Solve gives an incorrect answer

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Cond11[B_, A_, En_] := 4 + 2 Sqrt[1 + 4 B] - A^2/(2 (-En)^(3/2))

Cond22[B_, A_, En_] := 2 - (A^2 (3 + 2 Sqrt[1 + 4 B]))/(2 (-En)^(3/2))

Solve[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0, B, En]

Putting the result back into Cond11 or Cond22 does not yield zero.

What is the problem?

edited Aug 16 at 17:11

Carl Woll

55.6k271144

asked Aug 16 at 16:49

Nasser Saad

211

add a commentÂ |Â

up vote
4
down vote

favorite

Cond11[B_, A_, En_] := 4 + 2 Sqrt[1 + 4 B] - A^2/(2 (-En)^(3/2))

Cond22[B_, A_, En_] := 2 - (A^2 (3 + 2 Sqrt[1 + 4 B]))/(2 (-En)^(3/2))

Solve[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0, B, En]

Putting the result back into Cond11 or Cond22 does not yield zero.

What is the problem?

edited Aug 16 at 17:11

Carl Woll

55.6k271144

asked Aug 16 at 16:49

Nasser Saad

211

add a commentÂ |Â

up vote
4
down vote

favorite

Cond11[B_, A_, En_] := 4 + 2 Sqrt[1 + 4 B] - A^2/(2 (-En)^(3/2))

Cond22[B_, A_, En_] := 2 - (A^2 (3 + 2 Sqrt[1 + 4 B]))/(2 (-En)^(3/2))

Solve[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0, B, En]

Putting the result back into Cond11 or Cond22 does not yield zero.

What is the problem?

edited Aug 16 at 17:11

Carl Woll

55.6k271144

asked Aug 16 at 16:49

Nasser Saad

211

Cond11[B_, A_, En_] := 4 + 2 Sqrt[1 + 4 B] - A^2/(2 (-En)^(3/2))

Cond22[B_, A_, En_] := 2 - (A^2 (3 + 2 Sqrt[1 + 4 B]))/(2 (-En)^(3/2))

Solve[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0, B, En]

Putting the result back into Cond11 or Cond22 does not yield zero.

What is the problem?

edited Aug 16 at 17:11

Carl Woll

55.6k271144

asked Aug 16 at 16:49

Nasser Saad

211

edited Aug 16 at 17:11

Carl Woll

55.6k271144

edited Aug 16 at 17:11

Carl Woll

55.6k271144

edited Aug 16 at 17:11

Carl Woll

55.6k271144

asked Aug 16 at 16:49

Nasser Saad

211

asked Aug 16 at 16:49

Nasser Saad

211

asked Aug 16 at 16:49

Nasser Saad

211

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
5
down vote

From the documentation:

Solve uses non-equivalent transformations to find solutions of transcendental equations and hence it may not find some solutions and may not establish exact conditions on the validity of the solutions found.

With Method->Reduce, Solve uses only equivalent transformations and finds all solutions.

So, the non-equivalent transformations used by Solve are introducing spurious solutions. To avoid this:

Solve[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0, B, En, Method->Reduce]

showing that your equations have no solutions.

answered Aug 16 at 17:09

Carl Woll

55.6k271144

How about the result of NSolve in my answer?
â€“Â user64494
Aug 16 at 18:55

@ user64494 NSolve is basically just invoking Solve in this case.
â€“Â Daniel Lichtblau
Aug 17 at 0:49

@Daniel Lichtblau: If you execute the codes of OP and me, you will see quite different results. Can you base your claim?
â€“Â user64494
Aug 17 at 5:05

@user64494 I checked by tracking it in a debugging kernel. The difference in results is probably due to an intermediate result factoring in the exact case, at the Roots step.
â€“Â Daniel Lichtblau
Aug 17 at 15:04

add a commentÂ |Â

up vote
0
down vote

Making use of NSolve instead of Solve, one obtains

Cond11[B_, A_, En_] := 4 + 2 Sqrt[1 + 4 B] - A^2/(2 (-En)^(3/2));
Cond22[B_, A_, En_] := 2 - (A^2 (3 + 2 Sqrt[1 + 4 B]))/(2 (-En)^(3/2));
NSolve[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0, B, En]

B -> 0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
1]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 1]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 1], B ->
0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
2]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 2]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 2], B ->
0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
3]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 3]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 3], B ->
0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
4]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 4]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 4], B ->
0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
5]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 5]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 5], B ->
0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
6]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 6]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 6]

Addition. Also

FindRoot[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0 /. A -> 1, B, 1, En, 1]

B -> -1.10628 - 1.69546*10^-15 I, En -> 0.944732 + 0.168628 I

edited Aug 16 at 20:00

answered Aug 16 at 18:54

user64494

2,6731917

It doesn't appear that the result of NSolve or FindRoot is close to a zero of the conditions, for any value of A. (Plug into the functions.)
â€“Â Michael E2
Aug 16 at 20:48

@Michael E2: I did try the verification for some values of A. Indeed, the result was wrong or unclear to me. So we deal with a serious bug in NSolve/FindRoot.
â€“Â user64494
Aug 16 at 21:34

FindRoot gives an error/warning, so I wouldn't call that a bug. But I expect more from NSolve, which gives no error; I agree that's a bug.
â€“Â Michael E2
Aug 16 at 23:36

@Michael E2: Did your verify FindRoot for all complex values of A?
â€“Â user64494
Aug 17 at 5:07

Obviously not for all values of A, just a few thousand, like this Cond11[B, A, En], Cond22[B, A, En] /. A -> # /. FindRoot[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0 /. A -> #, B, 1 + 0. I, En, 1] & /@ RandomComplex[1/10 -5 - 5 I, 5 + 5 I, 1000] // Chop[#, 1*^-3] & // Unitize // Flatten // DeleteDuplicates // Quiet, with different domains for RandomComplex. It seems to me that the result of Reduce[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0] indicates one should consider why there is no solution.
â€“Â Michael E2
Aug 17 at 11:25

add a commentÂ |Â

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2 Answers
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active

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2 Answers
2

active

oldest

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up vote
5
down vote

From the documentation:

Solve uses non-equivalent transformations to find solutions of transcendental equations and hence it may not find some solutions and may not establish exact conditions on the validity of the solutions found.

With Method->Reduce, Solve uses only equivalent transformations and finds all solutions.

So, the non-equivalent transformations used by Solve are introducing spurious solutions. To avoid this:

Solve[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0, B, En, Method->Reduce]

showing that your equations have no solutions.

answered Aug 16 at 17:09

Carl Woll

55.6k271144

How about the result of NSolve in my answer?
â€“Â user64494
Aug 16 at 18:55

@ user64494 NSolve is basically just invoking Solve in this case.
â€“Â Daniel Lichtblau
Aug 17 at 0:49

@Daniel Lichtblau: If you execute the codes of OP and me, you will see quite different results. Can you base your claim?
â€“Â user64494
Aug 17 at 5:05

@user64494 I checked by tracking it in a debugging kernel. The difference in results is probably due to an intermediate result factoring in the exact case, at the Roots step.
â€“Â Daniel Lichtblau
Aug 17 at 15:04

add a commentÂ |Â

up vote
5
down vote

From the documentation:

Solve uses non-equivalent transformations to find solutions of transcendental equations and hence it may not find some solutions and may not establish exact conditions on the validity of the solutions found.

With Method->Reduce, Solve uses only equivalent transformations and finds all solutions.

So, the non-equivalent transformations used by Solve are introducing spurious solutions. To avoid this:

Solve[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0, B, En, Method->Reduce]

showing that your equations have no solutions.

answered Aug 16 at 17:09

Carl Woll

55.6k271144

How about the result of NSolve in my answer?
â€“Â user64494
Aug 16 at 18:55

@ user64494 NSolve is basically just invoking Solve in this case.
â€“Â Daniel Lichtblau
Aug 17 at 0:49

@Daniel Lichtblau: If you execute the codes of OP and me, you will see quite different results. Can you base your claim?
â€“Â user64494
Aug 17 at 5:05

@user64494 I checked by tracking it in a debugging kernel. The difference in results is probably due to an intermediate result factoring in the exact case, at the Roots step.
â€“Â Daniel Lichtblau
Aug 17 at 15:04

add a commentÂ |Â

up vote
5
down vote

From the documentation:

Solve uses non-equivalent transformations to find solutions of transcendental equations and hence it may not find some solutions and may not establish exact conditions on the validity of the solutions found.

With Method->Reduce, Solve uses only equivalent transformations and finds all solutions.

So, the non-equivalent transformations used by Solve are introducing spurious solutions. To avoid this:

Solve[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0, B, En, Method->Reduce]

showing that your equations have no solutions.

answered Aug 16 at 17:09

Carl Woll

55.6k271144

From the documentation:

Solve uses non-equivalent transformations to find solutions of transcendental equations and hence it may not find some solutions and may not establish exact conditions on the validity of the solutions found.

With Method->Reduce, Solve uses only equivalent transformations and finds all solutions.

So, the non-equivalent transformations used by Solve are introducing spurious solutions. To avoid this:

Solve[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0, B, En, Method->Reduce]

showing that your equations have no solutions.

answered Aug 16 at 17:09

Carl Woll

55.6k271144

answered Aug 16 at 17:09

Carl Woll

55.6k271144

answered Aug 16 at 17:09

Carl Woll

55.6k271144

answered Aug 16 at 17:09

Carl Woll

55.6k271144

How about the result of NSolve in my answer?
â€“Â user64494
Aug 16 at 18:55

@ user64494 NSolve is basically just invoking Solve in this case.
â€“Â Daniel Lichtblau
Aug 17 at 0:49

@Daniel Lichtblau: If you execute the codes of OP and me, you will see quite different results. Can you base your claim?
â€“Â user64494
Aug 17 at 5:05

@user64494 I checked by tracking it in a debugging kernel. The difference in results is probably due to an intermediate result factoring in the exact case, at the Roots step.
â€“Â Daniel Lichtblau
Aug 17 at 15:04

add a commentÂ |Â

How about the result of NSolve in my answer?
â€“Â user64494
Aug 16 at 18:55

@ user64494 NSolve is basically just invoking Solve in this case.
â€“Â Daniel Lichtblau
Aug 17 at 0:49

@Daniel Lichtblau: If you execute the codes of OP and me, you will see quite different results. Can you base your claim?
â€“Â user64494
Aug 17 at 5:05

@user64494 I checked by tracking it in a debugging kernel. The difference in results is probably due to an intermediate result factoring in the exact case, at the Roots step.
â€“Â Daniel Lichtblau
Aug 17 at 15:04

How about the result of NSolve in my answer?
â€“Â user64494
Aug 16 at 18:55

@ user64494 NSolve is basically just invoking Solve in this case.
â€“Â Daniel Lichtblau
Aug 17 at 0:49

@Daniel Lichtblau: If you execute the codes of OP and me, you will see quite different results. Can you base your claim?
â€“Â user64494
Aug 17 at 5:05

@user64494 I checked by tracking it in a debugging kernel. The difference in results is probably due to an intermediate result factoring in the exact case, at the Roots step.
â€“Â Daniel Lichtblau
Aug 17 at 15:04

add a commentÂ |Â

up vote
0
down vote

Making use of NSolve instead of Solve, one obtains

Cond11[B_, A_, En_] := 4 + 2 Sqrt[1 + 4 B] - A^2/(2 (-En)^(3/2));
Cond22[B_, A_, En_] := 2 - (A^2 (3 + 2 Sqrt[1 + 4 B]))/(2 (-En)^(3/2));
NSolve[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0, B, En]

B -> 0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
1]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 1]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 1], B ->
0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
2]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 2]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 2], B ->
0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
3]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 3]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 3], B ->
0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
4]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 4]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 4], B ->
0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
5]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 5]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 5], B ->
0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
6]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 6]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 6]

Addition. Also

FindRoot[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0 /. A -> 1, B, 1, En, 1]

B -> -1.10628 - 1.69546*10^-15 I, En -> 0.944732 + 0.168628 I

edited Aug 16 at 20:00

answered Aug 16 at 18:54

user64494

2,6731917

It doesn't appear that the result of NSolve or FindRoot is close to a zero of the conditions, for any value of A. (Plug into the functions.)
â€“Â Michael E2
Aug 16 at 20:48

@Michael E2: I did try the verification for some values of A. Indeed, the result was wrong or unclear to me. So we deal with a serious bug in NSolve/FindRoot.
â€“Â user64494
Aug 16 at 21:34

FindRoot gives an error/warning, so I wouldn't call that a bug. But I expect more from NSolve, which gives no error; I agree that's a bug.
â€“Â Michael E2
Aug 16 at 23:36

@Michael E2: Did your verify FindRoot for all complex values of A?
â€“Â user64494
Aug 17 at 5:07

Obviously not for all values of A, just a few thousand, like this Cond11[B, A, En], Cond22[B, A, En] /. A -> # /. FindRoot[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0 /. A -> #, B, 1 + 0. I, En, 1] & /@ RandomComplex[1/10 -5 - 5 I, 5 + 5 I, 1000] // Chop[#, 1*^-3] & // Unitize // Flatten // DeleteDuplicates // Quiet, with different domains for RandomComplex. It seems to me that the result of Reduce[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0] indicates one should consider why there is no solution.
â€“Â Michael E2
Aug 17 at 11:25

add a commentÂ |Â

up vote
0
down vote

Making use of NSolve instead of Solve, one obtains

Cond11[B_, A_, En_] := 4 + 2 Sqrt[1 + 4 B] - A^2/(2 (-En)^(3/2));
Cond22[B_, A_, En_] := 2 - (A^2 (3 + 2 Sqrt[1 + 4 B]))/(2 (-En)^(3/2));
NSolve[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0, B, En]

B -> 0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
1]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 1]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 1], B ->
0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
2]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 2]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 2], B ->
0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
3]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 3]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 3], B ->
0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
4]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 4]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 4], B ->
0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
5]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 5]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 5], B ->
0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
6]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 6]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 6]

Addition. Also

FindRoot[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0 /. A -> 1, B, 1, En, 1]

B -> -1.10628 - 1.69546*10^-15 I, En -> 0.944732 + 0.168628 I

edited Aug 16 at 20:00

answered Aug 16 at 18:54

user64494

2,6731917

It doesn't appear that the result of NSolve or FindRoot is close to a zero of the conditions, for any value of A. (Plug into the functions.)
â€“Â Michael E2
Aug 16 at 20:48

@Michael E2: I did try the verification for some values of A. Indeed, the result was wrong or unclear to me. So we deal with a serious bug in NSolve/FindRoot.
â€“Â user64494
Aug 16 at 21:34

FindRoot gives an error/warning, so I wouldn't call that a bug. But I expect more from NSolve, which gives no error; I agree that's a bug.
â€“Â Michael E2
Aug 16 at 23:36

@Michael E2: Did your verify FindRoot for all complex values of A?
â€“Â user64494
Aug 17 at 5:07

Obviously not for all values of A, just a few thousand, like this Cond11[B, A, En], Cond22[B, A, En] /. A -> # /. FindRoot[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0 /. A -> #, B, 1 + 0. I, En, 1] & /@ RandomComplex[1/10 -5 - 5 I, 5 + 5 I, 1000] // Chop[#, 1*^-3] & // Unitize // Flatten // DeleteDuplicates // Quiet, with different domains for RandomComplex. It seems to me that the result of Reduce[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0] indicates one should consider why there is no solution.
â€“Â Michael E2
Aug 17 at 11:25

add a commentÂ |Â

up vote
0
down vote

Making use of NSolve instead of Solve, one obtains

Cond11[B_, A_, En_] := 4 + 2 Sqrt[1 + 4 B] - A^2/(2 (-En)^(3/2));
Cond22[B_, A_, En_] := 2 - (A^2 (3 + 2 Sqrt[1 + 4 B]))/(2 (-En)^(3/2));
NSolve[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0, B, En]

B -> 0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
1]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 1]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 1], B ->
0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
2]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 2]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 2], B ->
0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
3]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 3]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 3], B ->
0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
4]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 4]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 4], B ->
0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
5]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 5]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 5], B ->
0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
6]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 6]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 6]

Addition. Also

FindRoot[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0 /. A -> 1, B, 1, En, 1]

B -> -1.10628 - 1.69546*10^-15 I, En -> 0.944732 + 0.168628 I

edited Aug 16 at 20:00

answered Aug 16 at 18:54

user64494

2,6731917

Making use of NSolve instead of Solve, one obtains

Cond11[B_, A_, En_] := 4 + 2 Sqrt[1 + 4 B] - A^2/(2 (-En)^(3/2));
Cond22[B_, A_, En_] := 2 - (A^2 (3 + 2 Sqrt[1 + 4 B]))/(2 (-En)^(3/2));
NSolve[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0, B, En]

B -> 0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
1]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 1]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 1], B ->
0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
2]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 2]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 2], B ->
0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
3]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 3]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 3], B ->
0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
4]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 4]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 4], B ->
0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
5]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 5]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 5], B ->
0.875 - (
0.21875 A^2 Sqrt[-1. Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &,
6]])/Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 6]^2,
En -> Root[1. A^8 + 20. A^4 #1^3 + 64. #1^6 &, 6]

Addition. Also

FindRoot[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0 /. A -> 1, B, 1, En, 1]

B -> -1.10628 - 1.69546*10^-15 I, En -> 0.944732 + 0.168628 I

edited Aug 16 at 20:00

answered Aug 16 at 18:54

user64494

2,6731917

edited Aug 16 at 20:00

answered Aug 16 at 18:54

user64494

2,6731917

answered Aug 16 at 18:54

user64494

2,6731917

answered Aug 16 at 18:54

user64494

2,6731917

It doesn't appear that the result of NSolve or FindRoot is close to a zero of the conditions, for any value of A. (Plug into the functions.)
â€“Â Michael E2
Aug 16 at 20:48

@Michael E2: I did try the verification for some values of A. Indeed, the result was wrong or unclear to me. So we deal with a serious bug in NSolve/FindRoot.
â€“Â user64494
Aug 16 at 21:34

FindRoot gives an error/warning, so I wouldn't call that a bug. But I expect more from NSolve, which gives no error; I agree that's a bug.
â€“Â Michael E2
Aug 16 at 23:36

@Michael E2: Did your verify FindRoot for all complex values of A?
â€“Â user64494
Aug 17 at 5:07

Obviously not for all values of A, just a few thousand, like this Cond11[B, A, En], Cond22[B, A, En] /. A -> # /. FindRoot[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0 /. A -> #, B, 1 + 0. I, En, 1] & /@ RandomComplex[1/10 -5 - 5 I, 5 + 5 I, 1000] // Chop[#, 1*^-3] & // Unitize // Flatten // DeleteDuplicates // Quiet, with different domains for RandomComplex. It seems to me that the result of Reduce[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0] indicates one should consider why there is no solution.
â€“Â Michael E2
Aug 17 at 11:25

add a commentÂ |Â

It doesn't appear that the result of NSolve or FindRoot is close to a zero of the conditions, for any value of A. (Plug into the functions.)
â€“Â Michael E2
Aug 16 at 20:48

@Michael E2: I did try the verification for some values of A. Indeed, the result was wrong or unclear to me. So we deal with a serious bug in NSolve/FindRoot.
â€“Â user64494
Aug 16 at 21:34

FindRoot gives an error/warning, so I wouldn't call that a bug. But I expect more from NSolve, which gives no error; I agree that's a bug.
â€“Â Michael E2
Aug 16 at 23:36

@Michael E2: Did your verify FindRoot for all complex values of A?
â€“Â user64494
Aug 17 at 5:07

Obviously not for all values of A, just a few thousand, like this Cond11[B, A, En], Cond22[B, A, En] /. A -> # /. FindRoot[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0 /. A -> #, B, 1 + 0. I, En, 1] & /@ RandomComplex[1/10 -5 - 5 I, 5 + 5 I, 1000] // Chop[#, 1*^-3] & // Unitize // Flatten // DeleteDuplicates // Quiet, with different domains for RandomComplex. It seems to me that the result of Reduce[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0] indicates one should consider why there is no solution.
â€“Â Michael E2
Aug 17 at 11:25

It doesn't appear that the result of NSolve or FindRoot is close to a zero of the conditions, for any value of A. (Plug into the functions.)
â€“Â Michael E2
Aug 16 at 20:48

@Michael E2: I did try the verification for some values of A. Indeed, the result was wrong or unclear to me. So we deal with a serious bug in NSolve/FindRoot.
â€“Â user64494
Aug 16 at 21:34

FindRoot gives an error/warning, so I wouldn't call that a bug. But I expect more from NSolve, which gives no error; I agree that's a bug.
â€“Â Michael E2
Aug 16 at 23:36

@Michael E2: Did your verify FindRoot for all complex values of A?
â€“Â user64494
Aug 17 at 5:07

Obviously not for all values of A, just a few thousand, like this

Cond11[B, A, En], Cond22[B, A, En] /. A -> # /. FindRoot[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0 /. A -> #, B, 1 + 0. I, En, 1] & /@ RandomComplex[1/10 -5 - 5 I, 5 + 5 I, 1000] // Chop[#, 1*^-3] & // Unitize // Flatten // DeleteDuplicates // Quiet

, with different domains for RandomComplex. It seems to me that the result of Reduce[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0] indicates one should consider why there is no solution.
â€“Â Michael E2
Aug 17 at 11:25

Obviously not for all values of A, just a few thousand, like this

Cond11[B, A, En], Cond22[B, A, En] /. A -> # /. FindRoot[Cond11[B, A, En] == 0, Cond22[B, A, En] == 0 /. A -> #, B, 1 + 0. I, En, 1] & /@ RandomComplex[1/10 -5 - 5 I, 5 + 5 I, 1000] // Chop[#, 1*^-3] & // Unitize // Flatten // DeleteDuplicates // Quiet

add a commentÂ |Â

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