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?
  • $\begingroup$ According to the documentation of Series (under "Scope"), it does generate a Laurent series about poles. $\endgroup$ – bbgodfrey Mar 11 '15 at 0:22
  • 1
    $\begingroup$ You could explicitly define z[0]=0 and then take the Series. $\endgroup$ – 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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.