# Thread: Unknown real non-analytic smooth function

1. ## Unknown real non-analytic smooth function

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:

$
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.

2. ## An idea for an approach...

The structure of $f\circ g$ (while looking quite random and complex at first) actually yields some good clues:

Start with the original function: $f(x)=kx-g(x)\left[1-e^{-\frac{kx}{g(x)}}\right]$

Rewrite: $e^{-\frac{kx}{g(x)}} = 1-\frac{kx}{g(x)}+\frac{f(x)}{g(x)}$

Denote new functions $A(x)=\frac{f(x)}{g(x)}$ and $B(x)=-\frac{kx}{g(x)}$

The equation now becomes: $e^{B(x)}=1+B(x)+A(x)$

Deriving, $B'(x)e^{B(x)}=B'(x)+A'(x)$, or $e^{B(x)}=1+\frac{A'(x)}{B'(x)}$

Therefore $A(x)+B(x)=\frac{A'(x)}{B'(x)}$ ------(1) (The same result can be shown using Taylor Series)

Find such an A and B and convert them back in terms of f and g: $f(x)=-kx\frac{A(x)}{B(x)}$ and $g(x)=\frac{-kx}{B(x)}$

I have looked at a handful of candidate functions, A and B but have not found anything by trial. Find any two functions A and B whose sum equals the ratio of their first derivatives or prove that none exist. This is just a first order differential equation, so it should be possible to solve. I took the liberty of starting a thread for the DiffEq guys to look into it: http://www.mathhelpforum.com/math-he...96754-b-b.html

3. ## ...which still doesn't solve my problem, but thanks anyway.

I was on holiday, so that's why I'm replying so late.

First of all, thank you for your interest on my post/problem. Unfortunately, besides really nice gymnastics on the functions $f(x)$ and $g(x)$, there's not much helpful information in there in terms of making $h(x)$ smooth, and hence all derivatives of $f(x)$ zero in $x=0$.