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Inverse Iteration
#1
Composition: f(x) o c = f( c )
Iterated composition (steinix-ankh): ☉f(x)☥^n_x = f(x)of(x)o...of(x) n times
Iterator-ankh operator: It aOb ☥ = ☉ aOb ☥^n-1_b |(b=a)|(n=b)
It a+b ☥ = ☉ a+b ☥^n-1_b |(b=a)|(n=b) = a+(a+(a+...a+b))|(b=a)|(n=b) = a*b
It a*b ☥ = a^b
It a^b ☥ = a^^b (tetration)
...etc.
It FoG ☥ = ☉ F ☥^G
Now, we know how to iterate operators. But I do not know how to iterate functions, or my question: What is the inverse method of iteration?
Xorter Unizo
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#2
> What is the inverse method of iteration?       
        
The closest thing I've found is counting iterations of the inverse. As an algorithm:     
Start with the outcome of your iteration of a group element as input. Count how often you can apply the corresponding inverse group element until you reach the neutral element. The counted number gives you the original number N of iterations.
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#3
(01/02/2019, 10:04 PM)MrFrety Wrote: > What is the inverse method of iteration?       
        
The closest thing I've found is counting iterations of the inverse. As an algorithm:     
Start with the outcome of your iteration of a group element as input. Count how often you can apply the corresponding inverse group element until you reach the neutral element. The counted number gives you the original number N of iterations.

Let me use another notation for iteration as iterative multiplication:
a ( O Z 0 ) b = a O b
a ( O Z 1 ) b = a O a O ... O a, b times which is the first iteration of operator O
a ( O Z n ) b = a ( O Z n-1 ) a ( O Z n-1 ) ... ( O Z n-1 ) a, b times
So
a ( + Z n-1 ) b = a [n] b
which is the well-known n-ation, like tetration (n=4).
It is binary operator, it needs an operator (e.g. addition, composition, logical and) and a number which is the level of the iteration. So it has two inverses: 1st is a trivial uniteration: ( O Z n ) Z -n = O
2nd is the iterative logarithm you talked about. It is a good method to get the inverse of the operator O, and apply it times such you do not get the neutral element... but what if you fall through the neutral element like how at function gamma. Let us define gamma as a binary operator: gamma_b(b,x) = int from 0 to +oo b^t/t^x dt. As I see this operator is between the exponentiation and tetration because of the limits of these quationts.
If I sign iterative logarithm as the following:
Zlog(O) - Zlog(P) = n, so that P Z n = O,
then what is Zlog(gamma_b)?
Xorter Unizo
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#4
(02/04/2019, 10:27 PM)Xorter Wrote:
(01/02/2019, 10:04 PM)MrFrety Wrote: > What is the inverse method of iteration?       
        
The closest thing I've found is counting iterations of the inverse. As an algorithm:     
Start with the outcome of your iteration of a group element as input. Count how often you can apply the corresponding inverse group element until you reach the neutral element. The counted number gives you the original number N of iterations.

Let me use another notation for iteration as iterative multiplication:
a ( O Z 0 ) b = a O b
a ( O Z 1 ) b = a O a O ... O a, b times which is the first iteration of operator O
a ( O Z n ) b = a ( O Z n-1 ) a ( O Z n-1 ) ... ( O Z n-1 ) a, b times
So
a ( + Z n-1 ) b = a [n] b
which is the well-known n-ation, like tetration (n=4).
It is binary operator, it needs an operator (e.g. addition, composition, logical and) and a number which is the level of the iteration. So it has two inverses: 1st is a trivial uniteration: ( O Z n ) Z -n = O
2nd is the iterative logarithm you talked about. It is a good method to get the inverse of the operator O, and apply it times such you do not get the neutral element... but what if you fall through the neutral element like how at function gamma. Let us define gamma as a binary operator: gamma_b(b,x) = int from 0 to +oo b^t/t^x dt. As I see this operator is between the exponentiation and tetration because of the limits of these quationts.
If I sign iterative logarithm as the following:
Zlog(O) - Zlog(P) = n, so that P Z n = O,
then what is Zlog(gamma_b)?
I'm not sure whether I follow completely but my first intuition (after older thought) would be that we just have to accept it as a new irreducable number like log(2), 1/2 or 15/4 ... There you also fall through the neutral element... 15-4-4-4-4=-1 ...
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