# Solving equations by inversion formulae

1. Jan 31, 2010

### zetafunction

the idea is let us suppose i must solve

$$f(x)= 0$$ (1)

let us suppose that f(x) have SEVERAL (perhaps infinite ) inverses, that is there is a finite or infinite solutions to the equation

$$f(x)= y$$ by $$g(y)= x$$ with $$f^{-1}(x)=g(x)$$

then solution to equation (1) would be $$g(0)=x$$

my problem is what would happen for multi-valued functions (example $$x^{2}$$ having several 'branches' (is this the correct word ?? )

Using Lagrange inversion theorem $$g(x) = a + \sum_{n=1}^{\infty} \left( \lim_{w \to a}\left( \frac{\mathrm{d}^{\,n-1}}{\mathrm{d}w^{\,n-1}} \left( \frac{w-a}{f(w) - b} \right)^n\right) {\frac{(x - b)^n}{n!}} \right).$$

then simply set x=0 but this would only give an UNIQUE solution to (1)