Chapter 9.3, Problem 22E

$\text{(a)}$

To determine

To find: The bound based on alternating series error bound.

Answer to Problem 22E

The error is less than $\frac{{0.1}^3}{3!}$ .

Explanation of Solution

Given information:

The approximation $\sin x\propto x$ when $\left|x\right|<0.1$

Calculation:

Because the terms of the series meet all three of the characteristics indicated in Chapter 10—they alternate in sign, decrease in absolute value, and approach $0$ as a limit— establish a bound on the error based on the Alternating Series Error Bound method.

The truncation error is therefore smaller than the absolute value of the term at $x$ :

  $\text{error }\le \left|\frac{x^4}{4!}\right|$

For all $|x\mid <0.1$ , based on this,

the error this,

  $\text{error }<\left|\frac{x^3}{3!}\right|=\frac{|x{|}^3}{3!}\le \frac{{0.1}^3}{3!}$

Therefore, the required error is less than $\frac{{0.1}^3}{3!}$ .

$\text{(b)}$

To determine

To find: The Lagrange error formula.

Answer to Problem 22E

The error is less than $\frac{{0.1}^3}{3!}$ .

Explanation of Solution

Given information:

The approximation $\sin x\propto x$ when $\left|x\right|<0.1$

Calculation:

Since the fourth derivative of sine is zero at $x=0,$

And

  $P_2(x)=P_3(x)$

But

  $P_4(x)$ differ,

thus

  $n=2$

The $(n+1)$ st derivative of sine is its fifth one, i.e. $-\cos x$ hence - based on the Lagrange error formula - the error at $x\in (-1,1)$ for some $c$ between $0$ and $x$ is:

  $\left|R_3(x)\right|=\left|\frac{f^{(3)}(c)}{3!}{x}^3\right|=\frac{|-\cos c|}{3!}\left|x^3\right|\le \frac{|x{|}^3}{3!}$

Based on this, $|x\mid <0.1$

Since, the error

  $\frac{|x{|}^3}{3!}<\frac{{0.1}^3}{3!}$

Therefore, the required error is less than $\frac{{0.1}^3}{3!}$ .

$(c)$

To determine

To find: The values that uses $x<\sin x$ .

Answer to Problem 22E

The answer is: $x<\sin x$ if $0<x<0.1$ .

Explanation of Solution

Given information:

The approximation $\sin x\propto x$ when $\left|x\right|<0.1$

Calculation:

If $0<x<0.1$ then $x^3/3\text{ !}$ is a positive number. know that the partial sums oscillate back and forth on either side of the eventual value of $\sin x$ on this interval, and the approximation

  $\begin{array}{c}\sin x=x\\ =P_1(x)\\ =P_2(x)\end{array}$

underestimates this value. Based on these, $P_3(x)$ will overestimate the exact value, hence $P_1(x)=x<\sin x$ In the other case, if $-0.1<x<0$ then because of both sine and the function $x$ are odd functions, $\sin x<x$

Summarizing,

  $\sin x<x$ if $x\in (0,0.1)$ .

Therefore, the required values that uses $x<\sin x$ if $0<x<0.1$ .