On Wed, May 13, 2020 at 2:32 PM <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)

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). But then I’m stuck. I’d love to have discovered that this implies a_i=0 for all i/=n and a_i=1 for i=n and then Binet’s formula follows.

But it doesn't imply that; a_i = 0 for all except i = 7 is a solution, too. Or any number of the a_i can be nonzero, as long as each of the associated x_u are either phi or phi-bar (other solution of the same quadratic). Solving something with multiple solutions is going to be harder than solving something with 1 solution. Since the constraint F_n = F_(n-1) + F_(n-2) is linear, than given any two series that are solutions, their sum (and in fact any linear combination of them) is a solution too. So looking for a solution that's a sum of a bunch of terms is going to be hugely underdetermined. Better first to find solutions that are a single summand, and then if you've found all of those, then all the things you can make as linear combinations of those will also be a solution. Then you can use the initial conditions F(0) = F(1) = 1 to determine which linear combination you want. Is this helpful? Andy

But I can’t quite figure that out. When I take that summation and plug it into F_n = F_(n-1) + F_(n-2) the best I can come up with a_n x^n = F_(n-2), which doesn’t look promising. What am I missing?

I know this isn't much "fun" but for odd reasons I've been revisiting math stuff that seemed obvious to me 50+ years, but has lain dormant. Thanks /bernie\

Bernie Cosell bernie@fantasyfarm.com -- Too many people; too few sheep --

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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 --