牛顿迭代法的C++求解,急!原题是用牛顿迭代法求解X^3-X^2-1=0的根,误差不超过10的负5次方#include#include double s(double);int main(){using namespace std;double xn,xn1;xn=1;xn1=s(xn);while(abs(xn1-xn)>1e-5){cout

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牛顿迭代法的C++求解,急!原题是用牛顿迭代法求解X^3-X^2-1=0的根,误差不超过10的负5次方#include#include  double s(double);int main(){using namespace std;double xn,xn1;xn=1;xn1=s(xn);while(abs(xn1-xn)>1e-5){cout
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牛顿迭代法的C++求解,急!原题是用牛顿迭代法求解X^3-X^2-1=0的根,误差不超过10的负5次方#include#include double s(double);int main(){using namespace std;double xn,xn1;xn=1;xn1=s(xn);while(abs(xn1-xn)>1e-5){cout
牛顿迭代法的C++求解,急!
原题是用牛顿迭代法求解X^3-X^2-1=0的根,误差不超过10的负5次方
#include
#include
double s(double);
int main()
{using namespace std;
double xn,xn1;
xn=1;
xn1=s(xn);
while(abs(xn1-xn)>1e-5)
{cout

牛顿迭代法的C++求解,急!原题是用牛顿迭代法求解X^3-X^2-1=0的根,误差不超过10的负5次方#include#include double s(double);int main(){using namespace std;double xn,xn1;xn=1;xn1=s(xn);while(abs(xn1-xn)>1e-5){cout
while(abs(xn1-xn)>le-5);
里面的abs改为fabs