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\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.

it's look good,post

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