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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 $3^{2n}$?

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

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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.

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  • $\begingroup$ 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? $\endgroup$ – Bark Jr. Dec 6 '14 at 21:40
  • $\begingroup$ @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." $\endgroup$ – Ron Gordon Dec 6 '14 at 22:24

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