We continued to provide some extra material for chapters 16 and 17 and finished looking at Fibonacci numbers. I gave some examples of where these occur in nature and then we talked about the “golden ratio.” I said I’d put the calculation behind deriving the golden ratio online. Here it is.

Computing the Golding Ratio

We take a line of length 1 and we place an $x$ somewhere along that line. The idea is that this $x$ partitions the line. On one side, you have a length of $x$ and on the other you have a length of $1-x$. What we want to do is something a bit weird. We want to make sure that the ratio of the entire line’s length to $x$ is the same as the ratio of $x$ to $1-x$. In other words, we’re trying to find the specific $x$ satisfying the equality


To solve for $x$ here, we cross-multiply (clear denominators), and we get


So, this is equivalent to $x^2+x-1=0$ if we place everything on one side (equating it to zero). We can solve this quadratic in a number of ways. The way I decided to do it in class was with completing the square.

\[\begin{align}x^2+x-1&=0\cr x^2+x+\frac{1}{4}-\frac{1}{4}-0 &=0\cr \left(x^2+x+\frac{1}{4}\right)-\frac{1}{4}-1&=0\cr \left(x+\frac{1}{2}\right)^2-\frac{5}{4}&=0\cr \left(x+\frac{1}{2}\right)^2 &=\frac{5}{4}\cr x+\frac{1}{2}&=\frac{\sqrt{5}}{2}\cr x&=\frac{\sqrt{5}-1}{2}.\end{align}\]

Now, when we take $1/x$, we get the ratio that we’re looking for, which is somewhere around

\[\phi\approx 1.61803398875\]

We also looked at several examples of where this number pops up. There are quite a few surprising examples, both in nature and in the constructed world around us.


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