习题5-7 使用函数求余弦函数的近似值

本题要求实现一个函数，用下列公式求cos(x)的近似值，精确到最后一项的绝对值小于e：

cos(x)=x^0/0!-x^2/2!+x^4/4!-x^6/6!+......

函数接口定义：

 double funcos( double e, double x )  

其中用户传入的参数为误差上限e和自变量x；函数funcos应返回用给定公式计算出来、并且满足误差要求的cos(x)的近似值。输入输出均在双精度范围内

#include <stdio.h>
#include <math.h>

double funcos( double e, double x );

int main()
{    
    double e, x;

    scanf("%lf %lf", &e, &x);
    printf("cos(%.2f) = %.6f\n", x, funcos(e, x));

    return 0;
}
// 函数实现cos(x)的近似值
double funcos( double e, double x )
{
	double tmp1 = 1,tmp2 = 1,tmp3 = 1;
	double sum = 1;   // 0的阶乘是1,所以sum = 1 
	int i,j;
    j = -1;  //等式中第0项是正号，第一项是负号，第二项又是正号......
    for(i=2; tmp1>e; i+=2)   //tmp>1 这一项是判断精确度
    {
    	tmp2 = tmp2 * x * x;    // 每次比前一项都需要多乘x^2
    	tmp3 = tmp3 * i * (i-1);  //每次比前一项都需要多乘i*（i-1）
    	sum = sum + k*tmp2/tmp3;  //sum 就是cos(x)的近似值
    	tmp1 = tmp2/tmp3;   // tmp1是第1、2、3...项的近似值
		j = -j;
		 
	}
	return sum;
}

/*
试数：e = 0.01 x = -3.14
1> i=2; 1>0.01; 
   tmp2 = x^2; 
   tmp3 = 2!; 
   sum = 1+(-1)*(-3.14)^2/2!; 
   tmp1 = (-3.14)^2/2! = 4.9298; 
   j = 1; 
   i = 4;
2> 4.9298 > 0.01;
   tmp2 = x^2*x^2 = x^4; 
   tmp3 = 2!*4*3 = 4!; 
   sum = 1+(-1)*(-3.14)^2/2!+(1)*(-3.14)^4/4!;
   tmp1 = (-3.14)^4/4! = 4.0505;
   j = -1;
   i = 6;
3> 4.0505>0.01;
   tmp2 = x^4*x^2 = x^6; 
   tmp3 = 4!*6*5 = 6!; 
   sum = 1+(-1)*(-3.14)^2/2!+(1)*(-3.14)^4/4!+(-1)*(-3.14)^6/6!;
   tmp1 = (-3.14)^6/6! = 1.3312;
   j = 1;
   i = 8;
4> 1.3312>0.01;
   tmp2 = x^6*x^2 = x^8; 
   tmp3 = 6!*8*7 = 8!; 
   sum = 1+(-1)*(-3.14)^2/2!+(1)*(-3.14)^4/4!+(-1)*(-3.14)^6/6!+(1)*(-3.14)^8/8!;
   tmp1 = (-3.14)^8/8! = 0.2344;
   j = -1;
   i = 10;
5> 0.2344>0.01;
   tmp2 = x^8*x^2 = x^10; 
   tmp3 = 8!*10*9 = 10!; 
   sum = 1+(-1)*(-3.14)^2/2!+(1)*(-3.14)^4/4!+(-1)*(-3.14)^6/6!+(1)*(-3.14)^8/8!+(-3.14)^10/10!;
   tmp1 = (-3.14)^10/10! = 0.0257;
   j = 1;
   i = 12;
6> 0.0257>0.01;
   tmp2 = x^10*x^2 = x^12; 
   tmp3 = 10!*12*11 = 12!; 
   sum = 1+(-1)*(-3.14)^2/2!+(1)*(-3.14)^4/4!+(-1)*(-3.14)^6/6!+(1)*(-3.14)^8/8!+(-1)*(-3.14)^10/10!+(1)*(-3.14)^12/12!;
   tmp1 = (-3.14)^12/12! = 0.0019;
   j = -1;
   i = 14;
7> 0.0019>0.01 不成立  循环结束  返回sum = -0.9999(近似运算)
*/