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: 17


$$f(x)=\sinh x=\frac{e^{x}-e^{-x}}{2}= x+\frac{x^3}{3!}+\frac{x^5}{5!}+...$$ Convergent for any value of $x$.

Work Step by Step

By making use of Table 2, we have the Maclaurin series for $f(x)=e^x$ as follows $$f(x)=e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$$ Now by the comparison with the function $f(x)=\sinh x=\frac{e^{x}-e^{-x}}{2}$, we have the Maclaurin series as follows $$f(x)=\sinh x=\frac{e^{x}-e^{-x}}{2}\\ =\frac{1}{2}(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...)\\ -\frac{1}{2}(1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}+...)\\ =x+\frac{x^3}{3!}+\frac{x^5}{5!}+...$$ Moreover, again from Table 2, the Maclaurin series for $f(x)=e^x$ is convergent for any value of $x$. Hence, the Maclaurin series for $f(x)=\sinh x$ is also convergent for any value of $x$.
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