求根号2， 网易的一道面试题

# 二分法
def solve():
    l, r = 1, 2
    eps = 1e-7
    while l + eps < r:
        mid = (l + r) / 2
        if mid * mid < 2:
            l = mid
        else:
            r = mid
    return l

# print('{:.6f}'.format(solve()))




# 泰勒展开
# f(x) = x^{1 / 2}
# f(x) = f(1) + f'(1)(x-1)+f''(1)/ 2 (x-1)**2+f'''(1)/ 3! (x-1)**3+....
#      = 1 + f'(1) + f''(1) / 2 + f'''(1)/ 3! + ...
#      = 1 + 0.5 + 0.5*(0.5 - 1) / 2 + 0.5*(0.5 - 1) * (0.5 - 2) / 3! + ...
def solve2():
    def daoshu(x0, n):
        # a, b = 1, 0.5
        res_a, res_b = 1, 0.5
        res_a_div_jiechen = 1
        dx = 1
        for i in range(n):
            res_a_div_jiechen = res_a_div_jiechen * res_b / (i + 1)  # 0.5*(0.5 - 1) * (0.5 - 2) * ...*(0.5 - n + 1) / n!
            res_a *= res_b
            res_b = res_b - 1
            dx *= (2 - x0)
        return res_a_div_jiechen * x0 ** res_b * dx  # x=2, x0=dx =x-x0 =  1
    res = 0
    x0 = 1 #1.5
    cnt = 0
    eps = 1e-6
    while True:
        now = daoshu(x0, cnt)
        # print(now)
        if abs(now) < eps:
            break
        res += now
        cnt += 1

    return res

# print('{:.6f}'.format(solve2()))

# 牛顿法
# f(x) = x**2 - 2
# x_next = x - f(x) / f'(x)
def solve3():
    x0 = 2
    eps = 1e-7
    while True:
        x0 = x0 - (x0 * x0 - 2) / (2 * x0)
        if abs(x0 * x0 - 2) < eps:
            break
    return x0
# print('{:.6f}'.format(solve3()))

# 梯度下降法
# 定义loss:  (x*x - 2) * 2
def sgd():
    def grad(x):
        return 2 * (x*x -2) * 2 * x
    x0 = 2
    eps = 1e-7
    lr = 0.001
    while True:
        x0_grad = grad(x0)
        x0 = x0 - lr * x0_grad
        if abs(x0 * x0 - 2) < eps:
            break
    return x0
# print('{:.6f}'.format(sgd()))

# 连分数逼近法
# 根号2 = 1 + 1 / (2 + 1 / (2 + 1 / (2 + ...)))
def lianfengshu():
    # 根号2 = 1 + 1 / (2 + ...)
    now = 1
    eps = 1e-7
    while True:
        now = 1 / (2 + now)
        if abs((1 + now) ** 2 - 2) < eps:
            break
    return 1 + now
# print('{:.6f}'.format(lianfengshu()))