1

Let $\displaystyle f(x) = \begin{cases}e^{- \frac{1}{x^2}} &\text{for } x \neq 0 \\ 0 & \text{when } x=0 \end{cases}.$

Prove that $f$ has derivatives of all orders at $x=0$, and that $f^n(0)=0$ for $n=1,2,3,...$

I found this problem from Rudin's principles of Mathematical Analysis book (p.196). My problem is, can I use the formula $e^x=1+x+\frac{1}{2!}x^2+...$ and use directly Theorem 8.1 and Corollary on page 173 in the same book? If not can anyone please explain me the reason?

Extremal
  • 5,785

2 Answers2

2

Hint:

Use the definition

$$f^{(n+1)}(0)=\lim_{x\to 0}\frac{f^{(n)}(x)-f^{(n)}(0)}{x}$$

togeather with induction to show that $f^{(n)}(0) = 0$ implies $f^{(n+1)}(0) = 0$.

For this you also need to find $f^{(n)}(x)$ for $x\not =0$. To do this use induction to prove that

$$f^{(n)}(x) = \frac{P_n(x)}{x^{3n}}e^{-\frac{1}{x^2}}$$

where $P_n(x)$ is a polynomial in $x$ (of degree $2n-2)$.

Winther
  • 24,478
  • 1
    I'll note that it's also sufficient to prove that $f^{(n)}(x) = R_n(x) e^{-\frac{1}{x^2}}$, for $R_n$ some rational function. The fact that the denominator is $x^{3n}$ isn't hugely important. –  Feb 14 '15 at 22:48
  • You're not a new user. This is a very usual question, which in fact was answered not so many days ago. Please make an effort to look for duplicates. (I'm not saying I'm free of sin, but we need to make a collective effort about this!) – Pedro Feb 14 '15 at 22:49
  • @Strants That is correct and the proof of that statement is even easier. – Winther Feb 14 '15 at 22:49
  • @PedroTamaroff Agreed, I forgot about it. I did find a dublicate and voted to close. – Winther Feb 14 '15 at 22:52
  • 1
    For the record, I thought this a nicer and cleaner answer than those on the other thread. Thanks Winther. – Simon S Feb 14 '15 at 23:04
0

Show that there exist a sequence $(P_n)_n$ of polynomials of $\mathbf{R}[T]$ such that for each $n\in\mathbf{N}$ and each $x\in\mathbf{R}_{+}^{\times}$ you have $f^{(n)} (x) = x^{-3n} P_n(x) e^{-\frac{1}{x^2}}$. From this, by basic comparison limits, you'll conclude.

Olórin
  • 12,040
  • 1
    You're not a new user. This is a very usual question, which in fact was answered not so many days ago. Please make an effort to look for duplicates. (I'm not saying I'm free of sin, but we need to make a collective effort about this!) – Pedro Feb 14 '15 at 22:49