用C语言程序改错:用以下公式计算e的近似值: 精度要求为10-6,计算精度公式为
时间: 2023-05-30 14:05:09 浏览: 73
#include <stdio.h>
int main()
{
int i = 1;
double item = 1.0, sum = 1.0, eps = 1e-6;
while (item > eps)
{
item = item / i;
sum = sum + item;
i++;
}
printf("The approximate value of e is %f", sum);
return 0;
}
相关问题
C语言利用公式求 cosx 的近似值(精度为10-6)
#include <stdio.h>
#include <math.h>
int main()
{
double x, cosx = 1, term = 1;
int i = 2;
printf("Enter the value of x in radians: ");
scanf("%lf", &x);
while (fabs(term) >= 1e-6)
{
term *= (-1) * x * x / (i * (i - 1));
cosx += term;
i += 2;
}
printf("cos(%lf) = %lf", x, cosx);
return 0;
}
用C语言写程序利用级数展开式计算求cos(x) 的近似值(精度为10-6)
可以的,我可以回答这个问题。以下是示例代码:
```
#include <stdio.h>
#include <math.h>
double computeCos(double x) {
double precision = 1e-6;
double sum = 1;
double term = 1;
int i = 2;
while (fabs(term) >= precision) {
term *= -x * x / ((i - 1) * i);
sum += term;
i += 2;
}
return sum;
}
int main() {
double x = 3.14 / 4; // 以弧度为单位
double cosX = computeCos(x);
printf("cos(%lf) = %lf\n", x, cosX);
return 0;
}
```
该程序利用级数展开式计算 cos(x) 的近似值,并通过循环迭代不断逼近目标精度,直到求得结果。