作业帮C语言编程：计算S=-x+2x^2/3!-4x^4/5!+6x^6/7!-.,x属于[1,2],要求计算精度为第n项的值小于10^-5
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![C语言编程：计算S=-x+2x^2/3!-4x^4/5!+6x^6/7!-.,x属于[1,2],要求计算精度为第n项的值小于10^-5](/uploads/image/z/1639610-26-0.jpg?t=C%E8%AF%AD%E8%A8%80%E7%BC%96%E7%A8%8B%EF%BC%9A%E8%AE%A1%E7%AE%97S%3D-x%2B2x%5E2%2F3%21-4x%5E4%2F5%21%2B6x%5E6%2F7%21-.%2Cx%E5%B1%9E%E4%BA%8E%5B1%2C2%5D%2C%E8%A6%81%E6%B1%82%E8%AE%A1%E7%AE%97%E7%B2%BE%E5%BA%A6%E4%B8%BA%E7%AC%ACn%E9%A1%B9%E7%9A%84%E5%80%BC%E5%B0%8F%E4%BA%8E10%5E-5)
C语言编程：计算S=-x+2x^2/3!-4x^4/5!+6x^6/7!-.,x属于[1,2],要求计算精度为第n项的值小于10^-5
C语言编程：计算S=-x+2x^2/3!-4x^4/5!+6x^6/7!-.,x属于[1,2],要求计算精度为第n项的值小于10^-5
C语言编程：计算S=-x+2x^2/3!-4x^4/5!+6x^6/7!-.,x属于[1,2],要求计算精度为第n项的值小于10^-5
#include <stdio.h>
#include <math.h>
int main()
{
char sign = -1;
int n = 0, n_factorial = 1;
double x, xn, S;
do
{
printf("Please input x([1,2]): ");
scanf("%lf", &x);
}while(!(1 <= x && x <= 2));//限制输入范围必须[1,2]
S = -x;
do
{
sign = -sign;//计算符号
n++; //计算当前n,n=0,1,2,3,...
n_factorial = n_factorial *(2*n) * (2*n + 1); //计算阶乘(2n+1)!
xn = (2*n)*pow(x, 2*n)/n_factorial; //计算第n项值（不包括负号）2n*x^(2n)/(2n+1)!
S += sign * xn; // 计算S=S+符号*xn
}while(xn >= (1e-5));//第n项的值大于等于10^-5时,重复循环
printf("The result S is : %lf\n", S);
}