Chapter 8.8, Problem 23E

To determine

To Find: The range of values of x for which the given approximation is accurate to within the stated error

Answer to Problem 23E

Range of value of x

  $-1.0371<x<1.0371$

Explanation of Solution

Given information:

Use Taylor’s Inequality to determine The range of values of x for which the given approximation is accurate to within the stated error

  $\sin x\approx x-\frac{x^3}{6}\left(\left|error\right|<0.01\right)$

Formula Used:

  $T_n(x)\to f(x)$ ???? ??>&#??2192;>&#??221??;

And $T_n(x)$ can be used as an approximation to

  $f:f(x)=T_n(x)$

Taylor’s Inequality, which says that if

  $\left|f^{(n+1)}(x)\right|\le M$ , then

  $\left|R_n(x)\right|\le \frac{M}{(n+1)!}{\left|x-a\right|}^{n+1}$

Calculation:

Maclaurin series

Of $\sin x$

  $T_3\approx x-\frac{x^3}{6}$

Maximum value of $\sin x$ is $1$

So $M=1$

for Maclaurin series, $a=0$ and $n=3$

  $\begin{array}{l}\left|f^4(x)\right|\le M\\ f^4(x)=\sin x\\ f^4(0)=0\end{array}$

So, use $n=4$

  $\begin{array}{l}\left|R_n(x)\right|\le \frac{M}{(n+1)!}{\left|x-a\right|}^{n+1}\\ \left|R_4(x)\right|\le \frac{M}{5!}{\left|x\right|}^5\\ \left|error\right|\le 0.01\\ \frac{1}{120}{\left|x\right|}^5<0.01\\ -1.0371<x<1.0377\end{array}$

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