02/06/2023, 10:42 PM
It is perhaps interesting to describe how fast the 2sinh method works.
We consider "speed" here by considering changing bases.
The gaussian method changes the bases by exp(t(s)) = exp( (1+erf(s))/2 ).
F(s+1) = exp( t(s) F(s) )
So we get exp(t(v) exp( t(v-1) exp (t(v-2) exp(...
This is fast. But much slower than the 2sinh method.
Because the bases change as function of iterations of 2sinh(x).
***
well ln ln ln ... 2sinh^[n] exp exp ... gives most attention to the 2sinh iteration part,
we could also consider the exp iterations part ... all is relative
afterall
lim
ln ln ... 2sinh^[x] exp^[y]( exp exp ...
is essentially equivalent to
lim
ln ln ... 2sinh^[x+y]( exp exp ...
keeping x between zero and 1 gives the most practical way ofcourse.
But to keep things simple we consider 2sinh here.
As we will later see here , the 2sinh vs exp iterations details here are not so important afterall
***
SO suppose 2sinh(x) has base A(x) [ setting 2sinh(x) = A^x is equivalent ] then the method has " speed "
A(2sinh^[n](x_0)) , in other words it converges faster than superexponential !
Indeed the new base depends on the previous iteration sinh(x) so we get somewhat superexponential growth.
But perhaps a little more precise would be nice.
Well a closed form that is simpler than the super of sinh or tetration is probably not possible.
But we can describe the limiting behaviour very well.
It comes down by understanding A(x) at +infinity.
So
A(x) = ln( 2sinh(x) ) / x.
Now consider the illuminating limit ;
lim x to +oo
( ln( 2sinh(x) ) / x )^(C + x exp(2x)) = exp(-1)
Or equivalent
( A(x) )^(C + x exp(2x)) = exp(-1)
for a constant C.
This implies that A(x) in the limit behaves alot like
A(x) = 1 - 2/W(2 (x + D))
Where D is a constant and W is the Lambert-W function.
So we get some idea how fast this method works.
***
I hesitated to post this earlier because the asymptotic is not very precise.
Perturbation theory and taylor series might be a better way.
Maybe there are better ways.
All roads lead to Rome but some require more effort than others.
But I like limits as you know , so I go a step further :
lim x to +oo
[ ( ln( 2sinh(x) ) / x )^(x exp(2x)) - exp(-1) ] exp(2x) = - 1/(2e)
At this point I think using an integral transform is probably smart.
Laplace probably.
The reason is taylor series might not be so useful because of a finite radius.
A taylor series after an integral transforms might be useful though.
A step further we get
lim x to +oo
[ [ ( ln( 2sinh(x) ) / x )^(x exp(2x)) - exp(-1) ] exp(2x) + 1/(2e)] x = - 1/(2e)
These methods are probably not optimal but Im dropping it here anyway for those who care.
Also going further by hand becomes quite somewhat dangerous and alot of work.
It would require more paper.
***
Notice the comment 2sinh iterations vs exp iterations written before becomes more important when studying these more precise limits.
This is logical.
Although the details , I admit they are not completely understood by myself.
Anyway we focus on A(x) here which is much more clearly defined.
***
***
Upper and lower bounds on A(x) can also be computed and might be more insightful than the situation at infinity.
On the other hand convergeance to infinity is fast and good so maybe not that much more interesting , well maybe as an exercise.
***
regards
tommy1729
We consider "speed" here by considering changing bases.
The gaussian method changes the bases by exp(t(s)) = exp( (1+erf(s))/2 ).
F(s+1) = exp( t(s) F(s) )
So we get exp(t(v) exp( t(v-1) exp (t(v-2) exp(...
This is fast. But much slower than the 2sinh method.
Because the bases change as function of iterations of 2sinh(x).
***
well ln ln ln ... 2sinh^[n] exp exp ... gives most attention to the 2sinh iteration part,
we could also consider the exp iterations part ... all is relative
afterall
lim
ln ln ... 2sinh^[x] exp^[y]( exp exp ...
is essentially equivalent to
lim
ln ln ... 2sinh^[x+y]( exp exp ...
keeping x between zero and 1 gives the most practical way ofcourse.
But to keep things simple we consider 2sinh here.
As we will later see here , the 2sinh vs exp iterations details here are not so important afterall
***
SO suppose 2sinh(x) has base A(x) [ setting 2sinh(x) = A^x is equivalent ] then the method has " speed "
A(2sinh^[n](x_0)) , in other words it converges faster than superexponential !
Indeed the new base depends on the previous iteration sinh(x) so we get somewhat superexponential growth.
But perhaps a little more precise would be nice.
Well a closed form that is simpler than the super of sinh or tetration is probably not possible.
But we can describe the limiting behaviour very well.
It comes down by understanding A(x) at +infinity.
So
A(x) = ln( 2sinh(x) ) / x.
Now consider the illuminating limit ;
lim x to +oo
( ln( 2sinh(x) ) / x )^(C + x exp(2x)) = exp(-1)
Or equivalent
( A(x) )^(C + x exp(2x)) = exp(-1)
for a constant C.
This implies that A(x) in the limit behaves alot like
A(x) = 1 - 2/W(2 (x + D))
Where D is a constant and W is the Lambert-W function.
So we get some idea how fast this method works.
***
I hesitated to post this earlier because the asymptotic is not very precise.
Perturbation theory and taylor series might be a better way.
Maybe there are better ways.
All roads lead to Rome but some require more effort than others.
But I like limits as you know , so I go a step further :
lim x to +oo
[ ( ln( 2sinh(x) ) / x )^(x exp(2x)) - exp(-1) ] exp(2x) = - 1/(2e)
At this point I think using an integral transform is probably smart.
Laplace probably.
The reason is taylor series might not be so useful because of a finite radius.
A taylor series after an integral transforms might be useful though.
A step further we get
lim x to +oo
[ [ ( ln( 2sinh(x) ) / x )^(x exp(2x)) - exp(-1) ] exp(2x) + 1/(2e)] x = - 1/(2e)
These methods are probably not optimal but Im dropping it here anyway for those who care.
Also going further by hand becomes quite somewhat dangerous and alot of work.
It would require more paper.
***
Notice the comment 2sinh iterations vs exp iterations written before becomes more important when studying these more precise limits.
This is logical.
Although the details , I admit they are not completely understood by myself.
Anyway we focus on A(x) here which is much more clearly defined.
***
***
Upper and lower bounds on A(x) can also be computed and might be more insightful than the situation at infinity.
On the other hand convergeance to infinity is fast and good so maybe not that much more interesting , well maybe as an exercise.
***
regards
tommy1729