Question

Playing around with Fibonacci numbers and why 1/phi appeared?

Answer 1

What do you mean by "limit" here? What exactly were you doing?




Let f_n be the n^th fibonacci number. Try taking the limit of f_n/f_(n-1) as n approaches infinity (starting with 1/1, of course). What do you find?

Answer 2

Phi can be arrived at from many different places.

For example, the continued fraction 1+ 1/(1+1/(1+1/1+...)) . You can 'solve' this fraction by setting it equal to x.

x = 1+ 1/(1+1/(1+1/1+...))

Then subtracting 1 from each side,

x-1 = 1/(1+1/(1+1/1+...))

Then inverting to get

1/(x-1) =  1+ 1/(1+1/(1+1/1+...))

But that was just x. So, you have the quadratic equation

1/(x-1) = x   <=> 0 = x\^2 -x -1 .

This has two roots, (1 +/- sqrt(5) )/2. The '+' here would yield phi. Now look at the constant term in the quadratic x\^2 -x -1. It's '-1'. If you think about 'FOIL', you can see the two roots of this quadratic have to multiply to be -1.

So ... phi \* (other root) = -1  <=> other root = -1/phi. Those two values are solutions to the same quadratic equation.

You can also see that the ratio of the Fib sequence terms converges to phi. You can do this with fixed points or generating functions or probably a dozen other methods.