How does the recursion relation work in the solution to this differential equation (using series)?

Question

Sorry for the vague title but it would not let me post the first step and last step of this equation (too many characters!).

How does $$\dfrac{a_0}{3n(3n-1)(3n-3)(3n-4)\cdots 9 \cdot 8 \cdot 6 \cdot 5 \cdot 3 \cdot 2} = \dfrac{a_0\Gamma(\frac{2}{3})}{3^nn!3^n\Gamma(n+\frac{2}{3})}?$$

This is a text book example I'm having trouble following. Solve the differential equation $y'' = xy$ using series. This makes sense in general. But when we get down to solving for the coefficients of the series, it confuses me.

Based of the relations $$a_nn(n-1) = 0 \text{ for } n=0,1,2$$ and $$a_nn(n-1) = a_{n-3} \text{ for } n = 3,4,\ldots$$

we have

$$\begin{align} a_{3n} &= \dfrac{a_0}{3n(3n-1)(3n-3)(3n-4)\cdots 9 \cdot 8 \cdot 6 \cdot 5 \cdot 3 \cdot 2} \\ &= \dfrac{a_0}{3^n n! 3^n(n-\frac{1}{3})(n-1-\frac{1}{3})(n-2-\frac{1}{3})\cdots (\frac{5}{3}) (\frac{2}{3})}\\ &= \dfrac{a_0\Gamma(\frac{2}{3})}{3^nn!3^n\Gamma(n+\frac{2}{3})}\\ \end{align}$$ Particularly, that middle step there is throwing me off.

The book also provided $$\begin{align} a_{3n+1} &= \dfrac{a_1}{(3n+1)(3n)(3n-2)(3n-3)\cdots 10 \cdot 9 \cdot 7 \cdot 6 \cdot 4 \cdot 3} \\ &= \dfrac{a_1}{3^n n! 3^n(n+\frac{1}{3})(n-1+\frac{1}{3})(n-2+\frac{1}{3})\cdots (\frac{7}{3}) (\frac{4}{3})}\\ &= \dfrac{a_1\Gamma(\frac{4}{3})}{3^nn!3^n\Gamma(n+\frac{4}{3})}\\ \end{align}$$ and $$a_{3n+2} = 0,$$ but that last one makes sense.

As a side note, why do they put $3^n$ twice in the denominator? Why not $3^{2n}$?

Answer 1

From what I understand, the product I denote as $P$ $$(3n+1)(3n)(3n−2)(3n−3)⋯10⋅9⋅7⋅6⋅4⋅3$$ is really (by writing the product pairwise) $$P = (3\cdot 4)(6\cdot 7)(9\cdot 10)\cdots(3n\cdot(3n+1))$$ the pattern is multiplying a pair, one of which is a multiple of three and the other, the increment of that. First, notice there are $n$ pairs, we can count from $3$ to $3n$, we can factor a 3 from each of these pairs to get $$P = 3^n(1\cdot4)(2\cdot 7)(3\cdot10)\cdots(n\cdot(3n+1))$$ as we can see, each pair now contains $1,2,3,...,n$. We can factor that as well to obtain $$P = 3^nn!(4)(7)(10)\cdots\left(3n+1\right)$$

Now I think they also factor out a 3 from each to obtain another $3^n$: $$P = 3^nn!3^n(4/3)(7/3)(10/3)\cdots(n+1/3)$$

Also, they don't write $3^n3^n$ as $3^{2n}$ because they might want to make it clearer for students as to where each $3^n$ contribution came from.

Answer 2

Use the fact that $\Gamma{(x+1)} = x \Gamma{(x)} $:

$$\begin{align}\Gamma{\left (n+\frac{2}{3} \right )} &= \left (n-\frac{1}{3} \right ) \Gamma{\left (n-\frac{1}{3} \right )}\\ &= \left (n-\frac{1}{3} \right ) \left (n-\frac{4}{3} \right )\Gamma{\left (n-\frac{4}{3} \right )}\\ &=\cdots\\&=\left (n-\frac{1}{3} \right ) \left (n-\frac{4}{3} \right )\cdots \frac{2}{3}\Gamma{\left (\frac{2}{3} \right )}\end{align}$$

The two $3^n$'s come from the factorial and the gamma function, each.