In the imho very intresting post :
http://math.eretrandre.org/tetrationforu...hp?tid=499
I had the idea to " fix the problem ".
And by that I mean to avoid the knots.
How I intend to do this is by adding variables or dimensions if you like. Hence the title.
Let \( 1 , x , x_2 , x_3 , x_4 , x_5 , x_6 , x_7 \) be the cyclic group C8.
Now construct the group ring by extending with the real \( R \).
( example of an element is \( 2,5 - 3x + 5x_2 \) see http://en.wikipedia.org/wiki/Integral_group_ring)
Now we define a "complex absolute value" operator by \( C \).
\( C(a + b x + c x_2 + ... ) = a + b 1^{1/8} + c 1^{2/8} + \)...
Lets call this group ring T8.
Notice all functions on T8 are defined and computable apart from division by zerodivisors.
Now let our starting value be z.
Let f(z) map z to an element in T8 such that \( C \)(f(z)) = z.
There are many such f(z) and that is crucial here.
Also C is far from a bijection , that is also crucial here.
Let f*(z) be a suitable f(z) , where suitable is explained below.
Let q be a positive real. Let r be a positive real.
Let Q be the complex value of the q th iteration of the exp of z that is near a knot.
Let Q2 be the complex value of the q+r th iteration of the exp of z that is near the same(!) knot , hence Q ~ Q2.
Let Q' be a T8 value of the q th iteration of the exp of f*(z).
Now \( C \)Q' = Q = Q2
Let Q'' be a T8 value of the q+r th iteraton of f*(z).
Now \( C \)Q'' = Q = Q2
However suitable f*(z) means Q' =/= Q'' , which also mean that there is no knot (anymore in T8 space) ... and we have an invertible function ( continu!).
this then solves the (original) problem for this particular knot.
That is the basic idea.
This is not the holy grail yet , but I think a promising idea.
we thus need to work with sexp and slog on T8 space.
and we need to find a way to work to find f*(z).
And then there is ofcourse the fact of multiple knots rather than one.
However statistically there exists a f*(z) that avoids all knots.
statistically because if we consider the knots as collisions of a trajectory than adding dimensions means probability of hits goes down by the cardinality of the reals !!
Plz do not confuse this with fixed points.
There are fixpoints in both C and T8 but that is not directly related to this post.
I wonder what you guys think.
Regards
Tommy1729
ps : sorry for my long absence.
pps : where is bo ?
http://math.eretrandre.org/tetrationforu...hp?tid=499
I had the idea to " fix the problem ".
And by that I mean to avoid the knots.
How I intend to do this is by adding variables or dimensions if you like. Hence the title.
Let \( 1 , x , x_2 , x_3 , x_4 , x_5 , x_6 , x_7 \) be the cyclic group C8.
Now construct the group ring by extending with the real \( R \).
( example of an element is \( 2,5 - 3x + 5x_2 \) see http://en.wikipedia.org/wiki/Integral_group_ring)
Now we define a "complex absolute value" operator by \( C \).
\( C(a + b x + c x_2 + ... ) = a + b 1^{1/8} + c 1^{2/8} + \)...
Lets call this group ring T8.
Notice all functions on T8 are defined and computable apart from division by zerodivisors.
Now let our starting value be z.
Let f(z) map z to an element in T8 such that \( C \)(f(z)) = z.
There are many such f(z) and that is crucial here.
Also C is far from a bijection , that is also crucial here.
Let f*(z) be a suitable f(z) , where suitable is explained below.
Let q be a positive real. Let r be a positive real.
Let Q be the complex value of the q th iteration of the exp of z that is near a knot.
Let Q2 be the complex value of the q+r th iteration of the exp of z that is near the same(!) knot , hence Q ~ Q2.
Let Q' be a T8 value of the q th iteration of the exp of f*(z).
Now \( C \)Q' = Q = Q2
Let Q'' be a T8 value of the q+r th iteraton of f*(z).
Now \( C \)Q'' = Q = Q2
However suitable f*(z) means Q' =/= Q'' , which also mean that there is no knot (anymore in T8 space) ... and we have an invertible function ( continu!).
this then solves the (original) problem for this particular knot.
That is the basic idea.
This is not the holy grail yet , but I think a promising idea.
we thus need to work with sexp and slog on T8 space.
and we need to find a way to work to find f*(z).
And then there is ofcourse the fact of multiple knots rather than one.
However statistically there exists a f*(z) that avoids all knots.
statistically because if we consider the knots as collisions of a trajectory than adding dimensions means probability of hits goes down by the cardinality of the reals !!
Plz do not confuse this with fixed points.
There are fixpoints in both C and T8 but that is not directly related to this post.
I wonder what you guys think.
Regards
Tommy1729
ps : sorry for my long absence.
pps : where is bo ?