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Solving Tanh[x]=Tanh[a]Tanh[b x + c]
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| Yaroslav Bulatov 2007-11-20, 4:29 am |
| I'd like to use Mathematica to show that solution of Tanh[x] - Tanh[a]
Tanh[b x + c]=0 can be written as
1/2 Log (Root[c1 x^(1+b) + c2 x^b + c3 x -1]) for certain coefficients
c1,c2,c3 when b is a positive integer
Tanh[x] - Tanh[a] Tanh[b x + c]// TrigToExp // Together // Numerator
gives me almost what I need, except now I need to factor out Exp[2x]
as a separate variable. What's the best way of achieving it? Using
syntactic replacement rules like {Exp[a_+b_]->Exp[a]Exp[b],Exp[2x]->x}
seems like an uphill battle against the evaluator which automatically
simplifies Exp expressions
Yaroslav
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Hi Yaroslav,
you must prevent the "uphill battle" by e.g. temporarily write
Exp[a]Exp[b] as {Exp[a],Exp[b]}, then do what you want and finally
eliminate the braces.E.g:
.... //.{Exp[a__+b_]->{Exp[a],Exp[b]},Exp[2x]->x,{a_,b_}->a b}
hope this helps, Daniel
Yaroslav Bulatov wrote:
> I'd like to use Mathematica to show that solution of Tanh[x] - Tanh[a]
> Tanh[b x + c]=0 can be written as
> 1/2 Log (Root[c1 x^(1+b) + c2 x^b + c3 x -1]) for certain coefficients
> c1,c2,c3 when b is a positive integer
>
> Tanh[x] - Tanh[a] Tanh[b x + c]// TrigToExp // Together // Numerator
> gives me almost what I need, except now I need to factor out Exp[2x]
> as a separate variable. What's the best way of achieving it? Using
> syntactic replacement rules like {Exp[a_+b_]->Exp[a]Exp[b],Exp[2x]->x}
> seems like an uphill battle against the evaluator which automatically
> simplifies Exp expressions
>
> Yaroslav
>
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