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 fibonacci like tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 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: 1,491 Threads: 355 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: 1,395 Threads: 91 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: 787 Threads: 121 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: 1,491 Threads: 355 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: 1,395 Threads: 91 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: 1,491 Threads: 355 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 $f(z) = f(z - 1) + f(z + i)$ If we assume a solution of the form $f(z) = r^z$ exists, like in the Fibonacci numbers, we can get the equation $r^z = r^{z-1} + r^{z+i}$ Dividing both sides by $r^{z-1}$ gives $r = 1 + r^{1+i}$ or $r^{1+i} - r = -1$. There is a caveat, however: $r^{1+i}$ is multivalued. It is more useful, then, to recast this equation in terms of $u = \log( r )$, $e^{(1+i)u} - e^u = -1$ and then the solutions for the functional equation are given by $f(z) = e^{uz}$ for any $u$-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 $u$-values, we can even consider infinite sums $f(z) = \sum_{n=0}^{\infty} C_n e^{u_n z}$ with arbitrary $C_n$, provided this sum converges. Since there are infinitely many constants $C_n$, one could say the equation is like it has "infinitely many initial conditions". This graph shows the function $e^{(1+i)u} - e^u + 1$ on the complex plane. You can see the roots, the values of $u$ used to construct solutions of the functional equation. The scale runs between $\pm30$ on each axis (x = real, y = imag). One set of roots seems to lie along the line $t + it$, while the other seems to lie along a slight curve (curving not visible here) that is asymptotic to the imaginary axis $it$.     For example, we could take two terms with the roots given by $u_0 \approx 0.5397851608092811048455891598 + 0.5397851608092811048455891598i$ and $u_1 \approx -1.453673666461041618684343568 - 1.453673666461041618684343568i$ and coefficients $C_0 = C_1 = 1$. This function is plotted below at the same scale. Numerical calculation can be done to verify it really does solve $f(z) = f(z-1) + f(z+i)$. I do not believe there is a closed form solution for these $u$-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 $f(z) = f(z-1)$ 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: 787 Threads: 121 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:$e^{(1+i)u} - e^u = -1$ 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:$e^{(1+i)u} - e^u = -1$ but gave up then... What's the problem? This is just from setting $r = e^u$ or $u = \log( r )$, since the $\log$ 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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