fibonacci miles and kilometers

In my running, I have been trying to keep track in kilometers. This presents a couple problems: Somehow, my mind still thinks in miles, which is weird because I do not really have /that/ good of an idea exactly how far a mile is, either. Or someone wants to know how far a 5k is. Or, when I was running the 2011 Chicago Marathon, there would be kilometer postings between the mile markers. Anyway, it is handy to convert between them, but a bit of a pain.

It turns out that the ratio of adjacent Fibonacci numbers are a decent approximation to the unit conversion between miles and kilometers. I decided to look into it a bit.

First, in case you have never seen or do not remember what Fibonacci numbers are, they are a sequence of numbers where the first two fibonacci numbers are $0$ and $1$, and then each number after that is the sum of the immediately prior two. Mathematically, this is written as ( f_0 = 0, f_1 = 1, f_n = f_{n-1} + f_{n-2}, ) which gives rise to the sequence ( 0,1,1,2,3,5,8,13,21,34,55,89,… )

You can actually come up with a closed form solution for Fibonacci numbers, which is borderline incredible, using the recursion relationship given above. If you assume that $f_n=\lambda^n$, then you get ( f_{n+2} = f_{n+1} + f_n \implies \lambda^{n+2} = \lambda^{n+1} + \lambda^n ) Then canceling a $\lambda^n$ from both sides, and rearranging you get that ( \lambda^2 - \lambda - 1 = 0 ) You can use the trusty quadratic formula to solve this, it gives ( \lambda^*_{\pm} = \frac{-(-1)\pm \sqrt{(-1)^2 - 4(+1)(-1)}}{2(1)} = \frac{1\pm\sqrt{5}}{2} )

Clearly the $n^{th}$ Fibonacci number is not just this messy looking irrational fraction to the $n^{th}$ power, so we need one more trick. Introduce a couple of constants $c_1$ and $c_2$, and let ( f_n = c_1(\frac{1+\sqrt{5}}{2})^n + c_2(\frac{1-\sqrt{5}}{2})^n ) Now, since we want $f_0=0$ and any number to the zeroth power is 1, we have that $f_0=c_1+c_2=0$, or that $c_1=-c_2$. Then $f_n = c_1((\frac{1+\sqrt{5}}{2})^n - (\frac{1-\sqrt{5}}{2})^n)$. The final step is requiring that $f_1=1$. This gives that ( f_1 = c_1(\frac{1+\sqrt{5} - 1 + \sqrt{5}}{2}) = c_1 (\frac{2\sqrt{5}}{2})\equiv 1 \implies c_1 = \frac{1}{\sqrt{5}} ) Now, finally, we have that ( f_n = \frac{1}{\sqrt{5}} \left( \left(\frac{1+\sqrt{5}}{2}\right)^n + c_2\left(\frac{1-\sqrt{5}}{2} \right) \right) )

It seems I have run a little off track. What does all of this have to do with running? Well, the Fibonacci numbers eventually approach this fancy $\lambda_+$ we came up with earlier, and it turns out that $\lambda_+ = 1.618$. It also turns out that one mile is about 1.609 kilometers. That is just 0.56% error! Sweet!

There is a catch though. Some Fibonacci ratios are better than others:

It looks to me like a sweet spot is at the 8 kilometer/5 mile mark. So now you can see that 25 miles is 40 kilometers, or whatever distance you wanted to know about in the first place.