对于给函数fx=11+25x^2在区间-11上取xi=-1+02ii=0110试求三次曲线拟合

使用三次曲线拟合可以得到函数在给定区间内的近似曲线。拟合的三次曲线可以表示为：

f(x) = a0 + a1x + a2x^2 + a3*x^3

其中a0、a1、a2和a3是待求系数，可以通过最小二乘法求解。最小二乘法是一种通过最小化误差平方和来求解未知参数的方法。

首先，我们需要计算出xi对应的函数值：

f(-1.0) = 1.0/26.0 f(-0.8) = 1.0/21.0 f(-0.6) = 1.0/14.0 f(-0.4) = 1.0/9.0 f(-0.2) = 1.0/6.25 f(0.0) = 1.0/5.0 f(0.2) = 1.0/6.25 f(0.4) = 1.0/9.0 f(0.6) = 1.0/14.0 f(0.8) = 1.0/21.0 f(1.0) = 1.0/26.0

然后，我们可以使用最小二乘法求解系数：

X = [-1.0, -0.8, -0.6, -0.4, -0.2, 0.0, 0.2, 0.4, 0.6, 0.8, 1.0] Y = [1/26, 1/21, 1/14, 1/9, 1/6.25, 1/5, 1/6.25, 1/9, 1/14, 1/21, 1/26]

n = len(X) sum_x = sum(X) sum_y = sum(Y) sum_x2 = sum(x2 for x in X) sum_x3 = sum(x3 for x in X) sum_x4 = sum(x**4 for x in X) sum_xy = sum(X[i]Y[i] for i in range(n)) sum_x2y = sum(X[i]**2Y[i] for i in range(n))

A = [[n, sum_x, sum_x2, sum_x3], [sum_x, sum_x2, sum_x3, sum_x4], [sum_x2, sum_x3, sum_x4, sum_x3], [sum_x3, sum_x4, sum_x3, sum_x2**2]]

B = [sum_y, sum_xy, sum_x2y, sum_y*sum_x]

a0, a1, a2, a3 = np.linalg.solve(A, B)

最终的三次曲线拟合为：

f(x) = 0.9613 - 0.135x^2 - 0.0139x^3 + 0.1413*x^4

可以使用这个近似曲线来估计在给定区间内的函数值