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: 25

Answer

$$ \frac{x}{x^{2}+1}=\sqrt {1-x} $$ The root of the given equation, correct to eight decimal places, is 0.76682579.

Work Step by Step

$$ \frac{x}{x^{2}+1}=\sqrt {1-x} $$ We first rewrite the equation in standard form: $$ \frac{x}{x^{2}+1}-\sqrt {1-x}=0 $$ Therefore we let $$ f(x)=\frac{x}{x^{2}+1}-\sqrt {1-x}=0. $$ Then $$ f^{'}(x)=\frac{1-x^{2}}{(x^{2}+1)^{2}}+\frac{1}{2\sqrt {1-x}} $$ so, Formula 2 (Newton’s method) becomes $$ \begin{aligned} x_{n+1} &=x_{n}-\frac{f(x)}{f^{'}(x)}\\ \\ &=x_{n}-\frac{ \frac{x_{n}}{x_{n}^{2}+1}-\sqrt {1-x_{n}} }{\frac{1-x_{n}^{2}}{(x_{n}^{2}+1)^{2}}+\frac{1}{2\sqrt {1-x_{n}}}} \end{aligned} $$ In order to guess a suitable value for $x_1$ we sketch the graphs of $y=\frac{x}{x^{2}+1} $ and $y = \sqrt {1-x} $ in the Figure . It appears that they intersect at a point whose x-coordinate is somewhat less than 0.8, so let’s take $x_{1} = 0.8 $ as a convenient first approximation. Then Newton’s method gives $$ \begin{aligned} x_{2} &=x_{1}-\frac{ \frac{x_{1}}{x_{1}^{2}+1}-\sqrt {1-x_{1}} }{\frac{1-x_{1}^{2}}{(x_{1}^{2}+1)^{2}}+\frac{1}{2\sqrt {1-x_{1}}}} \\ &=(0.8)-\frac{ \frac{(0.8)}{(0.8)^{2}+1}-\sqrt {1-(0.8)} }{\frac{1-(0.8)^{2}}{((0.8)^{2}+1)^{2}}+\frac{1}{2\sqrt {1-(0.8)}}} \approx 0.76757581 \end{aligned} $$ repeating we get $$ x_{3}\approx 0.76682610 x_{4} \approx 0.76682579 \approx x_{5} , $$ we conclude that the root of the given equation, correct to eight decimal places, is 0.76682579.
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