A question on the Lagrange Inversion Formula I have to use the L.I.F. for 
\begin{align*}
s\left(x,y\right)=\frac{1}{2}\left(1-x-y-\sqrt{1-2x-2y-2xy+x^2+y^2}\right)
\end{align*}
to obtain that 
\begin{align*}
s\left(x,y\right) = \sum_{p,q\geq1}\frac{1}{p+q-1}\binom{p+q-1}{p}\binom{p+q-1}{q}x^py^q
\end{align*}
I've tried finding some resources on examples on how to do this but so far I am quite clueless. Any help or suggestions are appreciated.
 A: We seek to use Lagrange Inversion to show that
$$s(x,y) = \frac{1}{2}
\left(1-x-y-\sqrt{1-2x-2y-2xy+x^2+y^2}\right)$$
has the series expansion
$$\sum_{p,q\ge 1} \frac{1}{p+q-1}
{p+q-1\choose p} {p+q-1\choose q} x^p y^q.$$
On squaring we obtain
$$4 s(x,y)^2 = (1-x-y)^2 + 1-2x-2y-2xy+x^2+y^2
\\ - 2(1-x-y) (1-x-y-2s(x,y))
\\ = 2 (1-x-y)^2 - 4xy - 2(1-x-y) (1-x-y-2s(x,y))
\\ = - 4xy + 4 (1-x-y) s(x, y).$$
We finally get
$$s(x,y)^2 = -xy + (1-x-y) s(x,y).$$
This implies
$$x = \frac{s(x,y) (1 - y - s(x,y))}{y + s(x,y)}.$$
We get with $p\ge 1$
$$[x^p] s(x, y) = \frac{1}{p} [x^{p-1}] \frac{d}{dx} s(x, y)
= \frac{1}{p} \; \underset{x}{\mathrm{res}} \;
\frac{1}{x^p} \frac{d}{dx} s(x,y).$$
Now put $s(x,y) = u$ so that $\frac{d}{dx} s(x, y) \; dx = du$
$$\frac{1}{p} \; \underset{u}{\mathrm{res}} \;
\frac{(y+u)^p}{u^p (1-y-u)^p}
\\ = \frac{1}{p} \frac{1}{(1-y)^p} \; \underset{u}{\mathrm{res}} \;
\frac{(y+u)^p}{u^p (1-u/(1-y))^p}.$$
This is
$$\frac{1}{p} \frac{1}{(1-y)^p}
\sum_{r=0}^{p-1} {p\choose r} y^{p-r}
{2p-2-r\choose p-1} \frac{1}{(1-y)^{p-1-r}}.$$
Extracting the coefficient on $[y^q]$ where we see that $q\ge 1$:
$$\frac{1}{p}
\sum_{r=0}^{p-1} {p\choose r} [y^q] y^{p-r}
{2p-2-r\choose p-1} \frac{1}{(1-y)^{2p-1-r}}
\\ = \frac{1}{p}
\sum_{r=0}^{p-1} {p\choose r}
{2p-2-r\choose p-1} {q+p-2\choose 2p-2-r}.$$
Next observe that
$${2p-2-r\choose p-1} {q+p-2\choose 2p-2-r}
= \frac{(q+p-2)!}{(p-1)! \times (p-1-r)! \times (q+r-p)!}
\\ = {p+q-2\choose p-1} {q-1\choose p-1-r}.$$
We get for our sum
$$\frac{1}{p} {p+q-2\choose p-1}
\sum_{r=0}^{p-1} {p\choose r} {q-1\choose p-1-r}
\\ = \frac{1}{p} {p+q-2\choose p-1} {p+q-1\choose p-1}.$$
This was Vandermonde. Some binomial coefficient algebra now yields
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{p+q-1} {p+q-1\choose p} {p+q-1\choose q}}$$
as claimed. 
This  was from Egorychev's Combinatorial sums.
