递归求简单交错幂级数的部分和

 

 1 double fn(double x, int n)
 2 {
 3     if(n == 1)
 4         return x;
 5 
 6     if((n-1)%2 == 1)        //如果n-1是奇数
 7         return -1*pow(x,n) + fn(x,n-1);
 8     else
 9         return pow(x,n) + fn(x,n-1);
10 }

 

转载于:https://www.cnblogs.com/FengZeng666/p/9696415.html