[Analysis] Notes and cleanup

Notes for tangent space computation and green's formula.
Also includes fixes for the pagination as result of additions

semester3/analysis-ii/cheat-sheet-jh/parts/vectors/differentiation/02_differential.tex

@@ -49,12 +49,16 @@ Finally, evaluate $\frac{\partial g}{\partial \phi}$ at the required points and
 % ────────────────────────────────────────────────────────────────────
 \setLabelNumber{all}{11}
 \compactdef{Tangent space} The graph of the affine linear approximation $g(x) = f(x_0) + u(x - x_0)$, or the set
-\vspace{-0.75pc}
+\rmvspace
 \begin{align*}
     \{ (x, y) \in \R^n \times \R^m : y = f(x_0) + u(x - x_0) \}
 \end{align*}
 
-\dnrmvspace
+\drmvspace
+\shade{gray}{Computing the tangent space} Also called the \bi{Tangent plane} in 3D.
+We only need to compute $g(x) = f(x_0) + J_f(x_0) \cdot (x - x_0)$, where both $x$ and $x_0$ are vectors (and $x_0$ is the point at which we compute the tangent space).
+All there is left to do is state the space: $\{ (x, y, \ldots) \in \R^n | z = g() \}$
+
 % ────────────────────────────────────────────────────────────────────
 \stepLabelNumber{all}
 \compactdef{Directional derivative} $f$ has a directional derivative $w \in \R^m$ in the direction of $v \in \R^n$,
@@ -71,7 +75,7 @@ Suppose we know the dir. der. $w_1$ and $w_2$ in directions $v_1$ and $v_2$, the
 \shade{gray}{Computing a directional derivative} Always normalize the vector! We can compute a directional derivative using the differential $\limit{h}{0} \frac{f(x_0 + hv) - f(x_0)}{h}$
 or using a $1$-dimensional helper function $g: h \mapsto f(x_0 + hv)$, calculating the derivative of it and evaluating $g'(0)$. That corresponds to the directional derivative.
 E.g. for function $f: x, y \mapsto x^2 + y^2$, we have $g: h \mapsto (x_0 + h)^2 + (y_0 + h)^2$.
-A final option is to compute it using a matrix-vector product: $D_v f(x_0) = J_f(x_0) v$
+An \bi{easy option} is to use this property: $D_v f(x_0) = J_f(x_0) \cdot v = \nabla f \cdot v$
 
 \rmvspace
 \begin{center}

semester3/analysis-ii/cheat-sheet-jh/parts/vectors/differentiation/04_change_of_variable.tex

@@ -1,4 +1,4 @@
-\newsectionNoPB
+\newsection
 \subsection{Change of variable}
 The idea is to substitute variables for others that make the equation easier to solve.
 A common example is to switch to polar coordinates from cartesian coordinates, as already demonstrated with continuity checks

semester3/analysis-ii/cheat-sheet-jh/parts/vectors/differentiation/05_taylor_polynomials.tex

@@ -1,4 +1,4 @@
-\newsection
+\newsectionNoPB
 \subsection{Taylor polynomials}
 \compactdef{Taylor polynomials}
 Let $f : X \rightarrow \R$ with $f \in C^k(X, \R)$ and $y \in X$. The Taylor-Polynomial of order $k$ of $f$ at $y$ is:

semester3/analysis-ii/cheat-sheet-jh/parts/vectors/integration/04_green_formula.tex

@@ -27,7 +27,7 @@ $\gamma_i$ as above, then
 Thus, the sum of all line integrals is the same thing as the Riemann-Integral of the curl.
 
 We can use Green's Formula to compute integrals. For that we need the set of curves that define the set.
-For the unit circle, that is just one curve,
+For the \bi{unit circle}, that is just one curve,
 being $\gamma(t) = \begin{pmatrix}
         R \cdot \cos(t) \\
         R \cdot \sin(t)
@@ -53,9 +53,8 @@ That set is derived from the image that is given for the line.
 Be cognizant of what direction the integral goes, if the set is on the right hand side of the curve, the final result has to be negated to change the direction of the integral.
 If the curve doesn't fully enclose the set, then we can simply compute the line integrals of the missing sections and subtract them from the final result.
 
-We can also use known formulas to compute the area of discs, etc (like $r^2 * \pi$ for a circle).
-To calculate the area enclosed by a curve using Green's formua, we can use the vector field
-% TODO: Finish
+We can also use known formulas to compute the area of discs, etc (like $r^2 \cdot \pi$ for a circle).
+To calculate the area enclosed by a curve using Green's formua, if not given a vector field, we can use the vector field $F(x, y) = (0, x)$.
 
 \shade{gray}{Center of mass}
 The center of mass of an object $\cU$ is given by $\displaystyle \overline{x}_i = \frac{1}{\text{Vol}(\cU)} \int_{\cU} x_i \dx x$.