Every one here has heard of the ackermann function; we'll define it as:
\( \{x,\, y,\, u\, \epsilon\, R^{+}\} \)
\( A(x,\, y,\, u)\, =\, x\, \{u\}\, y \)
I'm interested in itss inverses though.
There would be three inverses therefore, the first one with relation to x:
\( R(x,\, y,\, u)\, =\, x\, \}u\{\, y \)
which means R for roots, and }u{ is taken to mean: }0{ is subtraction, }1{ division, }2{ roots, }3{ super roots, etc etc
and then are next inverse is taken with relation to y
\( L(x,\, y,\, u)\, =\, u;\log_x(y) \)
where 0;log_x(y) = y-x, 1;log_x(y) = y/x, and 2;log_x(y) = \( \log_x(y) \)
and now, the final inverse, and definitely the most interesting inverse is the inverse taken with relation to u
\( I(x, y, u) = ? \) if
\( A(x, y, I(x, y, u)) = u) \)
this question is the equivalent of asking
\( 3\,\{u\}\, 4\, =\, 9 \)
what is u?
it's the act of rearranging an equation to solve for u. It might even seem intuitively impossible.
I've only found one solution to this problem... and that's by using logarithmic semi operators, though I is only defined over domain [0, 2]
Does anyone know if anybody else has ever looked into this inverse function?
\( \{x,\, y,\, u\, \epsilon\, R^{+}\} \)
\( A(x,\, y,\, u)\, =\, x\, \{u\}\, y \)
I'm interested in itss inverses though.
There would be three inverses therefore, the first one with relation to x:
\( R(x,\, y,\, u)\, =\, x\, \}u\{\, y \)
which means R for roots, and }u{ is taken to mean: }0{ is subtraction, }1{ division, }2{ roots, }3{ super roots, etc etc
and then are next inverse is taken with relation to y
\( L(x,\, y,\, u)\, =\, u;\log_x(y) \)
where 0;log_x(y) = y-x, 1;log_x(y) = y/x, and 2;log_x(y) = \( \log_x(y) \)
and now, the final inverse, and definitely the most interesting inverse is the inverse taken with relation to u
\( I(x, y, u) = ? \) if
\( A(x, y, I(x, y, u)) = u) \)
this question is the equivalent of asking
\( 3\,\{u\}\, 4\, =\, 9 \)
what is u?
it's the act of rearranging an equation to solve for u. It might even seem intuitively impossible.
I've only found one solution to this problem... and that's by using logarithmic semi operators, though I is only defined over domain [0, 2]
Does anyone know if anybody else has ever looked into this inverse function?