diff --git a/Lectures_my/NumMet/Lecture4/lecture4.tex b/Lectures_my/NumMet/Lecture4/lecture4.tex
index d9c3352..7893275 100644
--- a/Lectures_my/NumMet/Lecture4/lecture4.tex
+++ b/Lectures_my/NumMet/Lecture4/lecture4.tex
@@ -256,7 +256,7 @@
                     & \approx \xi + \mathcal{O}((\xi - x_k)^3)
             \end{align*}
         \item if $|f(a_+)| < |f(a_-)|$, then $x_{k + 1} \leftarrow a_+$; otherwise $x_{k + 1} \leftarrow a_-$
-        \item if $f(x_{k + 1}) \leq t$, where $t$ is an a-prior threshold, then stop; otherwise
+        \item if $f(x_{k + 1}) \leq t$, where $t$ is an a-priori threshold, then stop; otherwise
             jump back to step \#2.
     \end{enumerate}
 \end{frame}
@@ -265,13 +265,13 @@
     Does this iterative procedure converge? If yes, how fast? How can we quantify the rate of convergence?
 
     \begin{itemize}
-        \item we have \alert{local} convergence of order $p \leq 1$, if for all $x \in U(\xi)$
+        \item we have \alert{local} convergence of order $p \geq 1$, if for all $x \in U(\xi)$
         \begin{equation*}
             || \Phi[f](x) - \xi|| \leq C \cdot || x - \xi||^p\,,\quad\text{with}\quad C \geq 0\,.
         \end{equation*}
         Note: if $p = 1$ then we must have $C < 1$.
         \item we have \alert{global} convergence if $U(\xi) = \mathbb{R}$
-        \item for $D=1$ we can calculate $p$ is $\Phi[f](x)$ differentiable to sufficient degree:
+        \item for $D=1$ we can calculate $p$ if $\Phi[f](x)$ is differentiable to sufficient degree:
         \begin{align*}
             \Phi(x) - \xi = \Phi(x) - \Phi(\xi) = \frac{(x - \xi)^p}{p!} + o(||x - \xi||^p)
         \end{align*}
@@ -295,6 +295,127 @@
     \alert{(*): only if $f'(\xi) \neq 0$}, which is equivalent to $\xi$ is a simple root of $f$.
 \end{frame}
 
+\begin{frame}{Pathological Example}
+    \begin{columns}
+    \begin{column}[T]{.5\textwidth}
+    \begin{overlayarea}{\textwidth}{\textheight}
+        $\bm{k=0}$:
+        \only<2->{%
+        \begin{equation*}
+            x_0 := 0.5
+        \end{equation*}
+        }
+        \only<3->{%
+            $\bm{k=1}$:
+            \begin{align*}
+                x_0     & = 0.5 &
+                f(x_0)  & = +0.707 \\
+                        &     &
+                f'(x_0) & = -0.707 \\
+            \end{align*}
+        }
+        \only<4->{%
+            \begin{equation*}
+                {\color{red} x_1 \leftarrow -0.5}
+            \end{equation*}
+        }
+        \only<6->{%
+            $\bm{k=2}$:
+            \begin{align*}
+                x_1     & = -0.5 &
+                f(x_0)  & = +0.707 \\
+                        &     &
+                f'(x_0) & = -0.707 \\
+            \end{align*}
+        }
+        \only<7->{%
+            \begin{equation*}
+                {\color{red} x_2 \leftarrow +0.5}
+            \end{equation*}
+        }
+    \end{overlayarea}
+    \end{column}
+    \begin{column}[T]{.5\textwidth}
+    \hspace{-1cm}
+    \resizebox{1.1\textwidth}{!}{
+    \begin{tikzpicture}
+        \begin{axis}[%
+            axis x line=center,
+            axis y line=center,
+            ymin=-0.1,ymax=0.9,
+            xmin=-0.7,xmax=0.7,
+            title=$\sqrt(|x|)$,
+            samples=160
+        ]
+            \addplot[thick,black,domain=-1:+0] {
+                sqrt(abs(x))
+            };
+            \addplot[thick,black,domain=-0:+1] {
+                sqrt(abs(x))
+            };
+            \only<2-3>{
+                \addplot[thick,blue,only marks,mark=+,mark size=5pt] coordinates {
+                    ( +0.5, 0.707 )
+                };
+            }
+            \only<3-4>{
+                \addplot+[thick,blue,domain=-1:1,mark=none] {
+                    0.707 + 0.707 * (x - 0.5)
+                };
+            }
+            \only<4-5>{
+                \addplot[thick,red,only marks,mark=o,mark size=5pt] coordinates {
+                    ( -0.5, 0.707 )
+                };
+                \draw[thick,red] (axis cs:-0.5, 0) -- (axis cs:-0.5, +0.707);
+            }
+            \only<5-6>{
+                \addplot[thick,blue,only marks,mark=+,mark size=5pt] coordinates {
+                    ( -0.5, 0.707 )
+                };
+            }
+            \only<6-7>{
+                \addplot+[thick,blue,domain=-1:1,mark=none] {
+                    0.707 - 0.707 * (x + 0.5)
+                };
+            }
+            \only<7->{
+                \addplot[thick,red,only marks,mark=o,mark size=5pt] coordinates {
+                    ( 0.5, 0.707 )
+                };
+                \draw[thick,red] (axis cs:+0.5, 0) -- (axis cs:+0.5, +0.707);
+            }
+        \end{axis}
+    \end{tikzpicture}
+    }
+    \vfill
+    \begin{overlayarea}{\textwidth}{.3\textheight}
+        ~\\\bigskip
+        endless loop
+    \end{overlayarea}
+    \end{column}
+    \end{columns}
+\end{frame}
+
+\begin{frame}{Pitfalls of Newton-Raphson}
+    We had previously discussed that Newton-Raphson is guaranteed to converge
+    in an environment $U(\xi)$ of the root $\xi$.\\ \medskip
+
+    However, it may fail to converge or converge only very slowly\dots
+    \begin{itemize}
+        \item if $U(\xi)$ is a small interval and the initial point lies outside of $U$,
+        \item if the algorithm encounters a stationary point $\chi$ of $f$ (i.e. $f'(\chi) = 0$),
+        \item if $\xi$ is a $m$-multiple root of $f$ (i.e. $f(x) \sim (x - \xi)^m \dots$)
+    \end{itemize}
+
+    The last point can be accomodated for: If you know the mulitplicity $m$ of the root,
+    modify the Newton-Raphson step to read:
+    \begin{equation*}
+        x_{k + 1} \leftarrow x_{k} - m \frac{f(x_k)}{f'(x_k)}
+    \end{equation*}
+    Basically, this rescales the derivative from $f'(x) \to f'(x) / m$.
+\end{frame}
+
 \begin{frame}{Honorable mention: Horner Scheme}
     If $f(x) \equiv p_n(x)$ is a polynomial of degree $n$ in $x$, there might be multiple roots of $f$, i.e.
     roots $\xi_{n}$ with $f'(\xi_n) = 0$.\\
@@ -302,13 +423,176 @@
     The Horner Scheme allows to efficiently calculate all the (multiple) roots
     of the polynomial $p_n(x)$.\\
     \vfill
-    However, this is of limited use: These polynomials usually only arise in computing
-    the characteric polynomial $\chi_M$ of a matrix $M$. In this case, the roots of $\chi_M$
+    However, this is of limited use: The polynomials dominantly arise in numerical computations
+    as the characteric polynomial $\chi_M$ of a matrix $M$. In these cases, the roots of $\chi_M$
     correspond to eigenvalue of $M$. However, the are better/more stables ways to \alert{numerically} compute
-    all eigenvalues of $M$.
+    all eigenvalues of $M$, instead of finding all root of $\chi_M$.
 \end{frame}
 
-\begin{frame}{Newton-Raphson for $D > 1$}
+\begin{frame}{Intermission: Derivatives}
+    Both the original and the modified Newton-Raphson methods involve taking
+    the derivative $f'$ of our target function $f$ at arbitrary points. Of
+    course, if we can analytically calculate the derivative, there is no problem.\\
+
+    \vfill
+
+    What about functions that we can only evaluate numerically? (Think: Experiments,
+    numerical calculations)\\
+
+    \vfill
+
+    Enter the field of \emph{numerical differentiation}!
+\end{frame}
+
+\begin{frame}{Intermission: Derivatives}
+    How can we obtain the derivative $f'(x)$ numerically? Let's start from an expansion of
+    $f(x)$ close to the point of interest $x_0$:
+    \begin{tikzpicture}
+        \node at (0,0) {
+        \begin{minipage}{\textwidth}
+        \begin{align*}
+            f(x_0 + 2         h)  & = f(x_0) + f'(x_0) \cdot 2         h + \frac{f''(x_0)}{2} \cdot 4         h^2 + \dots\\
+            f(x_0 + \phantom2 h)  & = f(x_0) + f'(x_0) \cdot \phantom2 h + \frac{f''(x_0)}{2} \cdot \phantom4 h^2 + \dots\\
+            f(x_0 {\color{white}+ 2h}) & = f(x_0)\\
+            f(x_0 - \phantom2 h)  & = f(x_0) - f'(x_0) \cdot \phantom2 h + \frac{f''(x_0)}{2} \cdot \phantom4 h^2 + \dots\\
+            f(x_0 - 2         h)  & = f(x_0) - f'(x_0) \cdot 2         h + \frac{f''(x_0)}{2} \cdot 4         h^2 + \dots
+        \end{align*}
+        \end{minipage}
+        };
+        %\draw[step=0.50,black,thin] (-4.5,-2.0) grid (4.5,2.0);
+        \draw[green!75!black,thick,rounded corners=5pt] (-4.5,+0.15) rectangle (+4.5,+1.05);
+        \only<1>{
+        \draw[red,           thick,rounded corners=5pt] (-4.5,-0.5) rectangle (+4.5,+0.1);
+        }
+        \only<2>{
+        \draw[red,           thick,rounded corners=5pt] (-4.5,-1.5) rectangle (+4.5,-0.5);
+        }
+    \end{tikzpicture}
+
+
+    \begin{overlayarea}{\textwidth}{.4\textheight}
+    \only<1>{
+    \begin{block}{Variant \#1: Forward difference quotient}
+        \begin{equation*}
+            \frac{{\color{green!75!black}f(x_0 + h)} - {\color{red!90!black}f(x_0)}}{h} = f'(x_0) + \frac{f''(x_0)}{2} \cdot h = f'(x_0) + o(h)
+        \end{equation*}
+    \end{block}
+    }
+    \only<2>{
+    \begin{block}{Variant \#2: Central difference quotient}
+        \vspace{-1.5\medskipamount}
+        \begin{equation*}
+            \frac{{\color{green!75!black}f(x_0 + h)} - {\color{red!90!black}f(x_0 - h)}}{2h} = f'(x_0) + \frac{f'''(x_0)}{3} h^2 = f'(x_0) + o(h^2)
+        \end{equation*}
+    \end{block}
+    }
+    \end{overlayarea}
+\end{frame}
+
+\begin{frame}{Intermission: How to choose $h$?}
+    \begin{columns}
+    \begin{column}{.5\textwidth}
+        \includegraphics[width=\textwidth]{eps.png}\\
+        {\tiny \hfill source: Wikipedia}
+        \begin{itemize}
+            \item two sources of numerical errors compete:
+            \begin{itemize}
+                \item formula errors, and
+                \item round-off (finite precision) errors
+            \end{itemize}
+        \end{itemize}
+    \end{column}
+    \begin{column}{.5\textwidth}
+        \begin{itemize}
+            \item formula error can be reduced through ``intelligent formulae''
+            \item round-off error can be reduced through (schematically):
+            \begin{align*}
+                x_\text{approx}
+                    & \leftarrow x + h\\
+                h_\text{approx}
+                    & \leftarrow x_\text{approx} - x\\
+                f'  & \leftarrow \frac{f(x_\text{approx}) - f(x)}{h_\text{approx}}
+            \end{align*}
+            \item optimal choice of $h$ will be close to
+            \begin{equation*}
+                h \sim \sqrt{\varepsilon} \cdot x
+            \end{equation*}
+            with $\varepsilon$: working precision
+        \end{itemize}
+    \end{column}
+    \end{columns}
+\end{frame}
+
+\begin{frame}{Intermission: Higher derivatives}
+    What about $f''(x_0)$, $f^{(3)}(x_0)$ or even higher derivatives?\\
+    \medskip
+    \begin{block}{Central difference quotient for $f^{(n)}$}
+        \begin{align*}
+            \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{f(x_0 + (n - 2 k) h)}{(2 h)^n}
+                & = f^{(n)}(x_0) + o(h^2)
+        \end{align*}
+    \end{block}
+\end{frame}
+
+\newcommand{\del}{\partial}
+\begin{frame}{Intermission: Multiple partial derivatives}
+    What about (multiple) partial derivatives $\del_x \del_y f(x, y)$?\\
+    \medskip
+    \begin{block}{Central difference ``stencil'' for $\del_x \del_y f(x, y)$}
+        Combine what we learned previously:
+        \begin{multline*}
+            \frac{\only<1>{a}\only<2>{+} f(+ h_x, + h_y) \only<1>{+b}\only<2>{-} f(+ h_x, - h_y) \only<1>{+c}\only<2>{-} f(- h_x, + h_y) \only<1>{+d}\only<2>{+} f(- h_x, - h_y)}{4 h_x h_y}\\
+            = \only<1>{\dots}\only<2>{\del x\del_y f(x,y)|_{x=0,y=0} + o(h_x^2) + o(h_y^2)}
+        \end{multline*}
+    \end{block}
+    \vfill
+    \only<2>{End of intermission.}
+\end{frame}
+
+\begin{frame}{Newton-Raphson for $D > 1$?}
+    What happens if we generalize to $D > 1$?\\
+    \medskip
+    Let's consider the case $D=2$ first, for a polynomial of degree $1$
+    \begin{equation*}
+        f(x_1, x_2) = (1 - x) \cdot (2 - y)
+    \end{equation*}
+    \medskip
+    $f$ has infinitely many roots:
+    \begin{itemize}
+        \item $x = 1$ and $y$ arbitrary
+        \item $y = 2$ and $x$ arbitrary
+    \end{itemize}
+    In order to pin-point only one point in $D=2$, we need to solve $2$
+    equations:
+    \begin{align*}
+        f_1(x_1, x_2) & = 0, &
+        f_2(x_1, x_2) & = 0
+    \end{align*}
+    In $D$ dimensions, we will need to solve $D$ simultaneous equations!
+\end{frame}
+
+\begin{frame}{Generalization to $D > 1$}
+    Substitute:
+    \begin{align*}
+        f(x) \in \mathbb{R} \quad  & \to \quad f(x) \in \mathbb{R}^D\\
+        x \in \mathbb{R} \quad     & \to \quad x \in \mathbb{R}^D\\
+        f'(x) \in \mathbb{R} \quad & \to \quad [D_f]_{ij} \equiv \frac{\del f_i(x)}{\del x_j} \in \mathbb{R}^{D\times D}
+    \end{align*}
+
+    Then the Newton-Raphson steps reads
+    \begin{equation*}
+        x^{(k+1)} \leftarrow x^{(k)} - \left[D_f(x^{(k)})\right]^{-1} f(x^{(k)})\,.
+    \end{equation*}
+    Problems:
+    \begin{itemize}
+        \item the functional matrix $D_f(x^{(k)})$ might be singular, or numerically close to being
+            singular
+        \item inverting $D_f$ is generally computationally expensive!
+    \end{itemize}
+
+    In the next two lectures, Marcin will discuss how to solve systems of linear equations.
+    These methods will also introduce you to methods to numerically compute the above $x^{(k + 1)}$
+    from $D_f(x^{(k)})$.
 \end{frame}
 
 \backupbegin