[MSE] f(f(z)) = z , f(exp(z)) = exp(f(z))

[tommy1729]

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

[tommy1729]

The gaussian method idea variant method :

see : https://math.eretrandre.org/tetrationfor...p?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

[tommy1729]

The gaussian method idea variant method :

see : https://math.eretrandre.org/tetrationfor...p?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

[tommy1729]

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