08/11/2010, 04:10 PM
there is a simple equation to find a real to real tetration.
( [0,oo] to [0,oo] , no negative reals of course )
it works for all solutions.
lets call the solution 'kneser'.
and set kneser(0) = 0 , sexp(0) = 0.
real(z) = x
im(kneser(z)) = 0
sexp(x) = kneser(z)
thats it !
however finding the solution is something else.
i assume that by riemann mapping theorem that 'this sexp' is coo if ' this kneser ' is.
finding the parametric real and im part of ' kneser ' might be as complicated or even equivalent to riemann mapping.
and i assume the coo prop of the parametric real and im part is equivalent to the coo prop of ' kneser ' or ' sexp '.
looking at it this way , its trivial to see that riemann mapping , fourier series and the 'wave' are strongly related , not to say equivalent.
the question is if the equations
set kneser(0) = 0 , sexp(0) = 0.
real(z) = x
im(kneser(z)) = 0
sexp(x) = kneser(z)
lead to new insights or alternative methods.
i propose that the curve of the 'knesers' ( the parametric wave of the real outputs of all known non R->R solutions ) get a special function name or special command name ( for general iterations ).
( command could be 'knesercurve(f(x))' and in the general case its very likely that the curve cannot be computed if the superfunction cant , but who knows ! )
i wanted to note that the 'wave' does have some limitations and does not have total freedom as sometimes believed :
it must be an analytic wave and it must be a bounded wave that converges on C ( luckily for four series ! and thus not double periodic or finite radius )
but thats not all.
since we have a smooth curve the equations presented might be solvable without riemann mapping !
since we have a smooth curve we might use newton iteration.
im convinced that an iterative system exists to solve such a problem that is hardly more complicated than computing a schroeder function.
that is however a numerical method not suited for closed form ( unless the algoritm itself ( limit ) but prob no sum or integral ) or properties or proofs , unless by the squeezing theorem and comparing different tetration solutions...
i guess we are pretty close to reunderstanding knesers solution and being able to compute it ( rather than just the non-constructive existance ).
i know im not the only one doing research on it ( e.g. sheldonison ) , but i didnt want this to be left behind.
and maybe i helped some people who didnt understand kneser or sheldon and do understand now.
but im not finished...
about that wave again
because of kneser(z+1) = exp(kneser(z))
it is clear that the kneser curve has period 1.
kneser(0) = 0 and kneser(1) = 1.
if we draw a straith line trough them , then i assume the curve must be above or below the line and never cross it.
is that true ?
regards
tommy1729
( [0,oo] to [0,oo] , no negative reals of course )
it works for all solutions.
lets call the solution 'kneser'.
and set kneser(0) = 0 , sexp(0) = 0.
real(z) = x
im(kneser(z)) = 0
sexp(x) = kneser(z)
thats it !
however finding the solution is something else.
i assume that by riemann mapping theorem that 'this sexp' is coo if ' this kneser ' is.
finding the parametric real and im part of ' kneser ' might be as complicated or even equivalent to riemann mapping.
and i assume the coo prop of the parametric real and im part is equivalent to the coo prop of ' kneser ' or ' sexp '.
looking at it this way , its trivial to see that riemann mapping , fourier series and the 'wave' are strongly related , not to say equivalent.
the question is if the equations
set kneser(0) = 0 , sexp(0) = 0.
real(z) = x
im(kneser(z)) = 0
sexp(x) = kneser(z)
lead to new insights or alternative methods.
i propose that the curve of the 'knesers' ( the parametric wave of the real outputs of all known non R->R solutions ) get a special function name or special command name ( for general iterations ).
( command could be 'knesercurve(f(x))' and in the general case its very likely that the curve cannot be computed if the superfunction cant , but who knows ! )
i wanted to note that the 'wave' does have some limitations and does not have total freedom as sometimes believed :
it must be an analytic wave and it must be a bounded wave that converges on C ( luckily for four series ! and thus not double periodic or finite radius )
but thats not all.
since we have a smooth curve the equations presented might be solvable without riemann mapping !
since we have a smooth curve we might use newton iteration.
im convinced that an iterative system exists to solve such a problem that is hardly more complicated than computing a schroeder function.
that is however a numerical method not suited for closed form ( unless the algoritm itself ( limit ) but prob no sum or integral ) or properties or proofs , unless by the squeezing theorem and comparing different tetration solutions...
i guess we are pretty close to reunderstanding knesers solution and being able to compute it ( rather than just the non-constructive existance ).
i know im not the only one doing research on it ( e.g. sheldonison ) , but i didnt want this to be left behind.
and maybe i helped some people who didnt understand kneser or sheldon and do understand now.
but im not finished...
about that wave again
because of kneser(z+1) = exp(kneser(z))
it is clear that the kneser curve has period 1.
kneser(0) = 0 and kneser(1) = 1.
if we draw a straith line trough them , then i assume the curve must be above or below the line and never cross it.
is that true ?
regards
tommy1729