The Golden Ratio: Another Link to the Fibonacci's

Here is the previous post concerning the Golden Ratio. For a collection of all the posts concerning the Golden Ratio, click #GoldenRatio below.


It's been a while since I've last posted. I apologize. School has been busy, as well as the recent math olympiad contests here in the US. So, to refresh your memory of the Golden Ratio, remember that it is the positive solution to the following equation:

ϕ2=ϕ+1\phi^2 = \phi + 1

and

ϕ=1+52\phi = \frac{1+\sqrt{5}}{2}

Also remember that the Fibonacci numbers are described by the recurrence relation

F1=1,F2=1,Fn+1=Fn+Fn1F_1 = 1, \, F_2 = 1, \, F_{n+1} = F_n + F_{n-1}

Now, the golden ratio is tied into the Fibonacci numbers in this way: the ratios of successive Fibonacci numbers approaches the golden ratio as nn approaches infinity. However, there is even more connecting the golden ratio to the Fibonacci numbers, as we will explore here.


Notice the first equation above. What happens if I multiply both sides by ϕ\phi? ϕ2\phi^2? ϕn1\phi^{n-1}? Let's see with the last one:

ϕn+1=ϕn+ϕn1\phi^{n+1} = \phi^n + \phi^{n-1}

Aha! This is exactly like that Fibonacci number formula, but with powers of the golden ratio instead. Since they look very much the same, they also act very much the same. For instance, what if I wanted to find F7F_7, the seventh Fibonacci number? Well, I would create the sequence:

1,1,2,3,5,8,131,\, 1,\, 2,\, 3,\, 5,\, 8,\, 13

and see that the seventh term is F7=13F_7=13 I can do the same thing with the golden ratio, but based off of different starting values: ϕ0=1\phi^0 = 1, ϕ1=ϕ\phi^1 = \phi. Then our sequence (starting at ϕ0\phi^0) becomes:

1,ϕ,1+ϕ,1+2ϕ,2+3ϕ,3+5ϕ,5+8ϕ,8+13ϕ1,\, \phi,\, 1+\phi,\, 1+2\phi,\, 2+3\phi,\, 3+5\phi,\, 5+8\phi,\, 8+13\phi

Thus, ϕ7=8+13ϕ\phi^7 = 8 + 13\phi. Any integral power of ϕ\phi can be written in a+bϕa + b\phi form, and you should see the pattern for the numbers aa and bb.

PROBLEM 1: Prove that ϕn=Fn1+Fnϕ\phi^n = F_{n-1} + F_{n} \phi for all nZn\in \mathbb{Z}

In fact, any sequence GnG_n that follows the Fibonacci recurrence relation will follow this more general formula that Gn=Fn1G0+FnG1G_n = F_{n-1} G_0 + F_{n} G_1. Now, using this fact with the golden ratio, we can expand that to the other root of the first equation above, which is 1ϕ\frac{-1}{\phi} (try and check that yourself). Since this value satisfies the same initial formula, it will satisfy our deduced formula, namely

(1ϕ)n=Fn1+Fn(1ϕ)\left( \frac{-1}{\phi}\right)^n = F_{n-1} + F_{n}\left( \frac{-1}{\phi}\right)

Subtracting this from the equation in Problem 1, we get

ϕn(1ϕ)n=Fn(ϕ(1ϕ))\phi^n - \left( \frac{-1}{\phi}\right)^n = F_n \left(\phi-\left( \frac{-1}{\phi}\right)\right)

Fn=ϕn(1ϕ)nϕ(1ϕ)\Rightarrow F_n = \frac{\phi^n - \left( \frac{-1}{\phi}\right)^n}{\phi-\left( \frac{-1}{\phi}\right)}

or

Fn=ϕn(ϕ)n5F_n = \frac{\phi^n - (-\phi)^{-n}}{\sqrt{5}}

This is known as Binet's formula, although the result was known to Euler and de Moivre before. Binet developed a more general form of the formula, which we may discuss later.

Note the implications of this formula. We know longer have to calculate F1,F2,,F99F_1,\, F_2,\,\cdots,\, F_{99} to find F100F_{100}, for if we have a calculator, we can simply use this formula. Also, because (ϕ)n<1|(-\phi)^{-n}|<1 for all positive values of nn, we can abbreviate the formula to

Fn=[ϕn5]F_n = \left[\frac{\phi^n}{\sqrt{5}}\right]

where [a][a] represents the nearest integer to aa.


Test your knowledge with these questions:

Problem 1

Problem 2

More to come

Note by Bob Krueger
5 years, 3 months ago

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Please start a post like this for teaching calculus, I don't know what it is.....

Satvik Golechha - 5 years, 2 months ago

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This is an awesome thread. Please continue with the notes and problems.

Sharky Kesa - 5 years, 3 months ago

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Thanks Bob for helping many people learn things.........Please keep the noble great awesome work up!

Satvik Golechha - 5 years, 3 months ago

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Nice Bob continue this type of thread.

Rajeev Giri - 5 years, 3 months ago

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awesome post bob, thanks

Ashu Dablo - 5 years, 3 months ago

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