Chapter 10.3, Problem 33E

To determine

To estimate: The magnitude of the approximation error for $\left|x\right|\le 0.1$

Answer to Problem 33E

The magnitude of the approximation error for $\left|x\right|\le 0.1$is $4.251\times {10}^{-6}$

Explanation of Solution

Given information:

  $e^x\approx 1+x+\frac{x^2}{2}+\frac{x^3}{6}$

Calculation:

Let the function is, $f(x)=e^x$

Then,

  $P_3(x)=1+x+\frac{x^2}{2}+\frac{x^3}{6}$

Now,

  $f^{(4)}(x)=e^x$

And therefore,

  $\left|f^{(4)}(x)\right|=\left|e^x\right|\le e^{0.1}$ for $\left|x\right|<0.1$

Use $M=e^{0.1},r=1$

This gives $\left|R_3(x)\right|\le \frac{e^{0.1}|x{|}^4}{4!}$

Since $\left|x\right|<0.1$

$$

\frac{e^{0.1}|x|^{4}}{4 !}=\frac{e^{0.1}|0.1|^{4}}{4 !}

$$

Therefore, maximum error is, $\frac{e^{0.1}|x{|}^4}{4!}=\frac{e^{0.1}|0.1{|}^4}{4!}$ Alternate method: Using graphing

  $\approx 4.605\times {10}^{-6}$

technique, find that when $\begin{array}{cc}\Rightarrow & \frac{\text{error}=4.605\times {10}^{-6}}{\text{lerror}\mid }=\left|e^x-\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}\right)\right|\end{array}$

  $\left|x\right|<\mathrm{0.1.}\le \left|e^{0.1}-\left(1+0.1+\frac{0.01}{2}+\frac{0.001}{6}\right)\right|$

  $\begin{array}{cc}& =4.251\times {10}^{-6}\\ \Rightarrow & \text{error}=4.251\times {10}^{-6}\end{array}$