How do I truncate this infinite series by choosing an odd l?

Given the odd solution to the Legendre differential equation: y_("odd")(x) = sum_(n=0)^(oo) {(-1)^n ([(l - (2n-1))cdots(l-3)(l-1)][(l+2)(l+4)cdots(l+2n)])/((2n+1)!) x^(2n+1)} How do you show that for the appropriate choice of $l$, this sum reduces down to, for example, the term ${y}_{1} \left(x\right) = x$? Supposely, an odd $l$ truncates this series solution, which sums over $n$. The indexing for $y$ follows what is shown here. I'm feeling kinda braindead after trying the derivations at the beginning of this problem for hours. Maybe this is simple, but I don't see it right now.

1 Answer
Aug 30, 2017

I asked my professor about this yesterday, and I actually think I figured this out, but I had to make the assumption that $l$ is nonnegative.

To truncate the series, the idea is that if one constant term ${a}_{n}$ in the series goes to zero, the remaining successive ${a}_{n + 2}$ must also go to zero.

The recursion relation I obtained for this for the odd solution can be seen in the ${y}_{\text{odd}} \left(x\right)$ I gave:

a_(n+2) = (-1)^(n) ([(l - (2n-1))cdots(l-3)(l-1)][(l+2)(l+4)cdots(l+2n)])/((2n+1)!) a_n

Since we know that ${a}_{n + 2} = 0$ would truncate the series, we set ${a}_{n + 2} = 0$, having defined ${a}_{n}$ for the first allowed $n$ to be nonzero. Furthermore, we know that ${\left(- 1\right)}^{n} \ne 0$ and that no factorial is equal to zero. So:

${a}_{n + 2} = 0$

$\implies \left[\left(l - \left(2 n - 1\right)\right) \cdots \left(l - 3\right) \left(l - 1\right)\right] \left[\left(l + 2\right) \left(l + 4\right) \cdots \left(l + 2 n\right)\right] = 0$

We assume that $l \ge 0$, and we are left with:

$\left(l - 1\right) \left(l - 3\right) \cdots \left(l - 2 n + 1\right) = 0$

This then forces $l$ to be odd for the odd solution:

$l = 1 , 3 , 5 , . . . , 2 n + 1$

When one does this for the even solution, it would then turn out that $l = 0 , 2 , 4 , . . . , 2 n$ truncates the even series solution. The analogous recursion relation for the even solution was:

a_(n+2) = (-1)^(n) ([(l - (2n-2))cdots(l-2)l][(l+3)(l+5)cdots(l+2n-1)])/((2n)!) a_n

Together, they then form the solution set:

${y}_{l} \left(x\right) = \left\{\begin{matrix}{y}_{\text{odd" & n = "0 & 1 & 2 & . . . " & l = "1 & 3 & 5 & . . . " \\ y_"even" & n = "0 & 1 & 2 & . . . " & l = "0 & 2 & 4 & . . . }}\end{matrix}\right.$