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maths/numerical_analysis/intersection.py
54 строки
2 KB
Shikhar Maheshwari
Add doctest to maths/numerical_analysis/intersection.py (#12148)
30 дек 2024, 13:51
Не верифицирован
30 дек 2024, 13:51
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import math from collections.abc import Callable def intersection(function: Callable[[float], float], x0: float, x1: float) -> float: """ function is the f we want to find its root x0 and x1 are two random starting points >>> intersection(lambda x: x ** 3 - 1, -5, 5) 0.9999999999954654 >>> intersection(lambda x: x ** 3 - 1, 5, 5) Traceback (most recent call last): ... ZeroDivisionError: float division by zero, could not find root >>> intersection(lambda x: x ** 3 - 1, 100, 200) 1.0000000000003888 >>> intersection(lambda x: x ** 2 - 4 * x + 3, 0, 2) 0.9999999998088019 >>> intersection(lambda x: x ** 2 - 4 * x + 3, 2, 4) 2.9999999998088023 >>> intersection(lambda x: x ** 2 - 4 * x + 3, 4, 1000) 3.0000000001786042 >>> intersection(math.sin, -math.pi, math.pi) 0.0 >>> intersection(math.cos, -math.pi, math.pi) Traceback (most recent call last): ... ZeroDivisionError: float division by zero, could not find root """ x_n: float = x0 x_n1: float = x1 while True: if x_n == x_n1 or function(x_n1) == function(x_n): raise ZeroDivisionError("float division by zero, could not find root") x_n2: float = x_n1 - ( function(x_n1) / ((function(x_n1) - function(x_n)) / (x_n1 - x_n)) ) if abs(x_n2 - x_n1) < 10**-5: return x_n2 x_n = x_n1 x_n1 = x_n2 def f(x: float) -> float: """ function is f(x) = x^3 - 2x - 5 >>> f(2) -1.0 """ return math.pow(x, 3) - (2 * x) - 5 if __name__ == "__main__": print(intersection(f, 3, 3.5))