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