# 6.1: Review of Power Series

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

Before we go on to solving differential equations using power series, it would behoove you to go back to you calculus notes and review power series. There is one topic that was a small detail in first year calculus, but will be a main issue for solving differential equations. This is the technique of changing the index.

Example \(\PageIndex{1}\)

Change the index and combine the power series

\[ \sum_{n=1}^\infty n\,a_n\,x^{n+1} + \sum_{n=0}^\infty a_n\,x^n.\nonumber \]

**Solution**

There are two issues here: The first is the the powers of \(x\) are different and the second is that the summations begin at different values. To make the powers of \(x\) the same we perform the substitution

[u = n + 1, \;\;\; n = u - 1. \nonumber\]

Notice that when \(n = 1\), \(u = 2\) and when \(n\) is infinity so is \(u\). We can write

\[ \sum_{n=1}^{\infty} {n\,a_n\,x^{n+1}} = \sum_{u=2}^{\infty} {(u-1)\,a_{u-1}\,x^u}. \nonumber\]

Since \(u\) is a dummy index, we can rename it \(n\) to get

\[ \sum_{u=2}^{\infty} (u-1)\,a_{u-1}\,x^u = \sum_{n=2}^{\infty} (n-1)\,a_{n-1}\,x^n. \nonumber\]

We now need to find

\[ \sum_{n=2}^{\infty} (n-1)\,a_{n-1}\,x^n + \sum_{n=0}^{\infty} a_n \,x^n. \nonumber\]

The trouble now is that the starting numbers are different for the two series. We can pull the first two terms out of the second series to get

\[ \sum_{n=0}^{\infty} a_{n}\,x^n = a_0 + a_1\,x + \sum_{n=2}^{\infty} a_n\, x^n \nonumber \]

putting this together we get

\[\begin{align*} \sum_{n=0}^{\infty} (n-1)\, a_{n-1}\,x^n + a_0+a_1\,x+ \sum_{n=2}^{\infty} a_n\, x^n \\[5pt] &= a_0+a_1\,x+ \sum_{n=2}^{\infty} \left[ (n-1)\, a_{n-1} +a_n \right]x^n. \end{align*} \]

### Contributors

- Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.