s=x-x3/3!+x5/5!－…… (-1)n+1 x2n-1 /(2n-1)!

来源：百度知道 编辑：UC知道 时间：2024/06/30 03:25:12
s=x-x3/3!+x5/5!－…… (-1)n+1 x2n-1 /(2n-1)!
终止条件：最后一项 (-1)n+1 x2n-1 /(2n-1)!的绝对值小于10-8

(用C语言编程）

#include <iostream>
using namespace std;

double calculate( double x ){
double result;
result = x;
double progression = x;
for ( int n=3; progression > 1e-8 || progression < -1e-8; n = n+2 ){
progression = x*x*progression/(n*(n-1));
if ( n%4 == 1 ){
result = result + progression;
}
if ( n%4 == 3 ){
result = result - progression;
}
}
return result;
}

int main(){
double x;
while ( cin >> x ){
cout.precision(8);
cout << calculate(x) << endl;
}
}

c++的，楼主可以参考一下