# Why doesn't the Taylor expansion at 0 around $e^{-1/x^2}$ converge to the function itself?

I know that if you take the Taylor expansion at $$x = 0$$ of $$e^{-1/x^2}$$, you get $$f(x) = 0$$. However, I was wondering if a rigorous proof could be shown of why the Taylor expansion and the actual function differ at every point (other than 0) in this case. I was thinking of using the Lagrange Remainder: $$R_n(x) = \frac{f^{(n)}(z)x^n}{n!}$$ where $$z \in (0, x)$$ for any $$x$$. I tried to show why $$\lim_{n \rightarrow \infty} R_n(x) \neq 0$$ but I was running into trouble. Specifically, it seems like the $$n!$$ term grows faster than the top, alongside the fact it was quite hard to characterize the value of $$f^{(n)}(z)$$. Any help would be appreciated.

• The absolute value of $f^{(n)}(z)$ will be dominated by the supremum of $g(z)=e^{-1/z^2}2^nz^{-3n}$, which is achieved by $g(2/\sqrt{3n})=(Cn^{3/2})^n=h(n)$ when $n$ is large enough, with $C=e^{-3/4}3\sqrt{3}/4$. The correct bound of $R_n$ is thus $\approx h(n)x^n/n!$, which blows up as $n\to\infty$. – Edward H Dec 3 '19 at 3:34
• You just showed that $f$ is bounded by a function that blows up as n gets large. In fact, the remainder term is just f itself. – Matematleta Dec 3 '19 at 3:59
• @Matematleta Yes. – Edward H Dec 3 '19 at 7:04

Since the Maclaurin polynomial is zero for each integer $$n$$, the remainder term is just $$f$$ itself, and since this does not vanish for $$x\neq 0,\ f$$ is not represented by a Taylor series ar $$x=0.$$
The Taylor expansion does converge - to the function that's everywhere $$0$$. For every $$n$$ the remainder term at $$x$$ (the error in the Taylor expansion) is $$\exp(-1/x^2)$$, the value of the function you started with.
• The remainder term is (by definition) the difference between the $n$th degree Taylor polynomial and the function. Since that polynomial is identically $0$, the remainder term is the value of the function. You could use that fact to work out what the value of $z$ is in the Lagrange formula for the remainder. – Ethan Bolker Dec 3 '19 at 3:49