# Tetration Forum

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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
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?
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)) => ??????

regards

tommy1729
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)) => ??????

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).
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)) => ??????

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.

???
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.
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.
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) ??
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.
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 $b>e^{1/e}$.
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