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: 32


$1+ \dfrac{1}{2}x-\dfrac{1}{8}x^2+\dfrac{1}{16}x^3-....+(-1)^{n-1} \dfrac{1 \cdot 3 \cdot 5...(2n-3)x^n }{n! 2^n}$

Work Step by Step

Differentiate the given function $f(x)$ as follows: $f'(x)=\dfrac{1}{2 (x+1)^{1/2}} \implies f'(0)=1 $; $f''(x)=-\dfrac{1}{4(x+1)^{3/2}} \implies f''(0)=\dfrac{1}{2} \\ f'''(x)=\dfrac{3}{8(x+1)^{5/2}}\implies f'''(0)=-\dfrac{1}{4}$ The Maclaurin series at $x=0$ can be written as: $f(x)=\sqrt {x+1}=1+ \dfrac{1}{2}x-\dfrac{1}{8}x^2+\dfrac{1}{16}x^3-....+(-1)^{n-1} \dfrac{1 \cdot 3 \cdot 5...(2n-3) x^n}{n! 2^n}$
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