- #1
zetafunction
- 391
- 0
the idea is let us suppose i must solve
[tex] f(x)= 0 [/tex] (1)
let us suppose that f(x) have SEVERAL (perhaps infinite ) inverses, that is there is a finite or infinite solutions to the equation
[tex] f(x)= y [/tex] by [tex] g(y)= x [/tex] with [tex] f^{-1}(x)=g(x) [/tex]
then solution to equation (1) would be [tex] g(0)=x [/tex]
my problem is what would happen for multi-valued functions (example [tex] x^{2} [/tex] having several 'branches' (is this the correct word ?? )
Using Lagrange inversion theorem [tex] 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).
[/tex]
then simply set x=0 but this would only give an UNIQUE solution to (1)
[tex] f(x)= 0 [/tex] (1)
let us suppose that f(x) have SEVERAL (perhaps infinite ) inverses, that is there is a finite or infinite solutions to the equation
[tex] f(x)= y [/tex] by [tex] g(y)= x [/tex] with [tex] f^{-1}(x)=g(x) [/tex]
then solution to equation (1) would be [tex] g(0)=x [/tex]
my problem is what would happen for multi-valued functions (example [tex] x^{2} [/tex] having several 'branches' (is this the correct word ?? )
Using Lagrange inversion theorem [tex] 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).
[/tex]
then simply set x=0 but this would only give an UNIQUE solution to (1)