03/03/2021, 12:57 AM

Thinking about my 2sinh method, I came up with an intuitive iteration that might be interesting.

let z0 = a0 + b0 i with e < a0 , 0 =< b0 < 1.

Now define exp*(z) as abs( exp(re(z))*cos(im(z)) ) + abs( exp(re(z))*sin(im(z)) ) i.

notice exp*(positive real) = exp(positive real).

Now define log* as the appropriate log branches.

Let z_n = z_n(z0) = exp*^[n](z0).

Now consider for various functions F ;

log*^[n](F(exp*^[n](z0)))

Looks alot like the 2sinh method and many others.

BUT PERHAPS BETTER BEHAVED ??

Also how fast does this sequence z_n grow ?

Is log*^[n](F(exp*^[n](z0))) analytic ? equal to the 2sinh method for F an iteration of 2sinh ??

The idea is to avoid the need for analytic continuations and just keep iterating without issues.

Is log*^[n](2sinh(exp*^[n](z0))) = exp(z0) , exp*(z0) or conjugate exp(z0) ??

Another old sketchy idea but worth reconsidering imo

regards

tommy1729

let z0 = a0 + b0 i with e < a0 , 0 =< b0 < 1.

Now define exp*(z) as abs( exp(re(z))*cos(im(z)) ) + abs( exp(re(z))*sin(im(z)) ) i.

notice exp*(positive real) = exp(positive real).

Now define log* as the appropriate log branches.

Let z_n = z_n(z0) = exp*^[n](z0).

Now consider for various functions F ;

log*^[n](F(exp*^[n](z0)))

Looks alot like the 2sinh method and many others.

BUT PERHAPS BETTER BEHAVED ??

Also how fast does this sequence z_n grow ?

Is log*^[n](F(exp*^[n](z0))) analytic ? equal to the 2sinh method for F an iteration of 2sinh ??

The idea is to avoid the need for analytic continuations and just keep iterating without issues.

Is log*^[n](2sinh(exp*^[n](z0))) = exp(z0) , exp*(z0) or conjugate exp(z0) ??

Another old sketchy idea but worth reconsidering imo

regards

tommy1729