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tommy's uniqueness conditions

i will present 2 uniqueness conditions here.

however i dont know if the first uniqueness condition is actually truely unique .. the first is more like a conjecture.

also i dont know if the 2 uniqueness conditions are equivalent ...

i hope both are equivalent.

for clarity im talking about uniqueness of sexp resp slog for bases larger than sqrt(e).

i will not distinguish between sexp(0) = a or sexp(0) = b if they are equivalent apart from a lineair transform ( x + c ).

also i assume for both the conditions that they are analytic almost everywhere and have their fixpoints ( x = base^x ) at +/- oo i.

i partially already mentioned the first uniqueness condition before :

d f^n / d x^n sexp(slog(x) + k) > 0 for all positive integer n and all positive real k.

and probably ( i.e. if im not mistaken because of the local heat wave )this is true if and only if the following is true :

( i.e. i assume " equivalent to " )

d f^n / d k^n sexp(slog(x) + k) > 0 for all positive integer n and all positive real k.

i havent investigated this a lot yet ... basicly because of the intresting second uniqueness condition.

the second uniqueness condition could be a simple proof for the uniqueness of kouznetsov and thus be equivalent to kouznetsov's solution.

and at the same time shed some new insight on tetration and kouznetsov's solution.

the second uniqueness condition is this :

we search a uniqueness condition for sexp.

indirectly by finding a uniqueness condition for its inverse : slog.

in some sense this is the analogue of kouznetsov's uniqueness for sexp but for slog.

ok , the basic equation for slog(z) :

slog(base^z) = slog(z) + 1

lets take a certain base to simply matters : base e

slog(e^z) = slog(z) + 1

now clearly slog has period 2pi i.

now we introduce "another slog"

" another slog " = slog(x + v(x))

v(x) is meromorphic on C and has period 1 and v(0) = 0.

( we will later see that v(x) must be meromorphic rather than just entire )

so

slog(exp(x + v(x))) = slog(x + v(x)) + 1

letting y = x + v(x) we see that this is also a solution.

to explain further , why period 1 ?

well : sexp( x + v2(x) + 1 ) = exp ( sexp(x + v2(x)) )

it follows that with q = x + v2(x) we have another solution ;
" another sexp ".

but only if v2(x) has period 1. ( necc condition , not sufficient )

x + v(x) is the functional inverse of x + v2(x) and hence v(x) must have the period 1 as well.

but now the cruxial part occurs :

slog(exp(x + v(x))) = slog(x + v(x)) + 1

we now know that v(x) needs period 1 , but it also requires period 2pi i !

( again this can be shown by setting y = x + v(x) )

so any valid v(x) satisfies :

v(x) is complex differentiable.
v(0) = 0
v'(x) > 0
v(real) = real
v(x+2pi i) = v(x)
v(x+1) = v(x)

so v(x*) = v(x)* with fixpoint at 0 and v(x) is a non-constant meromorphic doubly periodic function.

such a function must have at least a pole and thus be meromorphic and not just entire BECAUSE :

( terminology : a prototype parallelogram (PP) is the vector span of its 2 periods i.e. with sides z , z + a , z + b , z + a + b where a and b are the periods )

A non-constant meromorphic doubly periodic function cannot be bounded on the prototype parallelogram (PP). For if it were it would be bounded everywhere, and therefore constant by Liouville's theorem.

also by picard's little , we know that it takes almost any value in such a parallelogram ( all but 1 actually ).

note that some values may be on the edge/sides though.

so we have almost all values of v(x) within (PP).

and thus slog(v(x)) contains almost all values of slog(x) in PP , even slog(oo) because of the pole in PP.

by pythagoras the diagonal length of PP is + sqrt ( 1 + (2pi)^2 ).

this means that in a circle with radius RT = + sqrt ( 1 + 4 pi^2 ) ,

x + v(x) takes on every value - no matter where the center of the circle is in the finite complex plane.

so if the elliptic function v(x) is not = 0 then

slog(x + v(x)) takes on every value of slog(x) in any circle with radius RT.

conversely , if F(x) = slog(x + v(x)) THUS ANY (complex differentiable) SOLUTION - only F(x) is given !! - such that F(x) does not take on every value of F(x) in a circle with radius RT that does not cross a branch point and F(x*) = F(x)* , then v(x) = 0 and hence F(x) is UNIQUE up to a linear substitution of x.

thus we have a uniqueness condition for slog(x).

since sexp is the unique inverse of slog up to branches , this means we also have a uniqueness condition for sexp(x).

--------------------------------
(note) that the one value not allowed for sexp(x + v2(x)) is a fixed point of base^x = x which kouznetsov calls 'L' and 'L*'.

if it does happen , the story is over before the end of my condition ; it cannot longer be Coo in the neighbourhood of the finite value x that maps sexp(x + v2(x)) to L or L*.

this might translate to another condition to slog , but i will leave in the sexp condition form for simplicity , since it is not in contradition to my slog condition.

basicly this is just a condition of Coo.
--------------------------------

as a sidenote , i would like to say that some other uniqueness conditions will follow from this one , so that ' incompleteness ' or ' incompatibility ' is not neccessarily true.

i wont go into details , since most of *those* conditions are not so intresting and havent occcured in this forum yet.

i know the presentation of this post could be better , but its really hot today and i have little time.

i guess you guys - as tetration experts - can follow the main ideas in it anyways.

i assume my solution to tetration for real bases b > sqrt(e)

$\operatorname{TommySexp_b}(z,x)= \lim_{n \to \infty } \ln_b^{[n]} (\2sinh_b^{[z]}(exp_b^{[n]}(x)))$

- or at least its analytic continuation - will satisfy at least one of these 2 uniqueness conditions.

it needs to be emphased that the domain and image of slog(x) needs to be investigated , since ' all values of slog(x) ' in some region is then better understood.

basicly because slog(x) is not entire , im not sure if we get almost all values ... hmmm ...

nevertheless , regardless of that outcome , the result above remains indepent of that.

thinking about it , i think we do get almost all values because slog(x) when defined for all complex x , it has singularities and/or pseudo fractal like structures + all exponents of them ; thus 'many values'.

regards

tommy1729
(07/10/2010, 10:15 PM)tommy1729 Wrote: [ -> ]now we introduce "another slog"

" another slog " = slog(x + v(x))

v(x) is meromorphic on C and has period 1 and v(0) = 0.

( we will later see that v(x) must be meromorphic rather than just entire )

so

slog(exp(x + v(x))) = slog(x + v(x)) + 1

letting y = x + v(x) we see that this is also a solution.

Dont know exactly how you arrive at that equation.
say slog2(x)=slog(x+v(x))
then slog2(exp(x))=slog(exp(x) + v(exp(x)))
I dont see why slog2 should be another slog.

sexp2(x)=sexp(x+v2(x)) that works, because:
sexp2(x+1)=sexp(x+1+v2(x+1))=sexp(x+v(x)+1)=exp(sexp(x+v(x))=exp(sexp2(x)).

For the slog one would choose
slog3(x)=slog(x)+v(slog(x))
to get another slog.

It seems you based your conclusions on a wrong assumption.
how about slog2(x) = x + v(x) [ slog(x) ] ...

maybe that works ... ill think about it.

(edit) ok i give up the second uniqueness ... and the first is very likely false too. sorry for those bad ideas.

at the edge of giving up , i was dreaming about (sexp(x) - sexp2(x))/(sexp(x)-sexp(x-1)) = v3(x) but im not even sure if v3(x) can be made periodic nor how to proceed from there ...

so that might be a dead end or a bad idea as well ...
(maybe start a new thread , after this 'mess' ? )

ok , another attempt and a related question.

forgive me for mistakes or ignorance , its still hot here ...

quick vague sketch of the idea :

sexp(x) = a0 + a1/1! x + a2/2! x^2 + a3/3! x^3 + ...

sexp2(x) = sexp(x + v(x)) with v(x) a periodic entire function satisfying v(0) = 0 and v*(x) = v(x*)

proposed uniqueness by :

a0 = 1 , a0 < a1 < a2 < a3 < a4 < ...

a1/a0 > the base , a2/a1 > the base , ...

(assuming existance ! )

the idea is now to show that for any v(x) =/= 0 ( but still with v(0) = 0 ) this is violates for the 'new' sexp2.

v(x) = 0 + v1 x + v2/2! x^2 + ...

now we use the so-called general leibniz rule to get to a system of inequalities :

a1*(1+v1) > 0 => v1 > -1

a2*(1+v1) + a1(v2) > 0

... ( by leibniz rule ) ...

now for large n , a_n < n!

now something important : note that a_n must grow fast.

and thus a_n probably becomes the dominant factor of the inequalities , more specific of the n'th inequality.

unless v_n becomes dominantly small !

since v(x) is periodic , there must be oo n such that v_n is negative.

lets call those v'_m. ( the m'th ) and let v'm = m_n corresponding to v_n.

now dominantly small means appromimately :

lim n -> oo m_n = around = 1/ a_n = little less than = 1/(n-1)!

but is there a v(x) , such that lim n-> oo m_n is around (n-1)! ??

i doubt it.

but such a v(x) is necc to break the uniqueness condition.

i doubt such a v(x) exists because if those negative v_n shrink so fast , then the series seems to converge to fast to be periodic.

for instance bessel functions resemble such v_n , but bessel is not periodic but is 'brutely' approximated by sin(x)/sqrt(x) ...

but of course in math we need formal proof.

this could be trivial even high skool , but right now i cant remember such a thing ...

i guess such a thing isnt hard to prove or disproof , but im having a bad day ...

but maybe a bad day with a good idea ...

notice that my uniqueness criterion reduces to a_n > 1 once we proved that no non-violating entire v(x) exists.

but i could not have started this post with a_n > 1 because i needed to explain the importance and growth rate of a_n and 1/v_n.

i couldnt help noticing that taylor series seems more adequate than fourier series for periodic functions in this context ( for v(x) ) ...

taylor fourier 1-0.

high regards

tommy1729