Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 9 - Power Series - 9.1 Approximating Functions with Polynomials - 9.1 Exercises - Page 672: 1

Answer

$p_2(0)=f(0)$ $p'_2(0)=f'(0)$ $p"_2(0)=f"(0)$

Work Step by Step

We have to approximate $f(x)$ by a Taylor polynomial of degree $n=2$ for $a=0$. The Taylor polynomial approximating $f(x)$ is: $p_2(x)=f(0)+f'(0)x+\dfrac{f"(0)}{2!}x^2$ The conditions that $p_2(x)$ must check are: $p_2(0)=f(0)$ $p'_2(0)=f'(0)$ $p"_2(0)=f"(0)$
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