Mathematical Properties of the Golden Ratio

Mathematical Properties of the Golden Ratio#

Defining the Golden Ratio and its Rational Conjugate#

The golden ratio $\phi = \frac{1+\sqrt5}{2}$ and its rational conjugate $\pb = \frac{1-\sqrt5}{2}$ are the solutions of the equation $x^2 = x + 1 = 0$.

Basic Operations on ${\phi}$ and ${\pb}$#

${\phi}$ and ${\pb}$ relate in the following ways:

\[\begin{split}\begin{align} \phi + \pb &= 1 \\ \phi - \pb &= \sqrt5 \\ \phi \cdot \pb &= -1 \end{align} \end{split}\]

import sympy as sym
from IPython.display import display, Math
phiv = (1 + sym.sqrt(5))/2
phibarv = 1 - phiv
phi, phibar = sym.symbols(r'\phi \vphantom{}\overline{\smash{\phi}\vphantom{;}}')
print('The sum of the two ratios is 1')
display(sym.Eq(phi + phibar, phiv + phibarv))
print('Their difference is root 5')
display(sym.Eq(phi - phibar, phiv - phibarv))
print('Their product is -1')
display(sym.Eq(phi * phibar, (phiv * phibarv).simplify()))

The sum of the two ratios is 1

\[\displaystyle \phi + \vphantom{}\overline{\smash{\phi}\vphantom{;}} = 1\]

Their difference is root 5

\[\displaystyle \phi - \vphantom{}\overline{\smash{\phi}\vphantom{;}} = \sqrt{5}\]

Their product is -1

\[\displaystyle \phi \vphantom{}\overline{\smash{\phi}\vphantom{;}} = -1\]

Powers of ${\phi}$ and ${\pb}$#

Since $\phi$ solves $x^2 = x + 1 = 0$, it obeys the relation:

\[\phi^2 = \phi + 1\]

Multiply both sides by $\phi$ to yield:

\[\phi^3 = \phi^2 + \phi\]

Now take advantage of $\phi^2 = \phi + 1$ and substitute $\phi + 1$ for $\phi^2$:

\[\begin{split}\begin{align} \phi^3 &= \phi + 1 + \phi \\ \phi^3 &= 2 \phi + 1 \end{align}\end{split}\]

So we have $\phi^3$ and $\phi^2$ each expressed as a multiple of $\phi$ plus an integer. A pattern is emerging. We can repeat the same trick to yield a similar representation for $\phi^4$:

\[\begin{split}\begin{align} \phi^3 &= 2 \phi + 1 && \text{Start with the expression for } \phi^3. \\ \phi^4 &= 2 \phi^2 + \phi && \text{Multiply both sides by } \phi. \\ \phi^4 &= 2 (\phi + 1) + \phi && \text{Substitute } \phi + 1 \text{ for } \phi^2. \\ \phi^4 &= 3 \phi + 2 \end{align} \end{split}\]

Continuing for higher powers:

\[\begin{split}\begin{align} \phi^5 &= 5\phi + 3\\ \phi^6 &= 8\phi + 5\\ \phi^7 &= 13\phi + 8\\ \end{align} \end{split}\]

p = phi
for i in range(2, 9):
    p = sym.expand(p * phi).subs(phi**2, phi + 1)
    display(sym.Eq(phi ** i, p))

\[\displaystyle \phi^{2} = \phi + 1\]

\[\displaystyle \phi^{3} = 2 \phi + 1\]

\[\displaystyle \phi^{4} = 3 \phi + 2\]

\[\displaystyle \phi^{5} = 5 \phi + 3\]

\[\displaystyle \phi^{6} = 8 \phi + 5\]

\[\displaystyle \phi^{7} = 13 \phi + 8\]

\[\displaystyle \phi^{8} = 21 \phi + 13\]

Amazingly, the coefficient on $\phi$ and integer term are both Fibonacci numbers! It can be shown that in general

\[\begin{equation}\label{eqn:phin} \phi^n = F_n \cdot \phi + F_{n-1} \ \text{for any integer }n. \tag{1} \end{equation}\]

Since $\pb$ is the other solution of $x^2 = x + 1=0$, a similar result holds:

\[\begin{equation}\label{eqn:phibn} \pbn{n} = F_n \cdot \pb + F_{n-1} \ \text{for any integer }n. \tag{2} \end{equation}\]

Formulas to Directly Calculate $F_n$ and $L_n$#

Subtract $(\ref{eqn:phin})$ from $(\ref{eqn:phibn})$:

\[\begin{split}\begin{align} \phi^n-\pbn{n} &= F_n(\phi - \pb) \\ \frac{\phi^n-\pbn{n}}{\phi - \pb} &= F_n \\ \frac{\phi^n-\pbn{n}}{\sqrt 5} &= F_n \end{align} \end{split}\]

Hence we have arrived at Binet’s formula:

\[\begin{equation}\label{eq:3} F_n = \frac{\phi^n-\pbn{n}}{\sqrt 5} \tag{3} \end{equation}\]

Recall the Lucas sequence $L_n$ defined in terms of Fibonaccci numbers by the relation $L_n = F_{n+1} + F_{n-1}$. Taking the sum of $(\ref{eqn:phin})$ and $(\ref{eqn:phibn})$ leads to the formula for Lucas numbers:

\[\begin{split}\begin{align} \phi^n+\pbn{n} &= F_n(\phi + \pb) + 2 F_{n-1} && \text{Apply } \phi + \pb = 1.\\ &= F_n + 2 F_{n-1} && \text{Apply } F_{n-1} = F_{n+1} - F_{n}.\\ &= F_{n+1} + F_{n-1} && \text{Apply the Lucas definition } L_n=F_{n+1} + F_{n-1}\\ \phi^n+\pbn{n} &= L_n \end{align} \end{split}\]

Thus the Lucas number analog to Binet’s formula is:

\[\begin{equation}\label{eq:4} L_n = \phi^n-\pbn{n} \tag{4} \end{equation}\]

Other Formulations of $\phi^n$ and $\pbn{n}$#

Compare the Lucas formula $(\ref{eq:4})$ to a slightly rearranged Binet formula:

\[ \sqrt 5 F_n = \phi^n-\pbn{n} \tag{3'} \label{eq:3'} \]

Using $\ref{eq:4}$ and $\ref{eq:3'}$ to solve for $\phi^n$ and $\pbn{n}$ leads to another fascinating pair of results:

\[\phi^n = \frac{L_n + F_n \sqrt 5}{2} \quad \text{and} \quad \pbn{n} = \frac{L_n - F_n \sqrt 5}{2} \tag{5} \label{eq:5} \]

Reversal of the Lucas Sequence Definition#

One way to define the Lucas sequence is $L_n = F_{n+1} + F_{n-1}$. Then it’s natural to wonder if $L_{n+1}+L_{n-1}$ has any special meaning. Yes:

\[\begin{split} \begin{align} L_{n+1}+L_{n-1} &= F_{n+2} + F_n + F_n+ F_{n-2} && \text{Apply the Lucas definition twice.} \\ &= F_{n+2} + 2F_{n} + F_{n-2} && \text{Substitute } F_{n+1} + F_{n} \text{ for } F_{n+2} \text{ and } F_{n} - F_{n-1} \text{ for } F_{n-2}.\\ &= F_{n+1} + 4F_{n} - F_{n-1} && \text{Substitute } F_{n} \text{ for } F_{n+1} - F_{n-1}.\\ L_{n+1}+L_{n-1} &= 5F_n \tag{6} \label{eq:6} \end{align} \end{split}\]

So we have another nice equation pair:
$$ F_{n+1}+F_{n-1} = L_n \text{ and } L_{n+1}+L_{n-1} = 5F_n $$

$\phi^n \sqrt 5$ and $\pbn{n} \sqrt 5$#

$\phi^n \sqrt 5$ takes a special form when we apply the first equation of $(\ref{eq:5})$:

\[\begin{split} \begin{align} \phi^n \sqrt 5 &= && \text{Apply (\ref{eq:5}).} \\ & = \frac{L_n + F_n \sqrt 5}{2} \sqrt 5 && \text{Multiply.}\\ &= \frac{\sqrt 5 L_n + 5F_n}{2} && \text{Apply the Lucas definition } L_n = F_{n+1} + F_{n-1} \\ & && \text{ and its counterpart }(\ref{eq:6}). \\ &= \frac{\sqrt 5 (F_{n+1} + F_ {n-1}) + (L_{n+1} + L_{n-1})}{2} && \text{Rearrange terms.} \\ &= \frac{L_{n+1} + \sqrt 5 F_{n+1}}{2} + \frac{L_{n-1} + \sqrt 5 F_{n-1}}{2} && \text{Apply (\ref{eq:5}) twice.} \\ &= \phi^{n+1} + \phi^{n-1}. \end{align}\end{split}\]

A similar result holds for $\pb$, but recall that $\pb = \frac{1-\sqrt 5}{2}$ is negative so there is a sign change. Thus we discover another pair of results:

\[ \phi^n \sqrt 5 = \phi^{n+1} + \phi^{n-1} \text{ and } \pbn{n} \sqrt 5= -\left(\pbn{n+1} + \pbn{n-1} \right) \tag{7} \label{eq:7}\]

Mathematical Properties of the Golden Ratio

Contents

Mathematical Properties of the Golden Ratio#

Defining the Golden Ratio and its Rational Conjugate#

Basic Operations on \({\phi}\) and \({\pb}\)#

Powers of \({\phi}\) and \({\pb}\)#

Formulas to Directly Calculate \(F_n\) and \(L_n\)#

Other Formulations of \(\phi^n\) and \(\pbn{n}\)#

Reversal of the Lucas Sequence Definition#

\(\phi^n \sqrt 5\) and \(\pbn{n} \sqrt 5\)#