Q. 72 - 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.

 

${\tan }^{-1}(-0.6)$

Step-by-Step Solution

Answer

the approximate value of ${\tan }^{-1}(-0.6)$up to three decimal places is $0.851$

also, to approximate the quantity  ${\tan }^{-1}0.4$to with ${10}^{-6}$ of its value, the number of terms required is $18$

step 1: Given information

consider the function ${\tan }^{-1}\left(-0.6\right)$

also, $f\left(x\right)={\tan }^{-1}x$

step 2: calculation

The Maclaurin series for the function $f\left(x\right)={\tan }^{-1}x$ is

$f\left(x\right)=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{2k+1}{x}^{2k+1}$

So, to find the Maclaurin series for the function  \tan ^{-1}(-0.6) , put x=0.4

Therefore,

$f(x=-0.6)=\sum _{k=0}^x\frac{(-1{)}^k}{2k+1}(-0.6{)}^{2k+1}$

That is

${\tan }^{-1}(-0.6)=\sum _{k=0}^{\infty }\frac{1}{2k+1}(0.6{)}^{2k+1}$

step 3:

Now, to approximate the value of ${\tan }^{-1}(-0.6)$ up to three decimal places, let us first vrite its corresponding Maclaurin series in expanded form.

So,

${\tan }^{-1}(-0.6)=\sum _{k=0}^{\infty }\frac{1}{2k+1}(0.6{)}^k$

$=1-\frac{1}{3}(0.6{)}^1+\frac{1}{5}(0.6{)}^2-\frac{1}{7}(0.6{)}^3+\frac{1}{9}(0.6{)}^4-\frac{1}{11}(0.6{)}^5$

$+\frac{1}{13}(0.6{)}^6-\frac{1}{15}(0.6{)}^7+\frac{1}{17}(0.6{)}^8$

$=1-\frac{0.6}{3}+\frac{0.36}{5}-\frac{0.216}{7}+\frac{0.1296}{9}-\frac{0.07776}{11}$

$+\frac{0.0466}{13}-\frac{0.02799}{15}+\frac{0.01679}{17}$

$=1-0.2+0.072-0.0309+0.0144-0.00707$

$+0.00358-0.001866+0.0009878$

$=0.851$