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+--- Thread: [MSE] f(f(z)) = z , f(exp(z)) = exp(f(z)) (/showthread.php?tid=1726)
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RE: [MSE] f(f(z)) = z , f(exp(z)) = exp(f(z)) - tommy1729 - 04/03/2023
Ok not sure if this is valid.
Let F be the carleman matrix at some point p of f(x).
Let E be the carleman matrix at some point p of exp(x).
Let I be the carleman matrix at some point p of id(x).
Then
F^2 = I
F E = E F
We require that F and E are invertible ( pick p in a good way ).
Functional composition and matrix multiplication are both associative.
So F , E , I are commutative and associative with respect to products or functional compositions.
Therefore the log of these compositions or products is well defined ( remember F and E are invertable ).
log(F E) = log(E F) mod 2pi i
And as we know from matrix theory :
log(F E) = log(E F) => log(F E) = log(F) + log(E) mod 2pi i
We know
F^2 = I
log(F F E) = log(F E F) = log(E F F) = log(E) = 2*log(F) + log(E) mod 2pi i
so
2*log(F) + log(E) = log(E) mod 2pi i
Now lets diagonalize
E = A1 D1 B1
F = A2 D2 B2
Now A1 and B1 are inverses , so they have logs.
Same with A2 and B2.
naively one would say ( mod 2pi i )
2 * log(F) + log(E) = 2 log(A2) + 2 log(D2) -2 log(A2) + log(A1) + log(D1) - log(A1) = log(A1) - log(D1) - log(A1)
SO
A1 = A2
However here we do not have commutative and associative anymore.
Since E and F are invertible , the eigenvalues are nonzero.
so log(E), log(F) , log(D1) , log(D2) must all exist.
The correct way to continue is thus :
2 * log(F) + log(E) = 2 * (A1 * log(D1) * B1) + (A2 * log(D2) * B2) = (A2 * log(D2) * B2)
Thus
2 * log(F) + log(E) = (A1 * 2log(D1) * B1) + (A2 * log(D2) * B2) = (A2 * log(D2) * B2)
and it follows
(A1 * 2log(D1) * B1) = 0 mod 2 pi i
and
( left and right division by A1 and B1 (who are invertible) )
2log(D1) + A2/A1 * log(D2) * B2/B1 = A2/A1 * log(D2) * B2/B1
Now I am stuck, assuming that F is an iteration or power of E then
A2/A1 = B2/B1 = 1
Must hold.
It follows that
(A1 * 2log(D1) * B1) = 0 mod 2 pi i
and
( left and right division by A1 and B1 (who are invertible) )
2log(D1) = 0 mod 2 pi i
so the eigenvalues of F are all +/- 1.
This implies :
IF A2/A1 = B2/B1 = 1 is true , then F is ALMOST an iteration of E.
Because it may have similar diagionalization but the eigenvalues of E are not +/- 1 !!
And if A2/A1 = B2/B1 = 1 is NOT true then F is not an iteration of E at all !!
So :
F is not an iteration of E.
But the eigenvalues of F are all +/- 1.
But hold on, F^2 = 1 already implied that !
So we are left with
Conjecture
F exists and A2/A1 = B2/B1 = 1 is true.
OR equivalent
F exists and diagonalizes just like E.
***
But now somehow you guys say F does not exist.
What does that mean ?
The above contains a mistake ?
There is no point p ??
The radius is 0 ???
OR is the big problem the infinite matrix ?? Do we have no convergeance as the n*n carleman matrix grows ?
---
But then I got an aha moment or an anti-aha moment.
Involutions are diagonalizable over the reals ( and indeed any field of characteristic not 2 ), with +/-1 on the diagonal.
But diagonalization is not unique !
For starters diagonalization is not unique in general, but unique up to the permutation of the diagonal entries in its diagonal matrix ( the eigenvalues ) and multiples of columns of the matrix used for diagonalisation.
There are also other decompositions such as LDU.
But long story short :
I do not see why
Conjecture
F exists and diagonalizes just like E
Is problematic ??
AND I do not see what base change has to do with it.
*confused screaming* lol
regards
tommy1729
RE: [MSE] f(f(z)) = z , f(exp(z)) = exp(f(z)) - tommy1729 - 04/22/2023
The gaussian method idea variant method :
see : https://math.eretrandre.org/tetrationforum/showthread.php?tid=1339
R(s) = exp( tr(s) * R(s) )
and tr(s) strictly rises from 0 at s = -oo to 1 at s = + oo in a fast way.
The gaussian method had t(s) = ( 1 + erf(s) )/2.
R(s) uses tr(s) = ( 1 + erf(2sinh(c s)) )/2
For some real c that makes it analytic.
Now 2 sinh(c s) is periodic with period V.
So this V is isomorphic to a function fV(z) such that
fV(exp(x)) = exp(fV(x))
and fV(fV(x)) = x.
***
The interesting part is that this might relate to caleb ideas.
Maybe assuming the periodicity is continuation beyond a boundary, where boundary implies natural boundary ( not analytic ) , not converging or not satisfying the basis equations.
In that case one might get a " fake " solution.
( not to confuse with fake function theory )
regards
tommy1729
RE: [MSE] f(f(z)) = z , f(exp(z)) = exp(f(z)) - tommy1729 - 04/22/2023
(04/22/2023, 12:10 PM)tommy1729 Wrote: The gaussian method idea variant method :
see : https://math.eretrandre.org/tetrationforum/showthread.php?tid=1339
R(s) = exp( tr(s) * R(s) )
and tr(s) strictly rises from 0 at s = -oo to 1 at s = + oo in a fast way.
The gaussian method had t(s) = ( 1 + erf(s) )/2.
R(s) uses tr(s) = ( 1 + erf(2sinh(c s)) )/2
For some real c that makes it analytic.
Now 2 sinh(c s) is periodic with period V.
So this V is isomorphic to a function fV(z) such that
fV(exp(x)) = exp(fV(x))
and fV(fV(x)) = x.
***
The interesting part is that this might relate to caleb ideas.
Maybe assuming the periodicity is continuation beyond a boundary, where boundary implies natural boundary ( not analytic ) , not converging or not satisfying the basis equations.
In that case one might get a " fake " solution.
( not to confuse with fake function theory )
regards
tommy1729
If c = 1 works that is ;
not " fake "
and analytic.
then the 2sinh method might also give a valid solution if the 2sinh is analytic ( since we have the same period !)
the 2sinh method satisfies the semi-group iso and the other one might not ;
resulting in maybe 2 different valid solutions ...
But this would be equivalent to having 2 valid solutions for f(x).
Which bring the question :
apart from existance , do we have a proof of uniqueness for f(x) ??
Remember we only want a local solution for f(x) !!
regards
tommy1729
RE: [MSE] f(f(z)) = z , f(exp(z)) = exp(f(z)) - tommy1729 - 04/23/2023
Ok this is very sketchy, I ignore branches and validity and locality and complex analysis and such so take with a grain of salt.
f f = id
g h = h g = id
f = g( 1 - h(x) )
so
f(exp(x)) = exp(f(x))
g( 1 - h exp ) = exp g(1 - h)
g( 1 - h exp g ) = exp g(1 - x)
g( 1 - h exp g(1-x) ) = exp(g(x))
1 - x = x if x = 1/2
g( 1 - h(exp(g(1/2))) ) = exp(g(1/2))
exp(g(1/2)) = Q
g(1 - h(Q)) = Q
f(Q) = Q
so
f(exp(Q)) = exp(f(Q))
f(exp(Q)) = exp(Q)
so exp(Q) is another fixpoint for f ?
also
g(1 - h(Q)) = Q , g(h(Q)) = Q
so
1 - h(Q) = h(Q)
so h(Q) = 1/2 = h(exp(g(1/2))) = 1/2
but also
h(g(1/2)) = 1/2.
so maybe h(exp(x)) = h(x) ?
But then g( h(exp(x)) ) = g(h(x)) = x and = (gh) exp = exp(x)
contradiction.
I need a drink ...
regards
tommy1729