Exercise 3

Added Exercise 3 with solutions and README to Repository

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/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;
+  }
+}