Chapter 8: Q. 72 (page 693)

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

Short Answer

Expert verified

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 by step solution

Given information

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

also, $f (x) = \tan^{- 1} x$

calculation

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

$f (x) = \sum_{k = 0}^{\infty} \frac{(- 1)^{k}}{2 k + 1} x^{2 k + 1}$

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}}{2 k + 1} (- 0.6)^{2 k + 1}

That is

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

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,

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

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

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

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

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

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

$+ 0.00358 - 0.001866 + 0.0009878$

$= 0.851$

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