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

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

    \[ f(x) = 1 + \sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{(3n)!} x^{3n}. \]

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

    \[ y'' = x^a y + b. \]

Find a and b.


First, to determine the interval of convergence for f(x) we use the ratio test,

    \begin{align*}  \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{1 \cdot 4 \cdot 7 \cdots (3n+1) \cdot x^{3n+3}}{(3n+3)!} \right) \left( \frac{(3n)!}{1 \cdot 4 \cdot 7 \cdots (3n-2) \cdot x^{3n}} \right) \\[9pt]  &= \lim_{n \to \infty} \frac{(3n+1)x^3}{(3n+1)(3n+2)(3n+3)} \\[9pt]  &= \lim_{n \to \infty} \frac{x^3}{(3n+2)(3n+3)} \\[9pt]  &= 0. \end{align*}

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

    \begin{align*}  y &= 1 + \sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{(3n)!} x^{3n} \\[9pt]  y' &= \sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)\cdot (3n)}{(3n)!} x^{3n-1} = \sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{(3n-1)!} x^{3n-1} \\[9pt]  y'' &= \sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)\cdot (3n-1)}{(3n-1)!} x^{3n-2} = \sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{(3n-2)!}x^{3n-2}.  \end{align*}

Then, we have the differential equation y'' = x^a y + b,

    \begin{align*}  && y'' &= x^a y + b \\[9pt]  \implies && \sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{(3n-2)!}x^{3n-2} &= x^a\left( 1 + \sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{(3n)!} x^{3n} \right) + b \\[9pt]  \implies && \sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{(3n-2)!} x^{3n-2} &= b + x^a + \sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{(3n)!} x^{3n+a}. \end{align*}

But, for this equation to hold we must have b = 0 (since there is no constant term in y'' on the left) and we must also have a = 1 since there the coefficient of x^1 on the left is 1 and the only possible x^1 term on the right is if a = 1. Using these values of a and b we verify that the given differential equation is satisfied since we have

    \begin{align*}  y'' &= \sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{(3n-2)!} x^{3n-2} \\[9pt]  x^a y + b = xy &= x + \sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{(3n)!} x^{3n+1} \\[9pt]  &= x + \sum_{n=2}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3(n-1)-2)}{(3(n-1))!} x^{3(n-1)+1} &(\text{Reindexing})\\[9pt]  &= \frac{1}{(3\cdot 1 - 2)!} x^{3 \cdot 1 - 2} + \sum_{n=2}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-5)}{(3n-3)!} \cdot \frac{(3n-2)}{(3n-2)} \cdot x^{3n-2} \\[9pt]   &= \frac{1}{(3 \cdot 1 - 2)!} x^{3 \cdot 1 - 2} + \sum_{n=2}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{(3n-2)!} x^{3n-2} \\[9pt]  &= \sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{(3n-2)!} x^{3n-2}.  \end{align*}

Hence, we indeed have y'' = x^a y + b.

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