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:

\[\phi^2 = \phi + 1\]

and

\[\phi = \frac{1+\sqrt{5}}{2}\]

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

\[F_1 = 1, \, F_2 = 1, \, F_{n+1} = F_n + F_{n-1}\]

Now, the golden ratio is tied into the Fibonacci numbers in this way: the ratios of successive Fibonacci numbers approaches the golden ratio as \(n\) 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 \(\phi\)? \(\phi^2\)? \(\phi^{n-1}\)? Let's see with the last one:

\[\phi^{n+1} = \phi^n + \phi^{n-1}\]

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 \(F_7\), the seventh Fibonacci number? Well, I would create the sequence:

\[1,\, 1,\, 2,\, 3,\, 5,\, 8,\, 13\]

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

\[1,\, \phi,\, 1+\phi,\, 1+2\phi,\, 2+3\phi,\, 3+5\phi,\, 5+8\phi,\, 8+13\phi\]

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

PROBLEM 1: Prove that \(\phi^n = F_{n-1} + F_{n} \phi\) for all \(n\in \mathbb{Z}\)

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

\[\left( \frac{-1}{\phi}\right)^n = F_{n-1} + F_{n}\left( \frac{-1}{\phi}\right)\]

Subtracting this from the equation in Problem 1, we get

\[\phi^n - \left( \frac{-1}{\phi}\right)^n = F_n \left(\phi-\left( \frac{-1}{\phi}\right)\right)\]

\[\Rightarrow F_n = \frac{\phi^n - \left( \frac{-1}{\phi}\right)^n}{\phi-\left( \frac{-1}{\phi}\right)}\]

or

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

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 \(F_1,\, F_2,\,\cdots,\, F_{99}\) to find \(F_{100}\), for if we have a calculator, we can simply use this formula. Also, because \(|(-\phi)^{-n}|<1\) for all positive values of \(n\), we can abbreviate the formula to

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

where \([a]\) represents the nearest integer to \(a\).

Test your knowledge with these questions:

More to come

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

awesome post bob, thanks

Log in to reply

Nice Bob continue this type of thread.

Log in to reply

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

Log in to reply

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

Log in to reply