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tetration base > exp(2/5)
#1
Very similar to the 2sinh method we can go to bases between eta and exp(1/2).

For instance bases > exp(2/5).

Basicly we just replace 2sinh(x) with

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! + ...

Notice the heart of these ideas ( 2sinh , f(x) above ) is series multisection.
f(A x) is close to exp(A x) , just like sinh(A x).
But f(A x) has derivative at 0 : 5/2 A instead of sinh(A x) which has derivative at 0 : 2 A.

Therefore by using f(A x) we maintain a hyperbolic fixpoint at 0 for all bases > exp(2/5).

Notice f(A x) has all its derivatives >= 0 and 0 =< f(A x) < exp(A x).

The rest is analogue to the 2sinh method.

A little more advanced is the idea that we can VERY LIKELY arrive at a uniqueness criterion regarding these multisection methods.

Although we can express these multisections in terms of standard functions , Im still curious about their " behaviour " such as fixed points , zero's etc.

Some plots would be nice.

( some ideas relating to fake function theory come to mind , but thats way too advanced and speculative for now )

***

The main idea to generalize this towards bases arbitrary close to eta is the simple observation that the multisection is of type "(a,b)" where exp(a/b) is the approximation to eta.

eta ~ exp(a/b)

This naturally makes me wonder about the nicest proofs for the irrationality of eta and the irrationality measure of eta.

It is clear that this method cannot be used for eta itself since

eta =/= exp(a/b)

Unless perhaps some limiting ideas.
Or maybe not since this method is not analytic. Not sure.

***

(musing about the multisection)

HOWEVER

The sky may not be perfectly blue ?

Suppose we work with a multisection type function such as

g(x) = 0 + Q x + Q x^S/S! + ...

Then we need to make sure that

Q x + Q x^S/S! < exp(x)

for all x > 0.

Although Q and S are not indep , the main issue is Q here and a more relaxed equation is

Q x < exp(x)

Therefore we look for the equation

Qx = exp(x)

with a single positive real solution x ;

a tangent line equation.

BUT LUCKILY we find

ex = exp(x) => e = exp(1).

so our Q is exactly bounded by e.

Therefore our base is exactly bounded by eta = exp(1/e).

SOOO , the sky is blue afterall.

Smile

THis also proof that the method works for bases larger than eta.

***

Perhaps some intresting links :

http://mathworld.wolfram.com/SeriesMultisection.html

http://mathworld.wolfram.com/IrrationalityMeasure.html

***

I think its true that if

a/b is an approximation of 1/e

then a*/(2b+1) is ALMOST Always a better one.

(where a and a* are chosen optimal ).

this leads to the imho intresting sequence :

( where the denom is iterated under the map 2x+1 )

1/2

2/5

4/11

9/23

17/47

35/95 = 7/19

...

a_n/b_n

and I guess a_(n+1)/b_(n+1) is Always a better approximation then all previous ones.

Its been a while since I did this kind of math so forgive any blunders.


regards

tommy1729
Reply
#2
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
Reply
#3
It appears that 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! + ...

is the only multisection type function that satisfies for all x > 0 :

x =< f(x) < exp(x).

The other candidates a/b seem to violate already within the bounds

0 < x < 11b.

SO assuming this is correct , probably the method based on base exp(2/5) converges the best.
As expected in post 1 and this answers part 1 of post 2.

More investigation is desired but probably the fact that the only number b such that 2/b is a good appr to 1/e is b = 5 is fundamental.

Large a in a/b lead to problems for x >> 1 because of the many powersums needed.
This makes some sense because 1/e < 1/2.

***

As for the cousin question , proving true is as hard as the cousin prime conjecture and probably equally hard as the twin prime conjecture.
Both unsolved as of today.

However proving false might be easier.

///

Notice fake function theory can in a way be an approximate inverse multisection.

a1 x^2 + a2 x^4 + ... = multisect

fake ( a1 x^2 + a2 x^4 + ... ) ~ Original function.

Nice.

///

regards

tommy1729
Reply


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