# Unknown real non-analytic smooth function

1. Jul 20, 2009

### milovanovic

This is my first post on physics forums, 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:

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

where k is some real constant, and g(x) is an arbitrary unknown also real function. I want to make another function h(x) in a following way:

$$h(x)= \left\{ \begin{array}{lr} f(x)&x\le 0\\ 0&x\ge 0 \end{array} \right.$$

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

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

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 construct g(x) such that h(x) is smooth at every point (zero is the non-trivial one).

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

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

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 my f(x) with:

$$z(x)= \left\{ \begin{array}{lr} -\exp\left(\frac{1}{x}\right)&x\le 0\\ 0&x\ge 0 \end{array} \right.$$

Cheers.