This code calculates an approximation of the mathematical constant π...

This code calculates an approximation of the mathematical constant π (pi) using the Leibniz formula for π, which states:

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

Explanation of the code:

1. int terms = 100000;: Sets the number of terms to include in the approximation. More terms result in better accuracy.
2. double sum = 0.0;: Initializes the sum where the partial results of the formula will be stored.
3. int i = 0;: Initializes the counter to track the current term in the formula.
4. double sign = 1.0;: Keeps track of the alternating signs (+ and -).
5. While loop (while (i < terms) {...}):
   • Calculates each term of the series: sign / (2 * i + 1).
   • Adds the term to the running total sum.
   • Flips the sign (+ to - or - to +) using sign = -sign.
   • Increments the counter i to proceed to the next term.
6. After the loop:
   • double piApprox = 4 * sum;: Multiplies the final sum by 4 to approximate π.
   • System.out.println(piApprox);: Prints the computed value of π.

Simplified Code:

We can make the code shorter by combining the variable declarations and avoiding sign explicitly:

int terms = 100000;        // Number of terms
double sum = 0.0;

for (int i = 0; i < terms; i++) {
    sum += (i % 2 == 0 ? 1.0 : -1.0) / (2 * i + 1);  // Alternating signs
}

double piApprox = 4 * sum;  // Compute approximation of pi
System.out.println(piApprox);

In this version:

• The while loop is replaced by a for loop.
• The alternating sign is calculated directly using (i % 2 == 0 ? 1.0 : -1.0) instead of manually flipping the sign variable. This checks if i is even or odd to determine the sign.