\documentclass{article}


\title{General Maxima and Minima, \\ Quadratic Forms}
\author{Yosun Chang}

\begin{document}
\maketitle

The most general function of two variables can be expanded in the form of a Taylor series. Near the critical point of either a maxima, minima, or saddle-point, one can drop off the higher order terms in the expansion. One could move the origin arbitrarily, so that the critical point is renamed $f(0,0)$. One has, thus: 

\begin{equation}
f(x,y)=f(0,0)+\frac{1}{2}f_{xx}x^2+\frac{1}{2}f_{yy}y^2+\frac{1}{2}f_{xy}xy+\frac{1}{2}f_{yx}yx+\mathcal{O}(3).
\end{equation} 

$f(x,y)$ is the value of the function at any point $(x,y)$. Its deviation from the critical point, call it $\Delta f$, is $\Delta f=f(x,y)-f(0,0)$. 

The deviation from the critical point can be re-expressed in matrix form, 

\begin{eqnarray}
\Delta f=\frac{1}{2}
\left(\begin{array}{cc}
x & y
\end{array}\right)
\left(\begin{array}{cc}
f_{xx} &  f_{xy} \\
f_{yx} &  f_{yy}
\end{array}\right)\left(\begin{array}{c}
x \\
y
\end{array}\right), 
\end{eqnarray}
which is also known as a \emph{quadratic form}. (Marion and Thornton. Mechanics. Chapter 7) 
\par
Because $x^2, y^2 \geq 0$, if the cross-terms $f_{xy}$ aren't there, one would be able to tell whether the deviation $\Delta f$ is overall positive or negative when both $f_xx$ and $f_yy$ are of the same sign. 
\par 
Because the coefficients $f_{ij}$ can be expressed in the form of a matrix, as shown, one can transform the whole matrix quantity into principle coordinates, via the usual method (see the Review in the attached Appendix).
\par

Recast in the basis of its principle coordinate system, the matrix has only diagonal entries, viz., its eigenvalues. Thus, 

\begin{eqnarray}
\Delta f=\frac{1}{2}
\left(\begin{array}{cc}
x & y
\end{array}\right)
\left(\begin{array}{cc}
f_{xx} &  0 \\
0 &  f_{yy}
\end{array}\right)\left(\begin{array}{c}
x \\
y
\end{array}\right)=\frac{1}{2}\left(f_{xx} x^2 + f_{yy} y^2\right). 
\end{eqnarray}

Thus, the following signs in the coefficients $f_{ii}$ determines everything:
\begin{center}
\fbox{$f_{xx},f_{yy}$\textless$ 0 \Rightarrow f(x,y)<f(0,0)\Rightarrow f(0,0)$ is a local max}\par
\fbox{$f_{xx},f_{yy}$\textgreater$ 0 \Rightarrow f(x,y)>f(0,0)\Rightarrow f(0,0)$ is a local min}
\end{center}

\section{Example}
Usually, the critical point isn't $(0,0)$ (however, since we can move our ``origin" anywhere, so the above is just as general). You can find the critical point by taking the first derivative of your function and setting the resulting expression equal to 0. 

$f=x^3 -2x+ y^2 $\par
$ f_x=3x^2-2=0\Rightarrow x=\pm\sqrt{2/3}$ and $f_y=2y=0\Rightarrow y=0$

Now, the $f_{ij}$ matrix (also known as the Hessian matrix) looks like this: 

\begin{eqnarray}
f_{ij}=\left(\begin{array}{cc}
f_{xx}&f_{xy}\\
f_{yx}&f_{yy}
\end{array}\right)=\left(\begin{array}{cc}
6x & 0\\
0&2
\end{array}\right)=\left(\begin{array}{cc}
\pm 6\sqrt{2/3} & 0\\
0&2
\end{array}\right)
\end{eqnarray}

It just happens that the function I randomly conjured happens to already be in its principle coordinate system, and thus we immediately have our eigenvalues of $6x$ and $0$. The last step above comes from plugging in $x$, as determined by setting $f_x=0$.

So, this means that at the point $(\sqrt{3/2},0)$, we have a local minimum (since both $f_{ij}>0$). At the point $(-\sqrt{32},0)$, we have a max. 

But, if the function \emph{isn't already} in its principle coordinate system, then you'd have to find its eigenvalues and pop them into the diagonals of your final $f_{ij}$ matrix. Details below.

\section{Review: Transforming to Principle Coordinates}
The matrix $A$ may be any funky square matrix filled with ugly entries. In the basis of its principle coordinate system, however, it will have only diagonal elements, viz., its eigenvalues. 

\subsection{Finding Eigenvalues}
$\left|A-\lambda\right|=0$. Solve for $\lambda$. A $n\times n$ matrix has $n$ $\lambda$'s.

\subsection{Finding Eigenvectors}
Plug in the lambda's into the original equation: $A | x \rangle = \lambda | x \rangle$. $x$ is a simple linear equation. Pick numbers that obey the equation, and you get your eigenvector corresponding to $\lambda$. 



\section{References}
Shankar. Basic Training in Mathematics. p 274ff

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