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Determine the interval of convergence of ∑ x4n / (4n)! and show that it satisfies a given differential equation

Consider the function f(x) defined by the power series,

    \[ f(x) = \sum_{n=0}^{\infty} \frac{x^{4n}}{(4n)!}. \]

Determine the interval of convergence of f(x) and show that f(x) satisfies the differential equation

    \[ f^{(4)}(x) = f(x). \]


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*}

Therefore, f(x) converges for all x (equivalently, r = +\infty). Next, to show that f(x) satisfies the differential equation y^{(4)} = y 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*}

But, reindexing this expression for the fourth derivative we have

    \[ y^{(4)} = \sum_{n=1}^{\infty} \frac{x^{4n-4}}{(4n-4)!} = \sum_{n=0}^{\infty} \frac{x^{4(n+1)-4}}{(4(n+1)-4)!} = \sum_{n=0}^{\infty} \frac{x^{4n}}{(4n)!} = y. \]

Thus, f(x) satisfies the given differential equation.

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