Mathematical Mystery Tour: Preparing for Fibonacci Day: Fibonacci and the Golden Ratio

Monday, November 17, 2014

Preparing for Fibonacci Day: Fibonacci and the Golden Ratio

One of the more well know properties of the Fibonacci sequence is its relationship to the Golden Ratio (or Phi):

If you take the ratio of two consecutive Fibonacci number (put the bigger number on top, and the smaller one on the bottom) you will get a fraction that approximates the Golden Ratio. The larger the Fibonaccci numbers are, the closer your approximation will be.

This is true, but what most people don’t know is that this property is not unique to the Fibonacci sequence. If you start with numbers other than 1 and 1, you will the sequence will still begin to approximate the value of the golden ration closer and closer as the numbers get bigger. And you don’t have to pick integers – you can choose fractions, decimals, or even negative numbers.

For example let’s look at the Lucas sequence. The Lucas sequence begins with 2 and 1 (instead of 1 and 1), but otherwise works just like the Fibonacci sequence.

The 41^st Lucas number divided by the 40^th Lucas number is:

370248451 / 228826127 = 1.61803398874989480550007298773185983259682579865541...

The 41^st Fibonacci number divided by the 40^th Fibonacci number is:

165580141 / 102334155 =
1.61803398874989489090910068099941803398874989489090...

Both of these are the same up to 14 decimal places! I don’t think I will ever need to be more accurate, but If I do I can do it.

If I use the equation listed above for the definition of the Golden Ration (Phi) is:

1.61803398874989484820458683436563811772030917980576...

So why did I pick the Lucas sequence? Well, the Lucas sequence hides another secret about Phi that the Fibonacci sequence does not have. In the Lucas sequence the first two terms are 2 and 1. We will call them L₀ = 2, and L₁ = 1. L₂ = 3, which is equal to Phi² rounded to the nearest integer. L₃ = 4, which is Phi³ rounded to the nearest integer. In fact, this pattern continues to at least the 53^rd term. I suspect that it continues further, but my spreadsheet program is only accurate to 12 digits.