Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 772: 81



Work Step by Step

$p(x+1)=\Sigma_{i=0}^{\infty}\frac{p^{i}({a})}{i!}(x+1-a)^{i}$ $=\Sigma_{i=0}^{n}\frac{p^{i}({x})}{i!}(x+1-x)^{i}$ $=\Sigma_{i=0}^{n}\frac{p^{i}({x})}{i!}(1)^{i}$ $=\Sigma_{i=0}^{n}\frac{p^{i}({x})}{i!}$ Hence, $p(x+1)=\Sigma_{i=0}^{n}\frac{p^{i}({x})}{i!}$
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