zetafunction
Jan31-10, 04:51 AM
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)
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)