Hi everyone,
About a year ago, I reported a few problems with the "residue" function, in
particular where residue(expr, z, z0) returns an expression that depends on
z..
More details here:
http://www.math.utexas.edu/pipermail/maxima/2009/016877.html
At the time, I was using maxima 5.18.1. I have now updated to 5.21.1 and as
far as I can tell, the residue function still suffers from the same
problems.
But it seems there are now a few more issues:
In maxima 5.18.1 I used to observe the following behaviour (which is
correct):
(%i1) set_display('ascii) $
H : (a*q^2 + b*q) / (q^2 + 2*c*q + d^2) $
p : solve(denom(H) = 0, q) $
G : fullratsimp(a + 1/z * lsum(residue(H / (1 - q/z), q, qk), qk,
[rhs(p[1]), rhs(p[2]), 0]));
Is (d - c) (d + c) positive, negative, or zero?p;
2
a z + b z
(%o4) ---------------
2 2
z + 2 c z + d
In maxima 5.21.1 I now get:
set_display('ascii) $
H : (a*q^2 + b*q) / (q^2 + 2*c*q + d^2) $
p : solve(denom(H) = 0, q) $
G : fullratsimp(a + 1/z * lsum(residue(H / (1 - q/z), q, qk), qk,
[rhs(p[1]), rhs(p[2]), 0]));
(%o4) a
In fact the problem can be tracked down to here:
(%i5)residue(H / (1 - q/z), q, rhs(p[1]));
(%o5) 0
(%i6)residue(H / (1 - q/z), q, rhs(p[2]));
(%o6) 0
Therefore it looks like something has been broken between 5.18.1 and 5.21.1.
However, both versions also suffer from the further problem below:
(%i1)assume(T > 0) $
assume(RC > 0) $
declare(N, integer) $
assume(N > 0) $
assume(a >= 0 and a <= 1) $
assume(b >= a and b <= 1) $
m1 : 1 - a $
m2 : 1 - b $
exp(-(m1-m2)*T/RC) * residue(1/(q - q0) * q^(m2*N) * q/(q - 1), q, 1);
residue(exp(-(m1-m2)*T/RC) * 1/(q - q0) * q^(m2*N) * q/(q - 1), q, 1);
(a - b) T
---------
RC
%e
(%o9) - -----------
q0 - 1
(%o10) 0
Obviously both answers should be identical, since the factor
exp(-(m1-m2)*T/RC) does not depend on the variable q, therefore it should
have no influence, inside or outside of "residue". I have tried to
understand the cause of the problem, to no avail..
Beause of all these issues, I am a bit struggling to work with the "residue"
function.. Obviously residues can always be determined with "ad hoc"
solutions, on a case by case basis. But I'm using residues in a "black box"
where ad hoc solutions are just not an option, and all I can do is call the
generic "residue" function. Can anyone help ?
Many thanks,
David.