Re: Solving Tanh[x]=Tanh[a]Tanh[b x + c]
- To: mathgroup at smc.vnet.net
- Subject: [mg83471] Re: [mg83424] Solving Tanh[x]=Tanh[a]Tanh[b x + c]
- From: DrMajorBob <drmajorbob at bigfoot.com>
- Date: Wed, 21 Nov 2007 02:51:55 -0500 (EST)
- References: <31486144.1195561117246.JavaMail.root@m35>
- Reply-to: drmajorbob at bigfoot.com
step1 = Tanh[x] - Tanh[a] Tanh[b x + c] // TrigToExp // Together //
Numerator;
assume = {xx \[Element] Reals, {c1 | c3} \[Element] Reals,
b \[Element] Integers, b > 0};
step2 = Simplify[
step1 /. {x -> Log[xx]/2, c -> Log[c1]/2, a -> Log[c3]/2},
assume];
step3 = step2/2 // Expand;
pure = Evaluate[step3 /. xx -> #] &
-1 + c3 #1 - c1 c3 #1^b + c1 #1^(1 + b) &
Bobby
On Tue, 20 Nov 2007 02:49:22 -0600, Yaroslav Bulatov
<yaroslavvb at gmail.com> 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
>
>
--
DrMajorBob at bigfoot.com