Exercise 6.6.9

Question

[Cauchy's Remainder Theorem]
  Let $f$ be differentiable $N+1$ times on $(-R, R)$. For each $a \in(-R, R)$, let $S_{N}(x, a)$ be the partial sum of the Taylor series for $f$ centered at $a$; in other words, define
  $$
  S_{N}(x, a)=\sum_{n=0}^{N} c_{n}(x-a)^{n} \quad 	ext { where } \quad c_{n}=\frac{f^{(n)}(a)}{n !} .
  $$
  Let $E_{N}(x, a)=f(x)-S_{N}(x, a) .$ Now fix $x 
eq 0$ in $(-R, R)$ and consider $E_{N}(x, a)$ as a function of $a$.
  \begin{enumerate}[label=(\alph*)] 
  \item Find $E_{N}(x, x)$.
  \item Explain why $E_{N}(x, a)$ is differentiable with respect to $a$, and show
    $$
    E_{N}^{\prime}(x, a)=\frac{-f^{(N+1)}(a)}{N !}(x-a)^{N} .
    $$
  \item Show
    $$
    E_{N}(x)=E_{N}(x, 0)=\frac{f^{(N+1)}(c)}{N !}(x-c)^{N} x
    $$
    for some $c$ between 0 and $x$. This is Cauchy's form of the remainder for Taylor series centered at the origin.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
    \item $$E_N(x,x) = f(x) - S_N(x,x) = f(x) - c_0 = f(x) - f(x) = 0$$
    \item $$\begin{aligned}E_N'(x,a) &= -S_N'(x,a) = \sum^N_{n=1} \frac{f^{(n)}(a)}{(n-1)!} (x-a)^{n-1} - \sum^N_{n=0}\frac{f^{(n+1)}(a)}{n!} (x-a)^n \
    &=\sum^{N-1}_{n=0} \frac{f^{(n+1)}(a)}{(n)!} (x-a)^{n} - \sum^N_{n=0}\frac{f^{(n+1)}(a)}{n!} (x-a)^n= \frac{-f^{(N+1)}(a)}{N!}(x-a)^{N}
    \end{aligned}$$
    \item By the Mean Value Theorem
    $$\frac{E_N(x,x) - E_N(x,0)}{x} = E'_N(x,c)$$
    for some $c \in (0,x)$. Plugging in $E_n(x,x) = 0$ and the expression for $E_N'$ derived in part $(b)$ leaves us with the desired result.
 \end{enumerate}

Exercise 6.6.9

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Exercise 6.6.9

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