1
$\begingroup$

Let \begin{equation*} f(x) = \begin{cases} \hfill \text{exp}(-1/x^2) \hfill & \text{ if $x \neq 0$} \\ \hfill 0 \hfill & \text{ if $x = 0$} \\ \end{cases} \end{equation*} Prove $f^{(n)}(x)$ exists $\forall \ x, n$. Also show that $f^{(n)}(0) = 0$ for $\forall \ n$.

Things I've tried/considered:

For $x \neq 0$ we have $f'(x) =$ exp$(-1/x^2)*2/x^3$.

Then $f^{(2)}(x) = $exp$(-1/x^2)*(2/x^3)*(2/x^3) -$ exp$(-1/x^2)*6/x^4$ ... Which can be factored to exp$(-1/x^2)*p_2(1/x)$ where we can think about $p_2(1/x)$ being a polynomial with argument $1/x$.

In general \begin{equation} f^{(n)}(x) = \text{exp}(-1/x^2)*p_n(1/x) \hspace{0.5cm} \text{if} \ x \neq 0 \end{equation} I tried looking at the behavior of these derivatives as $x \to 0$. It looks like $\lim_{x\to\ 0} f^{(n)}(x) = 0 * \infty$. However, I think the exponential term approaches zero faster than the "polynomial" approaches infinity so the limit goes to zero (I'm not sure about this) - possibly a case of L' Hopital's Rule?

If we can prove the $lim \to 0$ as $x \to 0$ we can say $f^{(n)}(x)$ exists (I think).

I'm not sure how you would prove $f^{(n)}(0) = 0, \ \forall \ n$. Aren't we just defining it to always be zero in order for the function to be continuous.

  • 1
    You need a separate argument showing $f^{(n)}(0)=0$ indeed, as your formular for the derivatives are only for $x\neq 0$, and there's no presupposed relation between $f^{(n)}(x)$ near 0 and $f^{(n)}(0)$. – 2017-02-03
  • 0
    If you type "smooth non-analytic function example" on google and spend 5 minutes on the search you will almost surely ;) find a couple of ways to prove that... – 2017-02-03

1 Answers 1

0

Let $u=1/x$ so that we get

$$\exp(-u^2)p_n(u)<\frac{p_n(u)}{e^{u}}$$

As $u\to\infty$, it's easy to see we get the indefinite form $\frac\infty\infty$, but after $n$ applications of L'Hospital's rule, we end up with

$$\lim_{u\to\infty}\frac{au^0}{e^{u}}=0$$

Same argument for $u\to-\infty$.