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Solving equations by inversion formulae

  1. Jan 31, 2010 #1
    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)
     
  2. jcsd
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