[programming/exercises] minors and add example data file for exercise 4

programming/exercises/matrices.tex

@@ -159,7 +159,7 @@ following pattern:
       \code{x(randi(100, 50, 1)])}
     \end{solution}
 
-    \part Can you imagine an advantage of using linear indexing instead of subscript indexing?
+    \part Can you imagine an advantage of using linear instead of subscript indexing?
     \begin{solution}
       Die Matrize ist 2-dimensional. Wenn mit dem subscript index
       zugegriffen werden soll, dann muss auf die Dimensionen
@@ -215,7 +215,7 @@ following pattern:
   \question Create a 3-D matrix from two 2-D matrices. Use the
   function \code{cat} (check the help to learn its usage).
   \begin{parts}
-    \part Select all elements of the first ``page'' (index 1, 3. dimension).
+    \part Select all elements of the first ``page'' (index 1 in the 3. dimension).
     \begin{solution}
       \code{x = randn(5,5); \\y = randn(5, 5);\\ z = cat(3, x, y);\\disp(z(:,:,1))}
     \end{solution}

programming/exercises/variables_types.tex

@@ -1,6 +1,6 @@
 \documentclass[12pt,a4paper,pdftex]{exam}
 
-\usepackage[german]{babel}
+\usepackage[english]{babel}
 \usepackage{natbib}
 \usepackage{graphicx}
 \usepackage[small]{caption}
@@ -171,7 +171,7 @@ following pattern: ``variables\_datatypes\_\{lastname\}.m''
 
   \question Floating point numbers I: Limited precision during additions
   \begin{parts}
-    \part Create the variable \code{a} and assign an arbitrary floting point number.
+    \part Create the variable \code{a} and assign an arbitrary floating point number.
     \begin{solution}
       \code{a = 3.14;}
     \end{solution}
@@ -185,12 +185,17 @@ following pattern: ``variables\_datatypes\_\{lastname\}.m''
       Result is 0! Even with doubles there is a limited precision in
       the decimal part.
     \end{solution}
-    \part Calculate \verb=(2^52 + 1) - 2^52= and
+    \part Calculate \verb=(2^52 + 1) - 2^52=, 
     \verb=(2^53 + 1) - 2^53=.
     \begin{solution}
       First command results in = 1, in the second case = 0. With such
       high numbers, small differences (1!) cannot be resolved. 
     \end{solution}
+    \part Same as before, but add larger numbers (100 and 101, for example).
+    \begin{solution}
+      The difference between 100 and 101 cannot be resolved, so it is
+      the delta, not the absolute value that is important.
+    \end{solution}
     \part Calculate \code{sqrt(1+1e-16)-1}. Is the result correct? Why (not)?
     \begin{solution}
       Square root of something larger than 1 should not be 1 and thus