Exercise 8.3.10

Question

\begin{enumerate}[label=(\alph*)] 
\item Make a rough sketch of $1 / \sqrt{1-x}$ and $S_2(x)$ over the interval $-1, 1$, and compute $E_2(x)$ for $x = 1/2, 3/4$, and $8/9$.
\item For a general $x$ satisfying $-1 < x < 1$, show
$$E_2(x) = \frac{15}{16}\int_0^x \left(\frac{x-t}{1-t}\right)^2 \frac{1}{(1-t)^{3/2}}dt$$
\item Explain why the inequality
$$\left|\frac{x-t}{1-t}\right| \leq \left|x\right|$$
is valid, and use this to find an overestimate for $\left|E_2(x)\right|$ that no longer involves an integral. Note that this estimate will necessarily depend on $x$. Confirm that things are going well by checking that this overestimate is infact larger than $\left|E_2(x)\right|$ at the three computed values from part $(a)$.
\item Finally, show $E_N(x) \to 0$ as $N \to \infty$ for an arbitrary $x \in (-1,1)$.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
\item $$S_2(x) = 1 + \frac{1}{2}x + \frac{3}{8}x^2$$
A plot is best made using your favourite graphing calculator.
$$
\begin{array}{|c|c|c|c|}
    \hline
    x & 1/2 & 3/4 & 8/9 \ \hline
    1/\sqrt{1-x} & 1.414 & 2 & 3\ \hline
    S_2(x) & 1.344 & 1.586 & 1.741 \ \hline
    E_2(x) & 0.070 & 0.414& 1.259\ \hline
\end{array}
$$
\item From Theorem 8.3.1,
$$E_2(x) = \frac{1}{2} \int_0^x f^{(3)}(t) (x-t)^2 = \frac{1}{2}\int_0^x \frac{15}{8} \frac{(x-t)^2}{(1-t)^{7/2}}  = \frac{15}{16}\int_0^x \left(\frac{x-t}{1-t}\right)^2 \frac{1}{(1-t)^{3/2}} $$
\item We have that $|t| \leq |x| < 1$ and $t$ having the same sign as $x$, so
$$\left|\frac{x-t}{1-t}\right| \leq \left|x\right| \Longleftrightarrow \left|x-t\right| \leq \left|x(1-t)\right|$$
For $x \geq t > 0$,
$$\left|x-t\right| \leq \left|x(1-t)\right| \Longleftrightarrow x -t \leq x - xt \Longleftrightarrow t \geq xt$$
which is true since $x < 1$. Similarly for $x \leq t < 0$,
$$\left|x-t\right| \leq \left|x(1-t)\right| \Longleftrightarrow t- x \leq xt - x\Longleftrightarrow t \leq xt$$
again true since $xt \geq 0 \geq t$. So
$$\left|E_2(x)\right| \leq \frac{15}{16} \left|\int_0^x \left|x\right|^2 \frac{1}{(1-t)^{3/2}}dt\right| = \frac{15x^2}{16} \int_0^x (1-t)^{-3/2} dt = \frac{15x^2}{8}\left|1 - \frac{1}{\sqrt{1-x}}\right|$$
$$
\begin{array}{|c|c|c|c|}
    \hline
    x & 1/2 & 3/4 & 8/9 \ \hline
    E_2(x) & 0.070 & 0.414& 1.259\ \hline
    \text{Overestimate} & 1.406 & 15.820 & 118.519 \ \hline
\end{array}
$$

\item Recall that
$$f^{(n)}(x) = \left(\prod^n_{i=1} \frac{2i-1}{2(1-x)}\right) \sqrt{\frac{1}{1-x}}$$
so
$$
    \begin{aligned}
        \left|E_N(x)\right| &= \frac{1}{N!}\left|\int_0^x \left(\prod^{N+1}_{i=1} \frac{2i-1}{2(1-x)}\right) \frac{1}{(1-t)^{1/2}} dt \right|\
        &= \frac{1}{N!} \left| \int_0^x \left(\prod^N_{i=1} \frac{(2i-1)(x-t)}{2 (i-t)}\right)\frac{2N+1}{2(1-t)^{3/2}}dt \right| \
        &= \frac{2N+1}{2} \left|\int_0^x \left(\prod^N_{i=1} \frac{2i-1}{2i}\right) \left(\frac{x-t}{1-t}\right)^N (1-t)^{3/2}dt \right|\
        &\leq \left|x\right|^N \left(\frac{2N+1}{2}\right) \left|\int^x_0(1-t)^{-3/2} dt \right|\
        &= \left|x\right|^N (2N+1) \left|1 - \frac{1}{\sqrt{1-x}}\right|
    \end{aligned}
$$
For a fixed $x$, with $|x| < 1$, as $N$ increases $|x|^N$ will go to 0 exponentially, faster than $2N+1$ increases, and therefore $\lim_{n \to \infty} E_N(x) = 0$ for $x \in (-1,1)$.
 \end{enumerate}

Exercise 8.3.10

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Exercise 8.3.10

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