Newton's Method

BabbleDay posted @ 2015年6月29日 22:02 in 见过的PL们 with tags python , 738 阅读

寻找函数零点:Quickly finds accurate approximations to zeros of differenciable functions.

只要问题是可微的,则问题是可解的。

Wikipedia上操作演示

https://en.wikipedia.org/wiki/Newton's_method#/media/File:NewtonIteration_Ani.gif

 

实现指南:

高阶函数为的是抽象出功能,不要被参数混乱思路

python中整数相除会返回整数,所以注意传入浮点数做参数

 

def square_root(a):
    def f(x):
        return x * x - a

    def df(x):
        return 2 * x

    return find_zero(f, df)

def find_zero(f, df):
    def near_zero(x):
        return appro_equal(f(x), 0)
    return improve(update(f, df), near_zero, guess=1.0, max_update=100)

def update(f, df):
    def u(x):
        return x - f(x)/df(x)
    return u

def improve(update, close, guess, max_update):
    k = 0
    while not close(guess) and k < max_update:
        print guess, k
        guess = update(guess)
        k += 1
    return guess  # the guess improved

def appro_equal(x, y, tolerance=1e-15):
    print abs(x-y) < tolerance
    return abs(x-y) < tolerance

 

推广至求n次根

>>> def power(x, n):
        """Return x * x * x * ... * x for x repeated n times."""
        product, k = 1, 0
        while k < n:
            product, k = product * x, k + 1
        return product
>>> def nth_root_of_a(n, a):
        def f(x):
            return power(x, n) - a
        def df(x):
            return n * power(x, n-1)
        return find_zero(f, df)

 

实际应用 Function Decorator

 

def trace(fn):
    def t(x):
        print "Calling", fn, "on argument", x
        return fn(x)
    return t

@trace
def square(x):
    return x * x
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