Consider the function f(x) defined by the power series,
![\[ f(x) = \sum_{n=0}^{\infty} \frac{x^n}{(n!)^2}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-bb463074b7c7e8217383f348a79c8a71_l3.png "Rendered by QuickLaTeX.com")
Determine the interval of convergence of f(x) and show that f(x) satisfies the differential equation
![\[ xy'' + y' - y = 0. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-e3ccfb0bae4e26ec2fffd1e0b7837f12_l3.png "Rendered by QuickLaTeX.com")
First, to determine the radius of convergence we use the ratio test,
![\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{x^{n+1}}{(n+1)!^2} \right) \left( \frac{(n!)^2}{x^n} \right) \\[9pt] &= \lim_{n \to \infty} \frac{x}{(n+1)^2} \\[9pt] &= 0. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-8dd935993ce7f6d969b811c135199337_l3.png "Rendered by QuickLaTeX.com")
Therefore, the series converges for all 
(or the radius of convergence is 
). Next, to show that 
satisfies the given differential equation we take the first two derivatives,
![\begin{align*} && y &= \sum_{n=0}^{\infty} \frac{x^n}{(n!)^2} \\[9pt] \implies && y' &= \sum_{n=1}^{\infty} \frac{nx^{n-1}}{(n!)^2} = \sum_{n=1}^{\infty} \frac{x^{n-1}}{n ((n-1)!)^2} \\[9pt] \implies && y'' &= \sum_{n=2}^{\infty} \frac{(n-1)x^{n-2}}{n ((n-1)!)^2} = \sum_{n=2}^{\infty} \frac{x^{n-2}}{n(n-1)((n-2)!)^2}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-f8c00c650e8831d0bd7d9c09dde620db_l3.png "Rendered by QuickLaTeX.com")
Plugging this into the given differential equation we have
![\begin{align*} xy'' + y' - y &= x \left( \sum_{n=2}^{\infty} \frac{x^{n-2}}{n(n-1)((n-2)!)^2} \right) + \left( \sum_{n=1}^{\infty} \frac{x^{n-1}}{n ((n-1)!)^2} \right) - \left( \sum_{n=0}^{\infty} \frac{x^n}{(n!)^2} \right) \\[9pt] &= \sum_{n=2}^{\infty} \frac{x^{n-1}}{n(n-1)((n-2)!)^2} + \sum_{n=1}^{\infty} \frac{x^{n-1}}{n ((n-1)!)^2} - \sum_{n=0}^{\infty} \frac{x^n}{(n!)^2} \\[9pt] &= \sum_{n=1}^{\infty} \frac{x^n}{(n+1)n((n-1)!)^2} + \sum_{n=0}^{\infty} \frac{x^n}{(n+1)(n!)^2} -\sum_{n=0}^{\infty} \frac{x^n}{(n!)^2} \\[9pt] &= \left( \sum_{n=1}^{\infty} \frac{x^n}{(n+1)n((n-1)!)^2} + \frac{x^n}{(n+1)(n!)^2} - \frac{x^n}{(n!)^2} \right) + 1 - 1 \\[9pt] &= \sum_{n=1}^{\infty} \left( \frac{1}{(n+1)n((n-1)!)^2} + \frac{1}{(n+1)(n!)^2} - \frac{1}{(n!)^2} \right)x^n \\[9pt] &= \sum_{n=1}^{\infty} \frac{n + 1 - (n+1)}{(n+1)(n!)^2} x^n \\[9pt] &= \sum_{n=1}^{\infty} \frac{0}{(n+1)(n!)^2} x^n \\[9pt] &= 0. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-ab759828132c53aadd721ecd9dcb98e8_l3.png "Rendered by QuickLaTeX.com")
Hence, 
indeed satisfies the given differential equation.