Discussion
Tue Oct 10
17:01
Alan Michael Gómez Calderón: Thanks to all of you for your great job.
Discussion
Tue Oct 10
06:59
Joerg Arndt: (89+12*sqrt(55))^(1-n) = (89-12*sqrt(55))^(n-1), and the latter form is better
07:46
Alan Michael Gómez Calderón: In fact dear Joerg Arndt, I agree with you, is also a bit more elegant form!, could I modify again?
16:37
Kevin Ryde: Yes you can.
Discussion
Tue Oct 10
06:04
Kevin Ryde: (Big bit first.)
Discussion
Sat Oct 07
22:28
Alan Michael Gómez Calderón: Good night dear Kevyn Ryde, I didn't understand what was the fail... so I set precision at 100 so work in Pari. Let me show you: ? \p 100
? [floor(((89+12*sqrt(55))^(-n)+(89+12*sqrt(55))^(n-1))/(2*sqrt(55))) | n<-[1..13]]
%2 = [0, 12, 2136, 380196, 67672752, 12045369660, 2144008126728, 381621401187924, 67926465403323744, 12090529220390438508, 2152046274764094730680, 383052146378788471622532, 68181130009149583854080016]
Sun Oct 08
06:29
Kevin Ryde: Don't use floating point. It leads to wrong results. In pari, sqrt(55) can be exactly represented by w = quadgen(4*55).
09:35
Alan Michael Gómez Calderón: Hello Kevin Ryde, I must to say sorry, but I still didn't understand this algebraic simplification... please could you explain me how to do it?
17:52
Kevin Ryde: Your first formula was right, but can be simplified. The form you now have is fine, but you've made two mistakes in the simplification.
17:54
Kevin Ryde: (Incidentally, you do know why these kind of formulas are right, yes? If you've been doing anything with floating point approximations then that may suggest an answer, but it's not proof ...)
Mon Oct 09
02:55
Alan Michael Gómez Calderón: Well, no I don't know :-(
07:07
Kevin Ryde: There's a theory on how linear recurrences work. It's not too hard. In an order 2 like here the powers are of solutions to a quadratic equation. If you have an apparent answer then substituting it into the linear recurrence can show it right or wrong.
20:39
Alan Michael Gómez Calderón: Thank you very Much Kevin Ryde, It was very interesting to know how to test this kind of formulas, I will learn more about a second order linear recurrence relation theory
Discussion
Sat Oct 07
19:55
Kevin Ryde: Formula wrong.