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: 18

Answer

$${P_4}\left( x \right) = 5 + 10x + 10{x^2} + \frac{{20}}{3}{x^3} + \frac{{10}}{3}{x^4}$$

Work Step by Step

$$\eqalign{ & f\left( x \right) = 5{e^{2x}} \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[ {5{e^{2x}}} \right] = 10{e^{2x}} \cr & {f^{\left( 2 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ {10{e^{2x}}} \right] = 20{e^{2x}} \cr & {f^{\left( 3 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ {20{e^{2x}}} \right] = 40{e^{2x}} \cr & {f^{\left( 4 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ {40{e^{2x}}} \right] = 80{e^{2x}} \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) = 5{e^{2\left( 0 \right)}} = 5 \cr & {f^{\left( 1 \right)}}\left( 0 \right) = 10{e^{2\left( 0 \right)}} = 10 \cr & {f^{\left( 2 \right)}}\left( 0 \right) = 20{e^{2\left( 0 \right)}} = 20 \cr & {f^{\left( 3 \right)}}\left( 0 \right) = 40{e^{2\left( 0 \right)}} = 40 \cr & {f^{\left( 4 \right)}}\left( 0 \right) = 80{e^{2\left( 0 \right)}} = 80 \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) = 5 + \frac{{10}}{{1!}}x + \frac{{20}}{{2!}}{x^2} + \frac{{40}}{{3!}}{x^3} + \frac{{80}}{{4!}}{x^4} \cr & {\text{simplifying}} \cr & {P_4}\left( x \right) = 5 + 10x + 10{x^2} + \frac{{20}}{3}{x^3} + \frac{{10}}{3}{x^4} \cr} $$
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