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

Question

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.

Charles Anctil · Accepted Answer

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, . . . ") :}#

Emilia Aguilar · Answer

undefined To truncate an infinite series by choosing an odd $ l $, you can use the following steps:

1. Start with the infinite series you want to truncate, usually given in terms of $ l $:
$$ \sum_{l=0}^{\infty} a_l $$

2. Decide on an odd number for $ l $, let's say $ l = 2n + 1 $, where $ n $ is a non-negative integer.

3. Substitute $ l = 2n + 1 $ into the series:
$$ \sum_{n=0}^{\infty} a_{2n+1} $$

4. This series will include terms for $ a_1, a_3, a_5, a_7, \ldots $ which are the odd-indexed terms of the original series.

5. If you want to truncate this series at a certain point, say $ N $, then you will sum up to the $ N $th odd term:
$$ \sum_{n=0}^{N} a_{2n+1} $$

6. This truncated series will consist of the terms $ a_1, a_3, a_5, \ldots, a_{2N+1} $.

So, to truncate the infinite series by choosing an odd $ l $, you simply replace $ l $ with $ 2n+1 $ and sum up to the desired term $ a_{2N+1} $.