Taylor Series (Horner)

e^x = 1 + x / 1 + x^2 / 2! + x^3 / 3! + x^4 / 4! + ... n times

This method will be faster, by taking less number of multiplications. First, we can rewrite the formula as:

x = x / 1 + x² / (1 * 2) + x³ / 1 * 2 * 3 + x⁴ / 1 * 2 * 3 * 4

We can reduce from quadratic O(n²) to linear O(n)

Using iterative loop:

double e(int x, int n) {
  double s = 1;
  for (n > 0; n--) {
    s = 1 + x / n * s;
  }
  return s;
}

Recursive version:

double e(int x, int n) {
  static double s = 1;
  if (n == 0)
    return s;

  s = 1 + x / n * s;
  return e(x, n - 1);
}