Q. 70 - 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.

$\ln 1.3$

Step-by-Step Solution

Answer

Also, to approximate the quantity $\ln 1.3$ to within ${10}^{-6}$ of its value, the number of terms required is $14$

step 1: given information

 Consider the function $\ln 1.3$

step 2: calculation

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

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

So, to find the Maclaurin series for the function $\ln 1.3$, put $x=0.3$

Therefore,

$f(x=0.3)=\sum _{k=1}^{\infty }\frac{(-1{)}^{k+1}}{k}(0.3{)}^k$

That is

$\ln 1.3=\sum _{i=1}^{\infty }\frac{(-1{)}^{k+1}}{k}(0.3{)}^k$

Now, to approximate the value of $\ln 1.3$ up to three decimal places, let us first write its corresponding Maclaurin series in expanded form.

So,

$\ln 1.3=\sum _{k=1}^{\infty }\frac{(-1{)}^{k+1}}{k}(0.3{)}^4$

$$\begin{gathered} =0.3-\frac{1}{2}(0.3{)}^2-\frac{1}{3}(0.3{)}^3+\frac{1}{4}(0.3{)}^4-\frac{1}{5}(0.3{)}^5 \\ =0.3-\frac{0.09}{2}+\frac{0.027}{3}-\frac{0.0081}{4}+\frac{0.00243}{5} \\ =0.3-0.045+0.009-0.002025+0.000486 \end{gathered}$$

Therefore, the approximate the value of $\ln 1.3$  up to three decimal places is $0.262$

Also, to approximate the quantity $\ln 1.3$ to within ${10}^{-6}$ of its value, the number of terms required is $14$