## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 9 - Section 9.8 - Taylor and Maclaurin Series - Exercises - Page 536: 10

#### Answer

$p_{0}(x)=1\\ p_{1}(x)=1-\dfrac{x}{2} \\p_{2}(x)=1-\dfrac{x}{2}-\dfrac{x^2}{8}\\p_{3}(x)=1-\dfrac{x}{2}-\dfrac{x^2}{8}-\dfrac{x^3}{16}$

#### 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, $f(0)=1 \\ f'(0)=\dfrac{-1}{2}\\f''(0)=-\dfrac{1}{4}\\ f'''(0)=-\dfrac{3}{8}$ Thus, $p_{0}(x)=1\\ p_{1}(x)=1-\dfrac{x}{2} \\p_{2}(x)= 1-\dfrac{x}{2}+(-1/4)\dfrac{(x-0)^2}{2}=1-\dfrac{x}{2}-\dfrac{x^2}{8}\\p_{3}(x)=1-\dfrac{x}{2}-\dfrac{x^2}{8}+\dfrac{(-3/8)x^3}{6}=1-\dfrac{x}{2}-\dfrac{x^2}{8}-\dfrac{x^3}{16}$

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