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A flat function is a smooth function $ƒ : \mathbb{R} → \mathbb{R}$ all of whose derivatives vanish at a given point $x_0 \in \mathbb{R}$.

Can anybody suggest a non-trivial example of function flat at $x_0=0$ which is not an "obvious" variant of the following:

  1. $f(x) = \begin{cases} 0 &\mbox{if } x \leq0 \\ e^{-\frac{1}{x}} & \mbox{if }x>0. \end{cases} $
  2. $f(x)=e^{-\frac{1}{x^2}}$ on $\mathbb{R}$.
user123933
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user95731
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    Actually I want a flat function where no exponential term comes or if comes then that should be non-trivially and drastically new function not just simple modification of given two examples. – user95731 Dec 05 '13 at 19:00
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    I think $$f(x) = \begin{cases} 0 &, x \leqslant 0\ \dfrac{1}{\Gamma\left(\frac1x\right)} &, x > 0 \end{cases}$$ works. But I'm not in the mood to check that really all derivatives in $0$ exist and are $0$. – Daniel Fischer Dec 05 '13 at 19:12
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    It's not clear (to me) what you would admit as "non-trivial" or "drastically new". For example, if $f$ is positive except at $0$, then $f(x) = e^{\log f(x)}$, which "involves the exponential function". Are you seeking a function of some specific type (e.g., a function not obtained by differentiating or antidifferentiating a piecewise-elementary function)? Is there some motivation other than idle curiosity? – Andrew D. Hwang Dec 06 '13 at 00:08
  • You are looking for a function which has zeros of infinite order. If such a function exists and is analytic, you could try writing down a power series representation for it and working out conditions under which all the derivatives vanish at one point. I'm not sure you'll find anything that different than what has already been pointed out though. – A. Thomas Yerger Apr 29 '15 at 06:38
  • Does $\lvert x\rvert^{1/x^2}$ work? – Akiva Weinberger Jun 11 '15 at 02:14
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    @AlfredYerger If such a function is analytic, it is the constant zero function, just by taylor expanding about the point in question. – Steven Gubkin Jul 13 '15 at 13:15

3 Answers3

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Sure. Let $f(x)$ be the first of your functions, and let $g(x) = f(x-a),$ for ANY real $a>0$

Igor Rivin
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If $p$ is a polynomial, and $g$ is your second function, then $pg$ is also extremely flat (using Spivak's terminology). Why? Because $(pg)'(0) = p'(0) g(0) + p(0)g'(0)$, where each of the $g$-factors is zero; repeated derivatives similarly all have the form $\sum_{i,k} c_{i,k} p^{(i)} g^{(k-i)}$, in which all the $g$-factors are zero because $g$ is extremely flat (and $g(0) = 0$).

John Hughes
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For example:

$f(x) = \begin{cases} 0 &\mbox{if } x \leq0 \\ \operatorname{erfc}\left(\frac{1}{x}\right) & \mbox{if }x>0. \end{cases} $

(Although $e^{-\frac{1}{x^2}}$ still plays a role because it's in the derivative of $\operatorname{erfc}\left(\frac{1}{x}\right)$.)

Luca Citi
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