# Proving $e^{1/x}$ is infinitely differentiable

I came across this problem awhile ago: Proving a function is infinitely differentiable. It is about proving that $$f$$ is infinitely differentiable for $$f=0, x\leq 0$$ and $$f=e^{-1/x}$$ for $$x>0$$.

It is stated "Similarly, when x is greater than zero the function is infinitely differentiable, by the properties of the exponential function." I don't understand how this statement is proven. How does one use properties of $$e^x$$ to show this?

• What is $f(0)$? – MPW Nov 4 '19 at 17:13

Your $$f$$ is the composition of two infinitely differentiable functions ($$e^x$$ and $$-1/x$$, for $$x>0$$) and therefore infinitely differentiable itself.
Your question concerns the region $$x>0$$. I claim that $$f^{(n)}(x)= e^{-1/x}\>p_n(1/x)\qquad(x>0, \ n\geq0)\ ,\tag{1}$$ where $$p_n(1/x)$$ is a polynomial of degree $$2n$$ in $${1\over x}$$. This is true for $$n=0$$, by definition of $$f$$. We therefore have to compute $$f^{(n+1)}(x)=e^{-1/x}\>{1\over x^2}\bigl(p_n(1/x)-p_n'(1/x)\bigr)\>\qquad(x>0, \ n\geq0)\ .$$ Here the RHS is $$e^{-1/x}$$ times a polynomial in $${1\over x}$$ of degree $$2(n+1)$$.
This shows that $$f$$ has derivatives of all orders $$n\geq0$$ on $${\mathbb R}_{>0}$$.
• @nicomezi: The question was about "the $x$ that are greater than zero". – Christian Blatter Nov 4 '19 at 19:33