Docs: linear equation systems

MathNet.Numerics.sln

@@ -50,6 +50,7 @@ Project("{2150E333-8FDC-42A3-9474-1A3956D46DE8}") = "Docs", "Docs", "{039229DA-A
 		docs\content\IntegralTransforms.fsx = docs\content\IntegralTransforms.fsx
 		docs\content\Integration.fsx = docs\content\Integration.fsx
 		docs\content\Interpolation.fsx = docs\content\Interpolation.fsx
+		docs\content\LinearEquations.fsx = docs\content\LinearEquations.fsx
 		docs\content\MKL.fsx = docs\content\MKL.fsx
 		docs\content\Random.fsx = docs\content\Random.fsx
 		docs\tools\templates\template.cshtml = docs\tools\templates\template.cshtml

docs/content/LinearEquations.fsx

@@ -0,0 +1,131 @@
+(*** hide ***)
+#I "../../out/lib/net40"
+#r "MathNet.Numerics.dll"
+#r "MathNet.Numerics.FSharp.dll"
+open MathNet.Numerics.LinearAlgebra
+
+(**
+Linear Equation Systems
+=======================
+
+A system of linear equations is a collection of linear equations involving the same set of variables:
+
+$$$
+\begin{alignat}{7}
+3x &\; + \;& 2y             &\; - \;& z  &\; = \;& 1 & \\
+2x &\; - \;& 2y             &\; + \;& 4z &\; = \;& -2 & \\
+-x &\; + \;& \tfrac{1}{2} y &\; - \;& z  &\; = \;& 0 &
+\end{alignat}
+
+More generally, we can write
+
+$$$
+\begin{alignat}{7}
+a_{11} x_1 &&\; + \;&& a_{12} x_2   &&\; + \cdots + \;&& a_{1n} x_n &&\; = \;&&& b_1 \\
+a_{21} x_1 &&\; + \;&& a_{22} x_2   &&\; + \cdots + \;&& a_{2n} x_n &&\; = \;&&& b_2 \\
+\vdots\;\;\; &&     && \vdots\;\;\; &&                && \vdots\;\;\; &&     &&& \;\vdots \\
+a_{m1} x_1 &&\; + \;&& a_{m2} x_2   &&\; + \cdots + \;&& a_{mn} x_n &&\; = \;&&& b_m \\
+\end{alignat}
+
+where we all parameters $a_{ij}$ and $b_i$ are known and we would like to find $x_j$ that satisfy
+all these equations. If we have the same number $n$ of unknown variables $x_j$ as number of
+equations $m$, and all these equations are independent, then there is a unique solution.
+
+This is a fundamental problem in the domain of linear algebra, and we can use its power to find the solution.
+Accordingly we can write the equivalent problem with matrices and vectors:
+
+$$$
+\mathbf{A}=
+\begin{bmatrix}
+a_{11} & a_{12} & \cdots & a_{1n} \\
+a_{21} & a_{22} & \cdots & a_{2n} \\
+\vdots & \vdots & \ddots & \vdots \\
+a_{m1} & a_{m2} & \cdots & a_{mn}
+\end{bmatrix},\quad
+\mathbf{x}=\begin{bmatrix}x_1\\x_2\\ \vdots \\x_n\end{bmatrix},\quad
+\mathbf{b}=\begin{bmatrix}b_1\\b_2\\ \vdots \\b_m\end{bmatrix}
+
+such that
+
+$$$
+\mathbf{A}\mathbf{x}=\mathbf{b}
+
+The initial example system would then look like this:
+
+$$$
+\begin{bmatrix}3 & 2 & -1 \\2 & -2 & 4 \\-1 & \tfrac{1}{2} & -1\end{bmatrix}
+\begin{bmatrix}x\\y\\z\end{bmatrix}
+\;=\;
+\begin{bmatrix}1\\-2\\0\end{bmatrix}
+
+Which we can solve explicitly with the LU-decomposition, or simply by using the Solve method:
+
+    [lang=csharp]
+    var A = Matrix<double>.Build.DenseOfArray(new double[,] {
+        { 3, 2, -1 },
+        { 2, -2, 4 },
+        { -1, 0.5, -1 }
+    });
+    var b = Vector<double>.Build.Dense(new double[] { 1, -2, 0 });
+    var x = A.Solve(b);
+
+The resulting $\mathbf{x}$ is $[1,\;-2,\;-2]$, hence the solution $x=1,\;y=-2,\;z=-2$.
+
+In F# the syntax is a bit lighter:
+*)
+
+let A = matrix [[ 3.0; 2.0; -1.0 ]
+                [ 2.0; -2.0; 4.0 ]
+                [ -1.0; 0.5; -1.0 ]]
+let b = vector [ 1.0; -2.0; 0.0 ]
+let x = A.Solve(b) // 1;-2;-2
+
+(**
+Normalizing Equation Systems
+----------------------------
+
+In practice, a linear equation system to be solved is often not in the standard form required
+to use the linear algebra approach. For example, let's have a look at the following system:
+
+$$$
+\begin{bmatrix}1 & 2 & 3 & 4\\2 & 3 & 4 & 5\\3 & 4 & 5 & 6\\4 & 5 & 6 & 7\end{bmatrix}
+\begin{bmatrix}0\\0\\V\\T\end{bmatrix}
+\;=\;
+\begin{bmatrix}F\\M\\20\\0\end{bmatrix}
+
+The first two values of the solution vector $[0,\;0,\;V,\;T]$ are constant zero, so we can simplify
+the system to:
+
+$$$
+\begin{bmatrix}3 & 4\\4 & 5\\5 & 6\\6 & 7\end{bmatrix}
+\begin{bmatrix}V\\T\end{bmatrix}
+\;=\;
+\begin{bmatrix}F\\M\\20\\0\end{bmatrix}
+
+Then we need to subtract the two unknowns from the right side back from the left (so that they
+become zero on the right side), by introducing a new column each. First we subtract
+$[F,\;0,\;0,\;0]^T$ from both sides:
+
+$$$
+\begin{bmatrix}3 & 4 & -1\\4 & 5 & 0\\5 & 6 & 0\\6 & 7 & 0\end{bmatrix}
+\begin{bmatrix}V\\T\\F\end{bmatrix}
+\;=\;
+\begin{bmatrix}0\\M\\20\\0\end{bmatrix}
+
+Then we subtract $[0,\;M,\;0,\;0]^T$ from both sides the same way:
+
+$$$
+\begin{bmatrix}3 & 4 & -1 & 0\\4 & 5 & 0 & -1\\5 & 6 & 0 & 0\\6 & 7 & 0 & 0\end{bmatrix}
+\begin{bmatrix}V\\T\\F\\M\end{bmatrix}
+\;=\;
+\begin{bmatrix}0\\0\\20\\0\end{bmatrix}
+
+Which is in standard from, so we can solve normally:
+*)
+
+let A' = matrix [[ 3.0; 4.0; -1.0; 0.0 ]
+                 [ 4.0; 5.0; 0.0; -1.0 ]
+                 [ 5.0; 6.0; 0.0; 0.0; ]
+                 [ 6.0; 7.0; 0.0; 0.0 ]]
+let b' = vector [ 0.0; 0.0; 20.0; 0.0 ]
+let x' = A'.Solve(b') // -140; 120; 60; 40

docs/tools/templates/template.cshtml

@@ -79,7 +79,7 @@
                         <li>Window Functions</li>
 
                         <li class="nav-header">Solving Equations</li>
-                        <li>Linear Equation Systems</li>
+                        <li><a href="@Root/LinearEquations.html">Linear Equation Systems</a></li>
                         <li>Nonlinear Root Finding</li>
 
                         <li class="nav-header">Optimization</li>