Tetration Forum - using sinh(x) ?

Pages: 1 2 3 4 5 6 7

(06/29/2010 03:18 AM)sheldonison Wrote: [ -> ]How is the fixed point removed? Starting with your equation for TommySexp, only interested in the "z" component, then let x=0,
$\operatorname{TommySexp_e}(z)= \lim_{n \to \infty } \ln^{[n]} (\operatorname{2sinh}^{[z]}(\exp^{[n]}(0)))$

This function gives the exact same results for all values of n as the following equivalent equation:
$\operatorname{TommySexp_e}(z)= \lim_{n \to \infty } \ln^{[n]} (\operatorname{2sinh}^{[z+\operatorname{SuperFunction^{-1}(\exp^{[n]}(0))]}} )$

Let $k=\operatorname{SuperFunction^{-1}(\exp^{[n]}(0))-n$ , then the following equation is also exactly equivalent:
$\operatorname{TommySexp_e}(z)= \lim_{n \to \infty } \ln^{[n]} (\operatorname{SuperFunction(z+k+n))$ As n goes to infinity, k converges quickly to the approximate value, k=0.067838366 (for n=0: k=-0.0734181, for n=1: k=0.0663658, for n>=2: k=0.067838366)

it is clear that iterations of exp and 2sinh are not exactly equal and thus n and z are not interchangeable ...

regards

tommy1729

(06/29/2010 10:45 PM)tommy1729 Wrote: [ -> ]it is clear that iterations of exp and 2sinh are not exactly equal and thus n and z are not interchangeable ...

regards

tommy1729

I'm confident in my equations, so we'll need to be more precise to figure out where we agree, and where we disagree. I assume the first equation, for the limit of the SuperFunction of 2sinh is agreed upon, but one never knows. Using this equation, the value of SuperFunction(0)=1.058049043.... I think the two equations after that are also fairly obvious, and I assume you also agree with them.
$\operatorname{SuperFunction}(z) = \lim_{n \to \infty} \operatorname{2sinh}^{[n]}(2^{z-n})$
$\operator{2sinh}^{[z]}(\operatorname{SuperFunc}(0))= \operatorname{SuperFunction}(z)$
$\operator{2sinh}^{[z]}(\operatorname{SuperFunc}(n))= \operatorname{SuperFunction}(z+n)$

Then substitute n with the inverse superfunction of n. Here the intent is to normalize so that SuperFunction(0)=n. And finally substitute n with the iterated exponentiation of n. This is equation is the equation I used to generate TommySexp; the only remaining step is iterating the ln function, and the equation I used for k. Do you still disagree with this equation? If so, why is this equation invalid, and is there another equation would you use instead? And does that alternative equation give identical values? I don't read minds, and I would be extremely interested in an alternative equation if you have one, because I would like to compare the two equations! They're probably equal!
$\operator{2sinh}^{[z]}(n)= \operatorname{SuperFunction}(z+\operatorname{SuperFunction^{[-1]}(n))$
$\operator{2sinh}^{[z]}(e^{[n]})= \operatorname{SuperFunction}(z+\operatorname{SuperFunction^{[-1]}(e^{[n]}))$

The other thing I posted was the shorthand substitution equation for k, which quickly converges to k+n, where k converges to k=0.067838366... for values of n>=2. At that point, after applying the iterating ln function, we have the equation you disagreed with, which still seems completely correct to me.
$k=\operatorname{SuperFunction^{[-1]}(e^{[n]})-n$
$\operator{2sinh}^{[z]}(e^{[n]})= \operatorname{SuperFunction}(z+k+n))$
$\operatorname{TommySexp}(z)=\lim_{n \to \infty}\ln^{[n]}( \operator{2sinh}^{[z]}(e^{[n]}))= \lim_{n \to \infty}\ln^{[n]}(\operatorname{SuperFunction}(z+k+n))$
-Sheldon

(06/30/2010 01:12 AM)sheldonison Wrote: [ -> ]
(06/29/2010 10:45 PM)tommy1729 Wrote: [ -> ]it is clear that iterations of exp and 2sinh are not exactly equal and thus n and z are not interchangeable ...

regards

tommy1729

I'm confident in my equations, so we'll need to be more precise to figure out where we agree, and where we disagree. I assume the first equation, for the limit of the SuperFunction of sexp2 is agreed upon, but one never knows. Using this equation, the value of SuperFunction(0)=1.058049043.... I think the two equations after that are also fairly obvious, and I assume you also agree with them.
$\operatorname{SuperFunction}(z) = \lim_{n \to \infty} \operatorname{sexp2}^{[n]}(2^{z-n})$
$\operator{sexp2}^{[z]}(\operatorname{SuperFunc}(0))= \operatorname{SuperFunction}(z)$
$\operator{sexp2}^{[z]}(\operatorname{SuperFunc}(n))= \operatorname{SuperFunction}(z+n)$

Then substitute n with the inverse superfunction of n. Here the intent is to normalize so that SuperFunction(0)=n. And finally substitute n with the iterated exponentiation of n. This is equation is the equation I used to generate TommySexp; the only remaining step is iterating the ln function, and the equation I used for k. Do you still disagree with this equation? If so, why is this equation invalid, and is there another equation would you use instead? And does that alternative equation give identical values? I don't read minds, and I would be extremely interested in an alternative equation if you have one, because I would like to compare the two equations! They're probably equal!
$\operator{sexp2}^{[z]}(n)= \operatorname{SuperFunction}(z+\operatorname{SuperFunction^{[-1]}(n))$
$\operator{sexp2}^{[z]}(\exp^{[n]}(1))= \operatorname{SuperFunction}(z+\operatorname{SuperFunction^{[-1]}(\exp^{[n]}(1)))$

The other thing I posted was the shorthand substitution equation for k, which quickly converges to k+n, where k converges to k=0.067838366... for values of n>=2. At that point, after applying the iterating ln function, we have the equation you disagreed with, which still seems completely correct to me.
$\operatorname{SuperFunction^{[-1]}(\exp^{[n]}(1))=k+n$
$\operator{sexp2}^{[z]}(\exp^{[n]}(1))= \operatorname{SuperFunction}(z+k+n))$
$\operatorname{TommySexp}(z)=\lim_{n \to \infty}\ln^{[n]}( \operator{sexp2}^{[z]}(\exp^{[n]}(1)))= \lim_{n \to \infty}\ln^{[n]}(\operatorname{SuperFunction}(z+k+n))$
-Sheldon

k depends on a limit and it depends on n !

iterations of exp cannot simply be replaced by iterations of 2 sinh , your formula is almost equal to mine but only if k starts approaching its limit , that is for large z and large n.

the convergence of 'k' is merely due to that fact that 2 sinh and exp are close , but that is only for large values , not around the fixpoints of 2 sinh or exp.

so you cannot write k for small n and small z.

since tetration grows so fast , carefully selecting parameters can lead to numerically undetectible differences.

( and btw , you havent even shown that k converges for large n and large z , although it makes sense that it does ... but what if it grows like e.g. slog(slog(slog(n))) ... though unlikely yes )

regards

tommy1729

(06/30/2010 02:35 PM)tommy1729 Wrote: [ -> ]k depends on a limit and it depends on n !

iterations of exp cannot simply be replaced by iterations of 2 sinh , your formula is almost equal to mine but only if k starts approaching its limit , that is for large z and large n.

the convergence of 'k' is merely due to that fact that 2 sinh and exp are close , but that is only for large values , not around the fixpoints of 2 sinh or exp.

so you cannot write k for small n and small z.

since tetration grows so fast , carefully selecting parameters can lead to numerically undetectible differences.

( and btw , you havent even shown that k converges for large n and large z , although it makes sense that it does ... but what if it grows like e.g. slog(slog(slog(n))) ... though unlikely yes )

regards

tommy1729

By the way, I had some typos, I should've typed 2sinh where I accidentally typed sexp2. I agree that tetration can be very sensitive to input conditions, though this isn't always the case. I also agree that k has a unique value for each integer n. If one uses the exact value for the real number k, then you're ok with this equation? Or is there a different equation you would use?
$\operatorname{SuperFunction^{[-1]}(e^{[n]})=k+n$
$\operator{2sinh}^{[z]}(e^{[n]})= \operatorname{SuperFunction}(z+(k+n))$

By the way, it is very easy to get values of k accurate to 10 significant digits, which in many cases is not enough precision, but in some cases, it is. The slope of the SuperFunction at z=0 is approximately one, and the value of the Superfunction(0) is also approximately one. So generating the half-iterate of TommySexp(0) is not a problem. At the img(z)=i0.5*pi/ln(2), the SuperFunction is also pretty well behaved since there is an attracting fixed point, small slopes, small values, so the approximation is not a problem. At that contour, real(superfunction(z))=0, and img(superfunction(z)) varies between i0.8 and i2.0 (see the graph I posted), so a value of k accurate to 10 significant digits is more than enough.

There are places where the equations are much more difficult, and extremely high precision may be required. Where the SuperFunction has superexponential growth for a couple of iterations, and then the img(z)=0, so the function is going to the fixed point after that. In that case, having enough precision to do the iterated natural logarithms with branches for increasing values of n could be tricky.

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 $Smile$

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

tommy1729 Wrote: $\operatorname{TommySexp_e}(z,x)= \lim_{n \to \infty } \ln^{[n]} (\operatorname{2sinh}^{[z]}(\exp^{[n]}(x)))$

(07/18/2010 10:41 PM)tommy1729 Wrote: [ -> ].... 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.
....

Using x= the fixed points exp(e), L, L* is an interesting idea, but I doubt it really helps. The most immediate draw back is that we're looking for a real valued super function, with real values at the real axis, which is why I assumed x=0.
- Sheldon

its not a matter of " helping " its a matter of being consistant.

x is free to choose and so is z.

i think it is easiest to understand that for fixed x , we have a sexp kind of function starting at value x and for fixed z we have slog kind of function.

from my post it should be clear that L and L* lie at oo i for the sexp interpretation of my formula.

there is no drawback , with respect but i think you are confused.

regards

tommy1729

(07/19/2010 10:20 PM)tommy1729 Wrote: [ -> ]its not a matter of " helping " its a matter of being consistant.

x is free to choose and so is z.

i think it is easiest to understand that for fixed x , we have a sexp kind of function starting at value x and for fixed z we have slog kind of function.

from my post it should be clear that L and L* lie at oo i for the sexp interpretation of my formula.

there is no drawback , with respect but i think you are confused.

regards

tommy1729

L and L* are the fixed point of the superfunction of exp(x). That superfunction of exp(x) has limiting values of L, L* and +/- i*infinity respectively. But the superfunction of 2sinh is periodic in the imaginary direction with period z=i*2pi/ln(2), so I don't understand how L and L* lie at oo i for the sexp interpretation of your formula.

(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

(07/20/2010 03:13 PM)sheldonison Wrote: [ -> ]But the superfunction of 2sinh is periodic in the imaginary direction with period z=i*2pi/ln(2), so I don't understand how L and L* lie at oo i for the sexp interpretation of your formula.

how do you know the superfunction of 2sinh is periodic in the imaginary direction ?

that is intresting.

regards

tommy1729

Pages: 1 2 3 4 5 6 7