Exercise 5.11

Question

Exercise 11: Suppose $f$ is defined in a neighborhood of $x$, and suppose $f''(x)$ exists. Show that $$\lim_{h\rightarrow 0}\frac{f(x+h)-f(x-h)-2f(x)}{h^2}=f''(x).$$
    Show by an example that the limit may exist even if $f''(x)$ does not.

analambanomenos · Accepted Answer

\def\where{\,|\,}

Let $g(h)=f(x+h)+f(x-h)-2f(x)$. Then $g$ is differentiable in a neighborhood of 0, and $g(0)=0$. Applying Theorem 5.13, we get
    \begin{align*}
        \lim_{hightarrow 0}\frac{g(h)}{h^2} &= \lim_{hightarrow 0}\frac{f'(x+h)-f'(x-h)}{2h} \
        &=\lim_{hightarrow 0}\frac{f'(x+h)-f(x)}{2h}-\lim_{kightarrow 0}\frac{f(x+k)-f(x)}{-2k}\quad\hbox{(where $k=-h$)} \
        &=\frac{f''(x)}{2}+\frac{f''(x)}{2} = f''(x)
    \end{align*}

Letting $f(x)=x^2\sin(1/x)$, $f(0)=0$, then $f$ is continuous and differentiable at $x=0$, and since $f$ is an odd function, $f(h)-f(-h)-2f(0)=0$ for all $h
eq 0$ so that
    the limit above exists for $x=0$. However $f'$ is not continuous at 0, so $f''(0)$ does not exist.

Exercise 5.11

Answers

Comments

Exercise 5.11

Answers

Comments

Add answer