Obtaining a New Representation for the Golden Ratio by Solving a Biquadratic Equation
\textsc{Leonardo Mondaini}
small{Instituto de F\'{\isica, Universidade do Brasil}}
\hspace{-0.55cm} In the present work we show how different ways to solve biquadratic equations can lead us to different representations of its solutions. A particular equation which has the golden ratio and its reciprocal as solutions is shown as example.
math0208210.tex
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\title{Obtaining a New Representation for the Golden Ratio by Solving a Biquadratic Equation}
\author{\textsc{Leonardo Mondaini}
\\\small{\emph{Instituto de F\'{\i}sica, Universidade do Brasil}}}
\date{\small{Cx.Postal 68528, Rio de Janeiro RJ 21945-970, Brazil\\e-mail address: mondaini@if.ufrj.br}}
\maketitle

\begin{abstract}
\hspace{-0.55cm}
\textbf{In the present work we show how different ways to solve biquadratic equations can lead us to different representations of its solutions. A particular equation which has the golden ratio and its reciprocal as solutions is shown as example.} 
\end{abstract}

\vskip1.0pc
\section{Introduction}

The study of algebraic equations has occupied the brightest mathematical minds throughout many centuries. We must highlight among the results of the studies in this area, the attainment of the formula for resolution of the general quadratic equations, which together with the formula for resolution of the general cubic equations, achieved by Niccolo Fontana (Tartaglia)\footnote{Usually the formula for resolution of the general cubic equation it is attributed to Girolamo Cardano, and being thus it receives his name. A more detailed description about the dispute for the priority of this solution can be found in [ 1 ].}, led to the creation of complex numbers, since the application of these formulas led to a ``misterious'' thing: the square root of a negative number.
 
The solution of general quartic equations by Ludovico Ferrari (a pupil of Cardano) comes to complete this picture, once it was established that a solution by radicals for generic equations of degree equal to or greater than 5 cannot be achieved (a result shown by the prodigies Niels Abel and Evariste Galois). What follows, we will concentrate on an interesting aspect associated to a particular kind of quartic equation, the biquadratic one.  
 
\section{An interesting equation}
Let us consider the following equation 

\begin{eqnarray}
det(\cal {A}  -\lambda \cal {I})&=0,
\end{eqnarray}
where $\cal {A}$ is a $4 \times 4$ symmetric real matrix defined by 

\begin{equation}
\cal {A} \equiv \left( 
\begin{array}
{cccc} 0&1&0&0\\1&0&1&0\\0&1&0&1\\0&0&1&0 
\end{array}
\right), 
\end{equation}
and $\cal{I}$ is the correspondent $4 \times 4$ identity matrix. We can write equation (1) in the following way   
 
\begin{eqnarray}
\left|
\begin{array}
{cccc} -\lambda&1&0&0\\1&-\lambda&1&0\\0&1&-\lambda&1\\
0&0&1&-\lambda 
\end{array}
\right|&=0,
\end{eqnarray}
or

\begin{equation}
\lambda^4-3\lambda^2+1=0.
\end{equation}

This simple biquadratic equation displays an interesting feature. The form its solutions are expressed depends, apparently, on the algorithm used for solving it. Still more interesting is the fact that one of these algorithms yields the numbers 
$\frac{\sqrt{5}+1}{2}$ and $\frac{\sqrt{5}-1}{2}$ (as well as their respective symmetrical ones) as solutions, which are the most known representation of this ubiquitous mathematical phenomenon, the golden ratio, and its reciprocal. This famous number appears historically as the solution of the quadratic equation 

\begin{equation}
(x+a)x=a^2,
\end{equation}
which is related to the geometrical problem of dividing a given line segment $\overline{AB}$ into what is called the golden section, which the celebrated astronomer Johannes Kepler called ``one of the two Jewels of Geometry'' (the second is the theorem of Pythagoras). Translated into mathematical language, the golden section means that the segment $\overline{AB}=a$ is cut at a point $C$ so that the whole segment is in the same ratio to the larger part $\overline{AC}=x$ as $\overline{AC}$ is to the other part, $\overline{CB}=a-x$. Stated otherwise, it produces the relation

\begin{eqnarray}
\frac{a}{x}=\frac{x}{a-x},&x>a-x.
\end{eqnarray}

This, in turn, leads to the quadratic equation $x(x+a)=a^2$ already mentioned, the positive root of which is $x=\frac{1}{2}a(\sqrt{5}-1)$. When $a=1$, the value $x=\frac{1}{2}(\sqrt{5}-1)$ is the reciprocal of the golden ratio,
$0.6180339...$.

We now consider two algorithms to solve equation (3).

\subsection{Algorithm 1}

Firstly, we will solve equation (3) by an algorithm very similar with the one employed by Ludovico Ferrari in his solution of quartic equations (polynomial equations of the fourth degree)\footnote{The goal of Ferrari's algorithm for solving the general quartic is to have perfect squares in both sides of the equation.}[2]. For such, we start by observing that equation (3) can be re-written in the following way:

\begin{equation}
\lambda^4-3\lambda^2+1 = \lambda^4-2\lambda^2-\lambda^2+1=0,  
\end{equation}
or

\begin{equation}
\lambda^4-2\lambda^2+1=\lambda^2.
\end{equation}



The left side of this equation, a perfect square, can be trivially simplified as

\begin{equation}
(\lambda^2-1)^2 = \lambda^2.
\end{equation}
This implies that

\begin{equation}
\lambda^2-1 = \pm \lambda,
\end{equation}
or

\begin{equation}
\lambda^2\mp\lambda-1=0.
\end{equation}

In solving these equations by the quadratic formula, it is found that the solutions for the equation with -$\lambda$ are

\begin{equation}
\lambda_- =\cases{\frac{\sqrt{5}+1}{2}\cr
\; \cr
-(\frac{\sqrt {5}-1}{2})},
\end{equation}
while for the equation with $+\lambda$ (which is identical to equation (5) with $a=1$) the following solutions are found

\begin{equation}
\lambda_+=\cases{\frac{\sqrt{5}-1}{2}\cr
\; \cr
-(\frac{\sqrt{5}+1}{2})}.
\end{equation}

Thus the four solutions of the original biquadratic equation are

\begin{equation}
\lambda=\cases{\frac{\sqrt{5}+1}{2}\cr
\; \cr
-(\frac{\sqrt{5}+1}{2})\cr
\; \cr
\frac{\sqrt{5}-1}{2}\cr
\; \cr
-(\frac{\sqrt{5}-1}{2})},
\end{equation}
where we remind the reader that $\frac{\sqrt{5}+1}{2}$ is the usual representation for the golden ratio, and $\frac{\sqrt{5}-1}{2}$ is its reciprocal.

\subsection{Algorithm 2}

The second algorithm follows the conventional method to solve biquadratic equations. In such method we start by letting the following substitution

\begin{equation}
\eta=\lambda^2,
\end{equation}
which reduces equation (3) to the form

\begin{equation}
\eta^2-3\eta+1=0.
\end{equation}

An application of quadratic formula yields the solutions

\begin{equation}
\eta = \cases{\frac{\sqrt{5}+3}{2} \cr
\; \cr
-(\frac{\sqrt{5}-3}{2})}.
\end{equation}



Since $\lambda=\pm \sqrt{\eta}$, we have the following solutions for the original equation\footnote{The solutions are displayed so as to indicate a direct correspondence with the set of solutions obtained with the first algorithm.}

\begin{equation}
\lambda=\cases{\sqrt{\frac{\sqrt{5}+3}{2}}\cr
\; \cr
-\sqrt{\frac{\sqrt{5}+3}{2}}\cr
\; \cr
\sqrt{\frac{3-\sqrt{5}}{2}}\cr
\; \cr
-\sqrt{\frac{3-\sqrt{5}}{2}}}.
\end{equation}

\section{Concluding Remarks}

We see that apparently distinct solutions are obtained by solving equation (3) by two different algorithms. In fact, there are no algebraic manipulations to be done on number $\sqrt{\frac{\sqrt{5}+3}{2}}$ that yield the number $\frac{\sqrt{5}+1}{2}$, but we can verify, easily, that they are identical! Just compute the square of the last one and check it!

We have obtained, then, a new representation for the golden ratio

\begin{equation}
\rho = \frac{\sqrt{5} + 1}{2}=\sqrt{\frac{\sqrt{5}+3}{2}}.
\label{expfoton}
\end{equation}

We must still notice that other representations could be achieved for the golden ratio just by elevating the number $\frac{\sqrt{5} + 1}{2}$ to an integer $n$ and writing the $nth$-root of the result. 

\begin{thebibliography}{99}
\bibitem{ref1}
\small{Boyer, Carl - \emph{History of Mathematics} - John Wiley $\&$ Sons (1989)}
\bibitem{ref2}
\small{Cardano, Girolamo - \emph{Ars Magna or the Rules of Algebra} - Dover (1993)}
\end{thebibliography}

\end{document}
math0208210.extracted_bib
\begin{thebibliography}{99}
\bibitem{ref1}
\small{Boyer, Carl - \emph{History of Mathematics} - John Wiley $\&$ Sons (1989)}
\bibitem{ref2}
\small{Cardano, Girolamo - \emph{Ars Magna or the Rules of Algebra} - Dover (1993)}
\end{thebibliography}