用C语言编写代码来实现“牛顿迭代法”求解方程 y = x 3 + e x y=x 3 +e x 的根 X0=4.5
时间: 2024-10-22 14:19:02 浏览: 26
在C语言中,我们可以使用牛顿迭代法来逼近函数y = x^3 + ex的零点。给定初始猜测X0=4.5,算法的基本步骤如下:
1. 定义函数f(x) = x^3 + ex - 0 (因为我们要找的是零点)
2. 定义函数f'(x) = 3x^2 + ex (一阶导数,用于计算下一个近似值)
3. 写一个循环,直到满足停止条件(如绝对误差小于某个阈值)或者达到最大迭代次数
以下是简单的C代码实现:
```c
#include <stdio.h>
#include <math.h>
double f(double x) {
return pow(x, 3) + exp(x);
}
double df(double x) {
return 3 * pow(x, 2) + exp(x);
}
double newton_raphson(double initial_guess, double tolerance, int max_iterations) {
double current_guess = initial_guess;
int iteration = 0;
while (iteration < max_iterations && abs(f(current_guess)) > tolerance) {
double derivative = df(current_guess);
if (derivative == 0) {
printf("Warning: Division by zero occurred. Exiting.\n");
return current_guess;
}
current_guess -= f(current_guess) / derivative;
iteration++;
}
return current_guess;
}
int main() {
double x0 = 4.5; // 初始猜测值
double tolerance = 1e-6; // 阈值
int max_iterations = 100; // 最大迭代次数
double root = newton_raphson(x0, tolerance, max_iterations);
printf("The root of the equation is approximately: %lf\n", root);
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