Exercise 8.1.13

Question

\begin{enumerate}[label=(\alph*)] 
\item For a particular $c_k \in [x_{k-1}, x_k]$ of $P$, show that
$$\left|F(x_k) - F(c_k) - f(c_k)(x_k - c_k)\right| < \epsilon (x_k - c_k)$$
and
$$\left|F(c_k) - F(x_{k-1}) - f(c_k)(c_k - x_{k-1})\right| < \epsilon (c_k - x_{k-1})$$
\item Now, argue that
$$\left|F(x_k) - F(x_{k-1}) - f(x_k)(x_k - x_{k-1})\right| < \epsilon(x_k- x_{k-1})$$
and use this fact to complete the proof of the theorem.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
\item For the first inequality, evaluate
$$\left|\frac{F(x) - F(c)}{x-c} - f(c)\right| < \epsilon$$
at $x = x_k$, $c = c_k$ and multiply both sides by $|x_k-c_k|$, noting that $x_k \geq c_k$.

For the second inequality, rewrite
$$\left|\frac{F(x) - F(c)}{x-c} - f(c)\right| = \left|\frac{F(c) - F(x)}{c-x} - f(c)\right| < \epsilon,$$
evaluate at $x = x_{k-1}$, $c = c_k$ and multiply both sides by $|c_k - x_{k-1}|$, noting that $c_k \geq x_{k-1}$.

\item Add the two inequalities in part (a) together and apply the Triangle Inequality. Then summing over $k$ from 1 to $n$ gets us
$$\begin{aligned}
    \epsilon (x_n - x_0) &= \epsilon(b - a) \
    &> \sum^n_{k=1} \left|F(x_k) - F(x_{k-1}) - f(c_k) (x_k - x_{k-1})\right| \
    &\geq \left|F(b) - F(a) - R(f,P)\right|
\end{aligned}$$
To prove the theorem we need $\left|F(b) - F(a) - R(f,P)\right| < \epsilon$, which can be readily obtained by instead constructing the gauge $\delta(c)$ for $\epsilon / (b-a)$.
 \end{enumerate}

Exercise 8.1.13

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Exercise 8.1.13

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