University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.8 - Taylor and Maclaurin Series - Exercises - Page 537: 45


$L(x)=x$ and $Q(x)=x$

Work Step by Step

We have $f(x) = \sin x$ and $f'(x)= \cos x ; \\f''(x)= - \sin x$ $\implies f(0)=0; f'(0)=1; f''(0)=0$ Therefore, $L(x)=f(0)+xf'(0)=x$ and $Q(x) =f(0) x +xf'(0) +\dfrac{x^2}{2!}f''(0)=x$ So, $L(x)=x$ and $Q(x)=x$
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