Question

牛顿迭代法,x*e^x-1=0 迭代初值为0.5

Answer 1

C++的写法：

#include

#include

using namespace std;

int main(){

double x=0.5;

double change=(x*exp(x)-1)/(x*exp(x)+exp(x));

while(fabs(change)>=1e-7){

x=x-change;

change=(x*exp(x)-1)/(x*exp(x)+exp(x));

}

cout