The Golden Ratio (represented by \(\displaystyle \varphi , \phi , \Phi\)) is one of the greatest discoveries Math has made with applications in but not limited to Architecture (eg. Parthenon) and designs. We will use \(\varphi\) for the golden ratio at all times in this note.

To start off, It can be represented in terms of itself and with numerous \(1\)'s: \[\displaystyle \varphi = 1 + \frac{1}{\varphi}\] which makes it slip neatly in the continued fraction and with a set of square roots: \[\displaystyle \varphi = 1+ \cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots}}} \]

\[\displaystyle \varphi=\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}\]

If a sequence follows a Fibonacci sequence structure, The ratio between a set of 2 numbers is closer to \(\displaystyle \varphi\).

\[\displaystyle 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, \cdots \] We can divide it in sets like this (preferably) \[\displaystyle (1, 1), (2, 3), (5, 8), (13, 21), (34, 55), (89, 144) \cdots \]

The bigger number divided by the smaller number is getting closer to the \(\displaystyle \varphi\) while the smaller number divided by the bigger number is getting closer to \(\displaystyle \varphi-1\)

Try it out yourself!

But we can also **RANDOMLY** choose 2 numbers to start off with. Say, \(58\) and \(6\)

\[\displaystyle 58, 6, 64, 70, 134, 214 \cdots \]

\(214 \div 134 \approx 1.6\) which is near \(\varphi\)'s \(1.618 \)ish

Try it out and see if you got it!

Stay tuned for more updates in the future and please see the my other note about \(\pi\)!

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## Comments

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TopNewestI remember that \(\varphi=\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}\)

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That is in the wiki. I do not want to copy the wiki but I will update the note for your sake

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It's OK. (Though I think \(\varphi= 1+ \cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots}}} \) is also in wiki. )

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here. Very surprised to know that the golden ratio page didn't have the continued fraction form.

Hmm...I found itLog in to reply

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wishI can answer this...Log in to reply

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\(\phi\) is a convergent of the continued fraction; such behaviour is to be expected. A more interesting question would be why the convergent satisfies such relation, regardless of the starting points of the sequence. (HINT: Think generating functions.)

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