This code approximates the value of \( \pi \) using...

This code approximates the value of ( \pi ) using the Leibniz formula for (\pi). The formula for (\pi) is:

[ \pi = 4 \left( 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \dots \right) ]

This implementation sums up to terms iterations to compute the approximation. It alternates between adding and subtracting terms, controlled by the sign variable.

Step-by-step explanation:

1. terms defines how many terms in the series are used for the approximation. A higher number of terms gives more accuracy.
2. sum accumulates the running total of the series approximation.
3. i tracks the current term index.
4. sign alternates between (+1.0) and (-1.0) to handle the alternating positive and negative signs in the series.
5. In the while loop:
   • The formula (\frac{\text{sign}}{2i + 1}) adds or subtracts the current term to the sum.
   • sign = -sign flips the sign after each iteration.
   • i++ progresses to the next term.
6. After the loop completes, the approximation of (\pi) is calculated by multiplying the sum by 4.
7. The result (piApprox) is printed to the console.

Simplified Code:

The essential structure is already simple, but here's a slightly more streamlined version that retains the while loop:

int terms = 100000;        // More terms = better approximation
double sum = 0.0;
int i = 0;
double sign = 1.0;

while (i < terms) {
    sum += sign / (2 * i + 1);
    sign = -sign; // Flip sign
    i++;
}

double piApprox = 4 * sum;
System.out.println(piApprox);

This is essentially as simple as it gets while keeping the while loop structure intact.