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

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 $3^{2n}$?

• I should clarify. My trouble is the first step to the middle step. Not the middle step to the last step. – Bark Jr. Nov 21 '14 at 20:21

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.

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.

• How are you able to be sure that the above $\Gamma\left(n +\frac{2}{3}\right)$ will end at $\frac{2}{3}$? How are you able to control for that so that the last term in your multiplication "sequence" isn't $\frac{1}{3}$? Because in the example demonstrated above (and going by Caitlin Dempsey's solution below), we need the last term to be $\frac{4}{3}$. How do we control for this? – Bark Jr. Dec 6 '14 at 21:40
• @BarkJr.: I have little idea of what you are trying to ask me, so I will advise you to plug in an integer, any integer you want. You will find that the endpoint for positive numbers is $2/3$. There is no "controlling for this." – Ron Gordon Dec 6 '14 at 22:24