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I wanna show that for $f:(0,\infty)\rightarrow\mathbb R, x\mapsto\exp(-\frac1{x^2})$ the sum of the derivates of $f$, so $\sum\limits_{n=0}^\infty f^{(n)}(x)$, converges to $0$, so $$\lim\limits_{x\rightarrow0}\sum\limits_{n=0}^\infty f^{(n)}(x)=0$$

It's intuitively clear but I have some issues on writting it down. Can anybody help? Thanks!

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The individual terms go to $0$ as $x \to 0$, but I don't think it's at all clear that the infinite series converges for any $x > 0$. – Robert Israel Jun 20 '12 at 19:22
Formally, $\sum_{n=0}^\infty f^{(n)}(x) = \int_0^\infty e^{-t} f(x+t)\ dt$, and as $x \to 0+$ that should go to $\int_0^\infty e^{-t} e^{-1/t^2}\ dt$ which is not $0$. – Robert Israel Jun 20 '12 at 19:36
Note that, by Taylor's Theorem with Lagrange remainder, for any $\epsilon > 0$ and any positive integer $n$ there is $t_n$ with $0 < t_n < \epsilon$ and $f^{(n)}(t_n) = n! \epsilon^{-n} f(\epsilon)$. As $n \to \infty$, this goes to $+\infty$. – Robert Israel Jun 20 '12 at 19:53

The series $$ \sum_{n=0}^\infty f^{(n)}(x) $$ does not converge, at least for $x$ close to zero. Indeed, one shows by induction that $$ f^{(n)}(x)=P_n(1/x)f(x) \quad \forall \ x>0, \ n \in \mathbb{N}, $$ where $\{P_n\}_{n \ge 0}$ is a sequence of polynomials such that $\deg P_n=3n$ and $$ P_0(t)=1, \quad P_n(t)=2t^3 P_{n-1}(t)+P_{n-1}'(t) \quad \forall \ t \in \mathbb{R}, \ n \in \mathbb{N}. $$ For every $x>0$ and $n \in \mathbb{N}$ we have $$ q_n(x)=\frac{P_{n+1}(1/x)}{P_n(1/x)}=\frac{2}{x^3}+\frac{P_n'(1/x)}{P_n(1/x)}. $$ Thus $$ \lim_{x \to 0+}q_n(x)=\infty. $$

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You can prove the following: $$\eqalign{ & y = \exp \left( { - \frac{1}{{{x^2}}}} \right) \cr & \log y = - \frac{1}{{{x^2}}} \cr} $$

So that

$$y' = \frac{2}{{{x^3}}}\exp \left( { - \frac{1}{{{x^2}}}} \right)$$


$$y'' = \left( {\frac{2}{{{x^3}}} - \frac{3}{x}} \right)\left( {\frac{2}{{{x^3}}}\exp \left( { - \frac{1}{{{x^2}}}} \right)} \right)$$

Then, you can prove that, for any $k>0$

$$\mathop {\lim }\limits_{x \to 0} \frac{1}{{{x^k}}}\exp \left( { - \frac{1}{{{x^2}}}} \right) = 0$$

Then proving (maybe by induction) that the derivatives will be linear combinations of that expression will do the job - basically, let

$$F(x) = \exp \left(-\frac 1 {x^2}\right) \sum_{k=1}^r\frac{a_k}{x^k}$$

and prove that any $F^{(n)}$ will be of the same type (product rule and induction).

ADD: I didn't spot the $\infty$ in the series. The fact that

$$\lim_{x \to 0}\sum_{k=0}^n f^{(k)}(x)=0 $$

for finite $n$ does not mean that

$$\lim_{x \to 0}\sum_{k=0}^\infty f^{(k)}(x)=0$$

This is related to the notion of uniform convergence of series, which explains why the limit can't be evaluated termwise. As Robert pointed out, the limit might be interpreted as $$\int\limits_0^\infty {\exp \left( { - x - \frac{1}{{{x^2}}}} \right)dx} $$

which evaluates to $\approx 0.293$

One obtains the above by some Taylor series manipulation,

Expand the function as a Taylor series

$$f\left( {x + t} \right) = \sum\limits_{n = 0}^\infty {{f^{\left( n \right)}}\left( x \right)} \frac{{{t^n}}}{{n!}}$$

Multiply by $e^{-t}$ and integrate over $(0,\infty)$:

$$\eqalign{ & \int\limits_0^\infty {{e^{ - t}}f\left( {x + t} \right)dt} = \sum\limits_{n = 0}^\infty {{f^{\left( n \right)}}\left( x \right)\frac{1}{{n!}}\int\limits_0^\infty {{t^n}{e^{ - t}}dt} } \cr & \int\limits_0^\infty {{e^{ - t}}f\left( {x + t} \right)dt} = \sum\limits_{n = 0}^\infty {{f^{\left( n \right)}}\left( x \right)\frac{1}{{n!}}n!} = \sum\limits_{n = 0}^\infty {{f^{\left( n \right)}}\left( x \right)} \cr} $$

The integral function $$\int\limits_0^\infty {\exp \left( { - t - \frac{1}{{{{\left( {x + t} \right)}^2}}}} \right)dt} $$

converges for any $x$ and is continuous so it might be reasonable to expect that

$$\mathop {\lim }\limits_{x \to 0} \sum\limits_{n = 0}^\infty {{f^{\left( n \right)}}\left( x \right)} = \mathop {\lim }\limits_{x \to 0} \int\limits_0^\infty {\exp \left( { - t - \frac{1}{{{{\left( {x + t} \right)}^2}}}} \right)dt} = \int\limits_0^\infty {\exp \left( { - t - \frac{1}{{{t^2}}}} \right)dt} \approx 0.293$$

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