01/13/2023, 08:45 PM
Ok I had some ideas.
And it feels or looks like it is alot like the ideas from Leo W.
Which I realized only much later.
Lets redefine most things with new symbols to avoid confusion.
we want
f(x+y) g(f(x)f(y)) = f(x) + f(y)
and some additional properties.
The idea was that those look like tanh.
Why ? well we want uniqueness :
f(x + y) =/= f(x + y + 1)
( f(x + y + z) gives complications when z being 0 or 1 gives the same. )
or so,
hence we want a function that is strictly increasing.
Strictly increasing from -1 to 1.
so
f := R -> [-1,-1]
and having a formal group law.
So
f(x) = Q(tanh(x))
Q(S(x)) = S(Q(x)) = id(x) = x
Q(-x) = - Q(x)
Then we "define the addition formula" for Q(tanh(x)) ;
Q(tanh(arctanh(S(x)) + arctanh(S(y)))
Now we let
Q(tanh(arctanh(S(x)) + arctanh(S(y))) = H(S(x),S(y))
Q(tanh(arctanh(S(x)) + arctanh(S(y))) = (x + y)/(1 + J(x,y))
substitute x,y with Q(x),Q(y) :
Q(tanh(arctanh(x) + arctanh(y))) = (Q(x) + Q(y))/( 1 + K(Q(x),Q(y)) )
Q(tanh(arctanh(x) + arctanh(y))) = (Q(x) + Q(y))/( 1 + L(Q(x)*Q(y)) )
We can simplify further :
Q( (x+y)/(1 + xy) ) = (Q(x) + Q(y))/( 1 + L(Q(x)*Q(y)) )
This is a very nice equation and it puts strong restrictions on Q.
This Q must be a solution to the ( more general ) equation for (a usually non-unique ) W ; what is just the above but setting x = y ;
W( 2x/(1 + x^2) ) = 2 W(x)/(1 + L(W(x)^2) )
Now define U(x) := 2 x / ( 1 + L(x^2) )
Then we arrive at
W( 2x/(1 + x^2) ) = U(W(x))
which looks like a typical equation from continu iterations !
This relates to complex dynamics !
And the branches for this type of equation is what Leo talked about when generalizing the abel equation.
2x/(1 + x^2) has the fixpoints -1,0,1.
So we can use many methods.
The methods with 2 fixpoints also work by symmetry of f(-x) = - f(x).
Hence the 3rd fixpoint is not a problem.
So our main focus to solving
W( 2x/(1 + x^2) ) = U(W(x))
are the iterations of U ;
U^[z](x).
So we want
P(U(x)) = P(x)+1.
I conjecture that if this P satisfies the semi-group iso , and there is indeed a formal group law for the given Q , then it is exactly computable by THAT solution P as shown above ( up to branches and speaking locally ).
regards
tommy1729
And it feels or looks like it is alot like the ideas from Leo W.
Which I realized only much later.
Lets redefine most things with new symbols to avoid confusion.
we want
f(x+y) g(f(x)f(y)) = f(x) + f(y)
and some additional properties.
The idea was that those look like tanh.
Why ? well we want uniqueness :
f(x + y) =/= f(x + y + 1)
( f(x + y + z) gives complications when z being 0 or 1 gives the same. )
or so,
hence we want a function that is strictly increasing.
Strictly increasing from -1 to 1.
so
f := R -> [-1,-1]
and having a formal group law.
So
f(x) = Q(tanh(x))
Q(S(x)) = S(Q(x)) = id(x) = x
Q(-x) = - Q(x)
Then we "define the addition formula" for Q(tanh(x)) ;
Q(tanh(arctanh(S(x)) + arctanh(S(y)))
Now we let
Q(tanh(arctanh(S(x)) + arctanh(S(y))) = H(S(x),S(y))
Q(tanh(arctanh(S(x)) + arctanh(S(y))) = (x + y)/(1 + J(x,y))
substitute x,y with Q(x),Q(y) :
Q(tanh(arctanh(x) + arctanh(y))) = (Q(x) + Q(y))/( 1 + K(Q(x),Q(y)) )
Q(tanh(arctanh(x) + arctanh(y))) = (Q(x) + Q(y))/( 1 + L(Q(x)*Q(y)) )
We can simplify further :
Q( (x+y)/(1 + xy) ) = (Q(x) + Q(y))/( 1 + L(Q(x)*Q(y)) )
This is a very nice equation and it puts strong restrictions on Q.
This Q must be a solution to the ( more general ) equation for (a usually non-unique ) W ; what is just the above but setting x = y ;
W( 2x/(1 + x^2) ) = 2 W(x)/(1 + L(W(x)^2) )
Now define U(x) := 2 x / ( 1 + L(x^2) )
Then we arrive at
W( 2x/(1 + x^2) ) = U(W(x))
which looks like a typical equation from continu iterations !
This relates to complex dynamics !
And the branches for this type of equation is what Leo talked about when generalizing the abel equation.
2x/(1 + x^2) has the fixpoints -1,0,1.
So we can use many methods.
The methods with 2 fixpoints also work by symmetry of f(-x) = - f(x).
Hence the 3rd fixpoint is not a problem.
So our main focus to solving
W( 2x/(1 + x^2) ) = U(W(x))
are the iterations of U ;
U^[z](x).
So we want
P(U(x)) = P(x)+1.
I conjecture that if this P satisfies the semi-group iso , and there is indeed a formal group law for the given Q , then it is exactly computable by THAT solution P as shown above ( up to branches and speaking locally ).
regards
tommy1729
