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

Question

Define that $f(x) = e^{-1/x^2}$ when $x$ is not $0$, and $f(x) = 0$ when $x=0$.

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

Answer 1

Using induction, it is easy to see/prove for $x \ne 0$, $f^{(n)}(x)$ has the form $Q_n(\frac{1}{x}) e^{-\frac{1}{x^2}}$ for some polynomial $Q_n$ of degree $3n$ in $\frac{1}{x}$ $\color{blue}{^{[1]}}$.

By definition $f(0) = 0$. Assume we have shown $$f(0) = f'(0) = \ldots f^{(n-1)}(x) = 0,$$ then by definition again: $$f^{(n)}(0) := \lim_{h\to 0} \frac{f^{(n-1)}(h) - f^{(n-1)}(0)}{h} = \lim_{h\to 0} \left(\frac{1}{h}Q_{n-1}(\frac{1}{h})\right) e^{-\frac{1}{h^2}}\tag{*}$$

Being a polynomial of degree $3n$, for sufficiently small $h$ and hence large $\frac{1}{h}$, say $|h| < \epsilon$, we can find a $M > 0$ such that $\color{blue}{^{[2]}}$:

$$|Q_n(\frac{1}{x})| < \frac{M}{|x|^{3n}}\quad\text{ for } x \in (-\epsilon,\epsilon) \setminus \{0\}$$

As a result $\color{blue}{^{[3]}}$, $$\limsup_{h\to 0} \left| \left(\frac{1}{h}Q_{n-1}(\frac{1}{h})\right) e^{-\frac{1}{h^2}} \right| \le \lim_{h\to 0}\frac{M e^{-\frac{1}{h^2}} }{|h|^{3n+1}} =\lim_{r\to+\infty} M r^{3n+1} e^{-r^2} = 0$$

This implies the limit in R.H.S of $(*)$ exists and vanishes. i.e. $f^{(n)}(0)$ also vanishes.

By induction again, we get $f^{(n)}(0) = 0$ for all $n \in \mathbb{N}$.

Notes

$\color{blue}{[1]}$ $Q_n(t)$ can be computed using the recurrence relation: $$Q_n(t) = \begin{cases} 1, & n = 0\\ 2t^3 Q_{n-1}(t) - t^2 Q'_{n-1}(t), & n > 0 \end{cases}$$

$\color{blue}{[2]}$ Recall for any polynomial $\sum_{k=0}^{m} a_k x^k$ of degree $m$, we have the inequality: $$ \left|\sum_{k=0}^{m} a_k x^k\right| < \left( \sum_{k=0}^{m} |a_k| \right) \max( |x|^m, 1 )$$

$\color{blue}{[3]}$ Notice when $r$ is positive,

$$e^{r} > 1 + r > r, \forall r \quad\implies\quad e^{\frac{r}{3n+1}} > \frac{r}{3n+1}, \forall r \quad\implies\quad e^{r} > \left(\frac{r}{3n+1}\right)^{3n+1}, \forall r $$ When $r > 2$, we get: $$r^{3n+1} e^{-r^2} < (3n+1)^{3n+1} e^{r-r^2} < (3n+1)^{3n+1} e^{-r}$$ and the R.H.S tends to $0$ as $r \to \infty$.

Answer 2

Hint:

By the definition of the derivative: $$f'(0) = \lim_{h\to 0}\frac{e^{-1/h^2}-0}{h}$$ And by L'Hopital, this is zero. For higher orders, show that the derivative $f^{(n)}(x>0)$ is of the form: $$\sum_i A_i\cdot \frac{e^{-1/x^2}}{x^i}$$ and use the derivative definition again.

Answer 3

Obviously the left side of the function, being a constant zero, has infinitely many zero derivatives coming into zero from the left. So we have to show it has infinitely many and they are all zero coming in from the right. The hint nbubis gives is not so much a hint as an answer provided you know that the exponential function will always swamp out the polynomial.

Since the 1/x tends to confuse things, I think there is a clearer way to see this. That is to consider the function

g(x) = f(1/x) = e^(-x^2). 

Equivalent to f(x) → zero, we observe g(x) as x → +∞. These derivatives clearly exist and just as clearly are zero for the same reason as above -- the exponential goes to zero faster than the polynomials can rise.

This is the standard example used to demonstrate that although a function may have infinitely many derivatives at a point, it is not necessarily analytic there. Viewed simply as a function of a real variable, this means that its Taylor series will not converge; however proving that on the real line could be quite a pain.

Yet it can be shown very easily by going to the complex plane, where f'(z) simply does not exist at zero. To see that, evaluate f'(z) (or the corresponding limit) along the i-axis. f'(z) = 2z^(-3)*f(z). Since f(i) ≡ 1 (check it out), this gives f'(i)= 2*i^(-3) which certainly is not going to zero as i goes to zero. So there is no complex derivative at zero. Various theorems from the study of complex functions state that f(z) is analytic at a point iff it has a derivative there as a function of a complex variable.