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diff --git a/FuVar.tex b/FuVar.tex
index 61eead8..552bc54 100644
--- a/FuVar.tex
+++ b/FuVar.tex
@@ -31,6 +31,7 @@
%% Layout
\usepackage{enumitem}
\usepackage{booktabs}
+\usepackage{footmisc}
%% Nice drwaings
\usepackage{tikz}
@@ -91,16 +92,17 @@ These are just my personal notes of the \themodule{} course, and definitively
not a rigorously constructed mathematical text. The good looking \LaTeX{}
typesetting may trick you into thinking it is rigorous, but really, it is not.
-\section{Derivatives of vector valued functions}
+\section{Derivatives of vector valued scalar functions}
\begin{definition}[Partial derivative]
- A vector values function \(f: \mathbb{R}^m\to\mathbb{R}\), with
+ A vector valued function \(f: \mathbb{R}^m\to\mathbb{R}\), with
\(\vec{v}\in\mathbb{R}^m\), has a partial derivative with respect to \(v_i\)
defined as
\[
\partial_{v_i} f(\vec{v})
- = f_{v_i}(\vec{v})
- = \lim_{h\to 0} \frac{f(\vec{v} + h\vec{e}_j) - f(\vec{v})}{h}
+ % = f_{v_i}(\vec{v})
+ = \frac{\partial f}{\partial v_i}
+ = \lim_{h\to 0} \frac{f(\vec{v} + h\vec{e}_i) - f(\vec{v})}{h}
\]
\end{definition}
@@ -113,17 +115,41 @@ typesetting may trick you into thinking it is rigorous, but really, it is not.
\]
\end{theorem}
-\begin{definition}[Linearization]
+\begin{application}[Find the slope of an implicit curve]
+ Let \(f(x,y) = 0\) be an implicit curve. It's slope at any point where
+ \(\partial_y f \neq 0\) is \(m = - \partial_x f / \partial_y f\)
+\end{application}
+
+\begin{definition}[Total differential]
+ The total differential \(df\) of \(f:\mathbb{R}^m\to\mathbb{R}\) is
+ \[
+ df = \sum_{i=0}^m \partial_{x_i} f\cdot dx .
+ \]
+ That reads, the \emph{total} change is the sum of the change in each
+ direction. This implies
+ \[
+ \frac{df}{dx_k} = \frac{\partial f}{\partial x_k} +
+ \sum_{i \in \{1 \leq i \leq m : i \neq k\}}
+ \frac{\partial f}{\partial x_i} \cdot \frac{dx_i}{dx_k} ,
+ \]
+ i.e. the change in direction \(x_k\) is how \(f\) changes in \(x_k\)
+ (ignoring other directions) plus, how \(f\) changes with respect to each
+ other variable \(x_i\) times how it (\(x_i\)) changes with respect to \(x_k\).
+\end{definition}
+
+\begin{application}[Linearization]
A function \(f: \mathbb{R}^m\to\mathbb{R}\) has a linearization \(g\) at
\(\vec{x}_0\) given by
\[
g(\vec{x}) = f(\vec{x}_0)
+ \sum_{i=1}^m \partial_{x_i} f(\vec{x}_0)(x_i - x_{i,0}) ,
\]
- if all partial derviatives are defined at \(\vec{x}_0\).
-\end{definition}
+ if all partial derivatives are defined at \(\vec{x}_0\). With the gradient
+ (defined below) \(g(\vec{x}) = f(\vec{x}_0) + \grad f(\vec{x}_0) \dotp
+ (\vec{x} - \vec{x}_0)\).
+\end{application}
-\begin{theorem}[Propagation of uncertanty]
+\begin{application}[Propagation of uncertanty]
Given a measurement of \(m\) values in a vector \(\vec{x}\in\mathbb{R}^m\)
with values given in the form \(x_i = \bar{x}_i \pm \sigma_{x_i}\), a linear
approximation the error of a dependent variable \(y\) is computed with
@@ -132,7 +158,7 @@ typesetting may trick you into thinking it is rigorous, but really, it is not.
\pm \sqrt{\sum_{i=1}^m \left(
\partial_{x_i} f(\bar{\vec{x}}) \sigma_{x_i}\right)^2}
\]
-\end{theorem}
+\end{application}
\begin{definition}[Gradient vector]
The \emph{gradient} of a function \(f(\vec{x}), \vec{x}\in\mathbb{R}^m\) is a
@@ -150,20 +176,20 @@ typesetting may trick you into thinking it is rigorous, but really, it is not.
\]
\end{definition}
+\begin{theorem}
+ The gradient vector always points towards \emph{the direction of steepest
+ ascent}, and thus is always perpendicular to contour lines.
+\end{theorem}
+
\begin{definition}[Directional derivative]
A function \(f(\vec{x})\) has a directional derivative in direction
- \(\vec{r}\) (with \(|\vec{r}| = 1\)) given by
+ \(\vec{r}\) (with \(|\vec{r}|=1\)) of
\[
\frac{\partial f}{\partial\vec{r}}
= \nabla_\vec{r} f = \vec{r} \dotp \grad f
\]
\end{definition}
-\begin{theorem}
- The gradient vector always points towards \emph{the direction of steepest
- ascent}.
-\end{theorem}
-
\begin{definition}[Jacobian Matrix]
The \emph{Jacobian} \(\mx{J}_f\) (sometimes written as
\(\frac{\partial(f_1,\ldots f_m)}{\partial(x_1,\ldots,x_n)}\)) of a function
@@ -192,7 +218,7 @@ typesetting may trick you into thinking it is rigorous, but really, it is not.
\begin{definition}[Hessian matrix]
Given a function \(f: \mathbb{R}^m \to \mathbb{R}\), the square matrix whose
entry at the \(i\)-th row and \(j\)-th column is the second derivative of
- \(f\) first with respect to \(x_j\) and then to \(x_i\) is know as the
+ \(f\) first with respect to \(x_j\) and then to \(x_i\) is known as the
\emph{Hessian} matrix.
\(
\left(\mx{H}_f\right)_{i,j} = \partial_{x_i}\partial_{x_j} f
@@ -212,11 +238,14 @@ typesetting may trick you into thinking it is rigorous, but really, it is not.
\section{Methods for maximization and minimization problems}
+\subsection{Analytical methods}
+
\begin{method}[Find stationary points]
Given a function \(f: D \subseteq \mathbb{R}^m \to \mathbb{R}\), to
find its maxima and minima we shall consider the points
\begin{itemize}
- \item that are on the boundary of the domain \(\partial D\),
+ \item that are on the boundary\footnote{If it belongs to \(f\).
+ \label{ftn:boundary}} of the domain \(\partial D\),
\item where the gradient \(\grad f\) is not defined,
\item that are stationary, i.e. where \(\grad f = \vec{0}\).
\end{itemize}
@@ -243,14 +272,14 @@ typesetting may trick you into thinking it is rigorous, but really, it is not.
\begin{remark}
The previous method is obtained by studying the second directional derivative
\(\nabla_\vec{r}\nabla_\vec{r} f\) at the stationary point in direction of a
- vector \(\vec{r} = \vec{e}_1\cos(\alpha) + \vec{e}_2\sin(\alpha)\)
+ vector \(\vec{r} = \vec{e}_1\cos(\alpha) + \vec{e}_2\sin(\alpha)\).
\end{remark}
\begin{method}[Determine the type of stationary point in higher dimensions]
- Given a scalar function of two variables \(f(x,y)\) and a stationary point
- \(\vec{x}_s\) (where \(\grad f(\vec{x}_s) = \vec{0}\)), we compute the
- Hessian matrix \(\mx{H}_f(\vec{x}_s)\). Then we compute its eigenvalues
- \(\lambda_1, \ldots, \lambda_m\) and
+ Given a scalar function of multiple variables \(f(\vec{x})\) and a stationary
+ point \(\vec{x}_s\) (\(\grad f(\vec{x}_s) = \vec{0}\)), we compute the
+ Hessian matrix \(\mx{H}_f(\vec{x}_s)\) and its eigenvalues \(\lambda_1,
+ \ldots, \lambda_m\), then
\begin{itemize}
\item if all \(\lambda_i > 0\), the point is a minimum;
\item if all \(\lambda_i < 0\), the point is a maximum;
@@ -284,7 +313,8 @@ typesetting may trick you into thinking it is rigorous, but really, it is not.
\(f: D \subseteq \mathbb{R}^2 \to \mathbb{R}\). To find the extrema we look for
points
\begin{itemize}
- \item on the boundary \(\vec{u} \in \partial D\) where \(n(\vec{u}) = 0\);
+ \item on the boundary\footref{ftn:boundary} \(\vec{u} \in \partial D\)
+ where \(n(\vec{u}) = 0\);
\item \(\vec{u}\) where the gradient either does not exist or is
\(\vec{0}\), and satisfy \(n(\vec{u}) = 0\);
@@ -316,7 +346,7 @@ typesetting may trick you into thinking it is rigorous, but really, it is not.
under \(k < m\) constraints \(n_1 = 0, \cdots, n_k = 0\). To find the extrema
we consider the following points:
\begin{itemize}
- \item Points on the boundary \(\vec{u} \in \partial D\) that satisfy
+ \item Points on the boundary\footref{ftn:boundary} \(\vec{u} \in \partial D\) that satisfy
\(n_i(\vec{u}) = 0\) for all \(1 \leq i \leq k\),
\item Points \(\vec{u} \in D\) where either
@@ -336,18 +366,56 @@ typesetting may trick you into thinking it is rigorous, but really, it is not.
\]
The \(\lambda\) values are known as \emph{Lagrange multipliers}. The same
calculation can be written more compactly by defining the
- \(m+k\) dimensional \emph{Lagrangian}
+ \emph{Lagrangian}
\[
\mathcal{L}(\vec{u}, \vec{\lambda})
- = f(\vec{u}) - \sum_{i = 0}^k \lambda_i n_i(\vec{u})
+ = f(\vec{u}) - \sum_{i = 0}^k \lambda_i n_i(\vec{u}),
\]
where \(\vec{\lambda} = \lambda_1, \ldots, \lambda_k\) and then solving
- \(\grad \mathcal{L}(\vec{u}, \vec{\lambda}) = \vec{0}\). This is
- generally used in numerical computations and not very useful by hand.
+ the \(m+k\) dimensional equation \(\grad \mathcal{L}(\vec{u},
+ \vec{\lambda}) = \vec{0}\) (this is generally used in numerical
+ computations and not very useful by hand).
\end{itemize}
\end{method}
-\section{Integration of vector values scalar functions}
+\subsection{Numerical methods}
+
+\begin{method}[Newton's method]
+ For a function \(f:\mathbb{R}^m\to\mathbb{R}\) we wish to numerically find
+ its stationary points (where \(\grad f = \vec{0}\)).
+ \begin{enumerate}
+ \item Pick a starting point \(\vec{x}_0\)
+ \item Set the linearisation\footnote{The gradient becomes a hessian matrix.}
+ of \(\grad f\) at \(\vec{x}_k\) to zero and
+ solve for \(\vec{x}_{k+1}\)
+ \begin{gather*}
+ \grad f(\vec{x}_k) + \mx{H}_f (\vec{x}_k)
+ (\vec{x}_{k+1} - \vec{x}_k) = \vec{0} \\
+ \vec{x}_{k+1} = \vec{x}_k - \mx{H}_f^{-1} (\vec{x}_k) \grad f(\vec{x}_k)
+ \end{gather*}
+ \item Repeat the last step until the magnitude of the error
+ \(|\vec{\epsilon}| = |\mx{H}_f^{-1} (\vec{x}_k) \grad f(\vec{x}_k)|\) is
+ sufficiently small.
+ \end{enumerate}
+\end{method}
+
+\begin{method}[Gradient ascent / descent]
+ Given \(f:\mathbb{R}^m\to\mathbb{R}\) we wish to numerically find
+ the stationary points (where \(\grad f = \vec{0}\)).
+ \begin{enumerate}
+ \item Define an arbitrarily small length \(\eta\) and a starting point
+ \(\vec{x}_0\)
+ \item Compute \(\vec{v} = \pm\grad f(\vec{x}_k)\) (positive for ascent,
+ negative for descent), then \(\vec{x}_{k+1} = \vec{x}_k + \eta\vec{v}\)
+ if the rate of change \(\epsilon\) is acceptable (\(\epsilon = |\grad
+ f(\vec{x}_{k+1})| > 0\)) else recompute \(\vec{v} := \pm \grad
+ f(\vec{x}_{k+1})\).
+ \item Stop when the rate of change \(\epsilon\) stays small enough for many
+ iterations.
+ \end{enumerate}
+\end{method}
+
+\section{Integration of vector valued scalar functions}
\begin{figure}
\centering
@@ -385,7 +453,7 @@ typesetting may trick you into thinking it is rigorous, but really, it is not.
\begin{theorem}[Transformation of coordinates]
The generalization of theorem \ref{thm:transform-coords} is quite simple.
- For an \(n\)-integral of a function \(f:\mathbb{R}^m\to\mathbb{R}\) over a
+ For an \(m\)-integral of a function \(f:\mathbb{R}^m\to\mathbb{R}\) over a
region \(B\), we let \(\vec{x}(\vec{u})\) be ``nice'' functions that
transform the coordinate system. Then as before
\[
@@ -407,14 +475,153 @@ typesetting may trick you into thinking it is rigorous, but really, it is not.
Cylindrical & r\,dr\,d\phi\,dz & \uvec{z}r\,dr\,d\phi \\
& & \uvec{\phi}\,dr\,dz \\
& & \uvec{r}r\,d\phi\,dz \\
- Spherical & r^2\sin\theta\, dr\,d\theta\,d\phi & \uvec{r}r^2\sin\theta\,d\theta\,d\phi \\
+ Spherical & r^2\sin\theta\, dr\,d\theta\,d\phi &
+ \uvec{r}r^2\sin\theta\,d\theta\,d\phi \\
Curvilinear & |\mx{J}_f|\,du\,dv\,dw & - \\
\bottomrule
\end{tabular}
\caption{Differential elements for integration.}
\end{table}
-\section{Derivatives of curves}
+\begin{application}[Physics]
+ Given the mass \(m\) and density function \(\rho\) of an object,
+ its \emph{center of mass} is calculated with
+ \[
+ \vec{x}_c = \frac{1}{m}\int_V \vec{x}\rho(\vec{x}) \,dv
+ \stackrel{\rho\text{ const.}}{=} \frac{1}{V} \int_V \vec{x}\,dv .
+ \]
+ The (scalar) \emph{moment of inertia} \(J\) of an object is given by
+ \[
+ J = \int_V \rho(\vec{r}) r^2 \,dv .
+ \]
+ % and similarly the \emph{area moment of inertia} \(I\)
+\end{application}
+
+\section{Parametric curves and line integrals}
+
+\begin{definition}[Parametric curve]
+ A parametric curve is a vector function \(\mathcal{C} : \mathbb{R} \to W
+ \subseteq \mathbb{R}^n, t \mapsto \vec{f}(t)\), that takes a parameter \(t\).
+\end{definition}
+
+\begin{definition}[Multivariable chain rule]
+ Let \(\vec{x}: \mathbb{R} \to \mathbb{R}^m\) and \(f: \mathbb{R}^m \to
+ \mathbb{R}\), so that \(f\circ\vec{x}: \mathbb{R} \to \mathbb{R}\), then
+ the multivariable chain rule states:
+ \[
+ \frac{d}{dt}f(\vec{x}(t)) = \grad f (\vec{x}(t)) \dotp \vec{x}'(t)
+ = \nabla_{\vec{x}'(t)} f(\vec{x}(t))
+ \]
+\end{definition}
+
+\begin{theorem}[Signed area enclosed by a planar parametric curve]
+ A planar (2D) parametric curve \((x(t), y(t))^t\) with \(t\in[r,s]\) that does
+ not intersect itself encloses a surface with area
+ \[
+ A = \int_r^s x'(t)y(t) \,dt
+ = \int_r^s x(t)y'(t) \,dt
+ \]
+\end{theorem}
+
+\begin{theorem}[Derivative of a curve]
+ The derivative of a curve is
+ \begin{align*}
+ \vec{f}'(t) &= \lim_{h\to 0} \frac{\vec{f}(t + h) - \vec{f}(t)}{h} \\
+ &= \sum_{i=0}^n \left(\lim_{h\to 0} \frac{f_i(t+h) - f_i(t)}{h}\right) \vec{e}_i \\
+ &= \sum_{i=0}^n \frac{df_i}{dt}\vec{e}_i
+ = \left(\frac{df_1}{dt}, \ldots, \frac{df_m}{dt}\right)^t
+ \end{align*}
+\end{theorem}
+
+\begin{definition}[Line integral in a scalar field]
+ Let \(\mathcal{C}:[a,b]\to\mathbb{R}^n, t \mapsto \vec{x}(t)\) be a
+ parametric curve. The \emph{line integral} in a field \(f(\vec{x})\) is the
+ integral of the signed area under the curve traced in \(\mathbb{R}^n\), and
+ is computed with
+ \[
+ \int_\mathcal{C} f(\vec{x}) \,d\ell
+ = \int_\mathcal{C} f(\vec{x}) \,|d\vec{x}|
+ = \int_a^b f(\vec{x}(t)) |\vec{x}'(t)| \, dt
+ \]
+\end{definition}
+
+\begin{application}[Length of a parametric curve]
+ By computing the line integral of the function \(\vec{1}(t) = 1\) we get the
+ length of the parametric curve \(\mathcal{C}:[a,b]\to\mathbb{R}^n\).
+ \[
+ \int_\mathcal{C}d\ell
+ = \int_\mathcal{C} |d\vec{x}|
+ = \int_a^b \sqrt{\sum_{i=1}^n x'_i(t)^2} \,dt
+ \]
+ In the special case with the scalar function \(f(x)\) results in
+ \(\int_a^b\sqrt{1+f'(x)^2}\,dx\)
+\end{application}
+
+\begin{definition}[Line integral in a vector field]
+ The line integral in a vector field \(\vec{F}(\vec{x})\) is ``sum'' of the
+ projections of the field's vectors on the tangent of the parametric curve
+ \(\mathcal{C}\).
+ \[
+ \int_\mathcal{C} \vec{F}(\vec{r})\dotp d\vec{r}
+ = \int_a^b \vec{F}(\vec{r}(t))\dotp \vec{r}'(t) \,dt
+ \]
+\end{definition}
+
+\begin{theorem}[Line integral in the opposite direction]
+ By integrating while moving backwards (\(-t\)) on the parametric curve gives
+ \[
+ \int_{-\mathcal{C}} \vec{F}(\vec{r})\dotp d\vec{r}
+ = -\int_{\mathcal{C}} \vec{F}(\vec{r})\dotp d\vec{r}
+ \]
+\end{theorem}
+
+\begin{definition}[Conservative field]
+ A vector field is said to be \emph{conservative} the line integral over a
+ closed path is zero.
+ \[
+ \oint_\mathcal{C} \vec{F}(\vec{r})\cdot d\vec{r} = 0
+ \]
+\end{definition}
+
+\begin{theorem}
+ For a twice partially differentiable vector field \(\vec{F}(\vec{x})\) in
+ \(n\) dimensions without ``holes'', i.e. in which each closed curve can be
+ contracted to a point (simply connected open set), the following statements
+ are equivalent:
+ \begin{itemize}
+ \item \(\vec{F}\) is conservative
+ \item \(\vec{F}\) is path-independent
+ \item \(\vec{F}\) is a \emph{gradient field}, i.e. there is a
+ function \(\phi\) called \emph{potential} such that \(\vec{F} = \grad
+ \phi\)
+ \item \(\vec{F}\) satisfies the condition \(\partial_{x_j} F_i =
+ \partial_{x_i} F_j\) for all \(i,j \in \{1,2,\ldots,n\}\). In the 2D case
+ \(\partial_x F_y = \partial_y F_x\), and in 3D
+ \[
+ \begin{cases}
+ \partial_y F_x = \partial_x F_y \\
+ \partial_z F_y = \partial_y F_z \\
+ \partial_x F_z = \partial_z F_x \\
+ \end{cases}
+ \]
+ \end{itemize}
+\end{theorem}
+
+\begin{theorem}
+ In a conservative field \(\vec{F}\) with gradient \(\phi\), using the
+ multivariable the chain rule:
+ \begin{align*}
+ \int_\mathcal{C} \vec{F} \dotp d\vec{r}
+ &= \int_\mathcal{C} \vec{F}(\vec{r}(t)) \dotp \vec{r}'(t) \,dt \\
+ &= \int_\mathcal{C} \grad \phi(\vec{r}(t)) \cdot \vec{r}'(t) \,dt \\
+ &= \int_\mathcal{C} \frac{d\phi(\vec{r}(t))}{dt}\,dt
+ = \phi(\vec{r}(b)) - \phi(\vec{r}(a))
+ \end{align*}
+\end{theorem}
+
+\section{Surface integrals}
+
+\section{Vector analysis}
\section*{License}
\doclicenseText