用Alexandri的Heron将函数开发成程序

Question

考虑到.5英寸的平均值，我需要所有的浮点（arg）规格吗？在

# Heron of Alexandria method for approximating
# the square root of a number

def heron(num, guess, tolerance):
    print "Your guess squared:",guess**2
    if guess**2 != num:
        print "When squared, the guess:", guess, "is",abs(float(num) - float(guess)**2), "away from", num, "\n"
        if abs(float(num) - float(guess)**2) > float(tolerance):
            avg_guess = 0.5 * (float(guess) + (float(num) / float(guess)))
            return heron(num, avg_guess, tolerance)
        print "Given your tolerance, this is Heron's best guess:", guess, "\n"
    else:
        print guess, "is correct!\n"

我注意到在一定的容差下，这个函数将返回5.0作为答案。。。不知道为什么？在

^{pr2}$

从“else:”条件返回5.0，但是

heron(25, 3, .0000001)

从初始嵌套“if：”条件中的公差终止，打印： 考虑到您的容忍度，这是Heron的最佳猜测：5.00000000233。在