Consider the infinite sum $\sum_{n=0}^\infty a_n(z-1)^n$ for the function $\frac{1}{z(z-1)}$ where |z-1| is between 0 and 1. How would one find the 'formula' for $a_n$? Is this similar to taylor series where the expansion should be produced and then the coefficients picked out?


A trick of algebraic manipulations and use of known expansion series: $$\frac{1}{z(z-1)} = \frac{1}{z-1} - \frac{1}{z} = \frac{1}{z-1} - \frac{1}{(z-1)+1} = \frac{1}{z-1} - \sum_{n=0}^{\infty} (-1)^n (z-1)^n $$ So explicitly, its Laurent series is $$ \frac{1}{z(z-1)} = \frac{1}{z-1} - 1 + (z-1) - (z-1)^2 + (z-1)^3 + ... $$

  • $\begingroup$ So the key is simply to see a relevant algebraic manipulation to apply in hopes of obtaining an expansion? Is your expansion above a Laurent series? It appears to be a taylor series. $\endgroup$ – user73041 Apr 28 '13 at 14:14
  • $\begingroup$ Note that $\frac{1}{z(z-1)}$ has a negative power term, $1/(z-1)$ that comes before the sum expression so it is a proper Laurent series. In any case you can always think of a Taylor series as a degenerate Laurent series with $0 = a_{-1} = a_{-2} = ...$. $\endgroup$ – LinAlgMan Apr 28 '13 at 14:39
  • $\begingroup$ Interesting, so Laurent series are really a generalization, not a separate object? $\endgroup$ – user73041 Apr 28 '13 at 14:44
  • $\begingroup$ Yes. Every Taylor series is a Laurent series. $\endgroup$ – LinAlgMan Apr 28 '13 at 15:33

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