Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 771: 7

Answer

$2,~\frac{1}{12}(x-8),~-\frac{1}{288}(x-8)^2,~\frac{5}{20736}(x-8)^3$

Work Step by Step

Let's suppose $n=0,1,2,3$ then, $f^n(x)=f^n(8)$ $f(x)=\sqrt[3] x$ $=\sqrt[3] 8$ $=(8)^\frac{1}{3}$ $=(2^3)^\frac{1}{3}$ $f(x)=2$ $f'(x)=\frac{d}{dx}\sqrt[3] x$ $~~~~~~~~~=\frac{1}{3}(x)^{\frac{1}{3}-1}$ $~~~~~~~~~=\frac{1}{3}(x)^{\frac{1-3}{3}}$ $~~~~~~~~~=\frac{1}{3}(x)^{\frac{-2}{3}}$ $f'(x)=\frac{1}{3(x)^{\frac{2}{3}}}$ $f'(8)=\frac{1}{3(8)^{\frac{2}{3}}}$ $f'(8)=\frac{1}{12}$ $f''(x)=\frac{d}{dx}(\frac{1}{3(x)^{\frac{2}{3}}})$ $~~~~=\frac{1}{3}\frac{d}{dx}(x)^{\frac{-2}{3}}$ $~~~~=\frac{-2}{9}(x)^{\frac{-2}{3}-1}$ $~~~~=\frac{-2}{9}(x)^{\frac{-2-3}{3}}$ $~~~~=\frac{-2}{9}(x)^{\frac{-5}{3}}$ $f''(x)~=-\frac{2}{9(x)^{\frac{5}{3}}}$ $f''(8)=-\frac{2}{9(8)^{\frac{5}{3}}}$ $f''(8)=-\frac{2}{288}$ $f'''(x)=\frac{d}{dx}(-\frac{2}{9(x)^{\frac{5}{3}}})$ $~~~~~~~~~~~=\frac{10}{27}(x)^{\frac{-5}{3}-1}$ $~~~~~~~~~~~=\frac{10}{27}(x)^{\frac{-5-3}{3}}$ $~~~~~~~~~~~=\frac{10}{27}(x)^{\frac{-8}{3}}$ $f'''(x)=\frac{10}{27(x)^{\frac{8}{3}}}$ $f'''(8)=\frac{10}{27(8)^{\frac{8}{3}}}$ $f'''(8)=\frac{10}{6912}$ The $nth$ term of Taylor Series $=\frac{f^{n-1}(8)x^{n-1}}{(n-1)!}$ The $1st$ term of Taylor Series $=\frac{f(8)x^{0}}{0!}$ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\frac{2(1)}{0!}$ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\frac{2}{0!}$ The $1st$ term of Taylor Series $=2$ The $2nd$ term of Taylor Series $=\frac{f(8)x^{1}}{1!}$ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\frac{\frac{1}{12}}{1!}(x-8)^1$ The $2nd$ term of Taylor Series $=\frac{1}{12}(x-8)$ The $3rd$ term of Taylor Series $=\frac{f(8)x^{2}}{2!}$ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\frac{-\frac{2}{288}}{2!}(x-8)^2$ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=-\frac{1}{288}(x-8)^2$ The $3rd$ term of Taylor Series $=-\frac{1}{288}(x-8)^2$ The $4th$ term of Taylor Series $=\frac{f(8)x^{3}}{3!}$ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\frac{\frac{10}{6912}}{3!}(x-8)^3$ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\frac{5}{20736}(x-8)^3$ The $4th$ term of Taylor Series $=\frac{5}{20736}(x-8)^3$ Hence, the $4th$ non zero Taylor Series is : $2,~\frac{1}{12}(x-8),~-\frac{1}{288}(x-8)^2,~\frac{5}{20736}(x-8)^3$
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