Exercise 6.6.7

Question

Find an example of each of the following or explain why no such function exists.
  \begin{enumerate}[label=(\alph*)] 
  \item An infinitely differentiable function $g(x)$ on all of $\mathbf{R}$ with a Taylor series that converges to $g(x)$ only for $x \in(-1,1)$.
  \item An infinitely differentiable function $h(x)$ with the same Taylor series as $\sin (x)$ but such that $h(x) 
eq \sin (x)$ for all $x 
eq 0$.
  \item An infinitely differentiable function $f(x)$ on all of $\mathbf{R}$ with a Taylor series that converges to $f(x)$ if and only if $x \leq 0$.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
    \item $g(x) = 1/(1+x^2)$. We already have that the Taylor series for this is $\sum^\infty_{n=0}(-1)^n x^2n$, and that it converges to $g(x)$ for $|x| < 1$, but it does not converge at all at $\pm 1$.
    \item Let $a(x)$ be the counterexample function introduced at the end of this section. Then set $h(x) = \sin(x) + a(x)$; since the Taylor series of $a(x)$ is identically 0, $h(x)$ has the same Taylor series as $\sin(x)$.
    \item $$f(x) = \begin{cases}
        0 & x \leq 0 \
        e^{-1/x^2} & x > 0
    \end{cases}$$
    has Taylor series identically 0.
   \end{enumerate}

Exercise 6.6.7

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Exercise 6.6.7

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