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Determine the interval of convergence for a given power series and show that it satisfies a given differential equation

by RoRi

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

    \[ f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n} x^{2n}}{(2n)!}. \]

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

    \[ y'' + 4y = 0. \]

First, to determine the interval of convergence for the power series we use the ratio test

    \begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{(-1)^{n+1}2^{2n+2} x^{2n+2}}{(2n+2)!} \right) \left( \frac{(2n)!}{(-1)^n 2^{2n} x^{2n}} \right) \\[9pt] &= \lim_{n \to \infty} \frac{-4x^2}{(2n+1)(2n+2)} \\[9pt] &= 0. \end{align*}

Hence, the series converges for all x. Next, to show that it satisfies the given differential equation we take the first two derivatives,

    \begin{align*} f(x) &= \sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n} x^{2n}}{(2n)!} \\[9pt] f'(x) &= \sum_{n=1}^{\infty} \frac{(-1)^n 2^{2n} (2n) x^{2n-1}}{(2n)!} \\[9pt] &= \sum_{n=1}^{\infty} \frac{(-1)^n 2^{2n} x^{2n-1}}{(2n-1)!} \\[9pt] f''(x) &= \sum_{n=1}^{\infty} \frac{(-1)^n 2^{2n} (2n-1) x^{2n-2}}{(2n-1)!} \\[9pt] &= \sum_{n=1}^{\infty} \frac{(-1)^n 2^{2n} x^{2n-2}}{(2n-2)!}. \end{align*}

Then, we have

    \begin{align*} y'' + 4y &= \sum_{n=1}^{\infty} \frac{(-1)^n 2^{2n} x^{2n-2}}{(2n-2)!} + 4 \sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n} x^{2n}}{(2n)!} \\[9pt] &= \sum_{n=0}^{\infty} \frac{(-1)^{n+1} 2^{2n+2} x^{2n}}{(2n)!} + 4 \sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n} x^{2n}}{(2n)!} \\[9pt] &= 4 \sum_{n=0}^{\infty} \frac{(-1)^{n+1} 2^{2n} x^{2n}}{(2n)!} + 4 \sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n} x^{2n}}{(2n)!} \\[9pt] &= -4 \sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n} x^{2n}}{(2n)!} + 4\sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n} x^{2n}}{(2n)!} \\[9pt] &= 0. \end{align*}

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

Further, in a previous exercise (Section 8.14, Exercise #2) that the solution of the differential equation y'' + 4y = 0 are all of the form

    \[ y = c_1 \cos (2x) + c_2 \sin (2x). \]

For this problem we also have f(0) = 1 so

    \[ f(0) = c_1 \cos 0 + c_2 \sin 0 = c_1 = 1. \]

Finally, we know this function is an even function (since f(x) = f(-x) for all x because we have x^{2n} inside the sum is the only x term). This means we must have c_2 = 0. Hence, we must have

    \[ f(x) = \cos (2x). \]

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