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+--- Thread: uniqueness (/showthread.php?tid=257)
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uniqueness - tommy1729 - 03/24/2009
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
RE: uniqueness - bo198214 - 03/24/2009
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?
RE: uniqueness - tommy1729 - 03/26/2009
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
RE: uniqueness - bo198214 - 03/26/2009
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).
RE: uniqueness - tommy1729 - 03/27/2009
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.
???
RE: uniqueness - bo198214 - 03/29/2009
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.
RE: uniqueness - tommy1729 - 03/29/2009
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.
RE: uniqueness - tommy1729 - 03/29/2009
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) ??
RE: uniqueness - bo198214 - 03/29/2009
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.
RE: uniqueness - bo198214 - 03/29/2009
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 \( b>e^{1/e} \).