# Proving a property of Legendre Polynomials

I've poked around Mathematics Stack Exchange for a while, and while I'm sure this is an elementary problem to you guys, I cannot figure this out. I have found similar solved problem prompts on here, but the issue is that I cannot use Rodrigues' Formula or the generating function, I have to use the definition of a Legendre polynomial itself. What I'm trying to do is show the following (here is the original prompt):

$\text{From the definition of a Legendre polynomial show that}$ $$P_{2n}(0)=(-1)^{n}\frac{(2n)!}{2^{2n}(n!)^2}$$

Here was my attempt so far:

$$\text{Definition:} \ \ \ P_n(x)=\frac{1}{2^n} \sum_{k=0}^{M}\frac{(-1)^k(2n-2k)!}{k!(n-k)!(n-2k)!}x^{n-2k}$$ $\\$ $$P_{2n}(x)=\frac{1}{2^{2n}}\sum_{k=0}^{M}\frac{(-1)^k(4n-2k)!}{k!(2n-k)!(2n-2k)!}x^{2n-2k}$$ $$P_{2n}(0)=\frac{1}{2^{2n}}\sum_{k=0}^{M}\frac{(-1)^k(4n-2k)!}{k!(2n-k)!(2n-2k)!}(0)^{2n-2k}$$

Here my confusion is that it looks like evaluating x at zero would make every single term disappear ($x^{2n-2k}=0$). Have I made a mistake, or how do I deal with this?

Remember the definition of $0^0$ (in series): $0^0 = 1$. Thus your $k=n$ term doesn't vanish. (I'm presuming you meant to have an $n$ in your upper limit, not $M$.) Making the appropriate substitution of $k=n$ in the Legendre polynomial coefficient yields the desired result.

• Oh, no, sorry, M was the correct upper limit, but in the context of M being the upper limit, M=(n)/2 or M=(n-1)/2, whichever gives it an integer value. This is going to sound dumb, but how do we have 0^0? I see 0^{2n-2k}, but I do not understand how 2n-2k=0. Edit: Just saw your remark on letting k=n, that makes sense now. Thank you. – DUTCHBAT III Apr 3 '15 at 1:41
• @DUTCHBATIII Oh. Well in that case, it follows in much the same way since in this case $M = n$. – Cameron Williams Apr 3 '15 at 1:43
• (Sorry, first time user on the site, was having huge trouble just properly formatting and submitting my comment for a second there.) – DUTCHBAT III Apr 3 '15 at 1:44
• No worries! You're doing better than a lot of first time users. You asked a good question and even gave your own input (and showed your own work). – Cameron Williams Apr 3 '15 at 1:45
• This is another thing that I've kind of hugely struggled with in math and have somehow made it this far without sufficiently answering: What exactly are the rules of substitution in stuff like this? I constantly shy away from trying substitution on my own because I feel like additional constraints that weren't specified in the prompt are being written in, and as a result, dealing with dummy variables/index shifts and stuff like this, I really suffer. Why am I "allowed" to make the substitution k=n? – DUTCHBAT III Apr 3 '15 at 1:54

The best way to remember an expression for the Legendre polynomials is Rodrigues' form:

$$P_n(x) = \frac{1}{2^nn!}[(x^2-1)^n]^{(n)}$$

To see why this equals the form you give, see the derivation below.

$$\begin{split} \frac{1}{2^nn!}[(x^2-1)^n]^{(n)} &= \frac{1}{2^nn!}[\sum_{k=0}^{n}\binom{n}{k}(x^2)^{n-k} \cdot (-1)^k]^{(n)} \\ &= \frac{1}{2^nn!}\sum_{k=0}^{n}(-1)^k \frac{n!}{k!(n-k)!}[x^{2n-2k}]^{(n)} \\ &= \frac{1}{2^n}\sum_{k=0}^{n}(-1)^k \frac{1}{k!(n-k)!}\cdot (2n-2k)(2n-2k-1)...(n-2k+1)x^{n-2k} \\ &= \frac{1}{2^n}\sum_{k=0}^{n}(-1)^k \frac{1}{k!(n-k)!}\cdot \frac{(2n-2k)!}{(n-2k)!}x^{n-2k} \\ &= \frac{1}{2^n}\sum_{k=0}^{n}\frac{(-1)^k(2n-2k)!}{k!(n-k)!(n-2k)!}x^{n-2k} \end{split}$$

Using Rodrigues' from, we can further derive that

$$\begin{split} P_n(x) &= \frac{1}{2^nn!}[(x+1)^n\cdot(x-1)^n]^{(n)} \\ &= \frac{1}{2^nn!}\sum_{k=0}^{n}\binom{n}{k}[(x+1)^n]^{(k)} \cdot [(x-1)^n]^{(n-k)} \\ &= \frac{1}{2^nn!}\sum_{k=0}^{n}\binom{n}{k}\frac{n!}{(n-k)!}(x+1)^{n-k} \cdot \frac{n!}{k!}(x-1)^k \\ &= \frac{1}{2^n}\sum_{k=0}^{n}\binom{n}{k}^2(x+1)^{n-k}(x-1)^k \end{split}$$

so $$\begin{split} P_{2n}(0) &=\frac{1}{2^{2n}}\sum_{k=0}^{2n}\binom{2n}{k}^2(-1)^k = \frac{1}{2^{2n}}\binom{2n}{n}(-1)^n = \frac{(-1)^n(2n)!}{2^{2n}n!^2} \end{split}$$

For deriving the value of $\sum_{k=0}^m\binom{m}{k}^2(-1)^k$, see below:

From the fact that $(1-x)^m(1+x)^m=(1-x^2)^m$, equating the coefficients for $x^m$, we see that $$$$\sum_{j=0}^{m}\binom{m}{j}^2(-1)^j=\sum_{j=0}^{m}\binom{m}{j}(-1)^j\binom{m}{m-j}\cdot (+1)^j= \begin{cases}0 &\text{if m is odd}\\ \binom{m}{m/2}(-1)^{m/2} & \text{if m is even} \end{cases}$$$$

• The thing was, I wasn't allowed to use Rodrigues formula as I said above, but thank you nonetheless very much. – DUTCHBAT III Apr 13 '15 at 11:43