[Analysis] Add more notes

semester3/analysis-ii/cheat-sheet-jh/parts/vectors/differentiation/01_partial_derivatives.tex

@@ -40,5 +40,6 @@ $\nabla f(x_0) =
 and the
 
 \drmvspace\rmvspace
-trace of the Jacobi Matrix, $\text{div}(f)(x_0) = \text{Tr}(J_f(x_0)) = \sum_{i = 1}^{n} \partial_{x_i} f_i(x_0)$ is called the \bi{divergence} of $f$ at $x_0$.
+trace of the Jacobi Matrix, $\text{div}(f)(x_0) = \text{Tr}(J_f(x_0)) = \sum_{i = 1}^{n} \partial_{x_i} f_i(x_0)$ is called the \bi{divergence} of $f$ at $x_0$.\\
+The gradient is simply the transpose of the Jacobian and it points in the direction of the \bi{steepest ascent}.
 \rmvspace

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

@@ -1,4 +1,4 @@
-\newsectionNoPB
+\newsection
 \subsection{The differential}
 \setLabelNumber{all}{2}
 \compactdef{Differentiable function} We have function $f: X \rightarrow \R^m$, linear map $u : \R^n \rightarrow \R^m$ and $x_0 \in X$. $f$ is differentiable at $x_0$ with differential $u$ if
@@ -6,7 +6,6 @@ $\displaystyle \lim_{\elementstack{x \rightarrow x_0}{x \neq x_0}} \frac{f(x) -
 We denote $\dx f(x_0) = u$.
 If $f$ is differentiable at every $x_0 \in X$, then $f$ is differentiable on $X$.
 
-\newpage
 % ────────────────────────────────────────────────────────────────────
 \stepLabelNumber{all}
 \shortproposition

semester3/analysis-ii/cheat-sheet-jh/parts/vectors/differentiation/06_critical_points.tex

@@ -16,7 +16,8 @@ To determine the kind of critical point, we need to determine if $H_f(x_0)$ is d
 
 \dhrmvspace
 \setLabelNumber{all}{6}
-\compactdef{Non-degenerate critical point} If $\det(H_f(x_0)) \neq 0$ (if $H_f(x_0)$ is semi-definite, then $\det(H_f(x_0)) = 0$, thus degenerate)\\
+\compactdef{Non-degenerate critical point} If $\det(H_f(x_0)) \neq 0$ (if $H_f(x_0)$ is semi-definite, then $\det(H_f(x_0)) = 0$, thus degenerate)
+
 To figure out if a matrix is definite, we can compute the eigenvalues. $A$ is positive (negative) definite, if and only if all eigenvalues are greater (lower) than $0$.
 $A$ is indefinite if and only if it has both positive and negative eigenvalues.
 $A$ is positive (negative) semi-definite if and only if all eigenvalues are greater (lower) or equal to $0$.
@@ -48,3 +49,5 @@ For $2 \times 2$ matrices (i.e. 2D functions), we can use the following scheme (
         (tr1) edge node [right] {$0$} (zero);
     \end{tikzpicture}
 \end{center}
+As in Analysis I, it is important to also check the boundaries for maximums and minimums.
+For that, formulate formulas for the borders and check them for critical points.