已知函数表如下: x 0.4 0.55 0.8 0.9 1 y 0.41075 0.57515 0.88811 1.02652 1.17520写出用牛顿插值法求解0.5,0.7,0.85三处点的函数值，并输出差商表的c语言和c++代码

根据给定的函数表，我们可以得到以下差商表： | x | y | f[x, y] | f[x, y, z] | |--------|----------|------------------|-----------------------| | 0.4 | 0.41075 | | | | 0.55 | 0.57515 | 1.019375 | | | 0.8 | 0.88811 | 1.1897833333 | 1.1108695652173911 | | 0.9 | 1.02652 | 0.9384000000 | -0.2356521739130435 | | 1 | 1.1752 | 1.4880000000 | 0.9434782608695652 | | 0.5075 | | 0.5652041667 | 0.7313043478260868 | | | | 0.1447916667 | -0.2208695652173913 | | | | -0.0218749999999 | 0.3228260869565217 | 其中，f[x, y]表示一阶差商，f[x, y, z]表示二阶差商。根据差商表，我们可以得到牛顿插值多项式： f(x) ≈ 0.41075 + 1.019375(x - 0.4) + 1.1108695652173911(x - 0.4)(x - 0.55) - 0.2356521739130435(x - 0.4)(x - 0.55)(x - 0.8) + 0.9434782608695652(x - 0.4)(x - 0.55)(x - 0.8)(x - 0.9) 代入x=0.5，0.7，0.85，可以得到对应的函数值： f(0.5) ≈ 0.54046640625 f(0.7) ≈ 0.8053555208333333 f(0.85) ≈ 1.0195064687500001 C语言代码实现： ```c #include <stdio.h> double newton_interpolation(double x, double *xs, double *ys, int n) { double f[n][n]; int i, j; for (i = 0; i < n; i++) { f[i][0] = ys[i]; } for (j = 1; j < n; j++) { for (i = j; i < n; i++) { f[i][j] = (f[i][j - 1] - f[i - 1][j - 1]) / (xs[i] - xs[i - j]); } } double result = f[0][0]; double prod = 1.0; for (i = 1; i < n; i++) { prod *= (x - xs[i - 1]); result += prod * f[i][i]; } return result; } int main() { double xs[] = {0.4, 0.55, 0.8, 0.9, 1}; double ys[] = {0.41075, 0.57515, 0.88811, 1.02652, 1.1752}; double x1 = 0.5, x2 = 0.7, x3 = 0.85; printf("f(%g) ≈ %g\n", x1, newton_interpolation(x1, xs, ys, 5)); printf("f(%g) ≈ %g\n", x2, newton_interpolation(x2, xs, ys, 5)); printf("f(%g) ≈ %g\n", x3, newton_interpolation(x3, xs, ys, 5)); return 0; } ``` C++代码实现： ```cpp #include <iostream> double newton_interpolation(double x, double *xs, double *ys, int n) { double f[n][n]; int i, j; for (i = 0; i < n; i++) { f[i][0] = ys[i]; } for (j = 1; j < n; j++) { for (i = j; i < n; i++) { f[i][j] = (f[i][j - 1] - f[i - 1][j - 1]) / (xs[i] - xs[i - j]); } } double result = f[0][0]; double prod = 1.0; for (i = 1; i < n; i++) { prod *= (x - xs[i - 1]); result += prod * f[i][i]; } return result; } int main() { double xs[] = {0.4, 0.55, 0.8, 0.9, 1}; double ys[] = {0.41075, 0.57515, 0.88811, 1.02652, 1.1752}; double x1 = 0.5, x2 = 0.7, x3 = 0.85; std::cout << "f(" << x1 << ") ≈ " << newton_interpolation(x1, xs, ys, 5) << std::endl; std::cout << "f(" << x2 << ") ≈ " << newton_interpolation(x2, xs, ys, 5) << std::endl; std::cout << "f(" << x3 << ") ≈ " << newton_interpolation(x3, xs, ys, 5) << std::endl; return 0; } ```