I am stuck in finding a function $g(x)$ that is not even, so $g(x)\neq g(-x)$, can be differentiated infinity many times and has $$\frac{g^{(2n-1)}(0)}{(2n-1)!}=0.$$

I have tried with at lot of different combinations, fx. $\sin{(x)}+e^{-x}$, but can't seem to find any that fits the criteria. Shall I try to just guess more functions, and see whether one fits, or is there a system of a kind that I haven't seen?

  • $\begingroup$ $x \mapsto e^{-1/x^2}$ seems a likely candidate. See, for example, this post math.stackexchange.com/questions/491227/… $\endgroup$ – michaelhowes Jun 4 '18 at 12:35
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    $\begingroup$ @michaelhowes Perhaps with some modification (it is an even function after all!) $\endgroup$ – preferred_anon Jun 4 '18 at 12:38
  • $\begingroup$ @Daniel Littlewood My apologies. Not thinking cleary. I believe using the same function for positive $x$ but setting it equal to 0 for all $x\le 0$ should work. $\endgroup$ – michaelhowes Jun 4 '18 at 12:40
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    $\begingroup$ @michaelhowes I agree, I think if you glue it to $-e^{-1/x^{2}}$ then that works too (choosing a smooth non-analytic function is definitely the important step). $\endgroup$ – preferred_anon Jun 4 '18 at 12:41
  • $\begingroup$ It is really being creative, but I have tried with many different combinations, and it seems for me like that a function which has odd derivatives with 0 that gives 0, must be an even function. $\endgroup$ – Frederik Jun 4 '18 at 12:46

If $g$ has the property that it is analytic - i.e. that it converges to its Taylor series at every point - then you will be out of luck. Since for a power series, $$\sum_{k=0}^{\infty}c_{k}x^{k} \equiv \sum_{k=0}^{\infty}c_{k}(-x)^{k} \iff \text{for all odd $k$, }c_{k}=0.$$ Sadly, most of the 'simple' functions are analytic, so it is not so easy to think of an example which will work.

The function $f(x) = e^{-1/x^{2}}$ with $f(0)=0$ is a famous example of a function which is smooth (differentiable $k$-times for all $k$), and is not analytic at $0$. It can be shown by induction that $f^{(k)}(0)=0$ for all $k$, so the Taylor series at $0$ is identically $0$, while $f$ is not zero in a neighbourhood of $0$.

Sadly $f$ happens to be an even function, but this is not an obstacle. The following functions $$g_{1}(x) = \begin{cases}e^{-1/x^{2}}& x > 0\\-e^{-1/x^{2}}&x <0\end{cases}$$ and

$$g_{2}(x) = \begin{cases}e^{-1/x^{2}}& x > 0\\0&x <0\end{cases}$$

(both with $g_{i}(0)=0$) both have $g_{i}^{(k)}(0) = 0$ for all $k$, while $g_{1}$ is an odd function and $g_{2}$ is neither even nor odd (the second example is due to michaelhowes in the comments). The proof is completely analogous to this answer, but feel free to comment if you get stuck.


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