non-natural operation ranks bo198214 Administrator Posts: 1,616 Threads: 102 Joined: Aug 2007 05/02/2008, 10:11 PM (This post was last modified: 05/02/2008, 10:32 PM by bo198214.) 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 $f_{-1,b}(x)=E^{-1}(f_{0,b})(x)=f_{0,b}(f_{0,b}^{-1}(x)+1)=(x-1+1)+1=x+1$ ... Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 05/25/2008, 05:31 PM 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. bo198214 Administrator Posts: 1,616 Threads: 102 Joined: Aug 2007 05/25/2008, 07:30 PM 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. andydude Long Time Fellow Posts: 510 Threads: 44 Joined: Aug 2007 05/27/2008, 06:31 PM 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 « Next Oldest | Next Newest »

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