07/18/2010, 10:41 PM

ok , some remarks about my method.

( i wont type my formula here again , go look in this thread ! )

lets work in base e for convenience.

if the x in my formula is set to the fixpoints L or L* we have

tommysexp(z,x) = tommysexp(z,L) = L

tommysexp(z,x) = tommysexp(z,L*) = L*

as is needed. (easy proof btw)

i wonder what the value - if converging ! - of 2sinh^[+ oo i](x) is.

in the simplest case , it converges to L.

so the fixpoints probably cause no problems for my formula.

so what else needs to be done ?

well ,

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 ...

attacking my own ideas

but there is no other way , no way around these properties !

i have to be honest.

can anyone prove any of those 3 ( converge , continu , complex differentiable ) formally ?

has any of it been formally proven for other solutions ( base >= e ) apart from kneser ?

regards

tommy1729

( i wont type my formula here again , go look in this thread ! )

lets work in base e for convenience.

if the x in my formula is set to the fixpoints L or L* we have

tommysexp(z,x) = tommysexp(z,L) = L

tommysexp(z,x) = tommysexp(z,L*) = L*

as is needed. (easy proof btw)

i wonder what the value - if converging ! - of 2sinh^[+ oo i](x) is.

in the simplest case , it converges to L.

so the fixpoints probably cause no problems for my formula.

so what else needs to be done ?

well ,

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 ...

attacking my own ideas

but there is no other way , no way around these properties !

i have to be honest.

can anyone prove any of those 3 ( converge , continu , complex differentiable ) formally ?

has any of it been formally proven for other solutions ( base >= e ) apart from kneser ?

regards

tommy1729