Generalised 'Fibonacci' Series and Phi
Date: 02/10/2002 at 11:47:18 From: Stuart Donnan Subject: Generalised 'Fibonacci' series and phi I have searched but I can't find an answer to this. I have shown with a spreadsheet that a Fibonacci-style series that starts with any two numbers at all, and adds successive items, produces a ratio of successive items that converges to phi in about the same number of terms as for the 1 1 2 3 5 etc. basic Fibonacci series. I can't find reference to this - maybe I don't know how to look - surely this is well known? Is it - and is it provable? Thanks.
Date: 02/11/2002 at 07:01:52 From: Doctor Floor Subject: Re: Generalised 'Fibonacci' series and phi Hi, Stuart, Thanks for your question. To prove your conjecture we will delve into formulas of generalized Fibonacci sequences (sequences satisfying X(n) = X(n-1) + X(n-2) ). Let's first investigate two very special members of the family of generalized Fibonacci sequences, those of the form X(1)=g=g^1, X(2)=g^2, X(3)=g^3, ..., X(n)=g^n [g nonzero] Since X(n)=X(n-1)+X(n-2), we must have: g^n = g^(n-1) + g(n-2) g^n - g^(n-1) - g(n-2) = 0 g^(n-2) * (g^2 - g - 1) = 0 g^2 - g - 1 = 0 [we can divide by g^(n-2) becuase g is nonzero] So g must be one of the roots of g^2 - g - 1 = 0, so it must be one of the following numbers: Phi = (1+SQRT(5))/2 (the Golden Ratio) 1-Phi = (1-SQRT(5))/2 So we have two very special generalized Fibonacci sequences: Phi^1, Phi^2, Phi^3, ..., Phi^n, ... (1-Phi)^1, (1-Phi)^2, ..., (1-Phi)^n, ... I leave it for you to check that, when you have two generalized Fibonacci sequences A(n) and B(n), and two numbers a and b, then a*A(n) + b*B(n) is a generalized Fibonacci sequence. In fact, all generalized Fibonacci sequences can be calculated in this way from Phi^n and (1-Phi)^n. This can be seen from the fact that any two initial terms can be created by some a and b from two (independent) pairs of initial terms from A(n) and B(n), and thus also from Phi^n and (1-Phi)^n. So each generalized Fibonacci sequences can be written as X(n) = a*Phi^n + b*(1-Phi)^n. Now your statement follows for all sequences for which a<>0, when we realize that |1-Phi|<1, so that (1-Phi)^n --> 0 for n --> infinity. So we see that there are exceptions: sequences b*(1-Phi)^n. If you have more questions, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
Date: 02/12/2002 at 04:31:46 From: Stuart Donnan Subject: Generalised Dr. Floor, Many thanks for your clear response. I did keep looking, and thought that this answer by Dr. Rob in the Dr. Math archives was helpful: The Golden Ratio http://mathforum.org/dr.math/problems/miller2.23.98.html Generalising F(n+1) = F(n) + F(n-1) (without requiring F(0)=0, F(1)=1 etc.) and dividing both sides by F(n) and rearranging etc. Dr. Rob concludes, 'if the ratio converges to anything, it converges to the Golden Ratio'. This is neat but not quite the same sort of 'proof' as yours. I'm surprised that there isn't more to be found (at least easily) about 'generalised' Fibonacci series, but there you go. Thanks again. Stuart
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