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: 1

Answer

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

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, $f(0)=1 ; f'(x)=e^{2x} \implies f'(0)=2; f''(x)=2e^{2x} \implies f''(0)=4; f'''(x)=8e^{2x} \implies f'''(0)=8$ Now, $p_{3}(x)=\dfrac{1}{2}+(\dfrac{-1}{4})(x-0)+(\dfrac{1}{4})\dfrac{(x-0)^2}{2!}+(\dfrac{-3}{8})\dfrac{(x-0)^3}{3!}$ Thus, $p_{0}(x)=1 \\ p_{1}(x)=1+2x \\p_{2}(x)=1+2x+2x^2 \\p_{3}(x)=1+2x+\dfrac{4}{2}x^2+\dfrac{8}{6}x^3=1+2x+2x^2+\dfrac{4}{3}x^3$
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