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+--- Thread: non-natural operation ranks (/showthread.php?tid=157)
non-natural operation ranks - bo198214 - 05/02/2008
Some basic thoughts
If we want to summarize our operation sequence [n] we can perhaps write:
\( b[n]x=f_{n,b}(x) \), where
\( f_{1,b}(x) = b+x \)
\( f_{n+1,b} = t\to f_{n,b}^{\circ t}(1) \), \( n\ge 1 \)
So we have a certain operator \( E \) that assigns to a given function \( f \) the function \( E(f)=t\to f^{\circ t}(1) \), this operator may be based on the natural Abel method, the diagonalization method, or the regular Abel method (with restrictions of \( b \)) and we can write
\( f_{n+1,b} = E(f_{n,b}) \), moreover
\( f_{n+1,b} = E^{\circ n}(f_{1,b}) = E^{\circ n}(x\mapsto b+x) \)
\( b[n+1]x = E^{\circ n}(f_{1,b})(x) \).
Note: I original found it more appropriate to start with the addition as 0th instead of the 1st operation. However I adapted to the already established nomenclature. If we would stick to my original counting (addition as 0th operation) we had the better looking formula \( b[n]x=E^{\circ n}(f_{0,b})(x) \)
As we have some methods for real functions to switch from \( f^{\circ n} \) to non-natural iterations \( f^{\circ t} \) maybe there are also methods for the operator \( E \) to compute \( E^{\circ t} \) which then gives definition for real and complex iteration ranks:
\( b[t]x=E^{\circ t-1}(f_{1,b})(x) \).
In the moment however it is even unclear how to generally express such operators which map powerseries.
At least we can determine that \( E^{-1} \) is the inverse operator, i.e. \( E^{-1}(g) \) is a function \( f \) such that \( g(x+1)=f(g(x)) \), i.e. \( E^{-1}(g)(x)=g(g^{-1}(x)+1) \). So \( E^{-1} \) is unique (independent of the method of \( E \) and independent on the initial condition).
If we compute \( f_{0,b} \) from this view point we get also:
\( f_{0,b}(x)=E^{-1}(f_{1,b})(x)=f_{1,b}(f_{1,b}^{-1}(x)+1)=b+(x-b+1)=x+1 \) and
\( f_{-1,b}(x)=E^{-1}(f_{0,b})(x)=f_{0,b}(f_{0,b}^{-1}(x)+1)=(x-1+1)+1=x+1 \)
...
RE: non-natural operation ranks - Ivars - 05/25/2008
bo198214 Wrote:Some basic thoughts
If we want to summarize our operation sequence [n] we can perhaps write:
\( b[n]x=f_{n,b}(x) \), where
\( f_{1,b}(x) = b+x \)
\( f_{n+1,b} = t\to f_{n,b}^{\circ t}(1) \), \( n\ge 1 \)
I was thinking of a function like:
\( b[t]z=F(b,t,z) \) so that for every combination of 3 at least complex variables we get one as a result. The type of number that will be the result is not obvious - there has been a discussion that new number types may appear.
RE: non-natural operation ranks - bo198214 - 05/25/2008
Ivars Wrote:I was thinking of a function like:
\( b[t]z=F(b,t,z) \) so that for every combination of 3 at least complex variables we get one as a result.
Exactly, everybody thinks of such a function. Thatswhy I showed a direction how one can obtain such a function, i.e. by:
\( b[t]x=E^{\circ t-1}(f_{1,b})(x) \).
where \( E \) is not a function, but an operator that maps functions to functions. The idea is to apply non-natural iteration to such an operator in a similar way as we apply non-natural iteration to a function. If we can compute non-natural iterates of functions via matrices, perhaps we can compute non-natural iterates of operators by the by Andrew mentioned tensors, who knows.
RE: non-natural operation ranks - andydude - 05/27/2008
bo198214 Wrote:\( b[n+1]x = E^{\circ n}(f_{1,b})(x) \).
I must say, this is so much more beautiful than my "hyper-exponential" section of the most recent Reference (I mistakenly called "FAQ"). However, it seems that this suffers from similar vagueness of the "iterative logarithm" in that it has both a function-parameter and a value-parameter. Can regular iteration even be performed on this?
Andrew Robbins