[Analysis} Finish Taylor polynomials, start critical points

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

@@ -22,4 +22,13 @@ where $i$ is a \textit{multi-index}, so:
 \end{multicols}
 
 \drmvspace\rmvspace
-The concept this formula uses is that we iterate through all possible partial derivatives of $f$ and assigns each a multi-index $i$
+The concept this formula uses is that we iterate through all possible partial derivatives of $f$ and assigns each a multi-index $i$.
+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:
+\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}
+\end{align*}
+
+\drmvspace

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

@@ -1,2 +1,25 @@
-\newsectionNoPB
+\newsection
 \subsection{Critical points}
+\stepLabelNumber{all}
+\compactdef{Critical Point} For $f: X \rightarrow \R^n$ differentiable, $x_0 \in X$ is called a \bi{critical point} of $f$ if $\nabla f(x_0) = 0$
+\shortremark As in 1 dimensional case, check edges of the interval for the critical point.\\
+%
+To determine the kind of critical point, we need to determine if $H_f(x_0)$ is definite:
+\drmvspace
+\begin{multicols}{3}
+    \begin{itemize}[noitemsep]
+        \item positive definite $\Rightarrow x_0$ local max
+        \item negative definite $\Rightarrow x_0$ local min
+        \item indefinite $\Rightarrow x_0$ point of inflection
+    \end{itemize}
+\end{multicols}
+
+\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)\\
+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 $0$.
+(Compute Eigenvalues using $\det(A - \lambda I) = 0$)
+
+For $2 \times 2$ matrices, we can use the following scheme: