Converted everything to orgmode

converted everything to orgmode and added solution to the README files

Informatik_I/Exercise_3/Task_4a/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, that we call sum 1:
-
-$$\frac{\pi}{4} = \sum_{j=0}^{m-1} \frac{(-1)^j}{2j + 1} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + ...$$
-
-Note that $m$ is the number of the terms in the sum. For example, $m=2$ refers to the sum of the terms $1$ (for $j=0$) and $-\frac{1}{3}$ (for $j=1$). This examples yields a value of $4 \cdot (1-\frac{1}{3})$ for $\pi$.
-
-Write a program that computes and outputs an approximation of Pi, based on sum 1. The input for your program is the number of **terms** $m$ of sum 1 that should be considered in the calculation. The output is the approximation of $\pi$.
-
-## Input
-
-A number $m \geq 1$.
-
-## Output
-
-The approximation of $\pi$ given by $4 \sum_{j=0}^{m-1} \frac{(-1)^j}{2j + 1}$, rounded to 6 significant digits. Note that 6 significant digits is the default precision of C++ for printing floating-point values. Use a variable of type double to calculate the sum. Note that that $x^0$ is 1 (if $x \neq 0$).
-
-**Important**: the use of functions from the math library (e.g., pow) is prohibited.

Informatik_I/Exercise_3/Task_4a/README.org

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+#+TITLE: Task 4a: Approximation of Pi: Sum 1
+#+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, that we call sum 1:
+
+$$\frac{\pi}{4} = \sum_{j=0}^{m-1} \frac{(-1)^j}{2j + 1} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + ...$$
+
+Note that \(m\) is the number of the terms in the sum. For example, \(m=2\)$ refers to the sum of the terms \(1\) (for \(j=0\)) and \(-\frac{1}{3}\) (for \(j=1\)). This examples yields a value of \(4 \cdot (1-\frac{1}{3})\) for \(\pi\).
+
+Write a program that computes and outputs an approximation of Pi, based on sum 1. The input for your program is the number of **terms** $m$ of sum 1 that should be considered in the calculation. The output is the approximation of \(\pi\).
+
+** Input
+
+A number \(m \geq 1\).
+
+** Output
+
+The approximation of \(\pi\) given by \(4 \sum_{j=0}^{m-1} \frac{(-1)^j}{2j + 1}\), rounded to 6 significant digits. Note that 6 significant digits is the default precision of C++ for printing floating-point values. Use a variable of type double to calculate the sum. Note that \(x^0\) is 1 (if \(x \neq 0\)).
+
+**Important**: the use of functions from the math library (e.g., pow) is prohibited.
+
+* Solution
+
+#+begin_src cpp
+#include <iostream>
+
+double power(double n, double e) { // Self defined function for exponent
+  int res = 1;
+  for (; e > 0; --e) {
+    res *= n;
+  }
+  return res;
+}
+
+int main() {
+  double m;      // User Input
+  double pi = 0; // Pi
+  float res;     // Outputed result with 6 significant digits
+
+  std::cin >> m;
+
+  if (m < 1) { // Check if Input is greater than 1.
+    return 0;
+  } else {
+    for (int i = 0; i < m; ++i) {
+      pi = pi + 4 * ((power(-1, i)) / ((2 * i) + 1)); // calculate Pi
+    }
+
+    res = (float)(pi); // round to 6 significant digits
+    std::cout << res;
+
+    return 0;
+  }
+}
+#+end_src
+
+-----
+
+Made by JirR02 in Switzerland 🇨🇭

Informatik_I/Exercise_3/Task_4a/main.cpp

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-#include <iostream>
-
-double power(double n, double e) { // Self defined function for exponent
-  int res = 1;
-  for (; e > 0; --e) {
-    res *= n;
-  }
-  return res;
-}
-
-int main() {
-  double m;      // User Input
-  double pi = 0; // Pi
-  float res;     // Outputed result with 6 significant digits
-
-  std::cin >> m;
-
-  if (m < 1) { // Check if Input is greater than 1.
-    return 0;
-  } else {
-    for (int i = 0; i < m; ++i) {
-      pi = pi + 4 * ((power(-1, i)) / ((2 * i) + 1)); // calculate Pi
-    }
-
-    res = (float)(pi); // round to 6 significant digits
-    std::cout << res;
-
-    return 0;
-  }
-}