What happens to the inverse function at infinity?

[zetafunction]

let be a function y=f(x) with poles f(a_{i} ) = \infty for some real 'a'

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

g(\infty)=a_{i} my question is that it seems that g(x) would have several asymptotes as x-->oo how it can be ??

 

[Dickfore]

The function

<br /> f(x) = \frac{A}{x - a} + \frac{B}{x - b}<br />

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

<br /> x = \frac{A}{y - a} + \frac{B}{y - b}<br />

<br /> x (y - a) (y - b) = A (y - b) + B (y -a)<br />

<br /> x y^{2} - (a + b) x y + a b x = (A + B) y - A b - B a<br />

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

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 x = \infty, you need to make the substitution x \rightarrow 1/x

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

Multiply out with x

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

and take x = 0, and you get:

<br /> y^{2} - (a + b) y + a b = 0 \Rightarrow (y - a)(y - b) = 0<br />

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