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 uniqueness tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 03/24/2009, 02:50 AM here are the equations that make half-iterate of exp(x) unique : (under condition f(x) maps reals to reals and f(x) > x ) exp(x) = f(f(x)) = D f(f(x)) = f ' (f(x)) * f ' (x) = D^2 f(f(x)) = f '' (f(x)) * f ' (x)^2 + f ' (f(x)) * f '' (x) = D^3 f(f(x)) = D^4 f(f(x)) regards tommy1729 bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 03/24/2009, 09:39 AM tommy1729 Wrote:here are the equations that make half-iterate of exp(x) unique : (under condition f(x) maps reals to reals and f(x) > x ) exp(x) = f(f(x)) = D f(f(x)) = f ' (f(x)) * f ' (x) = D^2 f(f(x)) = f '' (f(x)) * f ' (x)^2 + f ' (f(x)) * f '' (x) = D^3 f(f(x)) = D^4 f(f(x)) Arent these equations valid for every half iterate of exp? tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 03/26/2009, 07:34 PM bo198214 Wrote:tommy1729 Wrote:here are the equations that make half-iterate of exp(x) unique : (under condition f(x) maps reals to reals and f(x) > x ) exp(x) = f(f(x)) = D f(f(x)) = f ' (f(x)) * f ' (x) = D^2 f(f(x)) = f '' (f(x)) * f ' (x)^2 + f ' (f(x)) * f '' (x) = D^3 f(f(x)) = D^4 f(f(x)) Arent these equations valid for every half iterate of exp? NO of course not. for example : the first case : exp(x) = f ' (f(x)) * f ' (x) now consider a solution that satisfies f(f(x)) = exp(x) and let assume f ' (x) = 0 has a finite real zero at x = r1. thus f ' (r1) = 0 exp( r1 ) = 0 * f ' (f(r1)) => ?????? you see , contradiction. regards tommy1729 bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 03/26/2009, 08:11 PM tommy1729 Wrote:NO of course not. for example : the first case : exp(x) = f ' (f(x)) * f ' (x) now consider a solution that satisfies f(f(x)) = exp(x) and let assume f ' (x) = 0 has a finite real zero at x = r1. thus f ' (r1) = 0 exp( r1 ) = 0 * f ' (f(r1)) => ?????? you see , contradiction. That just shows that there is no halfiterate with f'(x)=0 for some x. The equation $\exp(x)=f'(f(x))*f'(x)$ is just a consequence of $\exp(x)=f(f(x))$, so it is valid for *every* halfiterate f (which of course must be differentiable). tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 03/27/2009, 12:57 PM bo198214 Wrote:tommy1729 Wrote:NO of course not. for example : the first case : exp(x) = f ' (f(x)) * f ' (x) now consider a solution that satisfies f(f(x)) = exp(x) and let assume f ' (x) = 0 has a finite real zero at x = r1. thus f ' (r1) = 0 exp( r1 ) = 0 * f ' (f(r1)) => ?????? you see , contradiction. That just shows that there is no halfiterate with f'(x)=0 for some x. The equation $\exp(x)=f'(f(x))*f'(x)$ is just a consequence of $\exp(x)=f(f(x))$, so it is valid for *every* halfiterate f (which of course must be differentiable). yes. so basicly its about f(x) being Coo or at least C4. ( notice bo didnt first notice , or at least didnt mention , " must be differentiable " ) i used to identies of the OP because those equations show restrictions ... so lets restate : conjecture if f(f(x)) = exp(x) f(x) is real -> real and > x and f(x) is Coo then f(x) is the unique solution to the above. ??? bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 03/29/2009, 07:50 PM tommy1729 Wrote:conjecture if f(f(x)) = exp(x) f(x) is real -> real and > x and f(x) is Coo then f(x) is the unique solution to the above. No, it is not unique. We discussed that already here. Even if you demand that f is analytic. tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 03/29/2009, 08:29 PM Ansus Wrote:I dont think uniqueness is a big problem for us now. We have a number of formulas for tetration which, it turns out, are equal. turns out to be equal ? that is not proven as i recall it. tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 03/29/2009, 08:51 PM bo198214 Wrote:tommy1729 Wrote:conjecture if f(f(x)) = exp(x) f(x) is real -> real and > x and f(x) is Coo then f(x) is the unique solution to the above. No, it is not unique. We discussed that already here. Even if you demand that f is analytic. ok , lets see. f(f(x)) = x^4 f(x) = x^2 another solution : f(x) = (x + q(x)) ^2 = x^2 + 2q(x)x + q(x)^2 f(f(x)) = ( x^2 + 2q(x)x + q(x)^2 + q(x^2 + 2q(x)x + q(x)^2) ) ^2 => ( x^2 + 2q(x)x + q(x)^2 + q(x^2 + 2q(x)x + q(x)^2) ) ^2 = x^4 => x^2 + 2q(x)x + q(x)^2 + q(x^2 + 2q(x)x + q(x)^2) = x^2 => 2q(x)x + q(x)^2 + q(x^2 + 2q(x)x + q(x)^2) = 0 this seems very different from your 1-periodic condition of q(x) ?? bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 03/29/2009, 09:06 PM tommy1729 Wrote:f(x) = (x + q(x)) ^2 = x^2 + 2q(x)x + q(x)^2 f(f(x)) = ( x^2 + 2q(x)x + q(x)^2 + q(x^2 + 2q(x)x + q(x)^2) ) ^2 q must appear inside q. bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 03/29/2009, 09:08 PM Ansus Wrote:I dont think uniqueness is a big problem for us now. We have a number of formulas for tetration which, it turns out, are equal. We only know that regular, Newton and Lagrange method are equal. But thats only a small fraction of methods. Matrix power, intuitive Abel and Cauchy integral method are still unclear and are more important as they are applicable to $b>e^{1/e}$. « Next Oldest | Next Newest »

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