Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.8 - Newton''s Method - 4.8 Exercises - Page 349: 21

Answer

$$ x^{3}=\tan^{-1} x $$ The roots of equation, correct to six decimal places, are 0, -0.902025 and 0.902025.
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Work Step by Step

$$ x^{3}=\tan^{-1} x $$ We first rewrite the equation in standard form: $$ x^{3}-\tan^{-1} x=0 $$ Therefore we let $$ f(x)=x^{3}-\tan^{-1} x=0 . $$ Then $$ f^{'}(x)=3x^{2}-\frac{1}{1+ x^{2}} $$ so, Formula 2 (Newton’s method) becomes $$ \begin{aligned} x_{n+1} &=x_{n}-\frac{f(x)}{f^{'}(x)}\\ \\ &=x_{n}-\frac{x_{n}^{3}-\tan^{-1} x_{n}}{3x_{n}^{2}-\frac{1}{1+ x_{n}^{2}}} \end{aligned} $$ In order to guess a suitable value for $x_1$ we sketch the graphs of $y=x^{3} $ and $y = \tan^{-1} x $ in the Figure . (*) It appears that they intersect at a point whose x-coordinate is somewhat less than 1, so let’s take $x_{1} = 1$ as a convenient first approximation. Then Newton’s method gives $$ \begin{aligned} x_{2} &=x_{1}-\frac{x_{1}^{3}-\tan^{-1} x_{1}}{3x_{1}^{2}-\frac{1}{1+ x_{1}^{2}}}\\ &=(1)-\frac{(1)^{3}-\tan^{-1} (1)}{3(1)^{2}-\frac{1}{1+ (1)^{2}}} \approx 0.914159 \end{aligned} $$ repeating we get $$ x_{3}\approx 0.902251, x_{4}\approx 0.902026 , x_{5}\approx 0.902025 . $$ (**) It appears that they intersect at a point whose x-coordinate is somewhat greater than -1, so let’s take $x_{1} = -1 $ as a convenient first approximation. Then Newton’s method gives $$ \begin{aligned} x_{2} &=x_{1}-\frac{x_{1}^{3}-\tan^{-1} x_{1}}{3x_{1}^{2}-\frac{1}{1+ x_{1}^{2}}}\\ &=(-1)-\frac{(-1)^{3}-\tan^{-1} (-1)}{3(-1)^{2}-\frac{1}{1+ (-1)^{2}}} \approx -0.914159 \end{aligned} $$ repeating we get $$ x_{3}\approx - 0.902251, x_{4}\approx -0.902026 , x_{5}\approx -0.902025 . $$ we conclude that the roots of equation, correct to six decimal places, are 0, -0.902025 and 0.902025.
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