Taylor series, the n-term calculation

Question

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

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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?

Answer 1

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.

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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.

Answer 2

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 $t=-x^2$, $$e^{-x^2} = \sum_{n=0}^{\infty} (-1)^n \frac{1}{n!} x^{2n}$$