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: 8


$p_{0}(x)=1\\ p_{1}(x)=1+2(x-\dfrac{\pi}{4}) \\p_{2}(x)=1+2(x-\dfrac{\pi}{4})+(x-\dfrac{\pi}{4})^2 \\p_{3}(x)=1+2(x-\dfrac{\pi}{4})+(x-\dfrac{\pi}{4})^2+\dfrac{8}{3}(x-\dfrac{\pi}{4})^3$

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, $f(\dfrac{\pi}{4})=1 ; f'(x)=sec^2 x \implies f'(\dfrac{\pi}{4})=2; f''(x)=2 sec^2 x \tan x \implies f''(\dfrac{\pi}{4})=-4; f'''(x)=4 sec^2 x \tan^2 x+2 \sec^4 x \implies f'''(\dfrac{\pi}{4})=16$ Thus, $p_{0}(x)=1\\ p_{1}(x)=1+2(x-\dfrac{\pi}{4}) \\p_{2}(x)=1+2(x-\dfrac{\pi}{4})+(x-\dfrac{\pi}{4})^2 \\p_{3}(x)=1+2(x-\dfrac{\pi}{4})+(x-\dfrac{\pi}{4})^2 +(16)(\dfrac{1}{6})(x-\dfrac{\pi}{4})^3=1+2(x-\dfrac{\pi}{4})+(x-\dfrac{\pi}{4})^2+\dfrac{8}{3}(x-\dfrac{\pi}{4})^3$
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