Bugs item #738848, was opened at 20030516 12:15
Message generated for change (Comment added) made by rtoy
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Category: Share Libraries
Group: Fix for 5.9.0
>Status: Closed
>Resolution: Duplicate
Priority: 5
Submitted By: Nobody/Anonymous (nobody)
Assigned to: Nobody/Anonymous (nobody)
Summary: Poverseries error for 1/(1t)^2
Initial Comment:
It seems powerseries(1(ab*t)^n, t, 0) produces an
error for n=2, but is correct for n=1,3,4:
(C1) POWERSERIES(1/(ab*t)^2, t, 0);
INF
====
\ I1  I1  2 I1
(D1) > a b (I1 + 1) t
/
====
I1 = 0
[I believe that powers of a and b above have opposite
signs].

>Comment By: Raymond Toy (rtoy)
Date: 20030528 12:41
Message:
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Bug marked as duplicate. I'll apply your patch for bug 727542.

Comment By: Martin Rubey (kratt5)
Date: 20030519 04:32
Message:
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This is fixed with the fix for [727542] powerseries wrong/fix
Martin
(please, somebody merge it into cvs, it is IMPORTANT!)

Comment By: Nobody/Anonymous (nobody)
Date: 20030517 14:32
Message:
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(C1) verbose:true;
(D1) TRUE
(C2) POWERSERIES(1/(ab*t)^2, t, 0);
In the first simplification we have returned:
1

2 2 2
b t  2 a b t + a
trying to do a rational function expansion of
1

2 2 2
b t  2 a b t + a
Using a special rule for expressions of form
M  N
(A + C VAR )
Here we have
[N = 2, A =  a, C = b, M = 1]
INF
====
\ I1  I1  2 I1
(D2) > a b (I1 + 1) t
/
====
I1 = 0
(C3)

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