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residues and the stability of fixed points

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

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

is a quotient of two polynomials. If Q has lesser degree than P, division may be used to write
r(z) = $\displaystyle g(z) + \frac{p(z)}{q(z)}$ (174)

wherein p and q have the desired property.

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

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

which will have poles wherever r(z) has fixed points. As previously discussed, r(z)-z can be factored, allowing the writing of
R(z) = $\displaystyle \frac{1}{g(z)+\frac{p(z)}{q(z)}-z}$ (176)
  = $\displaystyle \frac{q(z)}{g(z)q(z)+p(z)-zq(z)}.$ (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)

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

for fixed points of r, namely zj. The aj's are residues of R, but we are interested in the stability of r(z).
r(z) = $\displaystyle \frac{1}{R(z)}+z$ (179)
r'(zf) = $\displaystyle \lim_{z\rightarrow z_f}
\frac{1}{z-z_f}\left(\frac{1}{R(z)}+z\right)$ (180)
  = $\displaystyle 1 + \frac{1}{a_j}.$ (181)

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


next up previous contents
Next: representation of a function Up: Contour Integrals Previous: Schwartz's lemma
Microcomputadoras
2001-04-05