Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.8 - Taylor and Maclaurin Series - Exercises 10.8 - Page 618: 4


$A_{0}(x)=0 \\ A_{1}(x)=x \\A_{2}(x)=x-(\dfrac{1}{2})x^2 \\A_{3}(x)=x-(\dfrac{1}{2})x^2 +(\dfrac{1}{3})x^3$

Work Step by Step

The Taylor polynomial is defined for the function $f(x)$ at the point $a$ of order $n$ as: $A_n(x)=f(a)+\dfrac{f'(a)}{1!}(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+....+\dfrac{f^{n}(a)}{n!}(x-a)^n$ Now, we get $f(0)=0 \\ f'(x)=(1+x)^{-1} \implies f'(0)=1\\ f''(x)=-(1+x)^{-2} \implies f''(0)=-1 \\ f'''(x)=2(1+x)^{-3}\implies f'''(0)=2$ This implies that $A_{0}(x)=0 \\ A_{1}(x)=x \\A_{2}(x)=x-(\dfrac{1}{2})x^2 \\A_{3}(x)=x-(\dfrac{1}{2})x^2+(\dfrac{2}{6})(x-0)^3 =x-(\dfrac{1}{2})x^2 +(\dfrac{1}{3})x^3$
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