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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?

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  • $\begingroup$ What is $f(0)$? $\endgroup$ – MPW Nov 4 '19 at 17:13
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Your $f$ is the composition of two infinitely differentiable functions ($e^x$ and $-1/x$, for $x>0$) and therefore infinitely differentiable itself.

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  • $\begingroup$ Obviously, but not required here. $\endgroup$ – Lorenzo Cecchi Nov 4 '19 at 17:16
  • $\begingroup$ The question seems to be asking differentiability for positive values. I agree with you that, anyway, the interesting part is another. $\endgroup$ – Lorenzo Cecchi Nov 4 '19 at 17:37
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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 ${\mathbb R}_{>0}$.

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  • $\begingroup$ @nicomezi: The question was about "the $x$ that are greater than zero". $\endgroup$ – Christian Blatter Nov 4 '19 at 19:33
  • $\begingroup$ You are right. The first paragraph seems just to be some background ... It is not clear at all to me what this question is about .... $\endgroup$ – nicomezi Nov 4 '19 at 19:42

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