Define the set of equations as eq
(with all numbers rationalized), and assume that all variables are real. Then,
eq /. k -> 0
(* {False, True, c1 == 1, c2 == 0, False, False, c1 == 11/10, c2 == 0} *)
So, k != 0
.
Reduce[eq[[2]] && k != 0, b]
(* (k != 0 && A^2 + B^2 != 0 && Sech[d] == 0) || (C[1] ∈ Integers &&
k != 0 && A^2 + B^2 != 0 && (b == 2 π C[1] || b == π + 2 π C[1])) *)
Since Sech[d] != 0
for real d
, it follows that b
is an integer multiple of π
. Apply this result to obtain
eq1 = DeleteCases[Simplify[eq /. b -> n Pi, n ∈ Integers], True]
(* {((-1)^n k Sech[d])/Sqrt[A^2 + B^2] == 1,
c1 + (A k Tanh[d])/(A^2 + B^2) == 1,
c2 + (B k Tanh[d])/(A^2 + B^2) == 0,
(k Cos[a + n π] Sech[d + k])/Sqrt[A^2 + B^2] == 1,
(10 k Sech[d + k] Sin[a + n π])/Sqrt[A^2 + B^2] == 1,
c1 + (A k Tanh[d + k])/(A^2 + B^2) == 11/10,
c2 + (B k Tanh[d + k])/(A^2 + B^2) == 0} *)
Factor[First@eq1[[7]] - First@eq1[[3]]]
(* -((B k (Tanh[d] - Tanh[d + k]))/(A^2 + B^2)) *)
Hence, B == 0
, and
eq1 /. B -> 0;
(* {((-1)^n k Sech[d])/Sqrt[A^2] == 1,
c1 + (k Tanh[d])/A == 1,
c2 == 0,
(k Cos[a + n π] Sech[d + k])/Sqrt[A^2] == 1,
(10 k Sech[d + k] Sin[a + n π])/Sqrt[A^2] == 1,
c1 + (k Tanh[d + k])/A == 11/10,
c2 == 0} *)
Hence, c2 == 0
, and
eq2 = DeleteCases[% /. c2 -> 0, True]
(* {((-1)^n k Sech[d])/Sqrt[A^2] == 1,
c1 + (k Tanh[d])/A == 1,
(k Cos[a + n π] Sech[d + k])/Sqrt[A^2] == 1,
(10 k Sech[d + k] Sin[a + n π])/Sqrt[A^2] == 1,
c1 + (k Tanh[d + k])/A == 11/10} *)
This allows a
to be calculated.
as = Simplify[Solve[First@eq2[[3]]/First@eq2[[4]] == 1, a][[1,1]]/.C[1] -> m, m ∈ Integers]
(* a -> m π - n π + ArcCot[10] *)
eq3 = Union[eq2 /. as]
(* {((-1)^n k Sech[d])/Sqrt[A^2] == 1,
(k Cos[m π + ArcCot[10]] Sech[d + k])/Sqrt[A^2] == 1,
(10 k Sech[d + k] Sin[m π + ArcCot[10]])/Sqrt[A^2] == 1,
c1 + (k Tanh[d])/A == 1,
c1 + (k Tanh[d + k])/A == 11/10} *)
Note that there now are more equations than variables, so one must be redundant. Now, assume that n
and m
are even integers.
Reduce[eq3 /. {m -> 0, n -> 0}, Reals] // Simplify
(* A == Log[19531250/(30100251 - 2278951 Sqrt[101])]/Sqrt[101] &&
10 c1 == 11 && d == Log[1/10 (-1 + Sqrt[101])] &&
10 k == Log[19531250/(30100251 - 2278951 Sqrt[101])] *)
along with the message,
Reduce::ztest1: Unable to decide whether numeric quantity ... is equal to zero. Assuming it is.
Unfortunately, it does not provide the complete expression that it assumes equal to zero, so we cannot evaluate it. Next, assume that n
and m
are odd integers.
Reduce[eq3 /. {m -> 1, n -> 1}, Reals] // Simplify
(* A == (-Log[2] - 10 Log[5] + Log[30100251 + 2278951 Sqrt[101]])/Sqrt[101] &&
10 c1 == 11 &&
d == Log[1/10 (1 + Sqrt[101])] &&
10 k + Log[30100251 + 2278951 Sqrt[101]] == Log[2] + 10 Log[5] *)
Along with a similar message. On the other hand,
Reduce[eq3 /. {m -> 0, n -> 1}, Reals]
(* False *)
Reduce[eq3 /. {m -> 1, n -> 0}, Reals]
(* False *)
Finally, the question remains of whether the unknown expressions that Reduce
assumes equal to zero actually are. To proceed, assume c2 == 11/10
, as given in both tentative answers, and also that d == -k`, which also is given in the tentative answers, although not so obviously.
FullSimplify[Log[1/10 (-1 + Sqrt[101])] ==
-Log[19531250/(30100251 - 2278951 Sqrt[101])]/10]
(* True *)
DeleteCases[Union[Simplify[eq3 /. c1 -> 11/10 /. d -> -k] /. {m -> 0, n -> 0}], True]
(* k == ArcCosh[Sqrt[101]/10] &&
A == 100 Sqrt[1/101 (-1 + Sqrt[101]/10) (1 + Sqrt[101]/10)] ArcCosh[Sqrt[101]/10] *)
along with the same warning message, but this time with the assumed expression given.
Reduce::ztest1: Unable to decide whether numeric quantity ArcCosh[Sqrt[101]/10]+Log[2]+Log[5]-Log[1+Sqrt[101]] is equal to zero. Assuming it is.
Test it.
FullSimplify[ArcCosh[Sqrt[101]/10] + Log[2] + Log[5] - Log[1 + Sqrt[101]]]
(* 0 *)
Why Reduce
did not do this is unclear. Likewise, for odd integers,
FullSimplify[Log[1/10 (1 + Sqrt[101])] ==
-(Log[2] + 10 Log[5] - Log[30100251 + 2278951 Sqrt[101]])/10]
(* True *)
DeleteCases[Union[Simplify[eq3 /. c1 -> 11/10 /. d -> -k] /. {m -> 1, n -> 1}], True]
(* {-((10 k)/(Sqrt[101] Sqrt[A^2])) == 1,
-((k Sech[k])/Sqrt[A^2]) == 1,
(10 k Tanh[k])/A == 1} *)
Reduce[%, {k, A}, Reals]
(* k == -ArcCosh[Sqrt[101]/10] &&
A == 100 Sqrt[1/101 (-1 + Sqrt[101]/10) (1 + Sqrt[101]/10)] ArcCosh[Sqrt[101]/10] *)
FullSimplify[-ArcCosh[Sqrt[101]/10] + Log[2] + Log[5] - Log[-1 + Sqrt[101]]]
(* 0 *)
All unknowns now are evaluated. By the way, I attempted to use Reduce[eq, Reals]
to obtain the complete solution directly, but it failed after several hours for lack of memory on my 16 GB PC.
c1
andc2
used in the equations? You can get rid of three of the divisions byA^2+B^2
where the right-hand side of the equation is zero. $\endgroup$