residues and the stability of fixed points

Residues can show up in defining fixed points for rational functions. Suppose that the function r,

$$\frac{P(z)}{Q(z)},$$ r(z) = (173)

is a quotient of two polynomials. If Q has lesser degree than P, division may be used to write

$$g(z) + \frac{p(z)}{q(z)}$$ r(z) = (174)

wherein p and q have the desired property.

Consider a ``resolvent'' R(z) defined by

$$\frac{1}{r(z)-z}$$ R(z) = (175)

which will have poles wherever r(z) has fixed points. As previously discussed, r(z)-z can be factored, allowing the writing of

$$\frac{1}{g(z)+\frac{p(z)}{q(z)}-z}$$ R(z) = (176)

$$\frac{q(z)}{g(z)q(z)+p(z)-zq(z)}.$$ = (177)

This denominator has greater degree than the numerator, so it can be factored and R can be written as a sum of partial fractions (due allowance should be made for repeated factors)

$$\sum \frac{a_j}{z - z_j}$$ R(z) = (178)

for fixed points of r, namely z_j. The a_j's are residues of R, but we are interested in the stability of r(z).

$$\frac{1}{R(z)}+z$$ r(z) = (179)

$$\lim_{z\rightarrow z_f} \frac{1}{z-z_f}\left(\frac{1}{R(z)}+z\right)$$ r'(z_f) = (180)

$$1 + \frac{1}{a_j}.$$ = (181)

So stability depends on negative (but not too negative) a_j's.