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 A(z) = exp( A(z-1) - exp(-z) ) ? tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 03/03/2021, 12:26 AM Consider purely iterative methods not based on fixpoints. Most of them work with asymptotics to exp(z). For instance f(z) = exp(f(z-1) + z) as in the recent phi method. But that is not such a good asymptotic. One of my main ideas(conjecture!) is that asymptotics should be good (in order to try and get analytic solutions). This makes sense once you consider the neccessary conditions for a sequence of functions to converge to an analytic one. g(z) = exp(g(z-1)) + z is a better asymptotic. But still very far from good. h(z) = 2sinh(h(z-1)) is a much better one. ( hence my 2sinh method ) Notice an approximation to h is h* h*(z) = exp( h*(z-1) - 1/exp(2 h*(z-1)) ) because ln(2sinh(x)) - x = ln(exp(x)*(1 - 1/exp(2x))) - x = ln(1 - 1/exp(2x)). clearly an even better one is : t(z) = exp( t(z-1) - 1/exp(5 * t(z-1)) ) I would like to point out that the convergeance of lim t -> oo : ln ln ln ln (t times) ... A(z + t) is faster if A(z+1) is a better asymtotic to exp(A(z)),  again ; This makes sense once you consider the neccessary conditions for a sequence of functions to converge to an analytic one. Also we want convergeance to be fast IN A LARGE SUBSET OF THE COMPLEX PLANE , not just the reals. It is conjectured that picking the right asymptotic will achieve this analytic solutions , well trivially I guess because else a purely iterative method might not work ... ( In one variable that is !). One of the problems we first encounter is that t(z) = exp( t(z-1) - 1/exp(5 * t(z-1)) ) does not trivially get an iteration of function compositions like in the recent phi method, because there is too much selfreference. One way out is searching for the analogue sinh method that satisfies that .. similar to how h* appproximates h and vice versa. And then generate the super of that analogue sinh. However I guess that is not easier than the sinh method , is harder to compute or prove things about ? ANOTHER SOLUTION is by noticing t(z) must grow a bit like phi(z) and then we arrive at another approximation t* :  t*(z) = exp( t*(z-1) - 1/(phi(z))^5 ). Im not sure how well this works out on the complex plane. But it clearly works for the reals in a C^oo way ! What worries me is when phi(z+1) = 0 for nonreal z. so I propose : T(z) = exp( T(z-1) - exp(-5*phi(z-1)) ). On the other hand if phi(z-1) is close to i or r i for large r, this might still be a problem. So again I think about another solution. phi grows nice but on the complex plane less nice. A(z) = exp( A(z-1) - exp(-z) ) B(z) = exp( B(z-1) - exp(-5*A(z-1)) ) That might work. Or it might have similar problems as phi ... ( afterall A and B are also 2 pi i periodic and the behavior is complex ) What do you guys think ? Regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 03/03/2021, 01:26 PM So for instance  A(z) = exp( A(z-1) - exp(-z) ) We use A(z) = exp( - exp(-z) + exp( - exp(-z+1) + exp( - exp(-z+2) +  ...  We define the functional inverse of A(z) := C(z). Now we can go  James/sheldon method : for lim ln(ln(...* t times * A(z+t)...) similar to the phi method or - as i prefer -  tommy method : tet(v + slog(z) + p) = lim ln^[t]( A( C(exp^[t](z)) +v) ) for some constant p and 0

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