Question

c语言程序 用泰勒级数求e的近似值，直到最后一项小于le-6为止。 e=1+1/1!+...+1/n!

Answer 1

#include<stdio.h>

int main()

{float i,e=1.0,t=1.0;

 for(i=1;t>=1.0e-6;i+=1)

 {t/=i;

  e+=t;

 }

 printf("e=%f\n",e);

 getch();

return 0;

} 

Answer 2

#include <stdio.h>
int main()
{
double e = 1.0,single = 1.0,n = 1.0;
single /= n; //single表示每个单项，即1/n!
while(single >= (1e-6) )
{
e += single;
n += 1;
single /= n;
}
printf("e = %lf",e);
return 0;
}