How to show that $e^{-\frac{1}{x^2}}$ does not have a Taylor-Maclaurin expansion

Question

I know that $x\neq0 \Rightarrow f(x) = e^{-\frac{1}{x^2}},f(0)=0$ does not have a good Taylor-Maclaurin expansion (By good I mean that the this expansion is not eonverging to $e^{-\frac{1}{x^2}}$) since all of its derivatives are $f^{(n)}(x)|_{x=0} = 0$.

But, I want to show it by showing that the lagrange remainder is not converging to $0$. How can I do that. I mean, isn't the lagrange remainder always $0$ in this case and therefore also converges to $0$?

Thanks

Answer 1

The Lagrange remainder is not $0$, because the remainder $R_n$ is, by definition, equal to $$R_n = f(x) - P_n(x)$$

where $P_n(x)$ is the $n$-th degree Maclauring polynomial for $f$. Note that $P_n(x)$ is equal to a very simple expression for all values of $n$, which should make $R_n$ easy to calculate.

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Now, of course one way to calculate $R_n$ is by using the formula

$$R_n = \frac{f^{(n+1)}(x^*)}{(n+1)!}\cdot (x-x_0)^{n+1}$$

however this formula is true for some $x^*\in(x_0, x)$ so I wouldn't suggest using this here. Much easier to actually calculate $R_n$.