Python中求解f（x）=0的割线法

如何使用Python中的割线方法来求解方程f（x）=0，给定两个初始猜测x0和x1

def secant(f,x0,x1,tol):

我需要用它来找到二次函数和更高的x因子的解，例如，给定区间的x^3-4x^2+1=0。 这只是一个在黑暗中的镜头，所以任何有用的网站链接也将受到赞赏

维基百科有一个

def secant(f, x0, x1, tol):
    # store initial values
    fx0 = f(x0)
    fx1 = f(x1)
    while abs(fx1) < tol:
        # do calculation
        x2 = (x0 * fx1 - x1 * fx0) / (fx1 - fx0)
        # shift variables (prepare for next loop)
        x0,  x1  = x1,  x2
        fx0, fx1 = fx1, f(x2)
    return x1

def secant(f, x0, x1, tol, max_iterations=100):
    # keep initial values for error reporting
    init_x0 = x0
    init_x1 = x1

    # store y values instead of recomputing them
    fx0 = f(x0)
    fx1 = f(x1)

    # iterate up to maximum number of times
    for _ in range(max_iterations):
        # see whether the answer has converged
        if abs(fx1) < tol:
            return x1

        # do calculation
        x2 = (x0 * fx1 - x1 * fx0) / (fx1 - fx0)
        # shift variables (prepare for next loop)

        x0,  x1  = x1,  x2
        fx0, fx1 = fx1, f(x2)

    # for loop has ended - failed to converge
    raise ValueError(
        "call to secant(f={}, x0={}, x1={}, tol={}," \
        " max_iterations={}) has failed to converge"
        .format(
            f.__name__,
            init_x0,
            init_x1,
            tol,
            max_iterations
        )
    )

将第一次迭代转换为Python看起来像

x2 = (x0 * f(x1) - x1 * f(x0)) / (f(x1) - f(x0))

与多次调用

f（x0）

和

f（x1）

不同，我们应该调用它们一次并存储值：

fx0 = f(x0)
fx1 = f(x1)
x2 = (x0 * fx1 - x1 * fx0) / (fx1 - fx0)

对于第二次迭代，我们必须计算

x3 = (x1 * fx2 - x2 * fx1) / (fx2 - fx1)

但我们不想每次都用新的变量重新写方程。相反，我们将这些值移动：让

x1

取

x2

的值，

x2

取

x3

的值，依此类推。我们可以在循环中重复这样做，就像

# store initial values
fx0 = f(x0)
fx1 = f(x1)
while (something):
    # do calculation
    x2 = (x0 * fx1 - x1 * fx0) / (fx1 - fx0)
    # shift variables (prepare for next loop)
    x0,  x1  = x1,  x2
    fx0, fx1 = fx1, f(x2)

现在我们只需要知道何时停止，当最后一个答案与0之间的差值小于

tol

：

abs（f（x2）-0）

，可以简化为

abs（fx1）

所以函数看起来像

def secant(f, x0, x1, tol):
    # store initial values
    fx0 = f(x0)
    fx1 = f(x1)
    while abs(fx1) < tol:
        # do calculation
        x2 = (x0 * fx1 - x1 * fx0) / (fx1 - fx0)
        # shift variables (prepare for next loop)
        x0,  x1  = x1,  x2
        fx0, fx1 = fx1, f(x2)
    return x1

def secant(f, x0, x1, tol, max_iterations=100):
    # keep initial values for error reporting
    init_x0 = x0
    init_x1 = x1

    # store y values instead of recomputing them
    fx0 = f(x0)
    fx1 = f(x1)

    # iterate up to maximum number of times
    for _ in range(max_iterations):
        # see whether the answer has converged
        if abs(fx1) < tol:
            return x1

        # do calculation
        x2 = (x0 * fx1 - x1 * fx0) / (fx1 - fx0)
        # shift variables (prepare for next loop)

        x0,  x1  = x1,  x2
        fx0, fx1 = fx1, f(x2)

    # for loop has ended - failed to converge
    raise ValueError(
        "call to secant(f={}, x0={}, x1={}, tol={}," \
        " max_iterations={}) has failed to converge"
        .format(
            f.__name__,
            init_x0,
            init_x1,
            tol,
            max_iterations
        )
    )

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