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

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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.

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

The indexing for

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

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

To truncate the series, the idea is that if one constant term

The **recursion relation** I obtained for this for the odd solution can be seen in the

#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#

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

We assume that

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

This then forces

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

When one does this for the even solution, it would then turn out that

#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, . . . ") :}#