$f(x)=e^{-\frac{1}{x^{2}}}$ for $x\neq 0$ and $f(0)=0$. Prove that $f^{(k)}(0)=0$ for all postitive integer $k$.

Asked May 12 '22 at 12:22

Active May 12 '22 at 12:22

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Problem says "Use L'Hopital's Rule and induction"

Suppose $f^{(n)}(0)=0$

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

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

Then how I know $f^{(n+1)}(0)=0$?

asked May 12 '22 at 12:22

user1048734

2

See: https://math.stackexchange.com/questions/1436856/showing-that-fn0-0-for-fx-e-1-x2-if-x-neq-0-and-f0-0?rq=1 – Kavi Rama Murthy May 12 '22 at 12:32
$f^{(k)}$ is a polynom of $\frac{1}{x}$ times $\exp{-1/x^2}$ (you can prove it, by induction for example) – niobium May 12 '22 at 12:35

$f(x)=e^{-\frac{1}{x^{2}}}$ for $x\neq 0$ and $f(0)=0$. Prove that $f^{(k)}(0)=0$ for all postitive integer $k$.

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