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 536: 7

Answer

$p_{0}(x)=\dfrac{\sqrt 2}{2} \\ p_{1}(x)=\dfrac{\sqrt 2}{2}+\dfrac{\sqrt 2}{2}(x-\dfrac{\pi}{4}) \\p_{2}(x)=\dfrac{\sqrt 2}{2}+\dfrac{\sqrt 2}{2}(x-\dfrac{\pi}{4})-\dfrac{\sqrt 2}{4}(x-\dfrac{\pi}{4})^2 \\p_{3}(x)=\dfrac{\sqrt 2}{2}+\dfrac{\sqrt 2}{2}(x-\dfrac{\pi}{4})-\dfrac{\sqrt 2}{4}(x-\dfrac{\pi}{4})^2-\dfrac{\sqrt 2}{12}(x-\dfrac{\pi}{4})^3$

Work Step by Step

Taylor polynomial of order $n$ for the function $f(x)$ at the point $k$ can be defined as: $p_n(x)=f(k)+\dfrac{f'(k)}{1!}(x-k)+\dfrac{f''(k)}{2!}(x-k)^2+....+\dfrac{f^{n}(k)}{n!}(x-k)^n$ Here, $f(\dfrac{\pi}{4})=\dfrac{\sqrt 2}{2} ; f'(x)=\cos x \implies f'(\dfrac{\pi}{4})=\dfrac{\sqrt 2}{2}; f''(x)=-\sin x \implies f''(\dfrac{\pi}{4})=-\dfrac{\sqrt 2}{2}; f'''(x)=-\cos x\implies f'''(\dfrac{\pi}{4})=-\dfrac{\sqrt 2}{2}$ Thus, $p_{0}(x)=\dfrac{\sqrt 2}{2} \\ p_{1}(x)=\dfrac{\sqrt 2}{2}+\dfrac{\sqrt 2}{2}(x-\dfrac{\pi}{4}) \\p_{2}(x)=\dfrac{\sqrt 2}{2}+\dfrac{\sqrt 2}{2}(x-\dfrac{\pi}{4})-\dfrac{\sqrt 2}{4}(x-\dfrac{\pi}{4})^2 \\p_{3}(x)=\dfrac{\sqrt 2}{2}+\dfrac{\sqrt 2}{2}(x-\dfrac{\pi}{4})-\dfrac{\sqrt 2}{4}(x-\dfrac{\pi}{4})^2+(-\dfrac{\sqrt 2}{2})(\dfrac{1}{6})(x-\dfrac{\pi}{4})^3=\dfrac{\sqrt 2}{2}+\dfrac{\sqrt 2}{2}(x-\dfrac{\pi}{4})-\dfrac{\sqrt 2}{4}(x-\dfrac{\pi}{4})^2-\dfrac{\sqrt 2}{12}(x-\dfrac{\pi}{4})^3$
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