Chapter 8: Q. 27 (page 692)

In Exercises 23–32 we ask you to give Lagrange’s form for the corresponding remainder, $R_{4} (x)$
$\tan^{- 1} x$

Short Answer

Expert verified

The required answer is $R_{4} (x) = \frac{\{1 - 10 c^{2} + 5 c^{4}\}}{5 {(1 + c^{2})}^{5}} x^{5}$

Step by step solution

Step 1. Given Information

The given function is $f (x) = \tan^{- 1} x$

Step 2. Explanation

Using the Lagrange form for the remainder, we have,

$R_{n + 1} (x) = \frac{f^{n + 1} (c)}{(n + 1)!} x^{n + 1} R_{4} (x) = \frac{f^{5} (c)}{5!} x^{5}$

Now, we will find the fifth derivative of the function,

localid="1649334706302" $f^{1} (x) = \frac{1}{1 + x^{2}} f^{2} (x) = - \frac{2 x}{{(1 + x^{2})}^{2}} f^{3} (x) = - \frac{2 (1 - 3 x^{2})}{{(1 + x^{2})}^{3}} f^{4} (x) = \frac{24 (x - x^{4})}{{(1 + x^{2})}^{4}} f^{5} (x) = \frac{24}{{(1 + x^{2})}^{5}} \{1 - 10 x^{2} + 5 x^{4}\}$

Thus, we get,

localid="1649334781444" role="math" $R_{4} (x) = \frac{\frac{24 \{1 - 10 c^{2} + 5 c^{4}\}}{{(1 + c^{2})}^{5}}}{5!} x^{5} R_{4} (x) = \frac{24 \{1 - 10 c^{2} + 5 c^{4}\}}{120 {(1 + c^{2})}^{5}} x^{5} R_{4} (x) = \frac{\{1 - 10 c^{2} + 5 c^{4}\}}{5 {(1 + c^{2})}^{5}} x^{5}$

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In Exercises 23–32 we ask you to give Lagrange’s form for the corresponding remainder, $R_{4} (x)$
$\tan^{- 1} x$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Explanation

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In Exercises 23–32 we ask you to give Lagrange’s form for the corresponding remainder, R4(x)tan-1x

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Explanation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

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Study anywhere. Anytime. Across all devices.

In Exercises 23–32 we ask you to give Lagrange’s form for the corresponding remainder, $R_{4} (x)$
$\tan^{- 1} x$