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\title{Session 11 -- The golden sequence}
\subtitle{European section -- Season 2}
\date{}

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\begin{frame}{The Golden sequence}

\large

The Golden sequence is the geometric sequence with first term $\frac{1}{\varphi}$ and common ratio $\varphi$.

The successive terms of this sequence are 
\begin{center}
\begin{tabular}{lll}
$b_1=\frac{1}{\varphi}\simeq0.618$ & $b_2=\varphi^0=1$ & $b_3=\varphi\simeq1.618$ \\[5pt]
$b_4=\varphi^2\simeq2.618$ & $b_5=\varphi^3\simeq4.236$ & $b_6=\varphi^4\simeq6.854$\\[5pt]
& $b_7=\varphi^5\simeq11.090$ & \\
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\begin{frame}{The Golden sequence in the set $\Q[\sqrt5]$}

\large

The set $\Q[\sqrt5]$ is the set of all numbers of the form $a+b\sqrt5$, where $a$ and $b$ are fractions.

The terms of the Golden sequence are all in this set. The expressions for the first seven numbers are :
\begin{center}
\begin{tabular}{ll}
$b_1=\frac{1}{\varphi}=-\frac{1}{2}+\frac{1}{2}\sqrt{5}$ & $b_2=1=1+0\sqrt{5}$ \\[5pt]
$b_3=\varphi=\frac{1}{2}+\frac{1}{2}\sqrt{5}$ & 
$b_4=\varphi^2=\frac32+\frac{1}{2}\sqrt{5}$\\[5pt]
$b_5=\varphi^3=2+1\sqrt{5}$ & 
$b_6=\varphi^4=\frac72+\frac{3}{2}\sqrt{5}$\\[5pt]
$b_7=\varphi^5=\frac{11}{2}+\frac{5}{2}\sqrt{5}$ & \\
\end{tabular}
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\begin{frame}{The Golden sequence in the set $\Q[\sqrt5]$}

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It's interesting to consider the sum of the two coefficients of each term :
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\begin{tabular}{lcl}
$b_1=\frac{1}{\varphi}-\frac{1}{2}+\frac{1}{2}\sqrt{5}$ &:& $-\frac12+\frac12=0$\\[5pt]
$b_2=1=1+0\sqrt{5}$ &:& $1+0=1$\\[5pt]
$b_3=\varphi=\frac{1}{2}+\frac{1}{2}\sqrt{5}$ &:& $\frac12+\frac12=1$\\[5pt]
$b_4=\varphi^2=\frac32+\frac{1}{2}\sqrt{5}$ &:& $\frac32+\frac12=2$\\[5pt]
$b_5=\varphi^3=2+1\sqrt{5}$ &:& $2+1=3$\\[5pt]
$b_6=\varphi^4=\frac72+\frac{3}{2}\sqrt{5}$ &:& $\frac72+\frac32=5$\\[5pt]
$b_7=\varphi^5=\frac{11}{2}+\frac{5}{2}\sqrt{5}$ &:& $\frac{11}{2}+\frac52=8$\\
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These are the terms of the Fibonacci sequence !

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\begin{frame}{The relations between the terms}

\large

Using the expressions in $\Q[\sqrt5]$, it's easy to see that
$$\varphi=1+\frac{1}{\varphi}.$$
Multiplying by $\varphi$ this equation, we get
$$\varphi^2=\varphi+1$$
which proves that the Golden Ratio satisfies the quadratic equation
$$x^2=x+1.$$

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\begin{frame}{The relations between the terms}

\large

Multiplying the equation $\varphi^2=\varphi+1$ by $\varphi^{n}$, we prove that for any integer $n$
$$\varphi^{n+2}=\varphi^{n+1}+\varphi^{n}.$$
Therefore, we can conclude that for any integer $n$, the terms of the sequence satisfy the equality
$$b_{n+2}=b_{n+1}+b_n.$$SA2,
Incidently, this is the same relation as in the Fibonacci sequence. This sequence is the only one that is a geometric sequence and a Fibonacci-like sequence.

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\begin{frame}{The sum of consecutive terms}

\large

Using the formula for the sum of consecutive terms in a geometric sequence we get
\begin{eqnarray*}
\sum_{i=1}^{n} b_i &=& b_1+b_2+\ldots+b_n \\
\sum_{i=1}^{n} b_i &=& \frac{1}{\varphi}+1+\ldots+\varphi^{n-2} \\
\sum_{i=1}^{n} b_i &=& \frac{1}{\varphi}\times \frac{1-\varphi^n}{1-\varphi}\\
\sum_{i=1}^{n} b_i &=& \frac{1}{\varphi}\times \frac{\varphi^n-1}{\varphi-1}
\end{eqnarray*}

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\begin{frame}{The sum of consecutive terms}

\large

But we know that $\varphi=1+\frac{1}{\varphi}$, so $\frac{1}{\varphi}=\varphi-1$ and

\begin{eqnarray*}
\sum_{i=1}^{n} b_i &=& (\varphi-1)\times \frac{\varphi^n-1}{\varphi-1} \\
\sum_{i=1}^{n} b_i &=& \varphi^n-1
\end{eqnarray*}

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\Large

And now for something completely different\ldots

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