On 13 May 2020 at 12:32, rcs@xmission.com wrote:
(from Bernie Cosell <bernie@fantasyfarm.com>) (The list filtering software intercepted this message, without telling me why. I've retyped the formulas. --rcs)
Thanks I didn't know the standard syntax for such so I tried to past in actual formulae.
Binet’s formula for the Fibonacci numbers follows fairly easily from the assumption that F_n=x^n, but I didn’t quite follow how/why one would make that guess. I tried to see if I could come up with Binet’s formula not “guessing” that form. Instead I tried a finite polynomial F_n = sum(i=0,n,a_i*x^i).
Thanks for the comments and I now see where I screwed up. *IF* instead of doing i=0,n I had started with the simpler case: i=n-1,n then things become much simpler: We're want to find the x such that F_n = ax^n + bx^(n-1), n>1 and F_0 = 0 F_1 = 1; so F_n = ax^n + bx^(n-1) = ax^(n-1) + bx^(n-2) + ax^(n-2) + bx^(n-3) You can factor it all out so you get ax^(n-2)*(x^2-1-1) + bx^(n-3)*(x^2-x-1) = 0 or (ax^(n-2) + bx^(n-3))* x^2-x-1 = 0 Obviously the thing on the left always positive, so can't pick "special' a's and b's to get you to zero, so the only values of x that works is just what we expected. and then we get right back to Binet's formula. You just can't get away from phi :o) Thanks.. /bernie\ Bernie Cosell bernie@fantasyfarm.com -- Too many people; too few sheep --