The Golden RatioDate: 02/23/98 at 11:57:04 From: karen miller Subject: The golden mean I know that the limit of the ratios of the Fibonnaci sequence is the golden mean, but I would like to see a proof of this. Thank you, Karen Miller Date: 02/23/98 at 16:50:14 From: Doctor Rob Subject: Re: The golden mean The Fibonacci numbers satisfy the recursion F(0) = 0, F(1) = 1, F(2) = 1, F(n+1) = F(n) + F(n-1), for all n > 1. Divide both sides by F(n), and rearranging: F(n+1)/F(n) = 1 + 1/[F(n)/F(n-1)]. Now suppose the ratio converges to some limit A. Take the limit of both sides as n -> infinity: A = 1 + 1/A, A^2 - A - 1 = 0, A = (1 + Sqrt[5])/2, the Golden Ratio, after we reject the negative root as not meaningful. This shows that if the ratio converges to anything, it converges to the Golden Ratio. For another approach to show that it converges to the Golden Ratio, you can use the Binet Formula for F(n): F (n) = (A^n - B^n)/(A - B), A = (1 + Sqrt[5])/2, B = (1 - Sqrt[5])/2, A + B = 1, A - B = Sqrt[5]. Since |B| < 1, as n gets very large, B^n gets very small, and we find that, as n -> infinity, lim F(n+1)/F(n) = lim (A^[n+1] - B^[n+1])/(A^n - B^n), = lim A^[n+1]/A^n, = lim A, = A. The Binet Formula can be proved by induction. It also leads to the very interesting F(n) = Round[A^n/Sqrt[5]], for all n >= 0, which tells us that the Fibonacci numbers are almost in geometric progression with common ratio A. -Doctor Rob, The Math Forum Check out our web site http://mathforum.org/dr.math/ |
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