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I have heard much about the numerous appearances of the ratio found in nature: 1.6180339887.

Are there any actual mathematical uses that have been found of this number? What are its advantages? Just curiosity, really.

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You might enjoy Conway's soldiers: en.wikipedia.org/wiki/Conway's_Soldiers. It's a very beautiful and ingenious application of the golden ratio. –  Alex B. Mar 14 '11 at 4:54
    
How very interesting. Thank you! –  BKaylor Mar 14 '11 at 19:15
    
Here is a link to the same question on physics.stackexchange.com, for all interested! physics.stackexchange.com/questions/6904/… –  BKaylor Mar 15 '11 at 16:07
    
The golden ratio appears in formulas related to the icosahedron and the dodecahedron. –  Rahul Mar 24 at 5:33
    
I'm not sure if the edit (man-made -> mathematical) changes the meaning of the question.... –  Hurkyl Mar 24 at 5:37

4 Answers 4

up vote 11 down vote accepted

I'm not sure if you mean "man-made" as in real world applications or as in mathematical applications, but one of my favorite things about the golden ratio is that it is somehow the hardest number to approximate by rationals.

For one thing, it has the slowest converging simple continued fraction expansion. This is reflected in the fact that if you apply Euclid's algorithm to two successive Fibonacci numbers, you get a quotient of 1 at every step.

There is also Hurwitz's theorem, which says that for any irrational number $\xi$, there are infinitely rationals $\frac{n}{m}$ such that$$\left| \xi - \frac{n}{m} \right| < \frac{1}{\sqrt{5}m^2}.$$ The constant on the RHS can't be improved since, if we replace $\sqrt{5}$ with some larger number, then the statement of the theorem doesn't work when we let $\xi$ be the golden ratio. (Informally, we can't get too close, in a number theoretic way, to the golden ratio.)

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Thank you for your response, and for showing its application in Hurwitz's theorem! I was thinking mathematical in my mind, it translated to man-made somewhere between my brain and my fingertips... Anyways! Are there any formulas or theorems in physics which involve the ratio as a constant? –  BKaylor Mar 14 '11 at 4:30
    
The golden ratio seems to appear when you look at the ratios of some particles' magnetic moments, but that's about all that I know. Maybe the people at physics.stackexchange would be willing to give you an answer –  Daenerys Naharis Mar 14 '11 at 20:36
    
Shouldn't it be any "irrational number $\xi$"? –  dimensio1n0 Jun 11 '13 at 3:30
    
@dimension10 Yes, thanks for the correction. –  Daenerys Naharis Jun 11 '13 at 5:04

This is one realm in which Wikipedia can be amazingly helpful. http://en.wikipedia.org/wiki/Golden_ratio lists all sorts of applications, with citations if you're picky (which is commendable).

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I should have been more specific: By man-made I should have said mathematical or physics related. I've read the wiki! –  BKaylor Mar 14 '11 at 4:27
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Ah, that makes more sense. Well, the golden ratio occurs as the unique positive solution to $x^2 - x - 1 = 0$. It is also fundamental for the construction of a number of regular polyhedra (and probably regular polytopes of any dimension, although I'm not certain of this. –  Alex Becker Mar 14 '11 at 4:38

One part that I can think of mathematically would be the use of golden ratio in fibonacci numbers.

The golden ratio is the part of series of well known fibonacci number series .

Here is the link since I am unable to write the actual series for the fibonacci here...... fibonacci

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First, I strongly recommend that you watch the following three Vihart videos:

  1. Doodling in Math: Spirals, Fibonacci, and Being a Plant [1 of 3]
  2. Doodling in Math Class: Spirals, Fibonacci, and Being a Plant [2 of 3]
  3. Doodling in Math: Spirals, Fibonacci, and Being a Plant [Part 3 of 3]

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Now, to add to what @user8250 said, the golden ratio is very closely related to the Fibonacci Numbers. The Fibonacci Sequence is defined by the following recurrence relation,

$$F_{n+1} = F_n + F_{n-1}$$

where

$$F_0=0 \ \ \text{and} \ \ F_1=1.$$

From this definition we see that the sequence starts as follows,

$$0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...$$ An explicit formula for $F_n$ that can be derived using linear algebra is known as Binet's Formula. This formula is

$$F_n = \frac{\phi^n -(-\phi)^{-n}}{\sqrt{5}}$$

where $\phi = \frac{1+\sqrt{5}}{2}$, the golden ratio. Even if you are not familiar with linear algebra, you can prove (but not derive) this formula using induction.

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Lets look at the ratio of adjacent Fibonacci Numbers as they get large. We see that,

$$\lim\limits_{n \to \infty} \frac{F_{n+1}}{F_n} = \lim\limits_{n \to \infty} \frac{\frac{\phi^{n+1} -(-\phi)^{-(n+1)}}{\sqrt{5}}}{\frac{\phi^n -(-\phi)^{-n}}{\sqrt{5}}} = \lim\limits_{n \to \infty} \frac{\phi^{n+1} -(-\phi)^{-(n+1)}}{\phi^n -(-\phi)^{-n}} = \lim\limits_{n \to \infty} \frac{\phi^{n+1}}{\phi^n} = \phi.$$

Interestingly, we could have derived this limit just as easily with our original definition of Fibonacci Numbers:

Suppose that the limit of the ratios converges and that $\lim\limits_{n \to \infty} \frac{F_{n+1}}{F_n} = L$.

Because $F_{n+1}=F_n+F_{n-1}$, we can say,

$$L = \lim\limits_{n \to \infty} \frac{F_{n+1}}{F_n} = \lim\limits_{n \to \infty} \frac{F_n+F_{n-1}}{F_n} = \lim\limits_{n \to \infty} \left( \frac{F_n}{F_n} + \frac{F_{n-1}}{F_n} \right) = 1 + \frac{1}{L}.$$

So we have that $$L = 1 +\frac{1}{L}.$$

This brings us back to the golden ratio. The above equation is often used to define the golden ratio: Namely, the golden ratio is the only positive solution to this equation. We can determine the value of limit, and in process the value of the golden ratio by solving,

$$L=1+\frac{1}{L}$$ $$L^2 = L + 1$$ $$L^2 - L - 1 = 0$$

Now, apply the quadratic formula to obtain the two roots,

$$L = \frac{1+\sqrt{5}}{2} \ \ \text{and} \ \ L = \frac{1-\sqrt{5}}{2}$$

We know that the negative root must be extenouous as we are concenred with the limit of two positive numbers. So, as we previously saw, the limit is $L = \frac{1+\sqrt{5}}{2} = \phi$. With this method, one must also show that the limit exists, which I will omit, but isn't too difficult.

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