[ipsin]

What's interesting to me is that you can demonstrate this without too much work, and it's true for starting numbers other than 1,1, such as Lucas numbers (1,3,4,7,11,...)

f(n+2)=f(n+1)+f(n)

If you posit that this number has a solution in f(n)=a^x, you see that

a^(n+2)=a^(n+1)+a^n

Dividing by a^n, you get a^2=a+1, a simple quadratic with two roots, (1+sqrt(5))/2 and (1-sqrt(5))/2, or the golden ratio and approximately -0.618.

Because any constant multiple of the function will also solve the equation, a solution to f(n) will be some linear combination of f(n)=c1(1.618..)^n+c2(-0.618)^n, for constants c1 and c2. You calculate c1 and c2 to match your initial conditions.

As n increases, the second term tends to 0, so f(n+1)/f(n) approaches the golden ratio.

[NAFV_P]

> What's interesting to me is that you can demonstrate this without too much work, and it's true for starting numbers other than 1,1, such as Lucas numbers (1,3,4,7,11,...)

Your comment reminded me of chaos theory, small changes in initial conditions can have a radical effect upon the final outcome. In this case it seems to be the opposite.

> Because any constant multiple of the function will also solve the equation, a solution to f(n) will be some linear combination of f(n)=c1(1.618..)^n+c2(-0.618)^n, for constants c1 and c2. You calculate c1 and c2 to match your initial conditions.

Bear with me a second, I have to fish out an old geometry book...

"Geometry", Roger Fenn, Springer-Verlag 2001, page 24:

I've got it, the equation you gave is very similar to Binet's Formula:

http://mathworld.wolfram.com/BinetsFibonacciNumberFormula.ht...