Hofstadter mentions the following recursive relation in his great book "GĂ¶del, Escher, Bach": \[

\begin{align}

g(0) &= 0;\\

g(n) &= n-g(g(n-1)).

\end{align}

\] I claim that \( g(n) = \left\lfloor \phi\cdot (n+1) \right\rfloor \), where \( \phi = \frac{-1+\sqrt{5}}{2}\), and I'll show this using a technique that makes proving many identities of this type nearly automatic.

Let \( \phi\cdot n = \left\lfloor \phi\cdot n \right\rfloor + e\), where \( 0 < e < 1\) as \( \phi \) is irrational, nor can \(e = 1-\phi\), and note that \(\phi\) satisfies \( {\phi}^2 + \phi - 1 = 0\). Some algebra gives \[

\begin{align}

n-\left\lfloor \left( \left\lfloor \phi\cdot n \right\rfloor + 1 \right) \cdot \phi \right\rfloor

&= n-\left\lfloor \left( n\cdot \phi - e + 1 \right) \cdot \phi \right\rfloor \\

&= n-\left\lfloor n\cdot {\phi}^2 - e\cdot \phi + \phi \right\rfloor \\

&= n-\left\lfloor n\cdot \left(1-\phi\right) - e\cdot \phi + \phi \right\rfloor \\

&= n-n-\left\lfloor -n\cdot \phi - e\cdot \phi + \phi \right\rfloor \\

&= -\left\lfloor -n\cdot \phi - e\cdot \phi + \phi \right\rfloor \\

&= -\left\lfloor -n \cdot \phi + e - e - e\cdot \phi + \phi \right\rfloor \\

&= \left\lfloor \phi\cdot n \right\rfloor -\left\lfloor - e - e\cdot \phi + \phi \right\rfloor.

\end{align}

\]

Now if \[

\begin{align}

0 < e < 1-\phi &\implies 0 < - e - e\cdot \phi + \phi < \phi;\\

1-\phi < e < 1 &\implies -1 < - e - e\cdot \phi + \phi < 0.

\end{align}

\]

This implies \[ n-\left\lfloor \left( \left\lfloor \phi\cdot n \right\rfloor + 1 \right) \cdot \phi \right\rfloor = \left\lfloor \phi\cdot (n+1) \right\rfloor .\] Since \( \left\lfloor \phi\cdot (0+1) \right\rfloor = 0\), we're done.

The point of the algebra was to move all terms involving \(n\) out, and then checking to see how the remaining term varied with \( e\). A simple idea, but very useful.

## Saturday, October 11, 2014

## Sunday, October 5, 2014

### Closed Under Means

Here's a nice little problem from the 13th All Soviet Union Mathematics Olympiad.

First observe that if \(x\in S\) then both \(\frac{x}{4}\) and \(\frac{3x}{4}\) are in \(S\); average \(\{0,x\}\) to get \(\frac{x}{2}\), average \(\{0, \frac{x}{2}\}\) to get \(\frac{x}{4}\), average \(\{\frac{x}{2}, x\}\) to get \(\frac{3x}{4}\).

We could show any rational number \(\frac{m}{n}\) with \(1\leq m < n\) is in \(S\) if we had \(n\) distinct elements from \(S\) that summed to \(m\). Let's exhibit one.

Start with an array with \(m\) 1s on the left, \(n-m\) 0s on the right. Repeatedly replace adjacent \(x,y\) values with \(\frac{3(x+y)}{4}, \frac{(x+y)}{4}\), where either \(x=1,y\neq1\) or \(x\neq 0, y=0\), until there is one 0 and one 1 left. We can do this in exactly \(n-2\) steps, each resulting number is guaranteed to be in \(S\) by the above note, and each number is guaranteed to be distinct by construction!

Examples:

\(\frac{1}{3}: \left[1,0,0\right] \to \left[\frac{3}{4},\frac{1}{4},0\right] \)

\(\frac{2}{5}: \left[1,1,0,0,0\right] \to \left[1,\frac{3}{4},\frac{1}{4},0,0\right] \to \left[1,\frac{3}{4},\frac{3}{16},\frac{1}{16},0\right] \)

Given a set of real numbers \(S\) containing 0 and 1 that's closed under finite means, show that it contains all rational numbers in the interval \(\left[0,1\right]\).This isn't a difficult problem, but there's a particularly nice solution.

First observe that if \(x\in S\) then both \(\frac{x}{4}\) and \(\frac{3x}{4}\) are in \(S\); average \(\{0,x\}\) to get \(\frac{x}{2}\), average \(\{0, \frac{x}{2}\}\) to get \(\frac{x}{4}\), average \(\{\frac{x}{2}, x\}\) to get \(\frac{3x}{4}\).

We could show any rational number \(\frac{m}{n}\) with \(1\leq m < n\) is in \(S\) if we had \(n\) distinct elements from \(S\) that summed to \(m\). Let's exhibit one.

Start with an array with \(m\) 1s on the left, \(n-m\) 0s on the right. Repeatedly replace adjacent \(x,y\) values with \(\frac{3(x+y)}{4}, \frac{(x+y)}{4}\), where either \(x=1,y\neq1\) or \(x\neq 0, y=0\), until there is one 0 and one 1 left. We can do this in exactly \(n-2\) steps, each resulting number is guaranteed to be in \(S\) by the above note, and each number is guaranteed to be distinct by construction!

Examples:

\(\frac{1}{3}: \left[1,0,0\right] \to \left[\frac{3}{4},\frac{1}{4},0\right] \)

\(\frac{2}{5}: \left[1,1,0,0,0\right] \to \left[1,\frac{3}{4},\frac{1}{4},0,0\right] \to \left[1,\frac{3}{4},\frac{3}{16},\frac{1}{16},0\right] \)

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