Re: Solving Tanh[x]=Tanh[a]Tanh[b x + c]
- To: mathgroup at smc.vnet.net
- Subject: [mg83545] Re: Solving Tanh[x]=Tanh[a]Tanh[b x + c]
- From: dh <dh at metrohm.ch>
- Date: Thu, 22 Nov 2007 04:58:10 -0500 (EST)
- References: <fhu7h8$79u$1@smc.vnet.net>
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
>