Exercise 6.6.6

Question

Review the proof that $g^{\prime}(0)=0$ for the function
  $$
  g(x)= \begin{cases}e^{-1 / x^{2}} & 	ext { for } x 
eq 0 \ 0 & 	ext { for } x=0\end{cases}
  $$
  introduced at the end of this section.
  \begin{enumerate}[label=(\alph*)] 
  \item Compute $g^{\prime}(x)$ for $x 
eq 0$. Then use the definition of the derivative to find $g^{\prime \prime}(0)$.
  \item Compute $g^{\prime \prime}(x)$ and $g^{\prime \prime \prime}(x)$ for $x 
eq 0$. Use these observations and invent whatever notation is needed to give a general description for the $n$th derivative $g^{(n)}(x)$ at points different from zero.
  \item Construct a general argument for why $g^{(n)}(0)=0$ for all $n \in \mathbf{N}$.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
    \item $g'(x) = 2x^{-3} e^{-x^{-2}}$. Repeatedly using L'Hospital's Rule,
    $$ g''(0) = \lim_{x \to 0} \frac{2x^{-3} e^{-x^{-2}}}{x} = \lim_{x \to 0}\frac{2x^{-4}}{e^{1/x^2}} = 2 \lim_{x \to 0} \frac{-4x^{-5} x^3}{-2e^{1/x^2}} =4 \lim_{x \to 0} \frac{-2x^{-3} x^3}{-2e^{1/x^2}}=  0 $$
    \item Explicitly computing things sounds rather tedious, so let's skip to the general form. We claim that $g^{(n)}(x)$ for $x \neq 0$ is of the form $P_n(x) e^{-1/x^2}$ where $P_n(x)$ is some polynomial in $x^{-1}$. We can prove this inductively, by noting $g^{(n+1)}(x) = [P_n'(x) + P_n(x) (-2x^{-3})] e^{-1/x^2}$. Differentiating a polynomial in $x^{-1}$ with respect to $x$ only increases the powers, and polynomials are closed under addition and multiplication, so clearly the term in square brackets continues to be a polynomial in $x^{-1}$.
    \item Let $P_n(x) = \sum a_n x^{-b_n}$ with $a_n$ constant and $b_n \geq 0$. We'll compute the formula for the derivative termwise, and use induction.
    $$\lim_{x \to 0} \frac{a_n x^{-b_n} e^{-1/x^2}}{x} = a_n \lim_{x \to 0} \frac{x^{-(b_n + 1)}}{e^{1/x^2}}$$
    It's easy to show that this is equal to 0, since every time we apply L'Hospital's rule, the denominator doesn't change (and continues to go to infinity) while the exponent in the numerator increases by 2 each time, eventually becoming non-negative and preventing the numerator from also going to infinity.
 \end{enumerate}

Exercise 6.6.6

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Exercise 6.6.6

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