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MiszaBot I (talk | contribs) m Archiving 2 thread(s) from Talk:Fibonacci number. |
MiszaBot I (talk | contribs) m Archiving 2 thread(s) from Talk:Fibonacci number. |
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:::The details of my calculation?... I entered (((1+sqrt 5)/2)^i-(-(1+sqrt 5)/2))^i)/sqrt 5 into Google calculator... I am no mathematician, sorry, but maybe if you can look into how Google calculator got to this you might be able to change that into some king of closed solution involving e and/or pi. [[User:Robo37|Robo37]] ([[User talk:Robo37|talk]]) 13:38, 11 July 2011 (UTC)
::::Oops, wrong formular. The i<sup>th</sup> Fibonacci number is ''0.379294534 + 0.215939518 i'', which I got by ((1+sqrt 5)/2)^i-((-1)^i/((1+sqrt 5)/2)^i)))/sqrt 5.
== Fibonacci root? ==
What is x when F (x) = n? [[User:Robo37|Robo37]] ([[User talk:Robo37|talk]]) 09:57, 14 July 2011 (UTC)
:See [[Fibonacci number#Recognizing Fibonacci numbers]]. [[User:PrimeHunter|PrimeHunter]] ([[User talk:PrimeHunter|talk]]) 11:45, 14 July 2011 (UTC)
::Thanks, is it at all expressable without using uncommon logs? [[User:Robo37|Robo37]] ([[User talk:Robo37|talk]]) 17:58, 15 July 2011 (UTC)
:::You can always change a logarithm base to any other with the formula at [[Logarithm#Change of base]]. [[User:PrimeHunter|PrimeHunter]] ([[User talk:PrimeHunter|talk]]) 02:08, 16 July 2011 (UTC)
::::Wait, how come when I enter log ((21*sqrt 5)+0.5)/log ((1+sqrt 5)/2) into Google calculator I get 8.02106857? [[User:Robo37|Robo37]] ([[User talk:Robo37|talk]]) 07:42, 16 July 2011 (UTC)
:::::Because that is the (approximate) value of that expression. Note that to find the Fibonacci index you have to take the [[Floor and ceiling functions|floor]] (nearest integer less than or equal to) this value, which gives you 8. [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 14:56, 16 July 2011 (UTC)
::::::The difference to the real value converges to 0. In [[PARI/GP]] for powers of two:
<pre>
? for(n=1,8,print(2^n": "log((fibonacci(2^n)*sqrt(5))+0.5)/log((1+sqrt(5))/2)))
2: 2.09163988209225188766928942555474042584550500130492803616758
4: 4.10467845877076686103432829279967108337857360119089528434855
8: 8.02106857224470714263573083533459921808575680298676770865263
16: 16.0004703147334956159652160508455984425808456893171492670951
32: 32.0000002133187473657459828226685085556081114874256292899634
64: 64.0000000000000437950208463788551755832830904781992970335194
128: 128.000000000000000000000000001845932267159675589494339813885
256: 256.000000000000000000000000000000000000000000000000000003279
</pre>
::::::[[User:PrimeHunter|PrimeHunter]] ([[User talk:PrimeHunter|talk]]) 16:25, 16 July 2011 (UTC)
:::::::Oh, right, thanks. What is the (closed) exact formular, then? [[User:Robo37|Robo37]] ([[User talk:Robo37|talk]]) 17:08, 16 July 2011 (UTC)
::::::::Well, if you want an expression that does not involve the floor function then it is
:::::::::<math>n = \log_\varphi\bigg(\frac{\sqrt{5F_n^2\pm4}+\sqrt{5}F_n}{2}\bigg)</math>
::::::::where you choose the sign (plus or minus) that makes <math>\sqrt{5F_n^2\pm4}</math> an integer. [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 19:47, 16 July 2011 (UTC)
:::::::::Thanks, but that get's to (log (((5*(((1+sqrt 5)/2)^21-((-1)^21/((1+sqrt 5)/2)^21)))/sqrt 5)))+4)+((sqrt 5)*((1+sqrt 5)/2)^21-((-1)^21/((1+sqrt 5)/2)^21)))/sqrt 5)))/2))/(log ((1+sqrt 5)/2)) and when I enter it into Goodle calculater I get 58567.8366. With an equation as massive as that I'm bound to make an error somewhere along the line but there must be a shorter expression than that I can use '''without''' using the Fibonacci function itself. [[User:Robo37|Robo37]] ([[User talk:Robo37|talk]]) 20:17, 16 July 2011 (UTC)
::::::::::F<sub>8</sub> = 21, so
:::::::::::<math>\log_\varphi\bigg(\frac{\sqrt{5 \times 21^2+4}+21\sqrt{5} }{2}\bigg)
=\log_\varphi\bigg(\frac{\sqrt{2209}+21\sqrt{5}}{2}\bigg)=\log_\varphi\bigg(\frac{47+21\sqrt{5}}{2}\bigg)=8</math>
::::::::::Without using the floor function, that's as simple as it gets. [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 21:33, 16 July 2011 (UTC)
:::::::::::Here is a valid ASCII expression to compute x when F (x) = n:
:::::::::::x = log((sqrt(5*n^2+4)+sqrt(5)*n)/2)/log(((1+sqrt(5))/2))
:::::::::::Replace +4 by -4 if required to get an integer (this happens when x is odd). [[User:PrimeHunter|PrimeHunter]] ([[User talk:PrimeHunter|talk]]) 23:02, 16 July 2011 (UTC)
::::::::::::Thanks for that, we got there eventually. So, as I don't see <math>\log_\varphi\bigg(\frac{\sqrt{5 \times n^2+4}+n\sqrt{5} }{2}\bigg)=x</math> and <math>\log_\varphi\bigg(\frac{\sqrt{5 \times n^2-4}+n\sqrt{5} }{2}\bigg)=x</math> anywhere on the artical, maybe it should be included somewhere? Anyway, thanks again, I appreciate the help. [[User:Robo37|Robo37]] ([[User talk:Robo37|talk]]) 09:06, 17 July 2011 (UTC)
:::::::::::::Sorry about this, but is there a universal function that for that that isn't dependant on whether x is odd or even? [[User:Robo37|Robo37]] ([[User talk:Robo37|talk]]) 22:38, 29 July 2011 (UTC)
== On entering decimals into the Fibonacci function.. ==
We have F(x)=((1+sqrt 5)/2)^(x)-((-1)^(x)/((1+sqrt 5)/2)^(x))))/sqrt 5, which indeed satifies any integar as:
* ((1+sqrt 5)/2)^(0)-((-1)^(0)/((1+sqrt 5)/2)^(0))))/sqrt 5=0
* ((1+sqrt 5)/2)^(1)-((-1)^(1)/((1+sqrt 5)/2)^(1))))/sqrt 5=1
* ((1+sqrt 5)/2)^(2)-((-1)^(2)/((1+sqrt 5)/2)^(2))))/sqrt 5=1
* ((1+sqrt 5)/2)^(3)-((-1)^(3)/((1+sqrt 5)/2)^(3))))/sqrt 5=2
* ((1+sqrt 5)/2)^(4)-((-1)^(4)/((1+sqrt 5)/2)^(4))))/sqrt 5=3
* ((1+sqrt 5)/2)^(5)-((-1)^(5)/((1+sqrt 5)/2)^(5))))/sqrt 5=5
* ((1+sqrt 5)/2)^(6)-((-1)^(6)/((1+sqrt 5)/2)^(6))))/sqrt 5=8...
...and so on, so it thereby would make sense to use the same forumular for decimals.
Using Google calculator, we have:
{| class="wikitable"
|-
! x !! ((1+sqrt 5)/2)^(1/x)-((-1)^(1/x)/((1+sqrt 5)/2)^(1/x))))/sqrt 5
|-
| 1 || 1
|-
| 2 || 0.568864481 - 0.351577584 i
|-
| 3 || 0.905958432
|-
| 4 || 0.224000793 - 0.280383911 i
|-
| 5 || 0.898572747
|-
| 6 || 0.127109339 - 0.206373405 i
|-
| 7 || 0.896541471
|-
| 8 || 0.0858883613 - 0.161150332 i
|-
| 9 || 0.895706001
|-
| 10 || 0.063917354 - 0.131703889 i
|-
| 11 || 0.895283186
|-
| 12 || 0.0505164733 - 0.111197646 i
|-
| 13 || 0.895040036
|-
| 14 || 0.0415832749 - 0.0961519693 i
|-
| 15 || 0.894887492
|-
| 16 || 0.0352431027 - 0.084662092 i
|-
| 17 || 0.89478555
|-
| 18 || 0.0305294065 - 0.0756092279 i
|-
| 19 || 0.894714073
|-
| 20 || 0.0268975698 - 0.0682964386 i
|-
| 21 || 0.894662029
|-
| 22 || 0.0240191066 - 0.0622681185 i
|-
| 23 || 0.894622962
|-
| 24 || 0.0216849331 - 0.0572143342 i
|-
| 25 || 0.89459289
|}
So we have when x is odd, the number is real and converges to 2/sqrt 5. This is rather interesting as it shows that i is not just the solution to the square root of -1, but also the solution to non-integar Fibonacci numbers, which keeps me wondering what the ith Fibonacci root is? I tried using the obove Fibonacci root function wich leaves 2 + 3.2642513 i when you assume i to be odd or 1 + 3.2642513 i when you assume i to be even, with 3.2642513 aparently being what the rank-1 [[Grothendieck constant]] is at most in the tripartite graph G (whatever that means), but when you enter either of those numbers into ((1+sqrt 5)/2)^(1/x)-((-1)^(1/x)/((1+sqrt 5)/2)^(1/x))))/sqrt 5 you do not get i, which suggests the i is neither odd nor even.
Any help will be greatly appreciated. Thanks. [[User:Robo37|Robo37]] ([[User talk:Robo37|talk]]) 11:15, 31 July 2011 (UTC)
:That should be ((1+sqrt 5)/2)^(1/x)-((-1)^(1/x)/(( '''−'''1 +sqrt 5)/2)^(1/x))))/sqrt 5. Your biggest problem when trying to extend the Binet formula to non-integer exponents is that when ''x'' is not an integer, (-1)^''x'' is not well defined. For example, in the second line of your table, you have used the value ''i'' for (-1)^(1/2) rather than -''i'' - this is an arbitrary choice. It is equally valid to say that FR(1/2) could be 0.568864481 '''+''' 0.351577584 i. [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 08:03, 9 August 2011 (UTC)
::Thanks for the help, but there must be some kind of value of x when ((1+sqrt 5)/2)^(x)-((-1)^(x)/((1+sqrt 5)/2)^(x))))/sqrt 5=i, right?... is there any way to compute this? [[User:Robo37|Robo37]] ([[User talk:Robo37|talk]]) 10:51, 11 August 2011 (UTC)
::Your function seems to get me to -18.1040617 + 24.5918696 i instead of i for some reason. [[User:Robo37|Robo37]] ([[User talk:Robo37|talk]]) 11:02, 11 August 2011 (UTC)
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