用迭代法求多项式的导数

求函数的导数。

// 对 3.25x^3 - 2.75x^2 + 5.16x + 1 求导 #include <stdio.h> #include <math.h> // 定义多项式的单项式结构体 typedef struct { double coeff; // 系数 int exp; // 指数 } Term; double ployn(double a[], double x, int degree); int main() { // 数组中记录多项式的各项(系数和指数) Term poly[] = { {3.75, 4}, {3.25, 3}, {-2.75, 2}, {5.16, 1}, {1.0, 0} }; int i, n = sizeof(poly) / sizeof(poly[0]); int degree = poly[0].exp - 1; double coeff[degree+1], x; for(i=0; i <= degree; i++) coeff[i] = 0; printf("3.75x^4 + 3.25x^3 - 2.75x^2 + 5.16x + 1的导数:\n"); for (i = 0; i < n; i++) { // 常数项，直接跳过 if (poly[i].exp == 0) continue; // 新系数，新指数 double new_coeff = poly[i].coeff * poly[i].exp; int new_exp = poly[i].exp - 1; coeff[new_exp] = new_coeff; // 格式化输出（处理正负号和首项） if (i && new_coeff > 0) printf("+"); if (i && new_coeff < 0) printf("-"); // 指数为0 if (new_exp == 0) printf(" %.2f ", fabs(new_coeff)); // 指数为1 else if (new_exp == 1) printf(" %.2fx ", fabs(new_coeff)); else printf(" %.2fx^%d ", fabs(new_coeff), new_exp); } printf("\n系数是:"); for(i = 0; i <= degree; i++) printf("%.2f ", coeff[i]); x = 1.25; printf("\nf(%.2f) = %.4f\n", x, ployn(coeff, x, degree)+1e-9); return 0; } double ployn(double a[], double x, int degree) { int i; int limit = degree - 1; double sum1 = a[0], sum2 = 0; double xpwr1 = x, xpwr2 = x * x; for(i = 1; i <= limit; i += 2) { sum1 += a[i] * xpwr1; xpwr1 *= x * x; sum2 += a[i+1] * xpwr2; xpwr2 *= x * x; } for( ; i <= degree; i++) sum1 += a[i] * xpwr1; return sum1 + sum2; }

3.75x^4 + 3.25x^3 - 2.75x^2 + 5.16x + 1的导数:
15.00x^3 + 9.75x^2 - 5.50x + 5.16
系数是:5.16 -5.50 9.75 15.00
f(1.25) = 42.8163