Golden Ratio

The golden ratio is a truly remarkable mathematical constant. While, it's not quite as famous or important as π or e, there's still a wealth of information to be discovered from studying it.



For any two numbers a and b that satisfy the equation. That is, the ratio of the a + b to a is equal to the ratio of a to b. To find the exact value of φ we use the fact that (a/b) = φ, meaning a = bφ. Substituting in the first equation gives



And solving for the positive solution yields

.

(Approximated).


One of the coolest thing's about φ is it's relation to the Fibonacci series. The Fibonacci series begins with 1, and the next term is determined by the sum of the previous two terms. The first 10 terms of the series are

1, 1, 2, 3, 5, 8, 13, 21, 34, 55...

So, you get the idea. The explicit definition of the Fibonacci series is below:



The first relationship that comes to mind is seen when you look at the following table

F(n) / F(n-1)	Value	Difference From φ
1/1	1	.618
2/1	2	-0.381
3/2	1.5	0.118
5/3	1.6667	-0.048
8/5	1.6	0.0180
13/8	1.625	-0.0070
21/13	1.6154	0.0026
34/21	1.6190	-0.00101
55/34	1.6176	0.000386
89/55	1.6182	-0.000148

As you can see, the ratio between the consecutive terms of the Fibonacci series approaches φ. This is only the beginning of the relationships these two mathematical phenomena share.

Now, comes the interesting part. Look back at the quadratic we used to calculate the value of &phi. Leave the squared term on the left-hand side of the equation and move the other two terms to the right-side. Now, if we multiply both sides of the equation by φ once, we have φ^3 = φ^2 + φ. If we multiply by φ again we have φ^4 = φ^3 + φ^2. Obviously, since we can multiply by φ any number of times, the following equation is valid.



The recursive nature of this equation leads tho the conclusion that any power of φ can be represented in terms of it's first power and another number. For example, φ^3 is equal to φ^2 + φ, but φ^2 is equal to φ + 1, so φ^3 = 2φ + 1. Any power of φ will go through a similar process until it's eventually expressed in terms of the first power of φ and a constant. The interesting part is the pattern that emerges as you express powers of φ in terms of it's first power.

φ^1 = φ
φ^2 = φ + 1.
φ^3 = φ^2 + φ = (φ + 1) + φ = 2φ + 1
φ^4 = φ^3 + φ^2 = (2φ + 1) + φ +1 = 3φ + 2
φ^5 = φ^4 + φ^3 = (3φ + 2) + (2φ +1) = 5φ + 3.


Observe the coefficients of φ. In order, they go 1, 1, 2, 3, 5. Recognize the pattern? If you don't, I'll give you a hint. It starts with 'F' and ends with 'ibonacci'. That's right! The coefficients of the φ-terms follow the Fibonacci series. That's not all. If you observe the constant term as well, you find that it follows the same pattern, only offset by a single index.



It's relationship with the Fibonacci series doesn't end there. Due to the fact that the formula for calculating the powers of φ is very similar to the formula for determining terms of the Fibonacci series, we can actually use φ to calculate any term of the series we want without bothering to generate the entire series up to that number. The formula to do this is as follows:



It's not immediately obvious where this came from. Though, the proof for it is really quite simple. Before beginning, you should note that φ only refers to the positive solution to the quadratic. The negative solution is simply ignored. But, notice that the negative solution to the quadratic also satisfies the relationship between it's powers that φ does. That is, it's nth power is equal to the sum of the (n-1)th power and the (n-2)th power. Notice also, that the second solution to the quadratic is equivalent to 1 - φ, because:



(Now you know where the (1-φ) came from.)

Now comes the proof. Let's take two real numbers, a and b, and form the expression .aφ^n + b(1-φ)^n. Because of the properties of φ and (1-φ), the value of the expression follows the patterns from before. That is, if we form a series by starting with n = 1, and incrementing n by one each time, it's value at n = k is equal to it's to the sum of it's value at n = k-1 and n = k-2. This can be proven through substitution from the formulas we already have.



Because of this, if this, if we have a function G(n) = aφ^n + b(1-φ)^n, we have the fact that G(n) = G(n-1) + G(n-2). Exactly like the Fibonacci series. Now, all that's left to do is make sure that it begins the same way as the Fibonacci series and the rest of the terms will follow automatically. That is, we must substitute values for a and b so it starts the same way as the series. We do this with a system of equations, using the first two terms of the series. (We could actually use any two terms, but that would make things unnecessarily complicated). Now, we could use the 1st, and 2nd term (1 and 1) but things are even simpler if we use the 0th term and the 1st term. Intuitively, the 0th term should be equal to 0, because 0 + 1 = 1 (the second term.). Now, we obtain the equations




Upon solving, we find



Substituting these values in aφ^n + b(1-φ)^n we obtain the formula we set out to prove.

~Q.E.D.


While this formula is great and all, it leaves a little to be desired. What if you had the problem of finding the first Fibonacci number to be over 1000? Or even determining whether or not a certain number is a Fibonacci number? This formula would be insufficient.


Look back at the formula for generating a Fibonacci number. We can re-express it as


We can use the fact (1-φ)^n / sqrt(5) is always less than one half (you can observe this yourself by punching a few values in your calculator), meaning, that the nth Fibonacci number is the number closets to φ^n / sqrt(5). That is, you plug in φ^n / sqrt(5) and merely round it to the nearest number. This is a very useful formula. Like answering the question from below. What is the first Fibonacci number over 1000? Now, it becomes a rather simple process:




You can use the change of base formula to solve for n. Since n is the point at which φ^n / sqrt(5) becomes 1000 we merely need to take the first integer greater than n and plug it into the equation for finding Fibonacci numbers. Easy peasy lemon squeasy.


Well, that's it for now. It's some pretty cool stuff. Also, for those of you that frequent Project Euler, you could use the final expression discussed here (φ^n / sqrt(5)) to solve Problem 25 with a few calculations on a scientific calculator. It's really quite epic.