This is my first post on math help forum, and I feel quiet enthusiastic about it, although the problem I have is pretty challenging.

So let me start. I have a real function (function of a real variable) in form:

$\displaystyle

f(x)=kx-g(x)\left(1-\exp\left({-\frac{kx}{g(x)}}\right)\right)

$

wherekis some real constant, andg(x)is an arbitrary unknown also real function. I want to make another functionh(x) in a following way:

$\displaystyle

h(x)=

\left\{

\begin{array}{lr}

f(x)&x\le 0\\

0&x\ge 0

\end{array}

\right.

$

which should be smooth (all derivatives exist) at every point. Of course iff(x)is smooth, by definitionh(x) will be also smooth in all points butx=0. To make it also smooth at zero the following condition should be satisfied.

$\displaystyle

\lim_{x\to0^-}\frac{d^nf(x)}{dx^n}=0, \forall n\in\mathbb{N}

$

If we consider its Taylor series with this condition,f(x) is either going to be a constant or non-analytic function. Yes, you guessed, I need this second one.

Finally we came to the question.

I should constructg(x) such thath(x) is smooth at every point (zero is the non-trivial one).

I started calculating higher order derivatives off(x). Based on obtained results I reckon that if I findg(x) that satisfies following two conditions, my problem will be solved.

$\displaystyle

\lim_{x\to0^-}g(x)\to\infty

$

$\displaystyle

\lim_{x\to0^-}\frac{x}{g(x)}\frac{dg(x)}{dx}=0

$

Polynomial functions do not work here, so some trick with exponential one perhaps should work. Anyhow, if somebody has any clue how to proceed with this, euphemism is that I would be very grateful. Of course, I would be happy if someone proves me that this what I am looking for is not possible. I'll simply forward your message to my supervisor , and multiply myf(x) with:

$\displaystyle

z(x)=

\left\{

\begin{array}{lr}

-\exp\left(\frac{1}{x}\right)&x\le 0\\

0&x\ge 0

\end{array}

\right.

$

Cheers.