I am self-learning Real Analysis from the text, Understanding Analysis, by Stephen Abbott. The exercise problem 6.6.6 asks to prove certain stylized facts about the piecewise defined function:
$$g(x) = \begin{cases}e^{-1/x^2}, & x \neq 0 \\ 0, & x = 0 \end{cases}$$
Do you have clue/hint (without giving away the entire solution) for part (b) of this problem.
[Abbott 6.6.6] Review the proof that $\displaystyle g'( 0) =0$.
(a) Compute $\displaystyle g'( x)$ for $\displaystyle x\neq 0$. Then use the definition of the derivative to find $\displaystyle g''( 0)$.
(b) Compute $\displaystyle g''( x)$ and $\displaystyle g'''( x)$ for $\displaystyle x\neq 0$. Use these observations and invent whatever notation is needed to give a general description for the $\displaystyle n$th derivative $\displaystyle g^{( n)}( x)$ at points different from zero.
Proof.
(a) We have:
\begin{equation*} g'( x) =\frac{2}{x^{3}} e^{-1/x^{2}} \end{equation*}
By definition,
\begin{equation*} \begin{array}{ c l c } g''( 0) & =\lim _{x\rightarrow 0}\frac{g'( x) -g'( 0)}{x-0} & \\ & =\lim _{x\rightarrow 0}\frac{g'( x)}{x} & \\ & =\lim _{x\rightarrow 0}\frac{\frac{1}{x^{4}}}{e^{1/x^{2}}} & \\ & =\lim _{x\rightarrow 0}\frac{x^{3} \cdot \left( -4/x^{5}\right)}{\left( -2e^{1/x^{2}}\right)} & \left\{\text{L'Hospital's rule} ,\infty /\infty \right\}\\ & =2\lim _{x\rightarrow 0}\frac{\frac{1}{x^{2}}}{e^{1/x^{2}}} & \\ & =2\lim _{x\rightarrow 0}\frac{-\frac{2}{x^{3}}}{-\frac{2}{x^{3}} e^{1/x^{2}}} & \left\{\text{L'Hospital's rule} ,\infty /\infty \right\}\\ & =2\lim _{x\rightarrow 0}\frac{1}{e^{1/x^{2}}} & \\ & =0 & \end{array} \end{equation*}
(b) If $\displaystyle x\neq 0$, we have:
\begin{equation*} \begin{array}{ c l } g'( x) & =2e^{-1/x^{2}}\left(\frac{1}{x^{3}}\right)\\ g''( x) & =2\left[\left( x^{-3}\right) '\left( e^{-1/x^{2}}\right) +\left( x^{-3}\right)\left( e^{-1/x^{2}}\right) '\right]\\ & =2e^{-1/x^{2}}\left[ -\frac{3}{x^{4}} +\frac{2}{x^{6}}\right]\\ g'''( x) & =2\left[\left( e^{-1/x^{2}}\right) '\cdot \left( -\frac{3}{x^{4}} +\frac{2}{x^{6}}\right) +e^{-1/x^{2}}\left( -\frac{3}{x^{4}} +\frac{2}{x^{6}}\right)^{'}\right]\\ & =2e^{-1/x^{2}}\left[\frac{2}{x^{3}}\left( -\frac{3}{x^{4}} +\frac{2}{x^{6}}\right) +\left(\frac{12}{x^{5}} -\frac{12}{x^{7}}\right)\right]\\ & =2e^{-1/x^{2}}\left[\frac{12}{x^{5}} -\frac{12}{x^{7}} -\frac{6}{x^{7}} +\frac{4}{x^{9}}\right] \end{array} \end{equation*}
I can't seem to find the general expression for the $n$th derivative, $g^{(n)}(x)$.
