11/14/2008, 01:20 AM
(This post was last modified: 11/14/2008, 01:22 AM by Kouznetsov.)
Hello. This topic is related to
http://math.eretrandre.org/tetrationforu...hp?tid=208
In that branch, I explained why the lemma (when we prove it) about "alomost identical finction"=z+f(z))
leads to the uniqueness of the holomorphic tetration,
assuming, that tetration has no singularities at>-2)
.
"almost udentical" means, that function
is 1-periodic and "small enough" so that >0 \forall x\in \mathbb{R})
Here, I present the weak version of that theorem. This weak version can be used to prove the uniqueness of tetration, assuming that this tetration is holomorphic outside the real axis. It can be applied to the case of base)
, considered in
http://www.ils.uec.ac.jp/~dima/PAPERS/2008analuxp99.pdf ;
in this case, all the singulatities of tetration are at the real axis.
My today's proof does not match exactly all the criteria suggested in
http://math.eretrandre.org/tetrationforu...hp?tid=208
(that range allows tetration to have singulatities at<-2 )
).
To avoid confusions, I open the new branch for this.
I call it "weak theorem", because it cannot be applied as is to the tetation at base
)
, reserving the name "strong theorem" for the more general case that will work for all 
.
I begin with so long introduction in order to make this piece readable without background about previous posts.
Weak theorem about almost identical functions.
Let
is entire 1-periodic function
\ne 0 )
=0 )
=h(z)^* )
 = z+h(z) ~ \forall z\in \mathbb{C} )
>0~\forall x\in \mathbb{R} )
Then
 \in M )
Vulgarization: if function is somehow "small", then, function 
looks as "almost identical". The almost identical function must have negative integer values in some points outside the real axis.
Proof.
Let\in M \} )
Hypothesis 0: For each integer
,
there exists only one=m)
.
Below, I make some deduction based on Hypothesis 0 and show that this hypothesis is not consistent with Theorem 0.
From Hypothesis (0) it follows that
=\frac{z-m}{J(z)-m})
is entire function of 
.
=\frac{z-m}{z-h(z)-m}=<br />
\frac{z-m}{z-m-h(z-m)})
Define=\frac{z}{z+h(z)})
Then=v(z-m))
Then=V_m(z+m))
Theredore
is also entire function.
Function grows up to infinity along some contours )
at
. Let contour 
provides the fastest growth, id est, realizes the maximal
\big)\right|}<br />
{\left|\mathrm{d}\zeta(t)\right|}=<br />
\frac{<br />
\frac{\mathrm{d}}<br />
{\mathrm{d} t} \left| v \big(\zeta(t) \big) \right|<br />
}<br />
{ \left|\zeta^{\prime}(t) \right| } )
While, the combined function ))
should approach :
\big)-\zeta(t)\Big)=0)
The limit )
is not allowed to exist because
)
would be singularity of function .
Hence, grows to infinity.
Hence,)
should have asymptotically linear growth at infinity.
Among entire functions, the only polynomials are allowed to have polynomial growth at infinity; all other entire funcitons grow faster. Hence, is linear function.
Then,=\frac{z}{v(z)-z})
is rational function.
This property contradicts the conditions of the Theorem, because is non-trivial periodic function.
In such a qay, Hypothesis 0 contradicts conditions of the Theorem.
Hence, there exist more than one point such that \in M)
.
Due to the monotonous growth of function [math]J[/math] along the real axis, such a point is outside of the real axis.
(end of proof).
The theorem above provides the uniqueness of tetration at least for base
)
; in this case, all the singulatities of tetration are along the real axis. It colleagues agree with the proof of the theorem abovr, we can copypast here the condinuation, id est, the proof of uniqueness of tetration for this case.
P.S. Although Henryk suggests word "superexponentiation" instead of "tetration", I keep here "tetration" because I hope to make also unique holomorphic pentation in the similar way; in the case of superexponentiation, the pentation would have to be called supersuperexponentiation which is too long.
http://math.eretrandre.org/tetrationforu...hp?tid=208
In that branch, I explained why the lemma (when we prove it) about "alomost identical finction"
assuming, that tetration has no singularities at
"almost udentical" means, that function
Here, I present the weak version of that theorem. This weak version can be used to prove the uniqueness of tetration, assuming that this tetration is holomorphic outside the real axis. It can be applied to the case of base
http://www.ils.uec.ac.jp/~dima/PAPERS/2008analuxp99.pdf ;
in this case, all the singulatities of tetration are at the real axis.
My today's proof does not match exactly all the criteria suggested in
http://math.eretrandre.org/tetrationforu...hp?tid=208
(that range allows tetration to have singulatities at
To avoid confusions, I open the new branch for this.
I call it "weak theorem", because it cannot be applied as is to the tetation at base
I begin with so long introduction in order to make this piece readable without background about previous posts.
Weak theorem about almost identical functions.
Let
Then
Vulgarization: if function
looks as "almost identical". The almost identical function
Proof.
Let
Hypothesis 0: For each integer
there exists only one
Below, I make some deduction based on Hypothesis 0 and show that this hypothesis is not consistent with Theorem 0.
From Hypothesis (0) it follows that
Define
Then
Then
Theredore
Function
While
The limit
Hence,
Hence,
Among entire functions, the only polynomials are allowed to have polynomial growth at infinity; all other entire funcitons grow faster. Hence,
Then,
This property contradicts the conditions of the Theorem, because
In such a qay, Hypothesis 0 contradicts conditions of the Theorem.
Hence, there exist more than one point
Due to the monotonous growth of function [math]J[/math] along the real axis, such a point is outside of the real axis.
(end of proof).
The theorem above provides the uniqueness of tetration at least for base
P.S. Although Henryk suggests word "superexponentiation" instead of "tetration", I keep here "tetration" because I hope to make also unique holomorphic pentation in the similar way; in the case of superexponentiation, the pentation would have to be called supersuperexponentiation which is too long.
