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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.

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    $\begingroup$ 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 $n\to\infty$. $\endgroup$ – Edward H Dec 3 '19 at 3:34
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    $\begingroup$ 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. $\endgroup$ – Matematleta Dec 3 '19 at 3:59
  • $\begingroup$ @Matematleta Yes. $\endgroup$ – Edward H Dec 3 '19 at 7:04
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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 $x=0.$

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  • $\begingroup$ Ah, that's actually quite straightforward. I feel kinda stupid now :/ $\endgroup$ – yizzlez Dec 3 '19 at 14:11
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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.

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  • $\begingroup$ I'm wondering why the Taylor expansion doesn't converge to the actual function itself. I think what I'm looking for is why the remainder term is nonzero. $\endgroup$ – yizzlez Dec 3 '19 at 3:41
  • $\begingroup$ The remainder term is nonzero because the error will always exist. $\endgroup$ – Spencer Kraisler Dec 3 '19 at 3:49
  • $\begingroup$ 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. $\endgroup$ – Ethan Bolker Dec 3 '19 at 3:49
  • $\begingroup$ This answer doesn't deserve the -1 $\endgroup$ – Calvin Khor Dec 3 '19 at 4:25
  • $\begingroup$ I agree with @CalvinKhor . $\endgroup$ – kimchi lover Dec 3 '19 at 4:43

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