This is my first post on physics forums, and I feel quiet enthusiastic about it, although the problem I have is pretty challenging.(adsbygoogle = window.adsbygoogle || []).push({});

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

[tex]

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

[/tex]

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

[tex]

h(x)=

\left\{

\begin{array}{lr}

f(x)&x\le 0\\

0&x\ge 0

\end{array}

\right.

[/tex]

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.

[tex]

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

[/tex]

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.

[tex]

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

[/tex]

[tex]

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

[/tex]

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:

[tex]

z(x)=

\left\{

\begin{array}{lr}

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

0&x\ge 0

\end{array}

\right.

[/tex]

Cheers.

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# Unknown real non-analytic smooth function

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