I'm told to use L'Hopital and induction. Seems a bit intimidating to solve. Does anyone have any good suggestions?
Also, is $f(x)$ 'analytic?'
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It is not clear as to what you mean by "analytic". – Pedro Nov 19 '14 at 00:45
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According to Pugh, "A function that can be expressed locally as a convergent power series is analytic." – Dilettanter Nov 19 '14 at 00:50
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Note that if locally at $0$ we have $f(x)=\sum a_k x^k$ then forcedfully $k!a_k = f^{(k)}(0)=0$, which means the powerseries must vanish. But $f$ doesn't vanish in a nbhd of $0$; so no, this function is not analytic at zero. – Pedro Nov 19 '14 at 00:53
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For the sake of argument, extend $f$ to be identically zero when $x \leqslant 0$. It should be clear that $f$ has derivative of all orders at any point different from zero. Now consider $x = 0$. Since all left sided derivative vanish, if suffices we show all right sided derivatives vanish at $0$. Note that at any point $x \neq 0$; $f'(x) = x^{-2}e^{-1/x} = p_1(x^{-1})e^{-1/x}$. Then $$f''(x) = x^{-4}e^{-1/x}-2x^{-3}e^{-1/x} = p_2(x^{-1})e^{-1/x}$$
Inductively, show that $f^{(k)}(x) = p_k(x^{-1})e^{-1/x}$ where $p_k$ is a polynomial of degree $2k$. It follows that, since $w^n e^{-w}\to 0$ for any $n>0$ when $w\to\infty$, that $f^{(k)}(0)=0$: it is a theorem that if $f'(x)$ exists in a neighborhood of $a$ and $\lim\limits_{x\to a}f'(x)$ exists, then $f'(a)$ exists and it this limit.
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You wrote $w^n e^{-w} \to 0$ as $n > 0$ and $w \to \infty$, can you proof such statement? – ProFatXuanAll Aug 08 '21 at 10:08
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