Chapter 10.3, Problem 22E

To determine

The maximum error when the value is $\left|x\right|<0.01$

Answer to Problem 22E

The error value is $1.26\times {10}^{-3}$

Explanation of Solution

Given information:

The approximation $\sqrt{1+x}\approx 1+\left(x/2\right)$ is used when x is small. Estimate the maximum error when $\left|x\right|<0.01$ .

Formula used:

  $P_1\left(x\right)=1+\frac{x}{2}$

Calculation:

The given function is $f\left(x\right)=\sqrt{1+x}$

Then,

  $P_1\left(x\right)=1+\frac{x}{2}$

Now,

  $f"\left(x\right)=-\frac{1}{4}{\left(1+x\right)}^{-3/2}$

Therefore,

  $\left|f"\left(x\right)\right|=\left|-\frac{1}{4}{\left(1+x\right)}^{-3/2}\right|\le 0.2538$ for $\left|x\right|<0.01$

Use $M=0.2538,r=1$

This gives $\left|R_1\left(x\right)\right|\le \frac{0.2538{\left|x\right|}^2}{2!}$

Where,

  $\left|x\right|<0.01$

Therefore,

Maximum error is

  $\begin{array}{l}\frac{0.2538{\left|x\right|}^2}{2!}=\frac{0.2538{\left|0.01\right|}^2}{2!}\\ \text{error}=1.26\times {10}^{-3}\end{array}$

Alternate method: Using graphing technique, find that when $\left|x\right|<{10}^{-3}$ .|error|

  $\begin{array}{l}=\left|\sqrt{1+x}-\left(1+\frac{x}{2}\right)\right|\\ \le \left|\sqrt{1-0.01}-\left(1-\frac{0.01}{2}\right)\right|\\ =1.26\times {10}^{-5}\\ \text{error}=1.26\times {10}^{-3}\end{array}$

Conclusion:

The error value is $1.26\times {10}^{-3}$