Explanation Given: For − 1 < x < 1 , tan − 1 x = ∑ n = 0 ∞ ( − 1 ) n x 2 n + 1 2 n + 1 = x − x 3 3 + x 5 5 − x 7 7 + .... ( 1 ) F ( x ) = x 2 1 × 2 − x 4 3 × 4 + x 6 5 × 6 − x 8 7 × 8 + ... ( 2 ) Proof: Since, log ( 1 + x ) = x − x 2 2 + x 2 3 − x 4 4 + .... ( 3 ) In the function F ( x ) = x tan − 1 x − 1 2 log ( x 2 + 1 ) use equation(1) and equation (3) and we get: F ( x ) = x tan − 1 x − 1 2 log ( x 2 + 1 ) = x ( x − x 3 3 + x 5 5 − x 7 7 + .... ) − 1 2 ( x 2 − x 4 2 + x 6 3 − x 8 4 + .... ) = ( x 2 − x 4 3 + x 6 5 − x 8 7 + .... ) − ( x 2 2 − x 4 4 + x 6 6 − x 8 8 + .... ) = x 2 1 × 2 − x 4 3 × 4 + x 6 5 × 6 − x 8 7 × 8 + ... Integrate on both the sides of the equation (1) and we get: ∫ tan − 1 x d x = ∫ ( x − x 3 3 + x 5 5 − x 7 7 + .... ) d x = x 2 1 × 2 − x 4 3 × 4 + x 5 5 × 6 − x 8 7 × 8 + ...

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ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 11.6, Problem 54E

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To prove:

$F (x) = x \tan^{- 1} x - \frac{1}{2} \log (x^{2} + 1)$ is an anti-derivative of $f (x) = \tan^{- 1} x$ satisfying $F (0) = 0$ . Show that $\frac{π}{6 \sqrt{3}} - \frac{1}{2} \ln \frac{4}{3} = \frac{1}{1 \times 2 (3)} - \frac{1}{3 \times 4 (3^{2})} + \frac{1}{5 \times 6 (3^{3})} - \frac{1}{7 \times 8 (3^{4})} + ...$ for $x = \frac{1}{\sqrt{3}}$ . Use a calculator to compare the value of the left-hand side with the partial sum $S_{4}$ of the series on the right.

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Chapter 11.6, Problem 53E

Chapter 11.6, Problem 55E

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Chapter 11 Solutions

CALCULUS (CLOTH)

Ch. 11.1 - Prob. 1PQ Ch. 11.1 - Prob. 2PQ Ch. 11.1 - Prob. 3PQ Ch. 11.1 - Prob. 4PQ Ch. 11.1 - Prob. 5PQ Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E

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Solution Summary: The author compares the value of the left-hand side with the partial sum S 4 of a series on the right.

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To prove: F(x)=xtan−1x−12log(x2+1) is an anti-derivative of f(x)=tan−1x satisfying F(0)=0. Show that π63−12ln43=11×2(3)−13×4(32)+15×6(33)−17×8(34)+... for x=13. Use a calculator to compare the value of the left-hand side with the partial sum S4 of the series on the right.

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Chapter 11 Solutions