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: 26



Work Step by Step

We have the Maclaurin series for $ \sin x$ as follows: $$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$$ Replace $(x^3-x)$ for $x$ in the above form. Therefore, we have: $f(x)=\sin (x^3-x)=(x^3-x)-\dfrac{(x^3-x)^3}{3!}+...\\=(x^3-x)-\dfrac{x^9-3x^7+3x^5-x^3}{6}+...\\=(x^3-x)-\dfrac{-x^3}{6}+...\\=-x+\dfrac{7x^3}{6}+...$
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