给定一组坐标，实现三次Hermite差值曲线的绘制

好的，根据您的要求，我可以为您提供一个实现三次Hermite差值曲线的Python程序，代码如下：

import matplotlib.pyplot as plt
import numpy as np

# 定义Hermite插值函数
def hermite_interpolation(x, y, dy):
    n = len(x)
    A = np.zeros((2*n, 2*n))
    b = np.zeros((2*n, 1))
    for i in range(n):
        A[2*i][0] = 1
        A[2*i+1][0] = 1
        b[2*i] = y[i]
        b[2*i+1] = y[i]
        if i != 0:
            A[2*i][2*i-1] = x[i]-x[i-1]
            A[2*i+1][2*i-1] = x[i]-x[i-1]
            A[2*i][2*i] = (x[i]-x[i-1])**2
            A[2*i+1][2*i] = (x[i]-x[i-1])**2
        if i != n-1:
            A[2*i][2*i+1] = x[i+1]-x[i]
            A[2*i+1][2*i+1] = x[i+1]-x[i]
            A[2*i][2*i+2] = -(x[i+1]-x[i])**2
            A[2*i+1][2*i+2] = -(x[i+1]-x[i])**2
    for i in range(1, n-1):
        A[2*i][2*n-2] = 3*(x[i]-x[i-1])**2
        A[2*i+1][2*n-2] = 3*(x[i]-x[i-1])**2
        A[2*i][2*n-1] = -3*(x[i+1]-x[i])**2
        A[2*i+1][2*n-1] = -3*(x[i+1]-x[i])**2
        b[2*i][0] = dy[i-1]
        b[2*i+1][0] = dy[i]
    A[2*n-2][1] = 1
    A[2*n-1][2*n-2] = 1
    return np.linalg.solve(A, b)

# 定义绘制函数
def plot_hermite(x, y, dy):
    a = hermite_interpolation(x, y, dy)
    n = len(x)
    x_new = np.linspace(x[0], x[n-1], 1000)
    y_new = np.zeros((1000, 1))
    for i in range(1000):
        for j in range(n-1):
            if x_new[i] >= x[j] and x_new[i] <= x[j+1]:
                y_new[i] = a[2*j] + a[2*j+1]*(x_new[i]-x[j]) + a[2*n-2]*(x_new[i]-x[j])**2 + a[2*n-1]*(x[j+1]-x_new[i])**2
                break
    plt.plot(x_new, y_new)
    plt.scatter(x, y)
    plt.show()

# 测试程序
x = [1, 2, 3, 4]
y = [8, 14, 15, 18]
dy = [3, 2, 5]
plot_hermite(x, y, dy)

您可以将自己的坐标点和一阶导数传入 `plot_hermite` 函数中进行绘制，例如：

x = [1, 2, 3, 4]
y = [8, 14, 15, 18]
dy = [3, 2, 5]
plot_hermite(x, y, dy)

其中 `x` 和 `y` 分别为坐标点的横纵坐标，`dy` 为坐标点的一阶导数。程序会自动绘制出三次Hermite差值曲线。