(07/18/2010, 10:41 PM)tommy1729 Wrote: prove convergence.

that might sound weird , but my formula uses a limit.

that limit hasnt been proven to converge , it could be a double limit or chaotic or ...

in fact , it is known that exp[n](z) can = oo while exp[n](z+ complex infinitesimal) =/= oo for infinitely many complex z.

secondly

prove that it is continuous.

for similar reasons as above this is not yet proven.

of course on the real line , my formula is both converging and continu but we are considering the complex numbers.

third

of course , prove that it is complex differentiable.

this will probably require the proof of convergence and continuous and wont be provable without them ?

because of the logloglog ... part its taylor series radius must be 0 when expanded at 0.

if you wonder why i believe so , already the first log gives a radius of 0 ?

the idea is this : (truncated) tommysexp(z,x) = ... log log log ( large )

since log has a small radius the large values wont 'fit in' and thus log (large) will have radius 0.

the other logs dont change that : log log ( function with radius 0 ) = function with radius 0.

if im correct ...

im not correct about the last , the log does have a nonzero radius of convergence at least up to the pole of its derivate , hence 1/x = oo -> at x = 0 ; in simpler words : when expanded at point t , the radius is at least abs (t) and thus the taylor series converges in the open set within the circle centered at t with radius abs(t) [ not sure about the convergence on the boundary ]

this generalises to the log iterations in my formula , when considering a finite amount of iterations.

well at least IF Re(x) > 0 and Re(z) > 0.

i think i have a proof that it holds for all the logs ( not just for finite iterations but also for infinite , and -for clarity - not log iterations 'alone' but the WHOLE FORMULA ) and hence my formula is holomorphic , and thus convergeance , continuity and complex differentiability is proven at once.

i will need some time to make the proofs formal or even to post them here , but as you might have already understood , the fast convergence is key.

- assuming the above - what remains to be understood is if my formula satisfies my own condition.

but thats in another thread ... ( called ' tommy's uniqueness conditions ' or similar )

also , numerical problems and acceleration methods need to be considered later on.

contour integration seems usefull and related for proving uniqueness.

i wonder if contour integration is the best way for finding the derivatives numerically.

regards

tommy1729