# [Maxima] Problems with residue

Leo Butler l.butler at ed.ac.uk
Thu Jul 1 08:44:57 CDT 2010

On Tue, 29 Jun 2010, David Chappaz wrote:

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< Hi everyone,
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< 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….
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< More details here:
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< http://www.math.utexas.edu/pipermail/maxima/2009/016877.html
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< 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.
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< But it seems there are now a few more issues:
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< In maxima 5.18.1 I used to observe the following behaviour (which is correct):
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< (%i1) set_display('ascii) \$
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< H : (a*q^2 + b*q) / (q^2 + 2*c*q + d^2) \$
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< p : solve(denom(H) = 0, q) \$
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< G : fullratsimp(a + 1/z * lsum(residue(H / (1 - q/z), q, qk), qk, [rhs(p[1]), rhs(p[2]), 0]));
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< Is  (d - c) (d + c)  positive, negative, or zero?p;
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<                                      2
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<                                   a z  + b z
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< (%o4)                           ---------------
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<                                  2            2
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<                                 z  + 2 c z + d
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< In maxima 5.21.1 I now get:
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< set_display('ascii) \$
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< H : (a*q^2 + b*q) / (q^2 + 2*c*q + d^2) \$
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< p : solve(denom(H) = 0, q) \$
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< G : fullratsimp(a + 1/z * lsum(residue(H / (1 - q/z), q, qk), qk, [rhs(p[1]), rhs(p[2]), 0]));
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< (%o4)                                  a
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< In fact the problem can be tracked down to here:
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< (%i5)residue(H / (1 - q/z), q, rhs(p[1]));
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< (%o5)                                  0
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< (%i6)residue(H / (1 - q/z), q, rhs(p[2]));
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< (%o6)                                  0
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< Therefore it looks like something has been broken between 5.18.1 and 5.21.1.
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< However, both versions also suffer from the further problem below:
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< (%i1)assume(T > 0) \$
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< assume(RC > 0) \$
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< declare(N, integer) \$
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< assume(N > 0) \$
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< assume(a >= 0 and a <= 1) \$
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< assume(b >= a and b <= 1) \$
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< m1 : 1 - a \$
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< m2 : 1 - b \$
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< exp(-(m1-m2)*T/RC) * residue(1/(q - q0) * q^(m2*N) * q/(q - 1), q, 1);
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< residue(exp(-(m1-m2)*T/RC) * 1/(q - q0) * q^(m2*N) * q/(q - 1), q, 1);
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<                                      (a - b) T
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<                                      ---------
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<                                         RC
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<                                    %e
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< (%o9)                            - -----------
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<                                      q0 - 1
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< (%o10)                                 0
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< 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….
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< 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 ?
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< Many thanks,
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< David.

David, since there hasn't been a response to your email, would
you please file a bug report at
http://sourceforge.net/tracker/?group_id=4933&atid=104933

Leo
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