Question

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(x) = 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.

Answer 1

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_"odd"(x)` 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 `(-1)^n ne 0` and that no factorial is equal to zero. So:

`a_(n+2) = 0`

`=> [(l - (2n-1))cdots(l-3)(l-1)][(l+2)(l+4)cdots(l+2n)] = 0`

We assume that `l >= 0`, and we are left with:

`(l-1)(l-3)cdots(l-2n+1) = 0`

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

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

When one does this for the even solution, it would then turn out that `l = 0, 2, 4, . . . , 2n` 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(x) = {(y_"odd", n = "0, 1, 2, . . . ", l = "1, 3, 5, . . . "), (y_"even", n = "0, 1, 2, . . . ", l = "0, 2, 4, . . . ") :}`