02/10/2015, 10:43 PM
For all clarity the method works for bases larger then eta.
But that does not imply x < M(x) < exp(x)
where M(x) is a suitable multisection type function.
( Linear combinations of Mittag-Leffler functions )
it is only required for large x that
x < M(x) < exp(x).
But this follows automatically from the asymptotic behaviour of the multisection type function and the method we use :
f(x)
ln( f( exp(x) ) )
...
ln^[n]( f ( exp^[n](x) ) )
the third last line implies
exp(x) < M(exp(x)) < exp(exp(x))
...
exp^[n](x) < M(exp^[n](x)) < exp^[n+1](x)
and hence the method works.
However it is intresting to consider exp(x) - M(x).
This quantity affects how many iterations we need to take to get a good numeric result.
And I find it intresting by itself.
The number of sign changes / zero's of exp(x) - M(x) can be estimated by fourier-budan for instance.
But there must be many tools for this.
And the question of closed form zero's occurs naturally.
at x = 1 we can relatively easy check the sign of exp(x) - M(x).
M(x) = f(x) = 0 + 5/2 x + 5/2 x^5/5! + 5/2 x^6/6! + 5/2 x^10/10!
+ 5/2 x^11/11! + ...
then
exp(1) - M(1) =
e - 5/2 (1 + 1/5! + 1/6! + 1/10! +1/11! + ... )
truncated (at ...) this gives us :
0.193976
Notice the number of sign changes of exp(x) - M(x) is Always bounded as function of b in a/b ~ 1/e.
More investigation is desired.
---
Another thing is the sequence given before
2,5,11,23,47,...
these are the iterations by the map 2x+1 starting at 2.
This has a closed form : 3 * 2^n - 1.
( trivial to prove )
A similar sequence is
1,7,19,...
which are the iterations by the map 2x+5.
This also has a closed form : 3 * 2^n - 5.
Numbers of the form 3 * 2^n - 1 are called Thâbit ibn Kurrah numbers.
and if those numbers are prime they are called Thâbit ibn Kurrah primes.
Now if for a fixed n , 3 * 2^n - 1 and 3 * 2^n - 5 are both prime then we have a Thâbit ibn Kurrah cousin prime.
Conjecture : there are infinitely many Thâbit ibn Kurrah cousin primes.
7,11
19,23
41,47
379,383
...
regards
tommy1729
But that does not imply x < M(x) < exp(x)
where M(x) is a suitable multisection type function.
( Linear combinations of Mittag-Leffler functions )
it is only required for large x that
x < M(x) < exp(x).
But this follows automatically from the asymptotic behaviour of the multisection type function and the method we use :
f(x)
ln( f( exp(x) ) )
...
ln^[n]( f ( exp^[n](x) ) )
the third last line implies
exp(x) < M(exp(x)) < exp(exp(x))
...
exp^[n](x) < M(exp^[n](x)) < exp^[n+1](x)
and hence the method works.
However it is intresting to consider exp(x) - M(x).
This quantity affects how many iterations we need to take to get a good numeric result.
And I find it intresting by itself.
The number of sign changes / zero's of exp(x) - M(x) can be estimated by fourier-budan for instance.
But there must be many tools for this.
And the question of closed form zero's occurs naturally.
at x = 1 we can relatively easy check the sign of exp(x) - M(x).
M(x) = f(x) = 0 + 5/2 x + 5/2 x^5/5! + 5/2 x^6/6! + 5/2 x^10/10!
+ 5/2 x^11/11! + ...
then
exp(1) - M(1) =
e - 5/2 (1 + 1/5! + 1/6! + 1/10! +1/11! + ... )
truncated (at ...) this gives us :
0.193976
Notice the number of sign changes of exp(x) - M(x) is Always bounded as function of b in a/b ~ 1/e.
More investigation is desired.
---
Another thing is the sequence given before
2,5,11,23,47,...
these are the iterations by the map 2x+1 starting at 2.
This has a closed form : 3 * 2^n - 1.
( trivial to prove )
A similar sequence is
1,7,19,...
which are the iterations by the map 2x+5.
This also has a closed form : 3 * 2^n - 5.
Numbers of the form 3 * 2^n - 1 are called Thâbit ibn Kurrah numbers.
and if those numbers are prime they are called Thâbit ibn Kurrah primes.
Now if for a fixed n , 3 * 2^n - 1 and 3 * 2^n - 5 are both prime then we have a Thâbit ibn Kurrah cousin prime.
Conjecture : there are infinitely many Thâbit ibn Kurrah cousin primes.
7,11
19,23
41,47
379,383
...
regards
tommy1729
