Playing with Phi

Column

Recall that \[\frac{1}{\frac{1}{x}} == x\]

And that \[\phi = 1 + \frac{1}{1+\frac{1}{1+\frac{1}{1+...}}}\]

So our our original question (see right): \[y = \frac{1}{1+\frac{1}{1+\frac{1}{1+...}}}\] is just phi - 1.

maybe if we invert this fraction then we get \[\frac{1}{y} = 1+\frac{1}{1+\frac{1}{1+...}}\]

we can probably get our original thing again if we move things around again: \[\frac{1}{y} - 1 = \frac{1}{1+\frac{1}{1+...}} = y\]

From there: \[\frac{1}{y} - 1 = y\]

\[1 - y = y^2\]

\[y^2 + y - 1 = 0\]

Well that’s almost a quadratic but it’s not quite the {-1,1} droids I was looking for.

\[y^2 + y - 1 = 0\]

Final Answer \[y = \pm\frac{1- \sqrt{5}}{2}\]

caption

At first glance we can just start doing the sequence n times and see what it approaches

1 1/2 2/3 … ? ow my brain hurts, use R

findY <- function(iterations) {
  if (iterations==0) {
    return(1)
  } else {
    return(1/(1 + findY(iterations-1)))
  }
}

findY(1)

[1] 0.5

findY(2)

[1] 0.6666667

findY(3)

[1] 0.6

findY(4)

[1] 0.625

findY(5)

[1] 0.6153846

findY(50)

[1] 0.618034

So that looks pretty close to \[\phi\] the golden ratio.

Note tha phi is usually 1.618… not 0.618… which is the question being asked here.

The 0.618… is basically the short over long segment of the golden ratio rather than the long over the short and is called

“Golden Ratio Conjugate”

\[\Phi = \phi^{-1}\]

caption