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 Infinite tetration giving I*Omega costant=I*0.567143... Ivars Long Time Fellow    Posts: 366 Threads: 26 Joined: Oct 2007 03/07/2008, 10:54 PM (This post was last modified: 03/07/2008, 11:06 PM by Ivars.) 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 bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 03/09/2008, 11:45 AM (This post was last modified: 03/09/2008, 11:47 AM by bo198214.) 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. Ivars Long Time Fellow    Posts: 366 Threads: 26 Joined: Oct 2007 03/09/2008, 12:16 PM (This post was last modified: 03/09/2008, 12:21 PM by Ivars.) 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 bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 03/09/2008, 01:16 PM 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. « Next Oldest | Next Newest »

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