# How to obtain Laurent series coefficients of an (almost) arbitrary function?

I have observed that Series in Mathematica assumes that the given function is smooth in the point around which one wants to perform series expansion.

For instance:

Series[1/z[x]^2, {x, 0, 1}]


results in familiar Taylor expansion.

SeriesData[x, 0, {z[0]^(-2), (-2) z[0]^(-3) Derivative[1][z][0]}, 0, 2, 1]


whereas if we inform Mathematica about existence of pole of given function:

Assuming[{z[0] == 0}, Series[1/z[x]^2, {x, 0, 1}]]


Mathematica gives us something which looks like Laurent series:

SeriesData[x, 0, {Derivative[1][z][0]^(-2), -Derivative[1][z][0]^(-3)
Derivative[2][z][0], Rational[1, 12] Derivative[1][z][0]^(-4) (9 Derivative[2][z][0]^2 -
4 Derivative[1][z][0] Derivative[3][z][0]), Rational[1, 12]
Derivative[1][z][0]^(-5) ((-6) Derivative[2][z][0]^3 + 6 Derivative[1][z][0]
Derivative[2][z][0] Derivative[3][z][0] - Derivative[1][z][0]^2 Derivative[4][z][0])}, -2, 2, 1]


Now, I wanted to convince myself that this is indeed the proper Laurent series, so I tried explicit integration:

1/(2 π I) Integrate[(1/z[Exp[I α]]^2)*(α)^2, {α, 0, 2 π}]


taking the contour to be unit circle. Unfortunately, Mathematica is not able to reproduce the coefficient form mentioned series. So my questions are:

1. How to compute such complex integral?
2. How does Series deal with such a problem?
• According to the documentation of Series (under "Scope"), it does generate a Laurent series about poles. – bbgodfrey Mar 11 '15 at 0:22
• You could explicitly define z[0]=0 and then take the Series. – Daniel Lichtblau Mar 11 '15 at 1:43

I don't know any way to tell Integrate that 1/z[x]^2 has a pole at zero. However, we can integrate your series expression:

series = Assuming[{z[0] == 0}, Series[1/z[x]^2, {x, 0, 1}]];
1/(2 Pi I) Integrate[Normal[series] * I x /. {x -> Exp[I t]}, {t, 0, 2 Pi}]


This gives us

-z''[0]/z'[0]^3


This makes sense, since Cauchy's integral formula is:

$$\oint_\gamma \frac{f(z)}{(z-a)^n} dz = \frac{2\pi i}{(n-1)!}f^{(n-1)}(a)$$

For the pole of order 1, $f(z) = -\frac{z''(0)}{z'(0)^3}$ is a constant, so the residue is equal to that expression (times $2\pi i$).

For the pole of order 2, $f(z) = \frac{1}{z'(0)^2}$, again a constant. However, since the value of this residue depends on $f^{(1)}(z)=0$, the residue is simply zero, and only the first residue contributes to the result.