>What happens if you try the second example with a clean maxima?
>When you run both examples, would you trace solve also?
>Thanks,
>Dan
The second example doesn't change when run in a clean maxima. Do you have
a really clean maxima (i.e. without any patches etc.) handy? I think we
should check first if it is a problem with *my* installation...
GCL (GNU Common Lisp) Version(2.5.0) Sun Nov 17 15:58:09 CET 2002
Licensed under GNU Library General Public License
Contains Enhancements by W. Schelter
Maxima 5.9.0rc3 http://maxima.sourceforge.net
Distributed under the GNU Public License. See the file COPYING.
Dedicated to the memory of William Schelter.
This is a development version of Maxima. The function bug_report()
provides bug reporting information.
(C1) EquationP(e):=if part(e,0)="=" then true else false$
(C1)
(C2) load("/usr/labri/rubey/maxima/src/binary-gcl/comm.o")$
(C3) load("/usr/labri/rubey/maxima/src/binary-gcl/mutils.o")$
(C4) load("/usr/labri/rubey/maxima/src/binary-gcl/set.o")$
(C5) load("algebra/solver/misc")$
(C6) load("algebra/solver/solver")$
(C7) trace(solve);
(D7) [SOLVE]
(C8) solver([u-t*(u^(k+1)+1)=0,1-(k+1)*t*u^k=0],[u,t],[k]);
k
1 Enter SOLVE [(- k - 1) t u + 1, t]
1
1 Exit SOLVE [t = ----------]
k
(k + 1) u
******* obviously, this is wrong here ******** who set k=2 ??? This might
have to do with dynamic scope - in valuationsolver k is used, and if I
leave away the parameter k, i get the following message:
Warning: GET(OP(EXPR),'Valuation) was declared mode FIXNUM, has value:
FALSE
similar trouble if I use i instead of k:
(C7) solver([u-t*(u^(i+1)+1)=0,1-(i+1)*t*u^i=0],[u,t],[i]);
Is i an integer?
y;
1
(D7) [[u = 1, t = -]]
2
(C8)
***********************************************
3
2 u - 1
1 Enter SOLVE [--------, u]
2
3 u
SQRT(3) %I - 1 SQRT(3) %I + 1 1
1 Exit SOLVE [u = --------------, u = - --------------, u = ----]
1/3 1/3 1/3
2 2 2 2 2
4 k
---
3
SQRT(3) %I - 1 2
(D8) [[u = --------------, t = ---------------------------------------],
1/3 k k
2 2 (SQRT(3) %I - 1) k + (SQRT(3) %I - 1)
4 k
---
3
SQRT(3) %I + 1 2
[u = - --------------, t = -------------------------------------------],
1/3 k k
2 2 (- SQRT(3) %I - 1) k + (- SQRT(3) %I - 1)
k/3
1 2
[u = ----, t = -----]]
1/3 k + 1
2
(C9) solver([u-t*(u^(l+1)+1)=0,1-(l+1)*t*u^l=0],[u,t],[l]);
l
1 Enter SOLVE [(- l - 1) t u + 1, t]
1
1 Exit SOLVE [t = ----------]
l
(l + 1) u
l + 1
l u - 1
1 Enter SOLVE [------------, u]
l
(l + 1) u
Is l an integer?
y;
1
1 Exit SOLVE [u = ------]
1
-----
l + 1
l
l
-----
l + 1
1 l
(D9) [[u = ------, t = ------]]
1 l + 1
-----
l + 1
l
(C10)