def calculate_pi(limit: int) -> str: """ https://en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80 Leibniz Formula for Pi The Leibniz formula is the special case arctan(1) = pi / 4. Leibniz's formula converges extremely slowly: it exhibits sublinear convergence. Convergence (https://en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80#Convergence) We cannot try to prove against an interrupted, uncompleted generation. https://en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80#Unusual_behaviour The errors can in fact be predicted, but those calculations also approach infinity for accuracy. Our output will be a string so that we can definitely store all digits. >>> import math >>> float(calculate_pi(15)) == math.pi True Since we cannot predict errors or interrupt any infinite alternating series generation since they approach infinity, or interrupt any alternating series, we'll need math.isclose() >>> math.isclose(float(calculate_pi(50)), math.pi) True >>> math.isclose(float(calculate_pi(100)), math.pi) True Since math.pi contains only 16 digits, here are some tests with known values: >>> calculate_pi(50) '3.14159265358979323846264338327950288419716939937510' >>> calculate_pi(80) '3.14159265358979323846264338327950288419716939937510582097494459230781640628620899' """ # Variables used for the iteration process q = 1 r = 0 t = 1 k = 1 n = 3 l = 3 decimal = limit counter = 0 result = "" # We can't compare against anything if we make a generator, # so we'll stick with plain return logic while counter != decimal + 1: if 4 * q + r - t < n * t: result += str(n) if counter == 0: result += "." if decimal == counter: break counter += 1 nr = 10 * (r - n * t) n = ((10 * (3 * q + r)) // t) - 10 * n q *= 10 r = nr else: nr = (2 * q + r) * l nn = (q * (7 * k) + 2 + (r * l)) // (t * l) q *= k t *= l l += 2 k += 1 n = nn r = nr return result def main() -> None: print(f"{calculate_pi(50) = }") import doctest doctest.testmod() if __name__ == "__main__": main()