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 fibonacci like tommy1729 Ultimate Fellow     Posts: 1,370 Threads: 335 Joined: Feb 2009 07/10/2010, 12:27 PM i was thinking about a fibonacci like recursion. f(x) = f(x - 1) + f(x + i) together with some initial conditions ( you choose ) perhaps old hat ... but i dont recall a closed form solution ( though its really hot here ) regards tommy1729 tommy1729 Ultimate Fellow     Posts: 1,370 Threads: 335 Joined: Feb 2009 07/14/2010, 01:55 PM (This post was last modified: 07/14/2010, 01:55 PM by tommy1729.) anyone ? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 07/15/2010, 02:29 AM (07/10/2010, 12:27 PM)tommy1729 Wrote: f(x) = f(x - 1) + f(x + i) i = imaginary unit? Gottfried Ultimate Fellow     Posts: 764 Threads: 118 Joined: Aug 2007 07/15/2010, 08:56 AM (07/15/2010, 02:29 AM)bo198214 Wrote: (07/10/2010, 12:27 PM)tommy1729 Wrote: f(x) = f(x - 1) + f(x + i) i = imaginary unit? if it is meant f(x) = f(x - 1) + f(x + 1) then with b = 1/2 *(1+sqrt(3)*I) // complex cuberoot of -1 f(x) = b^x is one solution. Gottfried Gottfried Helms, Kassel tommy1729 Ultimate Fellow     Posts: 1,370 Threads: 335 Joined: Feb 2009 07/15/2010, 12:06 PM (07/15/2010, 02:29 AM)bo198214 Wrote: (07/10/2010, 12:27 PM)tommy1729 Wrote: f(x) = f(x - 1) + f(x + i) i = imaginary unit? yes , i^2 = -1. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 07/16/2010, 07:03 AM (07/15/2010, 12:06 PM)tommy1729 Wrote: (07/15/2010, 02:29 AM)bo198214 Wrote: (07/10/2010, 12:27 PM)tommy1729 Wrote: f(x) = f(x - 1) + f(x + i) i = imaginary unit? yes , i^2 = -1. then its not a recursion, i.e. you dont have a unique solution for natural number arguments. I dont think the usual methods apply. tommy1729 Ultimate Fellow     Posts: 1,370 Threads: 335 Joined: Feb 2009 07/16/2010, 12:27 PM (07/16/2010, 07:03 AM)bo198214 Wrote: (07/15/2010, 12:06 PM)tommy1729 Wrote: (07/15/2010, 02:29 AM)bo198214 Wrote: (07/10/2010, 12:27 PM)tommy1729 Wrote: f(x) = f(x - 1) + f(x + i) i = imaginary unit? yes , i^2 = -1. then its not a recursion, i.e. you dont have a unique solution for natural number arguments. I dont think the usual methods apply. i didnt claim unique solutions. the usual method relates such equations with polynomials ; f(x+a) is associated with x^a , but since x - x^-1 - x^i isnt a polynomial , its trivially not solvable by the usual methods , at least not the usual methods " alone ". if it was solvable by the usual method , i would have done so and would not have posted it here. why isnt it a recursion ??? im looking for solutions that are analytic almost everywhere. tommy1729 mike3 Long Time Fellow    Posts: 368 Threads: 44 Joined: Sep 2009 07/17/2010, 01:50 AM Hmm. We have If we assume a solution of the form exists, like in the Fibonacci numbers, we can get the equation Dividing both sides by gives or . There is a caveat, however: is multivalued. It is more useful, then, to recast this equation in terms of , and then the solutions for the functional equation are given by for any -value satisfying the above exponential equation. The functional equation is linear, so any linear combination of such solutions will be another solution, and since there are infinitely many such -values, we can even consider infinite sums with arbitrary , provided this sum converges. Since there are infinitely many constants , one could say the equation is like it has "infinitely many initial conditions". This graph shows the function on the complex plane. You can see the roots, the values of used to construct solutions of the functional equation. The scale runs between on each axis (x = real, y = imag). One set of roots seems to lie along the line , while the other seems to lie along a slight curve (curving not visible here) that is asymptotic to the imaginary axis . For example, we could take two terms with the roots given by and and coefficients . This function is plotted below at the same scale. Numerical calculation can be done to verify it really does solve . I do not believe there is a closed form solution for these -values in terms of any conventional special functions, but I could be wrong (and if I am, I'd like to know what the closed solution is.). Note that these may not be the only possible solutions -- remember that the very simple case has all 1-periodic functions as solutions. Not sure what the appropriate analogy is here. But the above could be thought of as a sort of "canonical" solution like how Binet's formula solves the Fibonacci numbers. Gottfried Ultimate Fellow     Posts: 764 Threads: 118 Joined: Aug 2007 07/17/2010, 07:02 AM (07/17/2010, 01:50 AM)mike3 Wrote: ... Very nice! I had tried it up to Quote: but gave up then... Gottfried Gottfried Helms, Kassel mike3 Long Time Fellow    Posts: 368 Threads: 44 Joined: Sep 2009 07/17/2010, 08:45 AM (This post was last modified: 07/17/2010, 09:15 AM by mike3.) (07/17/2010, 07:02 AM)Gottfried Wrote: (07/17/2010, 01:50 AM)mike3 Wrote: ... Very nice! I had tried it up to Quote: but gave up then... What's the problem? This is just from setting or , since the needed to raise to the complex exponent is multivalued. I do not think it can be solved in closed form. The values tested were obtained by numerical root-finding methods (Newton's method, specifically). There could be some kind of infinite series formula or something, but I would not know what it is. « Next Oldest | Next Newest »

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