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Infinite tetration giving I*Omega costant=I*0.567143... - Printable Version

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RE: Infinite tetration giving I*Omega costant=I*0.567143... - Ivars - 03/07/2008

bo198214 Wrote:
Ivars Wrote:

What do you mean by ? Btw. has tex code
Code:
\lim_{n\to\infty}
.

Thanks. is n-th iteration of , starting with , when n=0 .

It is the same as in iteration formula (map) , while is next iteration result =angle n+1, I just did not know how to add indexes in tex.

This limit is called rotation or winding number.

In my simulation, when n=1829, =1825.78027070570366694723301.... so

/1829=0.998239623130510479468142710945


RE: Infinite tetration giving I*Omega costant=I*0.567143... - bo198214 - 03/09/2008

Ivars Wrote:Thanks. is n-th iteration of , starting with , when n=0 .

Ok that means where and .

I put some effort into investigating this sequence . For the case of and that you describe one can see numerically that in the limit for some constant . By this observation we put up:





(1)

If converges we can choose so that converges to 0. In this case the equation has to be satisfied in the limit too:

So that if then
, , where has to be satisfied. Let then all possible solutions are
and for integer . However only one of them is the actual value in .

For example in our case and .

If vice versa is given as above, what conditions are to be satisfied that ?

We put into equation (1), first:
, for simplicity let ,

where we get for solutions and for solutions .

now into (1)


The interesting cases is (a) if and and (b) if and together with c) . We assume in the following that and only consider the case .

If then and only positive values are subtracted from , so satisfies case (a).

If then and the condition is equivalent to .

In this case and as both side are positive we can quadrate getting equivalently:


, let




so case (b) is occurs exactly for

So in the case it is this satisified for all otherwise for
.


As a last step we give conditions for c) .

Let and . In the case that we have to assure that the subtractions of and do not lead below , i.e. at most .

In the case , and we showed already that , so that the subtraction of and the addition of increases . We shall ascertain that .

The condition is easy to check:


Now we know that for , hence the condition is satisfied if we demand:
.

The condition , let and :

.
Similarly to the previous case the condition is satisfied for:
.

We know that is striclty increasing for and (graph it!). Hence there is an such that (which is equivalent to ) for all .

Now going backwards we get

Proposition. Let , , , , and let be the solution of .
If there exists an and integer such that , then , particularly .



Verify the conditions for your choice:
, , , , .

The last condition that , or even the stronger condition that is quite probable for any sufficient chaotic choice of and .

So for your choice but also many other choices for and indeed .

PS: By the length of the derivations I can not exclude that somewhere got an error into the computations, so dont take the result as granted.


RE: Infinite tetration giving I*Omega costant=I*0.567143... - Ivars - 03/09/2008

bo198214 Wrote:Ok that means where and .

I put some effort into investigating this sequence . For the case of and that you describe one can see numerically that in the limit for some constant .

Thanks,

It will .... take me some time to digest. I do not quite understand the first assumption,

Quote: one can see numerically that in the limit for some constant .

As far as I did it (1850 terms), nothing is constant-You mean you replaced small difference of sin from -1 with an argument?
But as sin argument nears n*(3pi/2), can You do it?

As I see it since teta(n+1)-teta(n) = 1 when n-> infinity, there shall be no differences between teta(n) and n integer part.

There can not be one also in reals, since limit n->infinity is 1.

So this constant seems suspicious to me-or may be I misunderstood something.

Ivars


RE: Infinite tetration giving I*Omega costant=I*0.567143... - bo198214 - 03/09/2008

Ivars Wrote:
Quote: one can see numerically that in the limit for some constant .

As far as I did it (1850 terms), nothing is constant-You mean you replaced small difference of sin from -1 with an argument?
But as sin argument nears n*(3pi/2), can You do it?

As I see it since teta(n+1)-teta(n) = 1 when n-> infinity, there shall be no differences between teta(n) and n integer part.

There can not be one also in reals, since limit n->infinity is 1.

So this constant seems suspicious to me-or may be I misunderstood something.

Whats not clear about this?
Just compute with . You see numerically that and that means that in the limit.