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 fibonacci like Gottfried Ultimate Fellow Posts: 758 Threads: 117 Joined: Aug 2007 07/17/2010, 10:31 AM (07/17/2010, 08:45 AM)mike3 Wrote: (07/17/2010, 07:02 AM)Gottfried Wrote: I had tried it up to Quote:$e^{(1+i)u} - e^u = -1$ but gave up then... What's the problem? No - there is no problem. It only looked not easy enough for an idle afternoon so I didn't try to proceed further on my own. Your arguments make it actually easy to proceed and even more: you've done some nice graphs... So your msg was a nice and interesting contribution! Gottfried Gottfried Helms, Kassel tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 07/17/2010, 08:03 PM (This post was last modified: 07/17/2010, 08:08 PM by tommy1729.) (07/17/2010, 01:50 AM)mike3 Wrote: $f(z) = f(z - 1) + f(z + i)$ If we assume a solution of the form $f(z) = r^z$ exists, like in the Fibonacci numbers, we can get the equation $e^{(1+i)u} - e^u = -1$ 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". 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. nice post mike ! i conjecture that all non-periodic entire function solutions are of this form. note that i didnt say anything about the periodic solutions , not even sure they exist. i wonder if elliptic solutions exist. (once again flirting with double periodic functions ) beautiful memories ; as an early teenager i defined 2 classes of functions as $f(z) = \sum_{n=0}^{\infty} C_n e^{G z}$ $f(z) = \sum_{n=0}^{\infty} D_n e^{E z}$ and assumed them to be equivalent , where G are the gaussian integers and E are the eisenstein integers. these were my " pre - taylor " series before i learned about taylor or laurent or even kahn series. similarly my gaussian / eisenstein polynomials were : $f(z) = \sum_{n=0}^{A} C_n e^{G z}$ $f(z) = \sum_{n=0}^{B} D_n e^{E z}$ for positive integer A and B , before i learned about polynomials or signomials. i did some investigations which could be considered pre-galois theory , pre-abelian variety , multisections and searching for zero's. and a lot of modular arithmetic , which seemed related. ( and abelian groups of order p^2 of course ) even today , i still find all that intresting. maybe mike has a similar history ? later i switched to number theory , but partially never forgot that. ( 'partially' because apparantly i forgot the relation to " fibonacci like " , i did find that fibo equation in my old papers of " gaussian polynomials " ) sorry for the emo. regards tommy1729 mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 07/18/2010, 01:47 AM (This post was last modified: 07/18/2010, 01:53 AM by mike3.) Periodic ones exist; indeed all the "basis" solutions $e^{uz}$ are periodic with complex period $\frac{2\pi i}{u}$. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 07/18/2010, 07:44 AM (07/16/2010, 12:27 PM)tommy1729 Wrote: (07/16/2010, 07:03 AM)bo198214 Wrote: 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. ... why isnt it a recursion ??? A recursion needs to be determined on the natural numbers (or in other words unique), i.e. you can trace back every f(n) to some initial given f(0) or f(1) or so. Like in the fibonacci numbers f(n)=f(n-1)+f(n-2),f(0)=0,f(1)=1 Or even put it differently: when you write a computer program with the defining relation it must terminate. Yours is not a recursion just some functional equation. tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 07/18/2010, 09:54 PM (07/18/2010, 01:47 AM)mike3 Wrote: Periodic ones exist; indeed all the "basis" solutions $e^{uz}$ are periodic with complex period $\frac{2\pi i}{u}$. maybe there are solutions with other periods as well. playing around with the c_n coefficients we can construct another periodic solution i guess. i conjectured about non-periodic entire functions only , because it might be that there exist periodic solutions that are of a different form ? i was aware of the period of the "basis" solutions , i just didnt express myself very well. you are correct of course. i hope nobody minds this is not really about tetration. ( and apparantly recursion is reserved for the integers as bo explained ) « Next Oldest | Next Newest »

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