Residue in complex analysis

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Partial answer : your second question is not legible, and the third doesn't make sanse without the second.

Maxima has a residue function :

(%i2) ? residue


 -- Function: residue (<expr>, <z>, <z_0>)

     Computes the residue in the complex plane of the expression <expr>
     when the variable <z> assumes the value <z_0>.  The residue is the
     coefficient of '(<z> - <z_0>)^(-1)' in the Laurent series for
     <expr>.

          (%i1) residue (s/(s**2+a**2), s, a*%i);
                                          1
          (%o1)                           -
                                          2
          (%i2) residue (sin(a*x)/x**4, x, 0);
                                           3
                                          a
          (%o2)                         - --
                                          6

(\%o2) \mathbf{true}

which can be called via the maxima_methods() interface:

sage: (1/x).maxima_methods().residue(x,0)
1
sage: (1/x).maxima_methods().residue(x,1)
0
sage: tan(x).maxima_methods().residue(x,pi/2)
-1
sage: tan(x).maxima_methods().residue(x,0)
0

HTH,

There's a residue method on symbolic expressions which is documented to do the thing you want, but of course you should check that it does indeed do what you want.

I think you can use:

k=3
br(x)=sqrt(x)-sqrt(1/x)
q1,q2=var('q1,q2')
X=var('x_0,x_1,x_2')
rho=[1,q1,q2]
chi=1/prod([ br(X[j])*br(q1/X[j])*X[j]*br(q2/X[j]) for j in range(k)])
[chi.residue(xi==rhoi) for xi,rhoi in zip(X,rho)]

or, if as in the example that you posted as an answer to the question here, you really want to take the residues iteratively,

for xi,rhoi in zip(X,rho):
    chi = chi.residue(xi,rhoi)
print chi

it returns almost instantaneously for me. You may want to simplify the answer a bit, with something like chi.factor() or so.

I write here the function I had in mind, using the solution I found:

 def br(x):
 return sqrt(x)-sqrt(1/x)

 import sympy as sp
 k=3
 q1,q2 = var('q1,q2')
 X = sp.symbols('x_0:{}'.format(k))
 rho = [1,q1,q2]
 chi = 1/prod([ br(X[j])*br(q1/X[j])*X[j]*br(q2/X[j]) for j in range(k)])
 for i in range(k):
     chi = sp.residue(chi,X[i],rho[i])
 chi

The problem is that even for a stupid example such as the one above, it takes forever to compute it.

Here's the same (with the actual function I need), using the above suggested answer:

k=3
br(x)=sqrt(x)-sqrt(1/x)
q1,q2,q3,m=var('q1,q2,q3,m')
# X=var('x_0,x_1,x_2')
# X=var('x_0:{}'.format(k))
X = [var("x%d" % i) for i in range(k)]
rho=[1,q1,q2]
# rho=[1,q1]
chi1 = prod([ br(X[j]/m)/(br(X[j])*X[j]) for j in range(k)])
chi2 = prod([prod([br(q1*q2*X[i]/X[j])*br(q1*q3*X[i]/X[j])*br(q2*q3*X[i]/X[j]) for i in range(k) if i != j]) for j in range(k)])
chi3 = prod([prod([br(q1*X[i]/X[j])*br(q2*X[i]/X[j])*br(q3*X[i]/X[j])*br(q1*q2*q3*X[i]/X[j]) for i in range(k) if i != j]) for j in range (k)])
chi = chi1*chi2/chi3
for xi,rhoi in zip(X,rho):
    chi = chi.residue(xi==rhoi)
print chi.factor()

This gives zero (wrong) for k=3 (unless I made a stupid mistake) and it is still slow, but faster than sympy for k=2.