Wikipedia:Reference desk/Archives/Mathematics/2011 August 14

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August 14

Imaginary Fibonacci

What is the Fibonacci root of i? That is, what put into the function ((1+sqrt 5)/2)^(x)-((-1)^(x)/((1+sqrt 5)/2)^(x))))/sqrt 5 = i? Robo37 (talk) 10:57, 14 August 2011 (UTC)

So I think you're asking us to solve $F_{x}=i$, where $F_{x}$ is the expression for the Fibonacci series, which you have slightly wrong, $F_{x}={\frac {({\frac {1+{\sqrt {5}}}{2}})^{x}-({\frac {1-{\sqrt {5}}}{2}})^{x}}{\sqrt {5}}}$.

This is not an easy problem. --COVIZAPIBETEFOKY (talk) 12:56, 14 August 2011 (UTC)

Does Wolfram Alpha's answer make sense? (I was using COVIZAPIBETEFOKY's version of your question – but Wolfram's has several different things about it). Grandiose (me, talk, contribs) 13:43, 14 August 2011 (UTC)

Thanks a lot for that, I really appreciate the help. Yeah I got my equation from the second equation listed at [1] because when I tried the first one with real integars in Google calculator I got incorrect resaults for some reason while the second one got 1 when I entered 1, 1 when I entered 2, 2 when I entered 3, 3 when I entered 4, 5 when entered 5, 8 when I entered 6... ect. Now there's something else that baffles me, howcome when I enter those answers you gave into my function I don't get i? Could you possibly calulate the same thing using my equation, out of curiousity? Thanks again. Robo37 (talk) 14:26, 14 August 2011 (UTC)

Wait, I've got it. Thanks for linking me to that program, it looks like it could come in really handy. I still don't understand why both forumla's work differently to each other in the complex plane but the same in the real plane but that's not a burning issue of mine. Thanks a lot for the link. Robo37 (talk) 14:44, 14 August 2011 (UTC)

The problem with trying to generalise Binet's formula

$F_{x}={\frac {({\frac {1+{\sqrt {5}}}{2}})^{x}-({\frac {1-{\sqrt {5}}}{2}})^{x}}{\sqrt {5}}}$

to non-integer values of x is that ${\frac {1-{\sqrt {5}}}{2}}$ is negative, so its xth power is not well defined for non-integer x (see exponentiation). Even if you extract a factor of -1 to get

$F_{x}={\frac {({\frac {1+{\sqrt {5}}}{2}})^{x}-(-1)^{x}({\frac {-1+{\sqrt {5}}}{2}})^{x}}{\sqrt {5}}}$

(which is what I think you are attempting to do) then you have the same problem because $(-1)^{x}$ is not well defined for non-integer x. You have two possible values for $(-1)^{\frac {1}{2}}$, three possible values for $(-1)^{\frac {1}{3}}$ etc. If you allow x to take irrational or complex values the problem becomes even worse - the set of possible values becomes infinite. Gandalf61 (talk) 14:50, 14 August 2011 (UTC)

The equation is

$e^{ax}+e^{i\pi bx}-i{\sqrt {5}}=0\,$

where

$a=\log(1+{\sqrt {5}})-\log 2\approx 0.481212$

and

$b=\log(-1+{\sqrt {5}})-\log 2=-a.$

Truncate the taylor series of the exponentials, and solve the resulting algebraic equation numerically. Bo Jacoby (talk) 10:38, 15 August 2011 (UTC).

That gives one solution - but my point is that selecting $e^{i\pi x}$ as a value of $(-1)^{x}$ is arbitrary - why not use $e^{-i\pi x}$ or $e^{3i\pi x}$ etc. For almost all non-integer values of x you have an infinite number of possible values for $(-1)^{x}$. Therefore there are (almost certainly) an infinite number of "Fibonacci roots" of i, not just one. Gandalf61 (talk) 16:23, 15 August 2011 (UTC)

The equation

$f(x)=0\,$

where

$f(x)=e^{ax}+e^{-i\pi ax}-i{\sqrt {5}}\,$

has an infinite number of solutions. Not just one solution. $f\,$ is like a polynomial of infinite order. There has been taking care of your point. Bo Jacoby (talk) 17:57, 15 August 2011 (UTC).

The solution −1.206+0.512i was computed in J like this.

  a=.^.-:>:%:5
  f=.^@(a&*)+^@((-j.o.a)&*)-(j.%:5)"_
  {:>{:p.f t.i.14
_1.20636j0.51203

Bo Jacoby (talk) 11:18, 16 August 2011 (UTC).