Possible bug in Solve function?
Bug introduced in 10.0 and persisting through 11.3 or later
In 11.3.0 for Microsoft Windows (64-bit) (March 7, 2018)
writing:
f[w_, x_, y_, z_] := w*x^2*y^3 - z*(w^2 + x^2 + y^2 - 1)
eqn = {D[f[w, x, y, z], w] == 0,
D[f[w, x, y, z], x] == 0,
D[f[w, x, y, z], y] == 0,
D[f[w, x, y, z], z] == 0};
sol = Solve[eqn];
Table[eqn /. sol[[n]], {n, Length[sol]}]
I get:
{{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True}}
from which there are four wrong solutions.
Am I wrong or is it a Solve
bug?
EDIT: through the email address support@wolfram.com I contacted Wolfram Technical Support who in less than three working days have confirmed that it is a bug and have already proceeded to report to their developers.
equation-solving bugs
add a comment |
Bug introduced in 10.0 and persisting through 11.3 or later
In 11.3.0 for Microsoft Windows (64-bit) (March 7, 2018)
writing:
f[w_, x_, y_, z_] := w*x^2*y^3 - z*(w^2 + x^2 + y^2 - 1)
eqn = {D[f[w, x, y, z], w] == 0,
D[f[w, x, y, z], x] == 0,
D[f[w, x, y, z], y] == 0,
D[f[w, x, y, z], z] == 0};
sol = Solve[eqn];
Table[eqn /. sol[[n]], {n, Length[sol]}]
I get:
{{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True}}
from which there are four wrong solutions.
Am I wrong or is it a Solve
bug?
EDIT: through the email address support@wolfram.com I contacted Wolfram Technical Support who in less than three working days have confirmed that it is a bug and have already proceeded to report to their developers.
equation-solving bugs
1
You could useList@ToRules@Reduce[eqn, {x, y, z, w}]
to get all valid solutions. Filter for those that only have numeric values on the RHS of->
.
– Szabolcs
Jan 5 at 22:30
Select[sol, And @@ eqn /. # &]
– Bob Hanlon
Jan 5 at 23:53
5
Next time, please do not add the bugs tag yourself on a question. The tag is only supposed to be added after your observations have been confirmed by other users.
– J. M. is computer-less♦
Jan 6 at 2:45
add a comment |
Bug introduced in 10.0 and persisting through 11.3 or later
In 11.3.0 for Microsoft Windows (64-bit) (March 7, 2018)
writing:
f[w_, x_, y_, z_] := w*x^2*y^3 - z*(w^2 + x^2 + y^2 - 1)
eqn = {D[f[w, x, y, z], w] == 0,
D[f[w, x, y, z], x] == 0,
D[f[w, x, y, z], y] == 0,
D[f[w, x, y, z], z] == 0};
sol = Solve[eqn];
Table[eqn /. sol[[n]], {n, Length[sol]}]
I get:
{{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True}}
from which there are four wrong solutions.
Am I wrong or is it a Solve
bug?
EDIT: through the email address support@wolfram.com I contacted Wolfram Technical Support who in less than three working days have confirmed that it is a bug and have already proceeded to report to their developers.
equation-solving bugs
Bug introduced in 10.0 and persisting through 11.3 or later
In 11.3.0 for Microsoft Windows (64-bit) (March 7, 2018)
writing:
f[w_, x_, y_, z_] := w*x^2*y^3 - z*(w^2 + x^2 + y^2 - 1)
eqn = {D[f[w, x, y, z], w] == 0,
D[f[w, x, y, z], x] == 0,
D[f[w, x, y, z], y] == 0,
D[f[w, x, y, z], z] == 0};
sol = Solve[eqn];
Table[eqn /. sol[[n]], {n, Length[sol]}]
I get:
{{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True}}
from which there are four wrong solutions.
Am I wrong or is it a Solve
bug?
EDIT: through the email address support@wolfram.com I contacted Wolfram Technical Support who in less than three working days have confirmed that it is a bug and have already proceeded to report to their developers.
equation-solving bugs
equation-solving bugs
edited 15 hours ago
Szabolcs
158k13432928
158k13432928
asked Jan 5 at 21:50
TeMTeM
1,990621
1,990621
1
You could useList@ToRules@Reduce[eqn, {x, y, z, w}]
to get all valid solutions. Filter for those that only have numeric values on the RHS of->
.
– Szabolcs
Jan 5 at 22:30
Select[sol, And @@ eqn /. # &]
– Bob Hanlon
Jan 5 at 23:53
5
Next time, please do not add the bugs tag yourself on a question. The tag is only supposed to be added after your observations have been confirmed by other users.
– J. M. is computer-less♦
Jan 6 at 2:45
add a comment |
1
You could useList@ToRules@Reduce[eqn, {x, y, z, w}]
to get all valid solutions. Filter for those that only have numeric values on the RHS of->
.
– Szabolcs
Jan 5 at 22:30
Select[sol, And @@ eqn /. # &]
– Bob Hanlon
Jan 5 at 23:53
5
Next time, please do not add the bugs tag yourself on a question. The tag is only supposed to be added after your observations have been confirmed by other users.
– J. M. is computer-less♦
Jan 6 at 2:45
1
1
You could use
List@ToRules@Reduce[eqn, {x, y, z, w}]
to get all valid solutions. Filter for those that only have numeric values on the RHS of ->
.– Szabolcs
Jan 5 at 22:30
You could use
List@ToRules@Reduce[eqn, {x, y, z, w}]
to get all valid solutions. Filter for those that only have numeric values on the RHS of ->
.– Szabolcs
Jan 5 at 22:30
Select[sol, And @@ eqn /. # &]
– Bob Hanlon
Jan 5 at 23:53
Select[sol, And @@ eqn /. # &]
– Bob Hanlon
Jan 5 at 23:53
5
5
Next time, please do not add the bugs tag yourself on a question. The tag is only supposed to be added after your observations have been confirmed by other users.
– J. M. is computer-less♦
Jan 6 at 2:45
Next time, please do not add the bugs tag yourself on a question. The tag is only supposed to be added after your observations have been confirmed by other users.
– J. M. is computer-less♦
Jan 6 at 2:45
add a comment |
2 Answers
2
active
oldest
votes
Thanks for asking! In version 9.0 only 16 solutions are returned and they are all valid. In version 10.2 there are 20 solutions, with the extra 4 all being invalid. Contragulations! I think you found a bug. You may want to click "Help", then "Give Feedback...", and then fill out the form in your browser to report.
As another answer notes, you can always try Reduce
instead which may give better results in some cases, but Solve
is usually what you want.
Well, it will mean that the next version will be the first calculation I will perform. Thank you! ^_^
– TeM
Jan 5 at 22:11
2
@TeM You should report it to Wolfram first, otherwise there's no chance for it to get fixed.
– Szabolcs
Jan 5 at 22:31
@Szabolcs: Could you direct me where I can do it correctly?
– TeM
Jan 5 at 22:44
4
@TeM wolfram.com/support/contact
– Szabolcs
Jan 5 at 23:49
add a comment |
You can use Reduce
f[w_, x_, y_, z_] := w*x^2*y^3 - z*(w^2 + x^2 + y^2 - 1)
eqn = {D[f[w, x, y, z], w] == 0, D[f[w, x, y, z], x] == 0,
D[f[w, x, y, z], y] == 0, D[f[w, x, y, z], z] == 0};
red = Reduce[eqn, Backsubstitution -> True]
$left(z=0land x=0land w=-sqrt{1-y^2}right)lor left(z=0land x=0land w=sqrt{1-y^2}right)lor left(z=0land y=0land
w=-sqrt{1-x^2}right)lor left(z=0land y=0land w=sqrt{1-x^2}right)lor (z=0land y=-1land x=0land w=0)lor (z=0land y=0land x=0land
w=-1)lor (z=0land y=0land x=0land w=1)lor (z=0land y=1land x=0land w=0)lor left(z=-frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land
x=-frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)\
lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)$
First@eqn //. {ToRules[red]}
{True, True, True, True, True, True, True, True, True, True, True,
True, True, True, True, True}
Yes, of course, I had already tried. I was almost certain it was a bug fromSolve
, so I pointed out. Thank you very much anyway, always very kind!
– TeM
Jan 6 at 1:14
4
I believeSolve
use the functionReduce
under the hood. when you removeBacksubstitution -> True
, you'll find implicit solution, somehowSolve
messes up somewhere and it is definitely a bug..
– Okkes Dulgerci
Jan 6 at 1:18
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for asking! In version 9.0 only 16 solutions are returned and they are all valid. In version 10.2 there are 20 solutions, with the extra 4 all being invalid. Contragulations! I think you found a bug. You may want to click "Help", then "Give Feedback...", and then fill out the form in your browser to report.
As another answer notes, you can always try Reduce
instead which may give better results in some cases, but Solve
is usually what you want.
Well, it will mean that the next version will be the first calculation I will perform. Thank you! ^_^
– TeM
Jan 5 at 22:11
2
@TeM You should report it to Wolfram first, otherwise there's no chance for it to get fixed.
– Szabolcs
Jan 5 at 22:31
@Szabolcs: Could you direct me where I can do it correctly?
– TeM
Jan 5 at 22:44
4
@TeM wolfram.com/support/contact
– Szabolcs
Jan 5 at 23:49
add a comment |
Thanks for asking! In version 9.0 only 16 solutions are returned and they are all valid. In version 10.2 there are 20 solutions, with the extra 4 all being invalid. Contragulations! I think you found a bug. You may want to click "Help", then "Give Feedback...", and then fill out the form in your browser to report.
As another answer notes, you can always try Reduce
instead which may give better results in some cases, but Solve
is usually what you want.
Well, it will mean that the next version will be the first calculation I will perform. Thank you! ^_^
– TeM
Jan 5 at 22:11
2
@TeM You should report it to Wolfram first, otherwise there's no chance for it to get fixed.
– Szabolcs
Jan 5 at 22:31
@Szabolcs: Could you direct me where I can do it correctly?
– TeM
Jan 5 at 22:44
4
@TeM wolfram.com/support/contact
– Szabolcs
Jan 5 at 23:49
add a comment |
Thanks for asking! In version 9.0 only 16 solutions are returned and they are all valid. In version 10.2 there are 20 solutions, with the extra 4 all being invalid. Contragulations! I think you found a bug. You may want to click "Help", then "Give Feedback...", and then fill out the form in your browser to report.
As another answer notes, you can always try Reduce
instead which may give better results in some cases, but Solve
is usually what you want.
Thanks for asking! In version 9.0 only 16 solutions are returned and they are all valid. In version 10.2 there are 20 solutions, with the extra 4 all being invalid. Contragulations! I think you found a bug. You may want to click "Help", then "Give Feedback...", and then fill out the form in your browser to report.
As another answer notes, you can always try Reduce
instead which may give better results in some cases, but Solve
is usually what you want.
edited Jan 6 at 1:16
answered Jan 5 at 22:05
SomosSomos
3628
3628
Well, it will mean that the next version will be the first calculation I will perform. Thank you! ^_^
– TeM
Jan 5 at 22:11
2
@TeM You should report it to Wolfram first, otherwise there's no chance for it to get fixed.
– Szabolcs
Jan 5 at 22:31
@Szabolcs: Could you direct me where I can do it correctly?
– TeM
Jan 5 at 22:44
4
@TeM wolfram.com/support/contact
– Szabolcs
Jan 5 at 23:49
add a comment |
Well, it will mean that the next version will be the first calculation I will perform. Thank you! ^_^
– TeM
Jan 5 at 22:11
2
@TeM You should report it to Wolfram first, otherwise there's no chance for it to get fixed.
– Szabolcs
Jan 5 at 22:31
@Szabolcs: Could you direct me where I can do it correctly?
– TeM
Jan 5 at 22:44
4
@TeM wolfram.com/support/contact
– Szabolcs
Jan 5 at 23:49
Well, it will mean that the next version will be the first calculation I will perform. Thank you! ^_^
– TeM
Jan 5 at 22:11
Well, it will mean that the next version will be the first calculation I will perform. Thank you! ^_^
– TeM
Jan 5 at 22:11
2
2
@TeM You should report it to Wolfram first, otherwise there's no chance for it to get fixed.
– Szabolcs
Jan 5 at 22:31
@TeM You should report it to Wolfram first, otherwise there's no chance for it to get fixed.
– Szabolcs
Jan 5 at 22:31
@Szabolcs: Could you direct me where I can do it correctly?
– TeM
Jan 5 at 22:44
@Szabolcs: Could you direct me where I can do it correctly?
– TeM
Jan 5 at 22:44
4
4
@TeM wolfram.com/support/contact
– Szabolcs
Jan 5 at 23:49
@TeM wolfram.com/support/contact
– Szabolcs
Jan 5 at 23:49
add a comment |
You can use Reduce
f[w_, x_, y_, z_] := w*x^2*y^3 - z*(w^2 + x^2 + y^2 - 1)
eqn = {D[f[w, x, y, z], w] == 0, D[f[w, x, y, z], x] == 0,
D[f[w, x, y, z], y] == 0, D[f[w, x, y, z], z] == 0};
red = Reduce[eqn, Backsubstitution -> True]
$left(z=0land x=0land w=-sqrt{1-y^2}right)lor left(z=0land x=0land w=sqrt{1-y^2}right)lor left(z=0land y=0land
w=-sqrt{1-x^2}right)lor left(z=0land y=0land w=sqrt{1-x^2}right)lor (z=0land y=-1land x=0land w=0)lor (z=0land y=0land x=0land
w=-1)lor (z=0land y=0land x=0land w=1)lor (z=0land y=1land x=0land w=0)lor left(z=-frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land
x=-frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)\
lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)$
First@eqn //. {ToRules[red]}
{True, True, True, True, True, True, True, True, True, True, True,
True, True, True, True, True}
Yes, of course, I had already tried. I was almost certain it was a bug fromSolve
, so I pointed out. Thank you very much anyway, always very kind!
– TeM
Jan 6 at 1:14
4
I believeSolve
use the functionReduce
under the hood. when you removeBacksubstitution -> True
, you'll find implicit solution, somehowSolve
messes up somewhere and it is definitely a bug..
– Okkes Dulgerci
Jan 6 at 1:18
add a comment |
You can use Reduce
f[w_, x_, y_, z_] := w*x^2*y^3 - z*(w^2 + x^2 + y^2 - 1)
eqn = {D[f[w, x, y, z], w] == 0, D[f[w, x, y, z], x] == 0,
D[f[w, x, y, z], y] == 0, D[f[w, x, y, z], z] == 0};
red = Reduce[eqn, Backsubstitution -> True]
$left(z=0land x=0land w=-sqrt{1-y^2}right)lor left(z=0land x=0land w=sqrt{1-y^2}right)lor left(z=0land y=0land
w=-sqrt{1-x^2}right)lor left(z=0land y=0land w=sqrt{1-x^2}right)lor (z=0land y=-1land x=0land w=0)lor (z=0land y=0land x=0land
w=-1)lor (z=0land y=0land x=0land w=1)lor (z=0land y=1land x=0land w=0)lor left(z=-frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land
x=-frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)\
lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)$
First@eqn //. {ToRules[red]}
{True, True, True, True, True, True, True, True, True, True, True,
True, True, True, True, True}
Yes, of course, I had already tried. I was almost certain it was a bug fromSolve
, so I pointed out. Thank you very much anyway, always very kind!
– TeM
Jan 6 at 1:14
4
I believeSolve
use the functionReduce
under the hood. when you removeBacksubstitution -> True
, you'll find implicit solution, somehowSolve
messes up somewhere and it is definitely a bug..
– Okkes Dulgerci
Jan 6 at 1:18
add a comment |
You can use Reduce
f[w_, x_, y_, z_] := w*x^2*y^3 - z*(w^2 + x^2 + y^2 - 1)
eqn = {D[f[w, x, y, z], w] == 0, D[f[w, x, y, z], x] == 0,
D[f[w, x, y, z], y] == 0, D[f[w, x, y, z], z] == 0};
red = Reduce[eqn, Backsubstitution -> True]
$left(z=0land x=0land w=-sqrt{1-y^2}right)lor left(z=0land x=0land w=sqrt{1-y^2}right)lor left(z=0land y=0land
w=-sqrt{1-x^2}right)lor left(z=0land y=0land w=sqrt{1-x^2}right)lor (z=0land y=-1land x=0land w=0)lor (z=0land y=0land x=0land
w=-1)lor (z=0land y=0land x=0land w=1)lor (z=0land y=1land x=0land w=0)lor left(z=-frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land
x=-frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)\
lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)$
First@eqn //. {ToRules[red]}
{True, True, True, True, True, True, True, True, True, True, True,
True, True, True, True, True}
You can use Reduce
f[w_, x_, y_, z_] := w*x^2*y^3 - z*(w^2 + x^2 + y^2 - 1)
eqn = {D[f[w, x, y, z], w] == 0, D[f[w, x, y, z], x] == 0,
D[f[w, x, y, z], y] == 0, D[f[w, x, y, z], z] == 0};
red = Reduce[eqn, Backsubstitution -> True]
$left(z=0land x=0land w=-sqrt{1-y^2}right)lor left(z=0land x=0land w=sqrt{1-y^2}right)lor left(z=0land y=0land
w=-sqrt{1-x^2}right)lor left(z=0land y=0land w=sqrt{1-x^2}right)lor (z=0land y=-1land x=0land w=0)lor (z=0land y=0land x=0land
w=-1)lor (z=0land y=0land x=0land w=1)lor (z=0land y=1land x=0land w=0)lor left(z=-frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land
x=-frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)\
lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)$
First@eqn //. {ToRules[red]}
{True, True, True, True, True, True, True, True, True, True, True,
True, True, True, True, True}
answered Jan 6 at 1:09
Okkes DulgerciOkkes Dulgerci
4,1851816
4,1851816
Yes, of course, I had already tried. I was almost certain it was a bug fromSolve
, so I pointed out. Thank you very much anyway, always very kind!
– TeM
Jan 6 at 1:14
4
I believeSolve
use the functionReduce
under the hood. when you removeBacksubstitution -> True
, you'll find implicit solution, somehowSolve
messes up somewhere and it is definitely a bug..
– Okkes Dulgerci
Jan 6 at 1:18
add a comment |
Yes, of course, I had already tried. I was almost certain it was a bug fromSolve
, so I pointed out. Thank you very much anyway, always very kind!
– TeM
Jan 6 at 1:14
4
I believeSolve
use the functionReduce
under the hood. when you removeBacksubstitution -> True
, you'll find implicit solution, somehowSolve
messes up somewhere and it is definitely a bug..
– Okkes Dulgerci
Jan 6 at 1:18
Yes, of course, I had already tried. I was almost certain it was a bug from
Solve
, so I pointed out. Thank you very much anyway, always very kind!– TeM
Jan 6 at 1:14
Yes, of course, I had already tried. I was almost certain it was a bug from
Solve
, so I pointed out. Thank you very much anyway, always very kind!– TeM
Jan 6 at 1:14
4
4
I believe
Solve
use the function Reduce
under the hood. when you remove Backsubstitution -> True
, you'll find implicit solution, somehow Solve
messes up somewhere and it is definitely a bug..– Okkes Dulgerci
Jan 6 at 1:18
I believe
Solve
use the function Reduce
under the hood. when you remove Backsubstitution -> True
, you'll find implicit solution, somehow Solve
messes up somewhere and it is definitely a bug..– Okkes Dulgerci
Jan 6 at 1:18
add a comment |
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1
You could use
List@ToRules@Reduce[eqn, {x, y, z, w}]
to get all valid solutions. Filter for those that only have numeric values on the RHS of->
.– Szabolcs
Jan 5 at 22:30
Select[sol, And @@ eqn /. # &]
– Bob Hanlon
Jan 5 at 23:53
5
Next time, please do not add the bugs tag yourself on a question. The tag is only supposed to be added after your observations have been confirmed by other users.
– J. M. is computer-less♦
Jan 6 at 2:45