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\title{Numerical Solving of Third Degree Equation}
\author{Yotam Medini}

\begin{document}

\maketitle

\begin{abstract}
Dedicated to: \textsl{Hillel, Roey, Tomer and Yotam}.
\end{abstract}

Given the equation
\begin{equation} \label{eq:orig}
\alpha x^3 + \beta x^2 + \gamma x + \delta = 0 
\qquad (\textnormal{where}\; \alpha \neq 0)
\end{equation}
we want to find a numerical solution. Dividing by \(\alpha\) we get
\begin{equation*}
x^3 + (\beta/\alpha)x^2 + (\gamma/\alpha)x + \delta/\alpha = 0.
\end{equation*}
If we put
\begin{equation*}
b = \beta/\alpha \qquad
c = \gamma/\alpha \qquad
d = \delta/\alpha
\end{equation*}
we get a simplified equation:
\begin{equation}
p(x) = x^3 + bx^2 + cx + d = 0
\end{equation}

We want to find some large enough value $M$ such that 
\begin{equation} \label{eq:-M:M}
p(-M) = -M^3 + bM^2 - cM + d\;< \;0\; <\;  M^3 + bM^2 + cM + d = p(M).
\end{equation}

We will show that if
\begin{equation} \label{eq:M:req}
M > \max(3|b|, \sqrt[2]{|3c|}, \sqrt[3]{|3d|})
\end{equation}
Then \eqref{eq:-M:M} will be true.
We can simply pick 
\begin{equation*}
M = \max(3|b|, 2|c|, 2|d|) + 1 
  > \max\left(3|b|, \sqrt[2]{|3c|}, \sqrt[3]{|3d|}\right).
\end{equation*}
By the requirement \eqref{eq:M:req} we are assured that
\begin{equation*}
M/3   > |b|    \qquad
M^2/3 > |c|  \qquad
M^3/3 > |d|
\end{equation*}


Now
\begin{align*}
p(M) 
&= M^3 + bM^2 + cM + d \\
&\geq M^3 - |b|M^2 - |c|M - |d| \\
&= (M^3/3 - |b|M^2) + (M^3/3 - |c|M) + (M^3/3 - |d|) \\
&> 0.
\end{align*}

Similarly
\begin{align*}
p(-M) 
&= -M^3 + bM^2 - cM - d \\
&\leq -M^3 + |b|M^2 + |c|M + |d| \\
&= (|b|M^2 - M^3/3) + (|c|M - M^3/3) + (|d| - M^3/3) \\
&< 0.
\end{align*}





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