0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – DUTCHBAT III Apr 3 '15 at 1:41
  • $\begingroup$ @DUTCHBATIII Oh. Well in that case, it follows in much the same way since in this case $M = n$. $\endgroup$ – Cameron Williams Apr 3 '15 at 1:43
  • $\begingroup$ (Sorry, first time user on the site, was having huge trouble just properly formatting and submitting my comment for a second there.) $\endgroup$ – DUTCHBAT III Apr 3 '15 at 1:44
  • $\begingroup$ 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). $\endgroup$ – Cameron Williams Apr 3 '15 at 1:45
  • $\begingroup$ 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? $\endgroup$ – DUTCHBAT III Apr 3 '15 at 1:54
0
$\begingroup$

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{equation} \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} \end{equation}

Using Rodrigues' from, we can further derive that

\begin{equation} \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} \end{equation}

so \begin{equation} \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} \end{equation}

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 $$ \begin{equation} \sum_{j=0}^{m}\binom{m}{j}^2(-1)^j=\sum_{j=0}^{m}\binom{m}{j}(-1)^j\binom{m}{m-j}\cdot (+1)^j= \begin{cases}0 &\text{if m is odd}\\ \binom{m}{m/2}(-1)^{m/2} & \text{if m is even} \end{cases} \end{equation}$$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.