Can anybody tell me how I come can rearrange the following formulae into each other:
$$\sum_{k=0}^{\frac{l}{2}} (-1)^k {{2l-2k}\choose{l}} {l\choose{k}} x^{l-2k} = \sum_{k=0}^l {l\choose{k}}^2(x-1)^{l-k}(x+1)^k .$$
These are two different representations of the Legendre polynomials, but I can't figure out how to rearrange the left-hand side of the equation into the right-hand side.
Using Rodrigues' formula $$P_n(x) = \frac{1}{2^nn!}[(x^2-1)^n]^{(n)}$$ we derive \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} and \begin{equation} \begin{split} P_n(x) &= \frac{1}{2^nn!}[(x^2-1)^n]^{(n)} \\ &= \frac{1}{2^nn!}[\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}x^{2(n-k)}]^{(n)} \\ &= \frac{1}{2^nn!}\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}[x^{2(n-k)}]^{(n)} \\ &= \frac{1}{2^nn!}\sum_{k=0}^{\lfloor{n/2}\rfloor}(-1)^{k}\binom{n}{k}x^{n-2k} \prod_{l=1}^{n}(l+n-2k) \\ &= \frac{1}{2^nn!}\sum_{k=0}^{\lfloor{n/2}\rfloor}(-1)^{k}\binom{n}{k}x^{n-2k} \binom{2n-2k}{n-k}n! \\ &= \frac{1}{2^n}\sum_{k=0}^{\lfloor{n/2}\rfloor}(-1)^{k}\binom{2n-2k}{n-k}\binom{n}{k}x^{n-2k} \end{split} \end{equation}
