Task 4a: Approximation of Pi: Sum 1
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
#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;
}
}Made by JirR02 in Switzerland 🇨🇭