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Good afternoon. I need to claculate the general n-term Taylor's expansion at zero of 2 functions:

  1. $e^{-x^2}$
  2. $e^{-\frac{1}{x^2}}$ if $x \ne 0$ and $0$ otherwise

For the first function everything seems easy.

First derivative is $-2x e^{-x^2}$, which is $0$

Second derivative is $-2 e^{-x^2} + 4x^2 e^{-x^2}$, which is $-2$

Third derivative is zero again

4th is $12 e^{-x^2} - 48x^2 e^{-x^2} + 16x^4 e^{-x^2}$ which is $12$

In the end after a few more calculation I realised that odd derivatives are equal to zero and then the form of th answer is something like $\frac{(-x^2)^n}{n!}$

Now, for the second function I instantly ran into a problem of derivative being undefined at zero. How do I deal with that?

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    $\begingroup$ All the derivatives at $0$ for the second function are $0$. $\endgroup$ – Kavi Rama Murthy Apr 15 at 8:42
  • $\begingroup$ Thank you for your reply, but correct me if I'm wrong: the first derivative is $e^{-\frac{1}{x^2}}*\frac{2}{x^3}$ which is kinda not $0$ at $x = 0$, right? $\endgroup$ – Makina Apr 15 at 8:45
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The function $e^{-1/x^2}$ is a pathological one. All its derivatives are zero, so that all its Taylor polynomials are identically zero. This does not invalidate Taylor's theorem: the whole function is supplied by the remainder of the expansion.


The expressions of the derivatives at zero are indeed indeterminate forms. Anyway a derivative is a limit, which allows you to replace the indeterminate forms by their limits.

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Sometimes it is easier to work directly with known series rather than applying Taylor's Theorem. For your first problem, for example, note that $$e^t = \sum_{n=0}^{\infty} \frac{1}{n!} t^n$$ so substituting $t=-x^2$, $$e^{-x^2} = \sum_{n=0}^{\infty} (-1)^n \frac{1}{n!} x^{2n}$$

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  • $\begingroup$ You should complete the argument by saying that this is the Taylor series of $e^{-x^2}$, and not just some entire series. $\endgroup$ – Yves Daoust Apr 15 at 12:40

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