06/04/2020, 02:08 PM
oexp(x' × olog(y(x))) = y^ox' =: super(y, 1)
oexp(oexp(x'' × x' × olog(olog(y)))) = super(y, 2)
oexp^op ((prod k=1 to p x[k]) × olog^op(y)) = super(y, p)
What if there is a osexp and a oslog operators for which:
osexp(f(x',oslog(y))) = super(y, x')
Let f(a,b) be a+b, so:
osexp(x' + oslog(y)) = super(y, x')
E. g.:
osexp(0 + oslog(y)) = super(y, 0) = y
So osexp is inverse of oslog and vica versa.
osexp(1 + oslog(y)) = oexp(x' × olog(y)) = super(y, 1)
osexp(-1 + oslog(y)) = y o (1+x) o y^o-1 = super(y, -1)
We know that [ oexp(y) ] = exp [y], [ olog(y) ] = log [y], but what is about [ osexp(y) ] and [ oslog(y) ]?
If I knew what these functions are, then I could determine Hyper(a, x, b) = a[x]b = osexp(x + oslog(a+x') o b) and it would be a really new thing in mathematics.
Ofc, I checked and [ osexp(y) ] cannot be sexp [y]...
oexp(oexp(x'' × x' × olog(olog(y)))) = super(y, 2)
oexp^op ((prod k=1 to p x[k]) × olog^op(y)) = super(y, p)
What if there is a osexp and a oslog operators for which:
osexp(f(x',oslog(y))) = super(y, x')
Let f(a,b) be a+b, so:
osexp(x' + oslog(y)) = super(y, x')
E. g.:
osexp(0 + oslog(y)) = super(y, 0) = y
So osexp is inverse of oslog and vica versa.
osexp(1 + oslog(y)) = oexp(x' × olog(y)) = super(y, 1)
osexp(-1 + oslog(y)) = y o (1+x) o y^o-1 = super(y, -1)
We know that [ oexp(y) ] = exp [y], [ olog(y) ] = log [y], but what is about [ osexp(y) ] and [ oslog(y) ]?
If I knew what these functions are, then I could determine Hyper(a, x, b) = a[x]b = osexp(x + oslog(a+x') o b) and it would be a really new thing in mathematics.
Ofc, I checked and [ osexp(y) ] cannot be sexp [y]...
Xorter Unizo