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 March 5th, 2017, 11:28 AM #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, 12:21 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 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

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