Re: Help me solving: (f(x)+3*x )^[5/3]=f(x /(x+1))+3*x /(x+1) .

From: Alain Verghote (alainverghote_at_yahoo.fr)
Date: 09/20/04 

Date: Mon, 20 Sep 2004 12:13:45 +0000 (UTC)

On 16 Sep 2004, G. A. Edgar wrote:
>
>In article <ciccds$44l$1@dizzy.math.ohio-state.edu>, ALAIN Verghote
><alainverghote@yahoo.fr> wrote:
>
>> f is a continuous real function,g(x)^[r] means r iterate of g,
>> r is real.
>> Is it a known method to solve such equations,
>> Thanks for replying,
>>
>
>I would proceed like this. I ignore the fact that non-integer iterates
>need not be unique (and this may be a real problem...).
>
>The proposed problem
> (f(x)+3*x )^[5/3]=f(x /(x+1))+3*x /(x+1)
>I take to mean: let g(x) = f(x)+3*x, then solve g^[5/3] = g h,
>where h(x) = x/(x+1). { g h means composition, * means multiplication. }
>So this is g^[2/3] = h , thus g = h^[3/2].
>Now I happen to know this one-parameter group:
>h_t(x) = x/(t*x+1) where h_1 = h and h_(s+t) = h_s h_t,
>so (ignoring as stated above) h^[3/2](x) = x/((3/2)*x+1)
>so g(x) = x/((3/2)*x+1) so f(x) = x/((3/2)*x+1) - 3*x
>= (-4*x-9*x^2)/(3*x+2) .
>
>We can check that g g g g g = r r r, where r(x) = f(x/(x+1))+3*x/(x+1)
>and g(x) = f(x)+3*x { they are both x -> 2*x/(15*x+2) },
>and that is presumably the meaning of the assertion g^[5/3] = r.
>
>--
>Gerald A. Edgar edgar at math.ohio-state.edu
>Department of Mathematics telephone: 614-292-0395 (Office)
>The Ohio State University 614-292-4975 (Math. Dept.)
>Columbus, OH 43210 614-292-1479 (Dept. Fax)

Dear Edgar,

I've read your interesting solution.
So two steps are needed:
            1°)reducing both sides by f+3*Id (Id identity function)
            2°)powering ((f(x)+3x)^[2/3])^[3/2] to obtain
           (f(x)+3x)^[1] or (f(x)+3x). [r] real iterate power ;
            g(x)^[5/3] means fractional iterate power between 1 and 2:
            that is 'between' (g(x) and g(g(x)).
     Notice x/(x+1)=1/(1+1/x) and (x/(x+1))^[r]=(x/(r*x+1)),
     The given solution seems the only one...

 Sincerely,Alain.

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