Elementary Differential Equations and Boundary Value Problems 9th Edition

Published by Wiley
ISBN 10: 0-47038-334-8
ISBN 13: 978-0-47038-334-6

Chapter 5 - Series Solutions of Second Order Linear Equations - 5.1 Review of Power Series - Problems - Page 249: 19

Answer

$$\sum^{\infty}_{n=0}a_{n}(x-1)^{n+1}$$

Work Step by Step

1. Add 1 to each n in the second expression: $$\sum^{\infty}_{n+1=1}a_{n+1-1}(x-1)^{n+1} =\sum^{\infty}_{n=1-1}a_{n}(x-1)^{n+1} $$ $$\sum^{\infty}_{n=0}a_{n}(x-1)^{n+1}$$ - We get exactly the first expression, therefore, they are equivalent.
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