Converted everything to orgmode

converted everything to orgmode and added solution to the README files

Informatik_I/Exercise_3/Task_4b/README.md

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-# Task
-
-The number $\pi$ can be defined through various infinite sums. The accuracy increases with the number of terms. Considering the following sum, which we call sum 2:
-
-$$\frac{\pi}{2} = 1 + \sum_{j = 1}^{m - 1} \frac{\prod_{i=1}^j i}{\prod_{i=1}^j (2i + 1)} = 1 + \frac{1}{3} + \frac{1 \cdot 2}{3 \cdot 5} + \frac{1 \cdot 2 \cdot 3}{3 \cdot 5 \cdot 7} + ...$$
-
-Write a program that computes and outputs an approximation of Pi, based on sum 2. The input for your program is the number of **terms** $m$ (including 1) to be included in the calculation to be done. The output is the approximation of $\pi$.
-
-**Optional**: After you have solved this and the previous task, think about which formula gives a more precise approximation of
-, sum 1 or sum 2 ? What are the drawbacks? Write your thoughts in a comment below the code.
-
-## Input
-
-A natural number $m$ ($m \geq 1$).
-
-## Output
-
-The approximation of $\pi$ given by $m$ terms, rounded to 6 significant digits. Note that 6 significant digits is the default precision of C++ for printing floating-point values. Use a double variable to store the sum.
-
-**Important**: the use of functions from the math library (e.g., pow) is prohibited.

Informatik_I/Exercise_3/Task_4b/README.org

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+#+TITLE: Task 4b: Approximation of Pi: Sum 2
+#+AUTHOR: JirR02
+
+* Task
+
+The number \(\pi\) can be defined through various infinite sums. The accuracy increases with the number of terms. Considering the following sum, which we call sum 2:
+
+$$\frac{\pi}{2} = 1 + \sum_{j = 1}^{m - 1} \frac{\prod_{i=1}^j i}{\prod_{i=1}^j (2i + 1)} = 1 + \frac{1}{3} + \frac{1 \cdot 2}{3 \cdot 5} + \frac{1 \cdot 2 \cdot 3}{3 \cdot 5 \cdot 7} + ...$$
+
+Write a program that computes and outputs an approximation of Pi, based on sum 2. The input for your program is the number of **terms** $m$ (including 1) to be included in the calculation to be done. The output is the approximation of \(\pi\).
+
+**Optional**: After you have solved this and the previous task, think about which formula gives a more precise approximation of \(\pi\), sum 1 or sum 2 ? What are the drawbacks? Write your thoughts in a comment below the code.
+
+* Input
+
+A natural number \(m\) (\(m \geq 1\)).
+
+* Output
+
+The approximation of $\pi$ given by $m$ terms, rounded to 6 significant digits. Note that 6 significant digits is the default precision of C++ for printing floating-point values. Use a double variable to store the sum.
+
+**Important**: the use of functions from the math library (e.g., =pow=) is prohibited.
+
+* Solution
+
+#+begin_src cpp
+#include <iostream>
+
+double productsumnum(double n) {
+  double res = 1;
+  for (int j = 1; j <= n; ++j) {
+    res *= j;
+  }
+  return res;
+}
+
+double productsumden(double n) {
+  double res = 1;
+  for (int j = 1; j <= n; ++j) {
+    res *= (2 * j + 1);
+  }
+  return res;
+}
+
+int main() {
+
+  double m;      // User Input
+  double pi = 2; // Pi
+  float res;     // Outputed result with 6 significant digits
+
+  std::cin >> m;
+
+  for (int i = 1; i < m; ++i) {
+    pi += (2 * (productsumnum(i) / productsumden(i)));
+  }
+
+  res = (float)(pi);
+  std::cout << res;
+
+  return 0;
+}
+#+end_src
+
+-----
+
+Made by JirR02 in Switzerland 🇨🇭

Informatik_I/Exercise_3/Task_4b/main.cpp

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-#include <iostream>
-
-double productsumnum(double n) {
-  double res = 1;
-  for (int j = 1; j <= n; ++j) {
-    res *= j;
-  }
-  return res;
-}
-
-double productsumden(double n) {
-  double res = 1;
-  for (int j = 1; j <= n; ++j) {
-    res *= (2 * j + 1);
-  }
-  return res;
-}
-
-int main() {
-
-  double m;      // User Input
-  double pi = 2; // Pi
-  float res;     // Outputed result with 6 significant digits
-
-  std::cin >> m;
-
-  for (int i = 1; i < m; ++i) {
-    pi += (2 * (productsumnum(i) / productsumden(i)));
-  }
-
-  res = (float)(pi);
-  std::cout << res;
-
-  return 0;
-}