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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: $\displaystyle A_{n}=A_{n1}+A_{n2}$ The characteristic equation is: $\displaystyle r^2r1=0$ 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: $\displaystyle \frac{A_{n+1}}{A_{n}}\to \frac{k_1\left(\dfrac{1+\sqrt{5}}{2}\right)^{n+1}} {k_1\left(\dfrac{1+\sqrt{5}}{2}\right)^n}= \frac{1+\sqrt{5}}{2}=\varphi$ 

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approach, fibonacci sequence, golden, golden ratio, ratio 
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