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From: ram@zedat.fu-berlin.de (Stefan Ram)
Newsgroups: comp.lang.python
Subject: Re: Beazley's Problem
Date: 23 Sep 2024 12:26:38 GMT
Organization: Stefan Ram
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Expires: 1 Jul 2025 11:59:58 GMT
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Annada Behera <annada@tilde.green> wrote or quoted:
>The "next-level math trick" Newton-Raphson has nothing to do with
>functional programming. 

  Nobody up the thread was claiming it was functional. And you can
  totally implement anything in an imperative or functional style.

from typing import Callable

def newton_raphson(
    f: Callable[[float], float],
    f_prime: Callable[[float], float],
    x0: float,
    epsilon: float = 1e-7,
    max_iterations: int = 100
) -> float:
    def recurse(x: float, iteration: int) -> float:
        if iteration > max_iterations:
            raise ValueError("Maximum iterations reached. The method may not converge.")
        
        fx: float = f(x)
        if abs(fx) < epsilon:
            return x
        
        x_next: float = x - fx / f_prime(x)
        return recurse(x_next, iteration + 1)
    
    return recurse(x0, 0)

# Example application: find a root of f(x) = x^2 - 4

# Define the function and its derivative
def f(x: float) -> float:
    return x**2 - 4

def f_prime(x: float) -> float:
    return 2*x

# Initial guess
x0: float = 1.0

try:
    root: float = newton_raphson(f, f_prime, x0)
    print(f"The root is approximately: {root}")
    print(f"f({root}) = {f(root)}")
except ValueError as e:
    print(f"Error: {e}")