\le k^nn!$

1) Show that f is the function null from $]x_0-\frac{1}{k},x_0+\frac{1}{k}[$.

I think that we must use Taylor's inegalities which say that (I don't know if it's the same name ...) :

for $u \in [a,b]$
$|f(b)-\sum_{k=0}^n\frac{(x-u)^k}{k!}f^{(k)}(u)|\le M\frac{(b-u)^{n+1}}{(n+1)!}$

where M majors $f^{n+1}$ from [a,b].

so, here, we have:

for $x\in ]x_0-\frac{1}{k},x_0+\frac{1}{k}[$

$|f(x)|\le \frac{|x-x_0|^{n+1}}{(n+1)!}M$

howerver we can take $M=k^{n+1}(n+1)!$

and $|x-x_0|\le \frac{1}{k}$

so we have:

$|f(x)|\le 1$ ...

and I have no idea to keep on the demonstration.

thanks

do you understand what I said ?

2. if I am mistaken tell me (In first, I want improve my english and do maths ) ...

thanks

3. Originally Posted by J.R
hello,

f is a function $C^{\infty}$ from [a,b] to $\mathbb{R}$.

We suppose :

$\exists x_0 \in ]a,b[ \ and \ k \ real \ >0 | \forall n\in \mathbb{N},f^{(n)}(x_0)=0 \ \ and \ \ sup_{[a,b]}\{|f^{(n)}(x)|\}\le k^nn!$

1) Show that f is the function null from $]x_0-\frac{1}{k},x_0+\frac{1}{k}[$.

I think that we must use Taylor's inegalities which say that (I don't know if it's the same name ...) :

for $u \in [a,b]$
$|f(b)-\sum_{k=0}^n\frac{(x-u)^k}{k!}f^{(k)}(u)|\le M\frac{(b-u)^{n+1}}{(n+1)!}$

where M majors $f^{n+1}$ from [a,b].

so, here, we have:

for $x\in ]x_0-\frac{1}{k},x_0+\frac{1}{k}[$

$|f(x)|\le \frac{|x-x_0|^{n+1}}{(n+1)!}M$

howerver we can take $M=k^{n+1}(n+1)!$

and $|x-x_0|\le \frac{1}{k}$

so we have:

$|f(x)|\le 1$ ...

and I have no idea to keep on the demonstration.
Your argument is correct, and it shows that $|f(x)|\leqslant k^{n+1}|x-x_0|^{n+1}$. But you are told that x belongs to the open interval $(x_0-\tfrac1k,x_0+\tfrac1k)$ (or in the horrible French notation $]x_0-\tfrac1k,x_0+\tfrac1k[$ ), so you know that $k|x-x_0|$ is strictly less than 1, so its (n+1)th power can be made arbitrarily small and hence $f(x)=0$.

4. ## Problem solve

message removes

5. ## Problem solve

Sincerely,
thank you very much.

@+