Calculus with Applications (10th Edition)

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

Chapter 12 - Sequences and Series - 12.3 Taylor Polynomials at 0 - 12.3 Exercises - Page 631: 14

Answer

$${P_4}\left( x \right) = - {x^3}$$

Work Step by Step

$$\eqalign{ & f\left( x \right) = \ln \left( {1 - {x^3}} \right) \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[ {\ln \left( {1 - {x^3}} \right)} \right] = \frac{{ - 3{x^2}}}{{1 - {x^3}}} \cr & {f^{\left( 2 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{ - 3{x^2}}}{{1 - {x^3}}}} \right] \cr & {\text{by using the quotient rule}} \cr & {f^{\left( 2 \right)}}\left( x \right) = \frac{{\left( {1 - {x^3}} \right)\left( { - 6x} \right) + 3{x^2}\left( { - 3{x^2}} \right)}}{{{{\left( {1 - {x^3}} \right)}^2}}} = \frac{{ - 6x + 6{x^4} - 9{x^4}}}{{{{\left( {1 - {x^3}} \right)}^2}}} \cr & {f^{\left( 2 \right)}}\left( x \right) = \frac{{ - 6x - 3{x^4}}}{{{{\left( {1 - {x^3}} \right)}^2}}} \cr & {f^{\left( 3 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{ - 6x - 3{x^4}}}{{{{\left( {1 - {x^3}} \right)}^2}}}} \right] \cr & {\text{by using the quotient rule}} \cr & {f^{\left( 3 \right)}}\left( x \right) = \frac{{{{\left( {1 - {x^3}} \right)}^2}\left( { - 6 - 12{x^3}} \right) + \left( {6x + 3{x^4}} \right)\left( 2 \right)\left( {1 - {x^3}} \right)\left( { - 3{x^2}} \right)}}{{{{\left( {1 - {x^3}} \right)}^4}}} \cr & {f^{\left( 3 \right)}}\left( x \right) = \frac{{\left( {1 - {x^3}} \right)\left( { - 6 - 12{x^3}} \right) + \left( {6x + 3{x^4}} \right)\left( 2 \right)\left( { - 3{x^2}} \right)}}{{{{\left( {1 - {x^3}} \right)}^3}}} \cr & {f^{\left( 3 \right)}}\left( x \right) = \frac{{ - 6 - 12{x^3} + 6{x^3} + 12{x^6} - 36{x^3} - 18{x^6}}}{{{{\left( {1 - {x^3}} \right)}^3}}} \cr & {f^{\left( 3 \right)}}\left( x \right) = \frac{{ - 6{x^6} - 42{x^3} - 6}}{{{{\left( {1 - {x^3}} \right)}^3}}} \cr & {f^{\left( 4 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{ - 6{x^6} - 42{x^3} - 6}}{{{{\left( {1 - {x^3}} \right)}^3}}}} \right] \cr & {\text{by using the quotient rule}} \cr & {f^{\left( 4 \right)}}\left( x \right) = \frac{{{{\left( {1 - {x^3}} \right)}^3}\left( { - 36{x^5} - 126{x^2}} \right) + \left( {6{x^6} + 42{x^3} + 6} \right)\left( 3 \right){{\left( {1 - {x^3}} \right)}^2}\left( { - 3{x^2}} \right)}}{{{{\left( {1 - {x^3}} \right)}^6}}} \cr & {f^{\left( 4 \right)}}\left( x \right) = \frac{{\left( {1 - {x^3}} \right)\left( { - 36{x^5} - 126{x^2}} \right) + \left( {6{x^6} + 42{x^3} + 6} \right)\left( 3 \right)\left( { - 3{x^2}} \right)}}{{{{\left( {1 - {x^3}} \right)}^4}}} \cr & {f^{\left( 4 \right)}}\left( x \right) = \frac{{ - 36{x^5} - 126{x^2} + 36{x^8} + 126{x^5} - 54{x^8} - 378{x^5} - 54{x^2}}}{{{{\left( {1 - {x^3}} \right)}^4}}} \cr & {f^{\left( 4 \right)}}\left( x \right) = \frac{{ - 18{x^8} - 288{x^5} - 180{x^2}}}{{{{\left( {1 - {x^3}} \right)}^4}}} \cr & \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) = \ln \left( {1 - {{\left( 0 \right)}^3}} \right) = 0 \cr & {f^{\left( 1 \right)}}\left( 0 \right) = \frac{{ - 3{{\left( 0 \right)}^2}}}{{1 - {{\left( 0 \right)}^3}}} = 0 \cr & {f^{\left( 2 \right)}}\left( 0 \right) = \frac{{ - 6\left( 0 \right) - 3{{\left( 0 \right)}^4}}}{{{{\left( {1 - {{\left( 0 \right)}^3}} \right)}^2}}} = 0 \cr & {f^{\left( 3 \right)}}\left( 0 \right) = \frac{{ - 6{{\left( 0 \right)}^6} - 42{{\left( 0 \right)}^3} - 6}}{{{{\left( {1 - {{\left( 0 \right)}^3}} \right)}^3}}} = - 6 \cr & {f^{\left( 4 \right)}}\left( 0 \right) = \frac{{ - 18{{\left( 0 \right)}^8} - 288{{\left( 0 \right)}^5} - 180{{\left( 0 \right)}^2}}}{{{{\left( {1 - {x^3}} \right)}^4}}} = 0 \cr & {\text{Replace the found values into the definition of Taylor Polynomial of Degree }}n \cr & {\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) = 0 + \frac{0}{{1!}}x + \frac{0}{{2!}}{x^2} + \frac{{ - 6}}{{3!}}{x^3} + \frac{0}{{4!}}{x^4} \cr & {\text{simplify }} \cr & {P_4}\left( x \right) = - {x^3} \cr} $$
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