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: 32


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

Work Step by Step

We have $f'(x)=\dfrac{1}{2\sqrt {x+1}} \implies f'(0)=1 $; $f''(x)=-\dfrac{1}{4(x+1)^{3/2}} \implies f''(0)=\dfrac{1}{2} \\ f'''(x)=\dfrac{3}{8(x+1)^{5/2}}\implies f'''(0)=-\dfrac{1}{4}$ and $f^{4}(0)=-\dfrac{15}{16}$ and so on. Therefore, the Maclaurin's series at $x=0$ is as follows: $\sqrt {x+1}=1+ \dfrac{1}{2}x-\dfrac{1}{8}x^2+\dfrac{1}{16}x^3-........+(-1)^{n-1} \dfrac{1 \cdot 3 \cdot 5......(2n-3)}{n! 2^n}x^n$
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