08/20/2016, 01:19 PM
There are two really important operators: + (addition) and o (composition)
If we iterate +, we get Hyper-operators, like this:
Hyper(A;n;B) = A[n-1]B
E. g.:
A[0]B = A+B
A[1]B = A*B
A[2]B = A^B
A[3]B = A^^B (tetration)
...
But if we iterate o, we get Hot-operators, like this
Hot(F;n;G)
E. g.:
Hot(f;1;c) = f(x) o c = f( c)
Hot(f;2;n) = ☉ f(x) ☥^n = f(x) o f(x) o ... o f(x) (steinix-ankh operator)
Hot(f;3;n) = ☉ f(x) ☥^(☉ f(x) ☥^...) (super-steinix-ankh operator)
...
For example:
Hot(2+x;1;2) = 4
Hot(2+x;2;2) = 4+x
Hot(2+x;3;2) = 4+3x
Hot(2+x;4;2) = 4*2^x+(4*2^x - 1)*x
The question is what Hot(f;n;g) gives if n is real or complex?
If we iterate +, we get Hyper-operators, like this:
Hyper(A;n;B) = A[n-1]B
E. g.:
A[0]B = A+B
A[1]B = A*B
A[2]B = A^B
A[3]B = A^^B (tetration)
...
But if we iterate o, we get Hot-operators, like this
Hot(F;n;G)
E. g.:
Hot(f;1;c) = f(x) o c = f( c)
Hot(f;2;n) = ☉ f(x) ☥^n = f(x) o f(x) o ... o f(x) (steinix-ankh operator)
Hot(f;3;n) = ☉ f(x) ☥^(☉ f(x) ☥^...) (super-steinix-ankh operator)
...
For example:
Hot(2+x;1;2) = 4
Hot(2+x;2;2) = 4+x
Hot(2+x;3;2) = 4+3x
Hot(2+x;4;2) = 4*2^x+(4*2^x - 1)*x
The question is what Hot(f;n;g) gives if n is real or complex?
Xorter Unizo