Exercise 6.6.8

Question

Here is a weaker form of Lagrange's Remainder Theorem whose proof is arguably more illuminating than the one for the stronger result.
  \begin{enumerate}[label=(\alph*)] 
  \item First establish a lemma: If $g$ and $h$ are differentiable on $[0, x]$ with $g(0)=h(0)$ and $g^{\prime}(t) \leq h^{\prime}(t)$ for all $t \in[0, x]$, then $g(t) \leq h(t)$ for all $t \in[0, x] .$
  \item Let $f, S_{N}$, and $E_{N}$ be as Theorem 6.6.3, and take $0<x<R$. If $\left|f^{(N+1)}(t)\right| \leq M$ for all $t \in[0, x]$, show
    $$
    \left|E_{N}(x)\right| \leq \frac{M x^{N+1}}{(N+1) !}
    $$
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
    \item Let $e(t) = h(t) - g(t)$. If $g(t) > h(t)$ for some $t \in [0, x]$ then we could apply the Mean Value Theorem on $e(t)$ between $0$ and $t$ to find $c \in (0, t) \subseteq [0, x]$ with $e'(c) = h'(t) - g'(t) < 0$, a contradiction since $h'(t) \geq g'(t)$.
    \item Since $S_N^{(N+1)} = 0$, $f^{(N+1)}(t) = E^{(N+1)}(t)$. We then have $\left|E_N^{(N+1)}(t)\right| \leq E_N^{(N+1)}\leq M$, and by repeated application of the lemma in part (a) we have $E_N(x) \leq \frac{Mx^{N+1}}{(N+1)!}$. A similar argument holds for $E_N(x) \geq -\frac{Mx^{N+1}}{(N+1)!}$, and we can combine these succiently as
    $$ \left|E_{N}(x)\right| \leq \frac{M x^{N+1}}{(N+1) !}$$
 \end{enumerate}

Exercise 6.6.8

Answers

Comments

Exercise 6.6.8

Answers

Comments

Add answer