Bug Fixed in newton_raphson_method.py

arithmetic_analysis/newton_raphson.py

@@ -0,0 +1,40 @@
+# Implementing Newton Raphson method in Python
+# Author: Syed Haseeb Shah (github.com/QuantumNovice)
+# The Newton-Raphson method (also known as Newton's method) is a way to
+# quickly find a good approximation for the root of a real-valued function
+
+from decimal import Decimal
+from math import *  # noqa: F401, F403
+from sympy import diff
+
+
+def newton_raphson(func: str, a: int, precision: int=10 ** -10) -> float:
+    """ Finds root from the point 'a' onwards by Newton-Raphson method
+    >>> newton_raphson("sin(x)", 2)
+    3.1415926536808043
+    >>> newton_raphson("x**2 - 5*x +2", 0.4)
+    0.4384471871911695
+    >>> newton_raphson("x**2 - 5", 0.1)
+    2.23606797749979
+    >>> newton_raphson("log(x)- 1", 2)
+    2.718281828458938
+    """
+    x = a
+    while True:
+        x = Decimal(x) - (Decimal(eval(func)) / Decimal(eval(str(diff(func)))))
+        # This number dictates the accuracy of the answer
+        if abs(eval(func)) < precision:
+            return float(x)
+
+
+# Let's Execute
+if __name__ == "__main__":
+    # Find root of trigonometric function
+    # Find value of pi
+    print(f"The root of sin(x) = 0 is {newton_raphson('sin(x)', 2)}")
+    # Find root of polynomial
+    print(f"The root of x**2 - 5*x + 2 = 0 is {newton_raphson('x**2 - 5*x + 2', 0.4)}")
+    # Find Square Root of 5
+    print(f"The root of log(x) - 1 = 0 is {newton_raphson('log(x) - 1', 2)}")
+    # Exponential Roots
+    print(f"The root of exp(x) - 1 = 0 is {newton_raphson('exp(x) - 1', 0)}")