Exercise 8.1

Question

Exercise 1: 
    Define $$f(x)=
    \begin{cases}
        e^{1/x^2} & (x
e 0), \
        0 & (x=0).
    \end{cases}$$
    Prove that $f$ has derivatives of all orders at $x=0$, and that $f^{(n)}(0)=0$ for $n=1,2,3,\ldots.$

ghostofgarborg · Accepted Answer

\def\where{\,|\,}
(	extit{Matt ``Frito'' Lundy}) \
    First note that according to the limit definition of the derivative:
    \begin{align*}
        f'(0) &= \lim_{x 	o 0} \frac{f(x) - f(0)}{x - 0} \
        &= \lim_{x 	o 0}\frac{e^{-1/x^2}}{x} \
        & = 0
    \end{align*}
    by theorem 8.6(f).
    This establishes a base case and we proceed with induction by
    noticing that for $x 
eq 0$, $n \in \mathbf N$, we have:

$$f^{(n)}(x) = e^{-1/x^2}r_n(x)$$

where $r_n(x)$ is some rational function of $x$. 
    Then, according to the limit definition of the derivative and
    using the inductive hypothesis that $f^{(n)}(0) = 0$ we have:
    \begin{align*}
        f^{(n+1)}(0) &= \lim_{x 	o 0} \frac{f^{(n)}(x) -f^{(n)}(0)}{x - 0} \
        &= \lim_{x 	o 0} \frac{e^{-1/x^2}r_n(x)}{x} \
        &= 0
    \end{align*}
    again using theorem 8.6(f).  This establishes that $f(x)$ is infinitely differentiable at $x = 0$ and that $f^{(n)}(0) = 0$.

The point here is that even though $f(x)$ is infinitely differentiable at $x = 0$, it does not have a taylor series expansion about $x = 0$.

Exercise 8.1

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Exercise 8.1

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