# Thread: Cauchy's function and differentiability

1. ## Cauchy's function and differentiability

How can I show that $\displaystyle f(x) = \{ \begin{array}{ll} exp(\frac{-1}{x^2} & x>0\\ 0 & x\leq 0 \end{array}$
is infinitely differentiable?

2. Originally Posted by dannyboycurtis
How can I show that $\displaystyle f(x) = \{ \begin{array}{ll} exp(\frac{-1}{x^2}) & x>0\\ 0 & x\leq 0 \end{array}$
is infinitely differentiable?
$\displaystyle f(x)=e^{-x^{-2}}$

The first derivative is $\displaystyle x^{-3}(2)f(x)$

The second derivative is $\displaystyle x^{-6}(4-6x^2)f(x)$

The third derivative is $\displaystyle x^{-9}(8-36x^2+24x^4)f(x)$

The fourth derivative is $\displaystyle x^{-12}(16-144x^2+300x^4-140x^6)f(x)$

So $\displaystyle f^{(n)}(x)=x^{-3n}(\text{polynomial~with~degree}<3n)f(x)$

(You might want to prove this using induction, though it will likely be ugly.)

Next you'd have to do another induction proof. Like assume that $\displaystyle f^{(n)}(0)=0$ and then prove that:

$\displaystyle f^{(n+1)}(0)=\lim_{x\to0^+}\frac{f^{(n)}(x)-f^{(n)}(0)}{x}=0$

Since you assumed $\displaystyle f^{(n)}(0)=0$, you need to prove that

$\displaystyle \lim_{x\to0^+}\frac{x^{-3n}(\text{polynomial})f(x)}{x}=\lim_{x\to0^+}x^{-3n-1}(\text{polynomial})e^{-x^{-2}}=0$.

There could be a cleaner way to do this using some theorems or something, but this should work.