Nonanalytic function with Taylor series of all orders

Asked Jan 31 '24 at 18:01

Active Jan 31 '24 at 18:16

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I am revising Calculus main concepts, I am with analytic functions. It is a well known fact that a function being analytic implies that admits a Taylor polynomial of arbitrary large order. However, the converse is not true. The prototypical example is the function given by $f(x) = e^{-\frac{1}{x^2}}$ if $x\ne 0$, $f(0) = 0$. My goal is to show that all derivatices vanish at $0$, so all Taylor polynomials are 0, and so is the corresponding infinite series. However, the function is only $0$ at $x=0$, so it can be analytical.

The first derivative would be given by $$ \lim_{x\to 0} \frac{-e^{-1/x^2}}{x} $$

What is the "elegant" way to compute this limit, avoiding L'Hôpital?

edited Jan 31 '24 at 18:16

asked Jan 31 '24 at 18:01

user210089

Try to substitute $t=-x^{-2}$ – Egor Ivanov Jan 31 '24 at 18:06
@EgorIvanov Is an acceptable change of variable? – user210089 Jan 31 '24 at 18:09
Why do you think it is unacceptable? – Egor Ivanov Jan 31 '24 at 18:10
@EgorIvanov Is not invertible. – user210089 Jan 31 '24 at 18:16
@MarianoSuárez-Álvarez thanks – user210089 Jan 31 '24 at 18:17
@user210089 But for limits it is unnecessary for substitution to be invertable. The only thing required is being continuous. Because the main idea is that $\lim_{x} f(x)={x=g(t)}=\lim_{t} f(g(t))$, where $g$ is continuous – Egor Ivanov Jan 31 '24 at 18:19
@user210089 But if something bothers you, you can find both $0+0$ and $0-0$ limits with invertible substitutions and check that they are equal – Egor Ivanov Jan 31 '24 at 18:21
@MarianoSuárez-Álvarez yeah, sure, because we need $x=g(t)$ and when $-t=x^{-2}$ we need to say that $x=\pm(\sqrt{-t})^{-1}$ and check both variants. Or another approach is it say that the function is odd and if one-side limit exists and equals $0$ then the other side one is to be the same. – Egor Ivanov Jan 31 '24 at 18:30
@MarianoSuárez-Álvarez I have a question about this. Think of the hierarchy of infinities. For example, $exp x/x$ goes to infinity as $x$ goes to infinity. Can I change $x$for another function that goes to infinity as $x$ goes to $a$, and write that $exp(f(x))/f(x)$ goes to infinity as x goes to $a$? What do I need? – user210089 Jan 31 '24 at 19:09
See my answer to Examples of applying L'Hôpitals rule ( correctly ) leading back to the same state? – Dave L. Renfro Jan 31 '24 at 19:43

Nonanalytic function with Taylor series of all orders

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