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 March 5th, 2017, 12:28 PM #1 Newbie   Joined: Feb 2017 From: Netherlands Posts: 8 Thanks: 2 Math Focus: Trigonometry and complex numbers Why does this approach the golden ratio? Hi. If you pick any two number, say for instance 5 and 7 and you keep adding them together (much like the Fibonacci Sequence) you get this: $\displaystyle 5, 7, 12, 19, 31, 50, 81, 131, ...$ and so on. The ratio between the last two numbers always approaches the golden ratio, no matter which set of numbers you choose. But I just cannot understand why this happens... Can anybody provide some good sources and/or explanation? Thanks in advance, Xxmarijnw March 5th, 2017, 01:21 PM #2 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs Suppose we look at the recurrence: $\displaystyle A_{n}=A_{n-1}+A_{n-2}$ The characteristic equation is: $\displaystyle r^2-r-1=0$ which has roots: $\displaystyle r=\frac{1\pm\sqrt{5}}{2}$ And so the closed form is: $\displaystyle A_{n}=k_1\left(\frac{1+\sqrt{5}}{2}\right)^n+ k_1\left(\frac{1-\sqrt{5}}{2}\right)^n$ The parameters $k_i$ will depend in the initial values. Since: $\displaystyle \left|\frac{1-\sqrt{5}}{2}\right|<1$ Then as $n$ grows without bound, we will find: $\displaystyle A_n\to k_1\left(\frac{1+\sqrt{5}}{2}\right)^n$ And so: $\displaystyle \frac{A_{n+1}}{A_{n}}\to \frac{k_1\left(\dfrac{1+\sqrt{5}}{2}\right)^{n+1}} {k_1\left(\dfrac{1+\sqrt{5}}{2}\right)^n}= \frac{1+\sqrt{5}}{2}=\varphi$ Thanks from topsquark and Xxmarijnw Tags approach, fibonacci sequence, golden, golden ratio, ratio Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Agno Number Theory 7 November 15th, 2019 05:55 AM wrightarya Algebra 2 July 18th, 2014 08:36 PM gaussrelatz Algebra 2 February 12th, 2013 09:54 PM Voltman Advanced Statistics 9 March 13th, 2011 09:39 PM Daniel Derwent Algebra 4 June 25th, 2010 10:03 PM

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