Consider the function 
defined by the power series,
![\[ f(x) = \sum_{n=0}^{\infty} \frac{x^{4n}}{(4n)!}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-0539e0b91917a3e8fd5d9c0d431a4b69_l3.png "Rendered by QuickLaTeX.com")
Determine the interval of convergence of and show that satisfies the differential equation
![\[ f^{(4)}(x) = f(x). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-5eecd0600ca3df023a7e2c0db05438b7_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^{4n+4}}{(4n+4)!} \right) \left( \frac{(4n)!}{x^{4n}} \right) \\[9pt] &= \lim_{n \to \infty} \frac{x^4}{(4n+1)(4n+2)(4n+3)(4n+4)} \\[9pt] &= 0. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-3e45e905162e56cf52f253bacd9c7497_l3.png "Rendered by QuickLaTeX.com")
Therefore, converges for all 
(equivalently, 
). Next, to show that satisfies the differential equation 
we take the first four derivatives,
![\begin{align*} && y&= \sum_{n=0}^{\infty} \frac{x^{4n}}{(4n)!} \\[9pt] \implies && y' &= \sum_{n=1}^{\infty} \frac{4n x^{4n-1}}{(4n)!} = \sum_{n=1}^{\infty} \frac{x^{4n-1}}{(4n-1)!} \\[9pt] \implies && y'' &= \sum_{n=1}^{\infty} \frac{(4n-1)x^{4n-2}}{(4n-1)!} = \sum_{n=2}^{\infty} \frac{x^{4n-2}}{(4n-2)!} \\[9pt] \implies && y''' &= \sum_{n=1}^{\infty} \frac{(4n-2)x^{4n-3}}{(4n-2)!} = \sum_{n=3}^{\infty} \frac{x^{4n-3}}{(4n-3)!} \\[9pt] \implies && y^{(4)} &= \sum_{n=1}^{\infty} \frac{(4n-3)x^{4n-4}}{(4n-3)!} = \sum_{n=4}^{\infty} \frac{x^{4n-4}}{(4n-4)!}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-83ba7251cd026d31568353062e92f0de_l3.png "Rendered by QuickLaTeX.com")
But, reindexing this expression for the fourth derivative we have
![\[ y^{(4)} = \sum_{n=1}^{\infty} \frac{x^{4n-4}}{(4n-4)!} = \sum_{n=0}^{\infty} \frac{x^{4(n+1)-4}}{(4(n+1)-4)!} = \sum_{n=0}^{\infty} \frac{x^{4n}}{(4n)!} = y. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-48eebf1176ea542dae292e6bd908a576_l3.png "Rendered by QuickLaTeX.com")
Thus, satisfies the given differential equation.