[Analysis] Fixes

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

@@ -8,7 +8,6 @@ Let $f : X \rightarrow \R$ with $f \in C^k(X, \R)$ and $y \in X$. The Taylor-Pol
 \end{align*}
 
 \drmvspace\rmvspace
-% TODO: Find out what the \partial_1 notation means (likely TA notes 09)
 where $i$ is a \textit{multi-index}, so:
 \drmvspace
 \begin{multicols}{3}
@@ -22,14 +21,19 @@ where $i$ is a \textit{multi-index}, so:
 \end{multicols}
 
 \drmvspace\rmvspace
+In the input, we have the vector $y$, which is the evaluation point, as well as the vector $x - y$ (where $y$ is the evaluation point again and $x = (x_1, \ldots, x_n)$)
+
 The concept this formula uses is that we iterate through all possible partial derivatives of $f$ and assigns each a multi-index $i$.
+Do note that the formula expands to $f(y) + \ldots$, so also include the original function in the sum!
+
 To denote that we want to take the partial derivative $\partial_{112}$, we use $i = (2, 1, 0)$, since we take the derivative of the first variable twice,
 of the second variable once and never of the third variable.
-So the expression is thus now:
+This is also the explanation for what the $\partial_1^{i_1}$ means (we take the derivative regarding the first variable $i_1$ times, etc).
+
+One of the elements of the sum (element with $i = (2, 1, 0)$) is for example:
 \mrmvspace
 \begin{align*}
-    \frac{\partial_{112} f(y) (x_1 - y_1)^2 (x_2 - y_2)^1 (x_3 y_3)^0}{2!1!0!} = \frac{\partial_{112} f(y) (x_1 - y_1)^2 (x_2 - y_2)}{2}
+    \frac{\partial_{112} f(y) (x_1 - y_1)^2 (x_2 - y_2)^1 (x_3 - y_3)^0}{2!1!0!} = \frac{\partial_{112} f(y) (x_1 - y_1)^2 (x_2 - y_2)}{2}
 \end{align*}
 
 \drmvspace
-% TODO: Add example

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

@@ -20,9 +20,10 @@ To determine the kind of critical point, we need to determine if $H_f(x_0)$ is d
 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 $0$.
+It is positive (negative) definite if and only if all eigenvalues are greater (lower) than $0$
 (Compute Eigenvalues using $\det(A - \lambda I) = 0$)
 
-For $2 \times 2$ matrices, we can use the following scheme:
+For $2 \times 2$ matrices (i.e. 2D functions), we can use the following scheme (remember that the trace is the sum of the diagonal entries):
 \begin{center}
     \begin{tikzpicture}[node distance = 0.3cm and 2cm, >={Stealth[round]}]
         \node (det) {$\det(A)$};