Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 11 - Infinite Sequences and Series - 11.10 Exercises - Page 790: 72

Answer

$f^{58}(0)=0$

Work Step by Step

Maclaurin series of $f(x)=\Sigma_{n=0}^{\infty}\frac{f^{n}(0)x^{n}}{n!}$ Coefficient of $x^{58}=\frac{f^{58}(0)}{n!}$ Note that the powers of $x$ in the expansion of $(1+x^{3})^{30}$ must be multiples of $58$. But $58$ is not a multiple of $3$. Thus, $0=\frac{f^{58}(0)}{n!}$ Hence, $f^{58}(0)=0$
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