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:
- How to compute such complex integral?
- How does
Series
deal with such a problem?
Series
(under "Scope"), it does generate a Laurent series about poles. $\endgroup$ – bbgodfrey Mar 11 '15 at 0:22z[0]=0
and then take theSeries
. $\endgroup$ – Daniel Lichtblau Mar 11 '15 at 1:43