Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.8 Newton's Method - 4.8 Exercises - Page 316: 14

Answer

\[\boxed{\begin{array}{*{20}{c}} n&{{x_n}} \\ 0&{1.700000} \\ 1&{1.718220} \\ 2&{1.718282} \\ 3&{1.718282} \\ 4&{1.718282} \\ 5&{1.718282} \\ 6&{1.718282} \\ 7&{1.718282} \\ 8&{1.718282} \\ 9&{1.718282} \\ {10}&{1.718282} \end{array}}\]

Work Step by Step

\[\begin{gathered} {\text{Let }}f\left( x \right) = \ln \left( {x + 1} \right) - 1,{\text{ and }}{x_0} = 1.7 \hfill \\ {\text{Using the Newton's Method }}{x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}} \hfill \\ f'\left( x \right) = \frac{1}{{x + 1}} \hfill \\ {\text{Then evaluating }}f\left( {{x_n}} \right){\text{ and }}f'\left( {{x_n}} \right) \hfill \\ {x_{n + 1}} = {x_n} - \frac{{\ln \left( {{x_n} + 1} \right) - 1}}{{\frac{1}{{{x_n} + 1}}}} \hfill \\ \hfill \\ {\text{Taking }}{x_0} = 1.7 \hfill \\ {x_0} = 1.7 \hfill \\ {x_{0 + 1}} = {x_1} = 1.7 - \frac{{\ln \left( {1.7 + 1} \right) - 1}}{{\frac{1}{{1.7 + 1}}}} = 1.718220 \hfill \\ {x_{1 + 1}} = {x_2} = 1.718220 - \frac{{\ln \left( {1.718220 + 1} \right) - 1}}{{\frac{1}{{1.718220 + 1}}}} \approx 1.718282 \hfill \\ {x_{1 + 1}} = {x_2} = 1.718282 - \frac{{\ln \left( {1.718282 + 1} \right) - 1}}{{\frac{1}{{1.718282 + 1}}}} \approx 1.718282 \hfill \\ {x_{1 + 3}} = {x_4} = 1.718282 - \frac{{\ln \left( {1.718282 + 1} \right) - 1}}{{\frac{1}{{1.718282 + 1}}}} \approx 1.718282 \hfill \\ {x_{1 + 4}} = {x_5} = 1.718282 - \frac{{\ln \left( {1.718282 + 1} \right) - 1}}{{\frac{1}{{1.718282 + 1}}}} \approx 1.718282 \hfill \\ {x_{1 + 5}} = {x_6} = 1.718282 - \frac{{\ln \left( {1.718282 + 1} \right) - 1}}{{\frac{1}{{1.718282 + 1}}}} \approx 1.718282 \hfill \\ {x_{1 + 6}} = {x_7} = 1.718282 - \frac{{\ln \left( {1.718282 + 1} \right) - 1}}{{\frac{1}{{1.718282 + 1}}}} \approx 1.718282 \hfill \\ {x_{1 + 7}} = {x_8} = 1.718282 - \frac{{\ln \left( {1.718282 + 1} \right) - 1}}{{\frac{1}{{1.718282 + 1}}}} \approx 1.718282 \hfill \\ {x_{1 + 8}} = {x_9} = 1.718282 - \frac{{\ln \left( {1.718282 + 1} \right) - 1}}{{\frac{1}{{1.718282 + 1}}}} \approx 1.718282 \hfill \\ {x_{1 + 9}} = {x_{10}} = 1.718282 - \frac{{\ln \left( {1.718282 + 1} \right) - 1}}{{\frac{1}{{1.718282 + 1}}}} \approx 1.718282 \hfill \\ \hfill \\ {\text{Thus}}{\text{, we obtain}} \hfill \\ \boxed{\begin{array}{*{20}{c}} n&{{x_n}} \\ 0&{1.700000} \\ 1&{1.718220} \\ 2&{1.718282} \\ 3&{1.718282} \\ 4&{1.718282} \\ 5&{1.718282} \\ 6&{1.718282} \\ 7&{1.718282} \\ 8&{1.718282} \\ 9&{1.718282} \\ {10}&{1.718282} \end{array}} \hfill \\ \end{gathered} \]
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