8===D

diff --git a/rapport.tex b/rapport.tex
index b7ab5af4404147a4d31db026aeb3cd05b7550df5..30222c15f8ed76771200c3915e3f275c978ea924 100644 (file)
--- a/rapport.tex
+++ b/rapport.tex
@@ -46,18 +46,28 @@
 %                                %\r
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\r
 \abstract{\r
-ABSTRACTABSTBSSTRACTABSTRACTABSTRACT\r
+  In this project we explore how one can use the method of steepest descent and\r
+  Newtons method to approximate the solution to a minimization problem\r
+  numerically. The method of steepest descent converges more rapidly when\r
+  supplied with an optimal step size, but in most cases Newtons method converges\r
+  in even fewer iterations.\r
 }\r
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\r
 %                                %\r
 %        RESULTATER              %\r
 %                                %\r
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\r
-\section{Resultater}\r
-This project explores how one can approximate the solution to a certain type of\r
-minimization problems. The function considered takes in a symmetric positive\r
-definite matrix, and two vectors. For this purpose we used the following\r
-randomly generated data:\r
+\section{Report}\r
+The function analyzed in this project was\r
+\begin{equation}\r
+  \label{eq:func_g}\r
+  g(\mathbf{x}) := -\mathbf{b}^{T}\mathbf{x} +\r
+  \frac{1}{2}\mathbf{x}^{T}H\mathbf{x} +\r
+  \frac{1}{12}\mathbf{x}^{T}C(\mathbf{x})\mathbf{x},\r
+\end{equation}\r
+which takes as input one symmetric definite matrix $H$, one vector\r
+$\mathbf{c}$ with positive elements, and one non-zero vector $\mathbf{b}$. For\r
+this purpose we used the following randomly generated data:\r
 \begin{equation}\r
   H = \r
   \begin{bmatrix}\r
@@ -86,13 +96,6 @@ in the uniform distribution over $[-3, 3]$, while the numbers on the diagonal
 are square roots of positive numbers from the same distribution. The matrix $H\r
 = U^{T}U$ is then symmetric positive definite.\r
 \r
-The function analyzed in this project was\r
-\begin{equation}\r
-  \label{eq:func_g}\r
-  g(\mathbf{x}) := -\mathbf{b}^{T}\mathbf{x} +\r
-  \frac{1}{2}\mathbf{x}^{T}H\mathbf{x} +\r
-  \frac{1}{12}\mathbf{x}^{T}C(\mathbf{x})\mathbf{x}.\r
-\end{equation}\r
 From equation \eqref{eq:func_g} we get\r
 \begin{eqnarray}\r
   \label{eq:grad_g}\r
@@ -107,7 +110,6 @@ implemented in the Python code below.
 \r
 Consider the Hessian $\nabla^{2}g(\mathbf{x})$, and any vector $\mathbf{v} \in\r
 \mathbb{R}^{n}_{*}$.  Then we have\r
-\r
 \begin{eqnarray}\r
   \mathbf{v}^{T}(\nabla^{2}g(\mathbf{x}))\mathbf{v} &=& \mathbf{v}^{T}(H+C(\mathbf{x}))\mathbf{v} \nonumber \\\r
   &=& \mathbf{v}^{T}H\mathbf{v}+\mathbf{v}^{T}C(\mathbf{x})\mathbf{v}.\r
@@ -128,7 +130,7 @@ that the Hessian, $\nabla^{2}g(\mathbf{x})$, is positive definite.
   \label{fig:fn_resid}\r
 \end{figure}\r
 \r
-\begin{figure}[htb]\r
+\begin{figure}[b]\r
   \centering\r
   \includegraphics[width=0.9\textwidth]{surface}\r
   \caption{\sf Surface plot of function with points showing iteration of steepest\r
@@ -136,6 +138,12 @@ that the Hessian, $\nabla^{2}g(\mathbf{x})$, is positive definite.
   \label{fig:surface}\r
 \end{figure}\r
 \r
+The steepest descent method works by always taking a step in the direction\r
+opposite the gradient in the point considered (i.e. the direction of the\r
+steepest descent). The step size can be set to a constant or chosen in a more\r
+optimal way. In the case of a constant step size, the method is equivalent to\r
+the Euler method.\r
+\r
 Consider the equation\r
 \begin{equation}\r
   \label{eq:forward}\r
@@ -153,7 +161,9 @@ vector equation, we get the following iteration:
 where $t_{n+1} = t_n + \alpha$. Looking at this result, we see that it is\r
 equivalent to the steepest descent iteration with constant step size $\alpha$. \r
 \r
-To find the optimal step length, $\alpha^*$, we solve the equation\r
+To find the optimal step length, $\alpha^*$, we walk along the line of steepest\r
+descent until we get to the lowest point. This point can be found analytically\r
+by solving the following equation for $\alpha$:\r
 \begin{eqnarray}\r
   \label{eq:poly}\r
   \frac{\partial}{\partial\r
@@ -162,7 +172,7 @@ To find the optimal step length, $\alpha^*$, we solve the equation
   &\Downarrow& \nonumber \\\r
   A\alpha^3 + B\alpha^2 + C\alpha + D &=& 0,\r
 \end{eqnarray}\r
-for $\alpha$. Where $A$, $B$, $C$, and $D$ are given by\r
+Where $A$, $B$, $C$, and $D$ are given by\r
 \begin{eqnarray}\r
   A &=& \frac{1}{3}\mathbf{u}^TC(\mathbf{u})\mathbf{u} \nonumber \\\r
   B &=& \mathbf{x}^TC(\mathbf{u}) \mathbf{u} \nonumber \\\r
@@ -208,7 +218,7 @@ The Newton method was also implemented to solve $\nabla g\left(\mathbf{x}^*\righ
 We see from table \ref{tab:time} that newtons method uses the least time to\r
 solve the problem, even though figure \ref{fig:fn_resid} shows that the\r
 combination method results in fewer iterations. This comes from the fact that\r
-computing the optimal step size requires solving a qubic equation and is\r
+computing the optimal step size requires solving a cubic equation and is\r
 computationally expensive.\r
 \r
 \end{document}\r