Q.69 - Calculus

Question

Use appropriate Maclaurin series to express the quantities in Exercises 67-76 as alternating series. Then use Theorem $7.38$ to approximate the value of the specified quantities to within 0.001 of their actual value. How many terms in each series would be needed to approximate the given quantity to within ${10}^{-6}$ of its value? In Exercises $73-76$ be sure to convert to radian measure first.

$\ln 1.5$

Step-by-Step Solution

Answer

Also, to approximate the quantity $\ln 1.5 to within {10}^{-6}$of its value, the number of terms required is $14$

Step 1: given information

 Consider the function $\ln 1.5$

Step 2: calculation

The Maclaurin series for the function  is$f\left(x\right)=\ln (1+x)$

$f\left(x\right)=\sum _{k=1}^{\infty }\frac{(-1{)}^{k+1}}{k}{x}^k$

So, to find the Maclaurin series for the function $\ln 1.5$, put $x=0.5$ Therefore,

$f(x=0.5)=\sum _{k=1}^{\infty }\frac{(-1{)}^{k+1}}{k}(0.5{)}^k$

That is

$\ln 1.5=\sum _{k=1}^{\infty }\frac{(-1{)}^{k+1}}{k}(0.5{)}^k$

Now, to approximate the value of $\ln 1.5$ up to three decimal places, let us first vrite its corresponding Maclaurin series in expanded form.

So,

$\ln 1.5=\sum _{k=1}^{\infty }\frac{(-1{)}^{k+1}}{k}(0.5{)}^k$

$$\begin{gathered} =0.5-\frac{1}{2}(0.5{)}^2-\frac{1}{3}(0.5{)}^3+\frac{1}{4}(0.5{)}^4-\frac{1}{5}(0.5{)}^5 \\ =0.5-\frac{0.25}{2}+\frac{0.125}{3}-\frac{0.0625}{4}+\frac{0.03125}{5} \\ =0.5-0.125+0.0416-0.015625+0.00625 \end{gathered}$$

 Therefore, the approximate the value of $\ln 1.5$  up to three decimal places is $0.407$

Also, to approximate the quantity $\ln 1.5 to within {10}^{-6}$ of its value, the number of terms required is $14$