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Inverse function and poles

  1. Jul 5, 2010 #1
    let be a function [tex] y=f(x) [/tex] with poles [tex] f(a_{i} ) = \infty [/tex] for some real 'a'

    my question is if we define the inverse function g(x) so [tex] g(f(x))=x [/tex] ,then is this true

    [tex] g(\infty)=a_{i} [/tex] my question is that it seems that g(x) would have several asymptotes as x-->oo how it can be ??
     
  2. jcsd
  3. Jul 5, 2010 #2
    The function

    [tex]
    f(x) = \frac{A}{x - a} + \frac{B}{x - b}
    [/tex]

    has two simple poles at [itex]x = a, b[/itex]. However, if you try to solve for its inverse, you get:

    [tex]
    x = \frac{A}{y - a} + \frac{B}{y - b}
    [/tex]

    [tex]
    x (y - a) (y - b) = A (y - b) + B (y -a)
    [/tex]

    [tex]
    x y^{2} - (a + b) x y + a b x = (A + B) y - A b - B a
    [/tex]

    [tex]
    x y^{2} - [(a + b) x + (A + B)] y + (a b x + A b + B a) = 0
    [/tex]

    This is a quadratic equation that has two roots in the set of complex numbers, corresponding to two branches of the inverse function. For the behavior at [itex]x = \infty[/itex], you need to make the substitution [itex] x \rightarrow 1/x[/itex]

    [tex]
    \frac{1}{x} y^{2} - [\frac{a + b}{x} + (A + B)] y + (\frac{a b}{x} + A b + B a) = 0
    [/tex]

    Multiply out with [itex]x[/itex]

    [tex]
    y^{2} - [(a + b) + (A + B) x] y + [a b + (A b + B a) x] = 0
    [/tex]

    and take [itex]x = 0[/itex], and you get:

    [tex]
    y^{2} - (a + b) y + a b = 0 \Rightarrow (y - a)(y - b) = 0
    [/tex]

    As you can see, there are two solutions even at infinity.
     
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