Functional-Equation
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From: Alfred Flaßhaar on 1 Feb 2010 11:24 Hello, I have got as a step in between during solving an equation the following equation f(f(x))/x = f(f(y))/y (*) where f is the function I am looking for. Now I have no idea to solve it. Is it correct to set (*)=const.? Regards, Alfred Fla�haar
From: Dan Cass on 1 Feb 2010 02:32 > Hello, > > I have got as a step in between during solving an > equation the following > equation > > f(f(x))/x = f(f(y))/y (*) > > where f is the function I am looking for. Now I have > no idea to solve it. Is > it correct to set (*)=const.? > > Regards, Alfred Flaßhaar > There may be (probably are) solutions for which neither side of (*) is constant. On the other hand there are solutions for which both sides of (*) are 1, such as any self-inverse f. Example: consider f(x) = 1/x. Then f(f(x)) = x. So both sides of (*) come out 1.
From: Patrick Coilland on 1 Feb 2010 12:53 Alfred Fla�haar a �crit : > Hello, > > I have got as a step in between during solving an equation the following > equation > > f(f(x))/x = f(f(y))/y (*) > > where f is the function I am looking for. Now I have no idea to solve > it. Is it correct to set (*)=const.? > Sure : $f(f(x))/x=f(f(y))/y forall x,y =/=0 means f(f(x))/x=f(f(1))/1=Cste. And you have to solve f(f(x))=cx which has infinitely many solutions
From: Alfred Flaßhaar on 1 Feb 2010 13:13 Patrick Coilland wrote: > Alfred Fla�haar a �crit : (...) > Sure : $f(f(x))/x=f(f(y))/y forall x,y =/=0 means > f(f(x))/x=f(f(1))/1=Cste. > And you have to solve f(f(x))=cx which has infinitely many solutions ^^^^^^^^^^^^^^^^^^^^^^ From where do You know this? Looking only for continous functions with f(x)>0. The subject "functional equations" is new to me. I wonder what it will turn out like. In my higher age I want to learn about. It is not a homework. Alfred
From: Patrick Coilland on 2 Feb 2010 04:06
Alfred Fla�haar a �crit : > Patrick Coilland wrote: >> Alfred Fla�haar a �crit : > > (...) > >> Sure : $f(f(x))/x=f(f(y))/y forall x,y =/=0 means >> f(f(x))/x=f(f(1))/1=Cste. >> And you have to solve f(f(x))=cx which has infinitely many solutions > ^^^^^^^^^^^^^^^^^^^^^^ > > From where do You know this? Looking only for continous functions with > f(x)>0. The subject "functional equations" is new to me. I wonder what > it will turn out like. In my higher age I want to learn about. It is not > a homework. > > Alfred There are no solution if you add the constraint f(x)>0 since for example f(f(-1))=-1<0 If you forget this constraint, you can build infinitely many continuous solutions with the "piece per piece" method : In this example, I consider c>1 (it's easy to adapt for 1>=c>=0, and there is obviously no continuous solution for c<0) Let any function h(x) defined on [1,c] with the following constraints : h(x) continuous and increasing c>h(1)>1 h(h(1))=c Let f(x)=h(x) for any x in [1,h(1)). You can then define f^{-1}(x) from [h(1),c]-->[1,h(1)] you can then define f(x)=cf^{-1}(x) for any x in [h(1),h(h(1)=c)] And, step by step, you can build f(x) : first on [1,h(1)) then on [h(1),c) then on [c,ch(1)) then on [ch(1),c^2) .... And same on (0,1) Adding f(0)=0 and the same construction for x<0, you can build infinitely many continuous solutions. (notice that this method does not provide all continuous solutions) |