| Author |
Comment/Response |
Bill Simpson
|
 09/29/12 00:16am
Reduce attempts to return an exact analytic equation that is valid for all values of your variables. It appears that Mathematica and Reduce simply do not know of any such solution for your problem.
If you specify an integer value for n then Reduce is happy to give you solutions.
In[1]:= n=1;Reduce[2 n-Tanh[z] Tanh[2 n z]==0,z]
Out[1]= C[1] ∈ Integers && (z == (-I/2)*Pi + (2*I)*Pi*C[1] || z == (I/2)*Pi + (2*I)*Pi*C[1])
In[2]:= n=2;Reduce[2 n-Tanh[z] Tanh[2 n z]==0,z]
Out[2]= C[1] ∈ Integers && (z == (-I/2)*Pi + (2*I)*Pi*C[1] || z == (I/2)*Pi + (2*I)*Pi*C[1] || z == (2*I)*Pi*C[1] + Log[-Sqrt[2/3 - (I/3)*Sqrt[5]]] || z == (2*I)*Pi*C[1] + Log[Sqrt[2/3 - (I/3)*Sqrt[5]]] || z == (2*I)*Pi*C[1] + Log[-Sqrt[2/3 + (I/3)*Sqrt[5]]] ||
z == (2*I)*Pi*C[1] + Log[Sqrt[2/3 + (I/3)*Sqrt[5]]])
In[3]:= n=3;Reduce[2 n-Tanh[z] Tanh[2 n z]==0,z]
Out[3]= C[1] ∈ Integers && (z == (-I/2)*Pi + (2*I)*Pi*C[1] || z == (I/2)*Pi + (2*I)*Pi*C[1] ||
z == (2*I)*Pi*C[1] + Log[-Sqrt[Root[5 - 8*#1 + 9*#1^2 - 8*#1^3 + 5*#1^4 & , 1, 0]]] ||
z == (2*I)*Pi*C[1] + Log[Sqrt[Root[5 - 8*#1 + 9*#1^2 - 8*#1^3 + 5*#1^4 & , 1, 0]]] ||
z == (2*I)*Pi*C[1] + Log[-Sqrt[Root[5 - 8*#1 + 9*#1^2 - 8*#1^3 + 5*#1^4 & , 2, 0]]] ||
z == (2*I)*Pi*C[1] + Log[Sqrt[Root[5 - 8*#1 + 9*#1^2 - 8*#1^3 + 5*#1^4 & , 2, 0]]] ||
z == (2*I)*Pi*C[1] + Log[-Sqrt[Root[5 - 8*#1 + 9*#1^2 - 8*#1^3 + 5*#1^4 & , 3, 0]]] ||
z == (2*I)*Pi*C[1] + Log[Sqrt[Root[5 - 8*#1 + 9*#1^2 - 8*#1^3 + 5*#1^4 & , 3, 0]]] ||
z == (2*I)*Pi*C[1] + Log[-Sqrt[Root[5 - 8*#1 + 9*#1^2 - 8*#1^3 + 5*#1^4 & , 4, 0]]] ||
z == (2*I)*Pi*C[1] + Log[Sqrt[Root[5 - 8*#1 + 9*#1^2 - 8*#1^3 + 5*#1^4 & , 4, 0]]])
Your notebook contained an I in your equation which you did not have in your description, but this same method of specifying an integer value for n works equally well for that.
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