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Here's the problem (from little Spivak):

Define $f:\mathbb{R}\rightarrow\mathbb{R}$ by

$$f(x)=\begin{cases} e^{-x^{-2}} & x\ne 0 \\ 0 & x=0 \end{cases}$$

Show that $f$ is a $C^\infty$ function, and $f^{(i)}(0)=0$ for all $i$.

Anyhow, here's my proof. It gets pretty ugly at the end and I do a little handwaving (every term is of the form ... pretty much "because I say so", and the last step should be a better form of induction instead of "successive applications of the rule..." but the exact way to say it correctly escapes me.

For arbitrary $p,q:\mathbb{R}\rightarrow\mathbb{R}$,

$$(e^{q(x)}p(x))'=e^{q(x)} q'(x)p(x) + e^{q(x)} p'(x)=e^{q(x)}(q'(x)p(x)+p'(x))$$

Let $q(x)=-x^{-2}$, $p_0(x)=1$, and $p_{n+1}(x)=q'(x)p_{n}(x)+p_n'(x)=\frac{2}{x^3} p_n(x) + p_n'(x)$. Every term of $p_n(x)$ is therefore of the form $a x^{-n}$ for some $a,n\in\mathbb{R}$ and $n>0$.

Then $f^{(n)}(x)=e^{q(x)} p_n(x)=e^{-x^{-2}}p_n(x)$

By the definition of derivative,

\begin{align*} f^{(n+1)}(0) &= \lim_{h\rightarrow 0}\frac{f^{(n)}(h)}{h}\ &= \lim_{h\rightarrow 0}\frac{e^{-x^{-2}} p_n(h)}{h}\ &= \lim_{h\rightarrow 0}\frac{h^{-1} p_n(h)}{e^{h^{-2}}}\ \end{align*}

Since every term of $p_n(h)$ is of the form $a h^{-m}$ for some $a,m\in\mathbb{R},n>0$, it follows that every term of $h^{-1} p_n(h)$ is also of the same form. Thus if we can show that $\lim_{h\rightarrow 0}\frac{ah^{-m}}{e^{h^{-2}}}=0$, the result follows.

Applying L'Hopital's rule results in

$$\lim_{h\rightarrow 0}\frac{-m h^{-m-1}}{e^{h^{-2}} (-\frac{2}{h^3})}=\lim_{h\rightarrow 0}\frac{2m h^{-m+2}}{e^{h^{-2}}}$$

Successive applications of L'Hopital's rule result in multiplying numerator by some constant and increasing the exponent of $h$ by 2. Thus apply the rule $m/2+1$ times will give a positive exponent, and the limit will be clearly 0.

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The organization is sloppy. You should set out an inductive hypothesis about the form of f^{(n)}(x) from the start and then prove it. –  Qiaochu Yuan Sep 15 '10 at 4:32
    
So start out with something like "We proceed by induction. First we show that $f'(0)=0$, and then we prove that $f^{(n)}(0)$ for all $n$" to tell the reader what my overall plan is? –  Unkz Sep 15 '10 at 4:57
1  
No, you need to induct on a stronger statement (namely that f^{(n)} = e^{-1/x^2} p_n(x) where p_n(x) has such-and-such form...) –  Qiaochu Yuan Sep 15 '10 at 5:18
    
If you have questions about typing cases and other constructs, see meta. –  Larry Wang Sep 15 '10 at 18:22
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1 Answer

up vote 3 down vote accepted

You seem to have two main observations.

statement $A_{n}$ (for $n \ge 1$): $$f^{n}(x) = e^{q(x)}p_{n}(x), x \neq 0$$ where $$p_{n}(x) = \sum \frac{a_{r}}{x^r}, x \neq 0$$

and

that

statement $B_{n}$ (for $n \ge -1$): $$\lim_{h \to 0} \frac{1}{h^{n} e^{h^{-2}}} = 0$$

The result you have is that $A_{n-1}$ implies $A_{n}$ and that $B_{n-2}$ implies $B_{n}$.

Can you rewrite those formally using the right form of induction, with proper base cases?

Once you have those, you finally can prove the third statement $C_{n}$ (for $n \ge 1$) :

$$ f^{n}(0) = 0$$

You haven't explicitly spelled this out in your proof, but you are showing that $C_{n}$ (combined with the $A_{i}$ and $B_{j}$) implies $C_{n+1}$ which can again be formalized by induction.

For instance the statement:

$$f^{(n+1)}(0) = \lim_{h\rightarrow 0}\frac{f^{(n)}(h)}{h}$$ is implicitly using the fact that $f^{n}(0) = 0$.

Hope that helps.

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That was a great help. –  Unkz Sep 16 '10 at 6:07
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