Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.7 Taylor Series - Exercises - Page 588: 28

Answer

$e+ex+ex^2+\dfrac{5ex^3}{6}+\dfrac{5ex^4}{8}+...$

Work Step by Step

We have the Maclaurin series for $f(x)$ as follows: $$f(x)=f(0)+f^{\prime}(0) x+\dfrac{f^{\prime \prime}(0) x^2}{2!}+\dfrac{f^{\prime \prime \prime}(0) x^3}{3!}+\dfrac{f^{\prime \prime \prime \prime}(0) x^4}{4!}+...$$ We are given that $f(x)=e^{e^x}$ Our aim is to evaluate the coefficient of the series $f^4(0)$. Therefore, we have: $f(x)=f(0)+f^{\prime}(0) x+\dfrac{f^{\prime \prime}(0) x^2}{2!}+\dfrac{f^{\prime \prime \prime}(0) x^3}{3!}+\dfrac{f^{\prime \prime \prime \prime}(0) x^4}{4!}+...\\=e+ex+\dfrac{2ex^2}{2!}+\dfrac{5ex^3}{3!}+\dfrac{15ex^4}{4!}+...\\=e+ex+ex^2+\dfrac{5ex^3}{6}+\dfrac{5ex^4}{8}+...$
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