# Lagrange inversion formula in 2 variables?

I have the following implicit equation for the function $G(x,y)$ in 2 variables:

$$G \;=\; (1 + x\,G^2)\,(1 + y\,G^2)$$

I want to use a form of a multivariable Lagrange inversion formula but I'm not sure that it's possible. Some ideas?

Thanks for help. Gianfranco

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What is the range of $x$ and $y$ that you are interested in: near zero, near $\infty$, or something else? –  user53153 Jan 9 '13 at 1:13
Hi Pavel,Thanks for your answer. In fact I'm looking for a 2 variable Taylor (near the origin) or Formal serie for G(x,y) . To give you an example, if you try to ask the same question for implicit equations like G = 1 + xG^2 (beside the fact that you can solve directly this degree 2 equation) you apply the 1-variable Lagrange inversion and you get the expected solution i.e. the one you'll get solving directly the quadratic equation. –  Gianfranco OLDANI Jan 9 '13 at 18:10

The coefficients of $G(x,y)=\sum_{i,j=0}^\infty c_{ij}x^i y^j$ are given by $$c_{ij}=\frac{1}{2i+2j+1}\binom{2i+2j+1}{i} \binom{2i+2j+1}{j}$$ Very Catalan-like, indeed. I did not try to write down a proof -- my computer is a much better combinatorialist than me.

Suspiciously similar expressions appear at the end of these slides.

The rest of the post refers to experiments that led to the above formula.

You can find any number of coefficients of the Taylor expansion at the origin just by solving a system for the coefficients. I used Maple to get coefficients up to 9th order; I'm sure it can do more. The system is actually trivial to solve if one does it in the right order, which means Maple could be replaced by a Python script or something.

n:=9:
c[0,0]:=1:
G:=sum(sum(c[i,j]*x^i*y^j,j=0..n-i),i=0..n):
H:=(1+x*G^2)*(1+y*G^2):
eqs:=[seq(c[0,j]=coeff(eval(H,x=0),y^j),j=1..n),seq(c[i,0]=coeff(eval(H,y=0),x^i),i=1..n),seq(seq(c[i,j]=coeff(coeff(H,x^i),y^j),j=1..n-i),i=1..n)]:
solve(eqs);


The output was

{c[0, 1] = 1, c[0, 2] = 2, c[0, 3] = 5, c[0, 4] = 14, c[0, 5] = 42, c[0, 6] = 132, c[0, 7] = 429, c[0, 8] = 1430, c[0, 9] = 4862, c[1, 0] = 1, c[1, 1] = 5, c[1, 2] = 21, c[1, 3] = 84, c[1, 4] = 330, c[1, 5] = 1287, c[1, 6] = 5005, c[1, 7] = 19448, c[1, 8] = 75582, c[2, 0] = 2, c[2, 1] = 21, c[2, 2] = 144, c[2, 3] = 825, c[2, 4] = 4290, c[2, 5] = 21021, c[2, 6] = 99008, c[2, 7] = 453492, c[3, 0] = 5, c[3, 1] = 84, c[3, 2] = 825, c[3, 3] = 6292, c[3, 4] = 41405, c[3, 5] = 247520, c[3, 6] = 1383732, c[4, 0] = 14, c[4, 1] = 330, c[4, 2] = 4290, c[4, 3] = 41405, c[4, 4] = 333200, c[4, 5] = 2372112, c[5, 0] = 42, c[5, 1] = 1287, c[5, 2] = 21021, c[5, 3] = 247520, c[5, 4] = 2372112, c[6, 0] = 132, c[6, 1] = 5005, c[6, 2] = 99008, c[6, 3] = 1383732, c[7, 0] = 429, c[7, 1] = 19448, c[7, 2] = 453492, c[8, 0] = 1430, c[8, 1] = 75582, c[9, 0] = 4862}

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These are the first few Catalan numbers, because by looking at c[0,k] you are in effect setting $x=0$ (so that $x^0=1$ and all higher powers of $x$ vanish). –  Andreas Blass Jan 9 '13 at 19:15
@Andreas That's interesting; is there a combinatorial interpretation of c[i,j] with both indices nonzero? –  user53153 Jan 9 '13 at 19:19
Yes Andreas, This can be derived from the original equation because setting x= 0 gives you: G = 1 + y*G^2 and this equation gives you for G the generating function of the simple Catalan numbers. These Catalan numbers can be interpreted as the number of one dimensional simple random walks on Z starting from 0 and returning to 0 without going under the horizontal "time" line. Now you are right Pavel what I'm looking is effectively a combinatorial interpretation for the C(i,j) . Some generalization of Catalan numbers? –  Gianfranco OLDANI Jan 9 '13 at 19:58