Smiles Await You When You Rise

April 6, 2014

The golden ratio \(\phi = \frac{1+\sqrt{5}}{2}\approx 1.618\ldots\) can be represented by a beautiful repeated fraction and a beautiful repeated square root. This leads to the mystical-looking identity:

\[1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\ddots}}}} = \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}}}.\]

Let’s try to understand the origin of this identity.

The golden ratio is the solution to the equation

\[\phi^2 - \phi - 1 = 0\]

which can be rearranged in two different ways.

The more common, perhaps, is to rearrange it into

\[\phi = 1+\frac{1}{\phi}\]

This rearrangement gives us a method for generating approximations to \(\phi\):

  1. Make an initial guess.
  2. Plug that guess into the right-hand side and
  3. Use the result as your next guess for \(\phi\).

Notice that if your initial guess is perfect (meaning you guessed the true value of \(\phi\)), the next guess (and all subsequent guesses) are also perfect. We call “perfect guesses” fixed points because when you put them through this little procedure they come out unchanged.

So, if our initial guess for \(\phi\) is \(\phi_0\), our first improved guess would be

\[\phi_1 = 1+\frac{1}{\phi_0}.\]

Our second improved guess would be

\[\phi_2 = 1+\frac{1}{\phi_1} = 1+\frac{1}{1+\frac{1}{\phi_0}},\]

the third improved guess would be

\[\phi_3 = 1+\frac{1}{\phi_2} = 1+\frac{1}{1+\frac{1}{1+\frac{1}{\phi_0}}},\]

and so on. As we continue this process our initial guess becomes less and less important to our later guesses, and when we repeat this process infinitely many times the influence of our initial guess vanishes entirely, so that the “infinityth” guess must be some value independent of our initial guess. This means it must be a fixed point, a true solution to the defining equation, or equivalently, a perfect guess:

\[\phi = \phi_\infty = 1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\ddots}}}}\]

Another rearrangement of the equation that defines \(\phi\) is

\[\phi = \sqrt{1+\phi}.\]

We can use this rearrangement in the same fashion as we did the other: view it as a pattern for generating improved guesses. If our initial guess is again \(\phi_0\) then

\(\phi_1 = \sqrt{1+\phi_0},\) \(\phi_2 = \sqrt{1+\phi_1} = \sqrt{1+\sqrt{1+\phi_0}},\) \(\phi_3 = \sqrt{1+\phi_2} = \sqrt{1+\sqrt{1+\sqrt{1+\phi_0}}}\)

and so on, so that an infinite number of guesses later we wind up with

\[\phi = \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}}},\]

demonstrating the mystical-looking equality.

There is one loophole to this whole story: there are actually two “perfect guesses” that yield themselves when you try to improve them in the first iterative manner. The first is the golden ratio, the one we actually want. The other is \(-1/\phi\). If you use this value as your initial guess, you will get stuck. However, if your initial guess differs from this value in any miniscule amount, you are guaranteed to wind up with the golden ratio eventually. We express this by calling \(\phi\) a stable fixed point of this procedure and \(-1/\phi\) an unstable fixed point. The second iterative procedure has only one fixed point, (\(\phi\)), because we implicitly mean that we want the positive square root and not the negative square root.

April 6, 2014

Smiles Await You When You Rise - April 6, 2014 - {"name"=>"Evan Berkowitz", "twitter"=>"evanberkowitz", "email"=>"", "phone"=>"+1 917-692-5685", "inspire"=>"", "arxiv"=>"", "github"=>"", "linkedin"=>"", "google_scholar"=>"", "orcid"=>"", "research_gate"=>""}