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 - 11.11 Exercises - Page 798: 8

Answer

$$T_{3}(x) =x-\frac{1}{2} x^{3}$$

Work Step by Step

Given $$f(x) =x\cos x,\ \ a=0$$ Since \begin{align*} f(x) &=x\cos x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f(0) =0\\ f'(x) &=-x\sin x+ \cos x\ \ \ \ \ \ \ f'(0)=1\\ f''(x) &=-x\cos x-2\sin x \ \ \ \ \ f''(0)=0\\ f'''(x) &=x\sin x-3 \cos x\ \ \ \ \ \ \ \ f'''(0 )= -3 \end{align*} Then \begin{aligned} T_{3}(x) &=\sum_{n=0}^{3} \frac{f^{(n)}(0)}{n !} x^{n} \\ &=0+\frac{1}{1 !} x+0+\frac{-3}{3 !} x^{3}\\ &=x-\frac{1}{2} x^{3} \end{aligned}
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