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Math Help - Unknown real non-analytic smooth function

  1. #1
    Newbie
    Joined
    Jul 2009
    Posts
    3

    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:

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

    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:

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

    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.

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

    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.

    <br />
 \lim_{x\to0^-}g(x)\to\infty<br />
    <br />
\lim_{x\to0^-}\frac{x}{g(x)}\frac{dg(x)}{dx}=0<br />

    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:

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

    Cheers.
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  2. #2
    Senior Member
    Joined
    Apr 2009
    From
    Atlanta, GA
    Posts
    408

    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
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  3. #3
    Newbie
    Joined
    Jul 2009
    Posts
    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.

    However, I appreciate your help.
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