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+--- Thread: How fast are these type of sums ? (/showthread.php?tid=1707)
How fast are these type of sums ? - tommy1729 - 02/16/2023
We know that in the limit x to +oo , f(x,n,m) = exp^[n]( (ln^[n](x))^m ) for all n,m > 1 is smaller than exp(x).
In fact even any fixed amount of iterations of that grow smaller than exp(x) at least in the limit x to +oo.
This implies lim_x f(x,n,m) < exp^[v](x) for any 0 < v.
Now suppose we want a function that grows ( in the limit ) faster than f(x,n,m) but slower than exp^[v] for some 0 < v < 1.
Also suppose we want to avoid defining a sexp or slog , and we want to avoid using functions for which we can clearly see it is asymptotic to exp^[v] for an easy to see value of v such that 0 < v < 1. ( so no half-iterate of sinh(x) or exp(x)-1 , no fake function theory and so on )
Also no taylor , fourier , pade or typical integral transforms.
We want an infinite sum.
So I define the following function for x > 0 :
T_n(x) = ( 1/ (2n)! ) * sinh^[n]( (arcsinh^[n](x))^2 ) / sinh^[n]( (arcsinh^[n](n))^2 )
T(x) = T_1(x) + T_2(x) + T_3(x) + ...
T(x) = sum_n T_n(x)
where the sum is over the strict positive integers n.
Now this function T(x) eventually grows faster than exp^[n]( (ln^[n](x))^2 ) or even exp^[n]( (ln^[n](x))^m ) for all large x and all integer n and m > 1.
But how fast does T(x) grow ?
What is the smallest value of v such that 0 < v < 1 and
LIM x to +oo
T(x) < exp^[v](x)
??
***
Btw this series might be a good first guess in iterative or recursive methods to find the semi-exp(x) ...
***
Now T(x) is ofcourse just an instance of the more general idea.
T_g_n(x) = ( 1/ (2n)! ) * sinh^[n]( (arcsinh^[n](x))^2 ) / sinh^[n]( (arcsinh^[n](g(n)))^2 )
T_g(x) = T_g_1(x) + T_g_2(x) + T_g_3(x) + ...
T_g(x) = sum_n T_g_n(x)
where the sum is over the strict positive integers n.
for some strictly increasing integer sequence g(n).
And ofcourse we could also consider different coefficients than 1/ (2n)! .
But I wanted to start with a small idea and specific case.
Hence T(x).
Sorry for not using tex , im tired and need sleep.
regards
tommy1729
RE: How fast are these type of sums ? - tommy1729 - 02/17/2023
(02/16/2023, 11:29 PM)tommy1729 Wrote: We know that in the limit x to +oo , f(x,n,m) = exp^[n]( (ln^[n](x))^m ) for all n,m > 1 is smaller than exp(x).
In fact even any fixed amount of iterations of that grow smaller than exp(x) at least in the limit x to +oo.
This implies lim_x f(x,n,m) < exp^[v](x) for any 0 < v.
Now suppose we want a function that grows ( in the limit ) faster than f(x,n,m) but slower than exp^[v] for some 0 < v < 1.
Also suppose we want to avoid defining a sexp or slog , and we want to avoid using functions for which we can clearly see it is asymptotic to exp^[v] for an easy to see value of v such that 0 < v < 1. ( so no half-iterate of sinh(x) or exp(x)-1 , no fake function theory and so on )
Also no taylor , fourier , pade or typical integral transforms.
We want an infinite sum.
So I define the following function for x > 0 :
T_n(x) = ( 1/ (2n)! ) * sinh^[n]( (arcsinh^[n](x))^2 ) / sinh^[n]( (arcsinh^[n](n))^2 )
T(x) = T_1(x) + T_2(x) + T_3(x) + ...
T(x) = sum_n T_n(x)
where the sum is over the strict positive integers n.
Now this function T(x) eventually grows faster than exp^[n]( (ln^[n](x))^2 ) or even exp^[n]( (ln^[n](x))^m ) for all large x and all integer n and m > 1.
But how fast does T(x) grow ?
What is the smallest value of v such that 0 < v < 1 and
LIM x to +oo
T(x) < exp^[v](x)
??
Ofcourse if we understand the speed of sinh^[x]( (arcsinh^[x](x))^2 ) very well , that would already be a great help and might even resolve all problems.
The problem is that sinh^[n]( (arcsinh^[n](x))^2 ) grows very fast and reaches relatively high values for small x, before it starts to be relatively slow compared to exp(x).
So understanding a relative max (and maybe some kind of average) of sinh^[x]( (arcsinh^[x](x))^2 ) would be very enlightning.
regards
tommy1729
RE: How fast are these type of sums ? - tommy1729 - 02/17/2023
(02/17/2023, 10:19 PM)tommy1729 Wrote:(02/16/2023, 11:29 PM)tommy1729 Wrote: We know that in the limit x to +oo , f(x,n,m) = exp^[n]( (ln^[n](x))^m ) for all n,m > 1 is smaller than exp(x).
In fact even any fixed amount of iterations of that grow smaller than exp(x) at least in the limit x to +oo.
This implies lim_x f(x,n,m) < exp^[v](x) for any 0 < v.
Now suppose we want a function that grows ( in the limit ) faster than f(x,n,m) but slower than exp^[v] for some 0 < v < 1.
Also suppose we want to avoid defining a sexp or slog , and we want to avoid using functions for which we can clearly see it is asymptotic to exp^[v] for an easy to see value of v such that 0 < v < 1. ( so no half-iterate of sinh(x) or exp(x)-1 , no fake function theory and so on )
Also no taylor , fourier , pade or typical integral transforms.
We want an infinite sum.
So I define the following function for x > 0 :
T_n(x) = ( 1/ (2n)! ) * sinh^[n]( (arcsinh^[n](x))^2 ) / sinh^[n]( (arcsinh^[n](n))^2 )
T(x) = T_1(x) + T_2(x) + T_3(x) + ...
T(x) = sum_n T_n(x)
where the sum is over the strict positive integers n.
Now this function T(x) eventually grows faster than exp^[n]( (ln^[n](x))^2 ) or even exp^[n]( (ln^[n](x))^m ) for all large x and all integer n and m > 1.
But how fast does T(x) grow ?
What is the smallest value of v such that 0 < v < 1 and
LIM x to +oo
T(x) < exp^[v](x)
??
Ofcourse if we understand the speed of sinh^[x]( (arcsinh^[x](x))^2 ) very well , that would already be a great help and might even resolve all problems.
The problem is that sinh^[n]( (arcsinh^[n](x))^2 ) grows very fast and reaches relatively high values for small x, before it starts to be relatively slow compared to exp(x).
So understanding a relative max (and maybe some kind of average) of sinh^[x]( (arcsinh^[x](x))^2 ) would be very enlightning.
regards
tommy1729
ok here is an idea
sinh^[x]( (arcsinh^[x](x))^2 )
=
sinh^[x]( arcsinh^[x](x) * arcsinh^[x](x) )
keep that in mind ,
remember how sinh^[x]( 2 arcsinh^[x](x) ) was similar to sinh^[x+1] ( arcsinh^[x](x) ) and therefore
sinh^[x]( 2 arcsinh^[x](x) ) is close to sinh(x).
so returning to the previous equation :
sinh^[x]( arcsinh^[x](x) * arcsinh^[x](x) )
is close to
sinh^[x + ln(arcsinh^[x](x))/ln(2)] (arcsinh^[x](x) )
= close to
sinh^[ arcsinh[x+1](x) ] (x)
so in the limit as x goes to + oo we still have growth 0.
But this gives us a tool to work with.
sinh^[1/2](x)
=
sinh^[x +1/2]( arcsinh^[x](x) )
= close to
sinh^[x]^[ sqrt(2) arcsinh^[x](x) ]
which makes alot of sense !
This is the classic limit formula n to +oo
sinh^[n]^[ sqrt(2) arcsinh^[n](x) ]
where we replace oo with x.
Since that limit formula converges very fast this is a reasonable approximation for large x.
We should not jump to conclusions yet.
But it gives a way to attack the problem !
Conjecture for x > 3:
sinh^[1/2](x) - sinh^[x]^[ sqrt(2) arcsinh^[x](x) ] = O( x^( 1/n ) - x^( 1/(n-1) ) )
where O is big O notation and n is the next integer after x.
But I think we already know
sinh^[1/2](x) - sinh^[x]^[ sqrt(2) arcsinh^[x](x) ] = O( 1/(sqrt(2)^x) )
because the convergeance is exp type.
Then again that last equation sinh^[1/2](x) - sinh^[x]^[ sqrt(2) arcsinh^[x](x) ] = O( 1/(sqrt(2)^x) )
might only hold near the origin because that is where the iterations are exp in growth.
And ofcourse we actually want to understand exp^[1/2] but I like to use sinh^[1/2] as a tool.
Yes yes , this is still all very informal.
regards
tommy1729
RE: How fast are these type of sums ? - tommy1729 - 02/17/2023
Im excited about this !
Assuming my ideas are correct and combining with some basic calculus and fake function theory this really gives a deep insight into asymptotics of series expansions !
In fact if I push these ideas , I can even make a series expansion asymptotic to the half-iterate of tetration.
Yes.
f(x+1) = exp(f(x))
g(g(x)) = f(x)
without using a fixpoint !
regards
tommy1729
RE: How fast are these type of sums ? - tommy1729 - 02/17/2023
It might be conceivable that using these approximation functions as upper or lower bounds give uniqueness criterions for fractional iterations.
regards
tommy1729
RE: How fast are these type of sums ? - tommy1729 - 02/17/2023
im working with x > 0.
By using functional equations we can get anywhere.
But I would like a method that is a good approx for all real x ...
hmm
that is tricky.
regards
tommy1729