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Using Legendre polynomial generating function \begin{equation} \sum_{n=0}^\infty P_n (x) t^n = \frac{1}{\sqrt{(1-2xt+t^2)}} \end{equation} Or $$ P_n(x)=\frac{1}{2^n n!} \frac{d^n}{dx^n} [(x^2-1)^n] $$

Show$$ P_{2n}(0)=\frac{(-1)^n (2n)!}{(4)^n (n!)^2} $$ And $$ P_{2n+1}(0)=0$$

I expressed $$(x^2 -1)^{2n}= \sum_{k=0}^{2n} {2n \choose k}x^{4n-2k}(-1)^k$$ And using second formula given, the only term that remains after differentiating 2n times and substituting x=0 is where $$4n-2k=2n$$ so $$2n=2k$$ k=n i.e $$(-1)^n \frac{(2n)!}{(n!)^2} $$ but multiplying this by$$ \frac{1}{2^{2n} (2n)!} $$ doesn't give desired solution. Where have I gone wrong?

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\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} and \begin{equation} \begin{split} P_{2n+1}(0) &=\frac{1}{2^{2n+1}}\sum_{k=0}^{2n+1}\binom{2n+1}{k}^2(-1)^k = \frac{1}{2^{2n}}\cdot 0 = 0 \end{split} \end{equation} See legendre polynomial problem (pls help!) for the derivation of the value of $\sum_{k=0}^m\binom{m}{k}^2(-1)^k$.

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This one can also be done using complex variables.

Starting from the generating function $$\sum_{n\ge 0} P_n(x) t^n = \frac{1}{\sqrt{1-2xt+t^2}}$$

Call $P_n(0) = Q_n$ so that $$\sum_{n\ge 0} Q_n t^n = \frac{1}{\sqrt{1+t^2}}.$$

Now using Lagrange Inversion we get $$Q_{2n} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{2n+1}} \frac{1}{\sqrt{1+z^2}} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{2n+2}} \frac{1}{\sqrt{1+z^2}} \; z \; dz.$$

Put $1+z^2 = w^2$ so that $z\; dz = w\; dw$ to obtain $$2\times \frac{1}{2\pi i} \int_{|w-1|=\epsilon} \frac{1}{(w^2-1)^{n+1}} \frac{1}{w} \; w\; dw.$$

The scalar two in front is because when $z$ makes one turn around the origin, $w$ makes two, but the contour only counts one turn. Continuing we have $$\frac{2}{2\pi i} \int_{|w-1|=\epsilon} \frac{1}{(w-1)^{n+1}} \frac{1}{(w+1)^{n+1}} \; dw \\ = \frac{2}{2\pi i} \int_{|w-1|=\epsilon} \frac{1}{(w-1)^{n+1}} \frac{1}{(2+w-1)^{n+1}} \; dw \\ = \frac{1}{2^{n}} \frac{1}{2\pi i} \int_{|w-1|=\epsilon} \frac{1}{(w-1)^{n+1}} \frac{1}{(1+(w-1)/2)^{n+1}} \; dw \\ = \frac{1}{2^{n}} \frac{1}{2\pi i} \int_{|w-1|=\epsilon} \frac{1}{(w-1)^{n+1}} \sum_{q\ge 0} (-1)^q {q+n\choose n} \frac{(w-1)^q}{2^q} \; dw .$$ Extracting coefficients this gives $$\frac{1}{2^{n}} (-1)^n {2n\choose n} \frac{1}{2^n}.$$ This is $$\frac{(-1)^n}{4^n} \frac{(2n)!}{(n!)^2}$$ as claimed.

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