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