Tetration Forum
uniqueness - Printable Version

+- Tetration Forum (https://math.eretrandre.org/tetrationforum)
+-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1)
+--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3)
+--- Thread: uniqueness (/showthread.php?tid=257)

Pages: 1 2


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 is just a consequence of , 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 is just a consequence of , 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 .