Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 12 - Sequences and Series - Chapter Review - Review Exercises - Page 662: 23

Answer

$${P_4}\left( x \right) = 1 + \frac{2}{3}x - \frac{1}{9}{x^2} + \frac{4}{{81}}{x^3} - \frac{7}{{243}}{x^4}$$

Work Step by Step

$$\eqalign{ & f\left( x \right) = {\left( {1 + x} \right)^{2/3}} \cr & {\text{Use the definition of Taylor Polynomial of Degree }}n\,\,\,\left( {{\text{see page 629}}} \right) \cr & {\text{Let }}f{\text{ be a function that can be differentiated }}n{\text{ times at 0}}{\text{. The Taylor }} \cr & {\text{polynomial of degree }}n{\text{ for }}f{\text{ at 0 is }} \cr & {P_n}\left( x \right) = f\left( 0 \right) + \frac{{{f^{\left( 1 \right)}}\left( 0 \right)}}{{1!}}x + \frac{{{f^{\left( 2 \right)}}\left( 0 \right)}}{{2!}}{x^2} + \cdots + \frac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}{x^n} = \sum\limits_{i = 0}^n {\frac{{{f^{\left( n \right)}}\left( 0 \right)}}{{i!}}} {x^i} \cr & {\text{Find the Taylor polynomials of degree 4 at 0}}{\text{. }} \cr & {\text{then }}n = 4. \cr & {\text{The }}n{\text{ - th derivatives are}} \cr & {f^{\left( 1 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ {{{\left( {1 + x} \right)}^{2/3}}} \right] = \frac{2}{3}{\left( {1 + x} \right)^{ - 1/3}} \cr & {f^{\left( 2 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ {\frac{2}{3}{{\left( {1 + x} \right)}^{ - 1/3}}} \right] = - \frac{2}{9}{\left( {1 + x} \right)^{ - 4/3}} \cr & {f^{\left( 3 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ { - \frac{2}{9}{{\left( {1 + x} \right)}^{ - 4/3}}} \right] = \frac{8}{{27}}{\left( {1 + x} \right)^{ - 7/3}} \cr & {f^{\left( 4 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ {\frac{8}{{27}}{{\left( {1 + x} \right)}^{ - 7/3}}} \right] = - \frac{{56}}{{81}}{\left( {1 + x} \right)^{ - 10/3}} \cr & {\text{evaluate }}f\left( 0 \right),{f^{\left( 1 \right)}}\left( 0 \right),{f^{\left( 2 \right)}}\left( 0 \right),{f^{\left( 3 \right)}}\left( 0 \right),{f^{\left( 4 \right)}}\left( 0 \right) \cr & f\left( 0 \right) = {\left( {1 + 0} \right)^{2/3}} = 1 \cr & {f^{\left( 1 \right)}}\left( 0 \right) = \frac{2}{3}{\left( {1 + 0} \right)^{ - 1/3}} = \frac{2}{3} \cr & {f^{\left( 2 \right)}}\left( 0 \right) = - \frac{2}{9}{\left( {1 + 0} \right)^{ - 4/3}} = - \frac{2}{9} \cr & {f^{\left( 3 \right)}}\left( 0 \right) = \frac{8}{{27}}{\left( {1 + 0} \right)^{ - 7/3}} = \frac{8}{{27}} \cr & {f^{\left( 4 \right)}}\left( 0 \right) = - \frac{{56}}{{81}}{\left( {1 + 0} \right)^{ - 10/3}} = - \frac{{56}}{{81}} \cr & {\text{Replace the found values into the definition of Taylor}} \cr & {\text{ Polynomial of Degree }}n{\text{ for }}n = 4 \cr & {P_4}\left( x \right) = f\left( 0 \right) + \frac{{{f^{\left( 1 \right)}}\left( 0 \right)}}{{1!}}x + \frac{{{f^{\left( 2 \right)}}\left( 0 \right)}}{{2!}}{x^2} + \frac{{{f^{\left( 3 \right)}}\left( 0 \right)}}{{3!}}{x^3} + \frac{{{f^{\left( 4 \right)}}\left( 0 \right)}}{{4!}}{x^4} \cr & {P_4}\left( x \right) = 1 + \frac{{2/3}}{{1!}}x + \frac{{ - 2/9}}{{2!}}{x^2} + \frac{{8/27}}{{3!}}{x^3} + \frac{{ - 56/81}}{{4!}}{x^4} \cr & {\text{simplify }} \cr & {P_4}\left( x \right) = 1 + \frac{2}{3}x - \frac{1}{9}{x^2} + \frac{4}{{81}}{x^3} - \frac{7}{{243}}{x^4} \cr} $$
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