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

= 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 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 &minus;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).