# Math Help - Real-valued function

1. ## Real-valued function

Hi ! Do you know how to solve this problem:

Find a real-valued function on R possessing derivatives of all orders whose Taylor series at a certain point converges to the function only at that point.

2. Originally Posted by Milus
Hi ! Do you know how to solve this problem:

Find a real-valued function on R possessing derivatives of all orders whose Taylor series at a certain point converges to the function only at that point.
Consider $x^{x^x}$ at $x=0$

3. Thank,but still I do not know what means that up side down L inside the log parentasice.
Also, do you have any idea how to prove it.

4. Originally Posted by Milus
Thank,but still I do not know what means that up side down L inside the log parentasice.
Also, do you have any idea how to prove it.
That is my signature, that is not part of the post. And for the proof try finding the general form for the n-th derivative of x.

5. Can you help me a little more?
I am still strugling with general form of the derivative

6. Originally Posted by Milus
Hi ! Do you know how to solve this problem:

Find a real-valued function on R possessing derivatives of all orders whose Taylor series at a certain point converges to the function only at that point.
I didn't know MathStud28's example. The example that I would have called the "classical" one is: $f(x)=e^{-\frac{1}{|x|}}$ for $x\neq 0$ and $f(0)=0$. I can't remember which famous mathematician introduced it, perhaps Cauchy, anyway it has a long history. And it seems simpler to deal with.

For $x>0$, $f(x)=e^{-\frac{1}{x}}$. First notice that $f(x)\to_{x\to 0^+} 0$. Notice as well $\frac{e^{-\frac{1}{x}}}{x^n}\to_{x\to 0^+}0$ for any integer $n$. To convince yourself of this, substitute $u=\frac{1}{x}$, so that the limit is $\lim_{u\to\infty} u^n e^{-u}$.
Now, the $n$-th derivative of $f$ is (easily) seen to be a rational function of $x$ (a quotient of two polynomials) times $e^{-\frac{1}{x}}$. This is proved by induction (suppose there is a rational function $R$ such that the $n$-th derivative is, for $x>0$, $R(x)e^{-\frac{1}{n}}$, and prove that the next derivative is again of the same kind).
Because of the previously mentioned limit, you obtain that the n-th derivative converges to 0 as $x$ tends to 0 from the right. By symmetry, the same holds from the left.
Finally, there is a theorem (that you probably know if you're asked this problem) telling you that, as a consequence, $f$ is indefinitely differentiable at 0 and the derivatives at 0 are the limits of the derivatives at $x$ when $x$ tends to 0. That is to say, all derivatives at 0 exist and are 0. Yet, the function is only zero at zero...