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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

> 




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