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

{| width = "100%"

|- ! colspan="3" align="center" | Mathematics desk |- ! width="20%" align="left" | < August 13 ! width="25%" align="center"|<< Jul | August | Sep >> ! width="20%" align="right" |Current desk > |}

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

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).