关于#泰勒级数计算sinx的值#的问题，如何解决？

Question

求看一下哪里错了

#include <stdio.h>
#include <math.h>
double Fact(double n);
double pow_n(double x,double n);
void sinx(double x);

int main()
{
    double x;
    printf("Input x:\n");
    scanf("%lf",&x);
    sinx(x);
    return 0;
}

//计算sin
void sinx(double x) {
    int sign = -1;//正负号标识符
    int count=0;//用于计数
    double fraction,sinx=x;
    do{
        int i=1;
        i = i+2;
        sign = -sign;//正负号切换
        fraction = pow_n(x, i) / Fact(i);//计算每个分母
        fraction = fraction * sign;
        count++;  //计数
        sinx=sinx + fraction;
    }while(fabs(fraction) < 1e-5);
    printf("sin(x)=%.3f,count=%d\n", sinx, count);
}

//阶乘
double Fact(double n)
{
    int result = 1;
    for (int i = 1; i <= n; i++)
    {
        result = result * i;
    }
    return result;
}

////自定义pow函数x的n次幂
double pow_n(double x,double n)
{
    double m=1;
    if(n==0)
        return 1;
    else {
        for (int i = 1; i <= n; i++) {
            m = m * x;
        }
        return m;
    }
}

Answer 1

int i=1;放到循环前面

#include <stdio.h>
#include <math.h>
double Fact(double n);
double pow_n(double x,double n);
void sinx(double x);
int main()
{
    double x;
    printf("Input x:\n");
    scanf("%lf",&x);
    sinx(x);
    return 0;
}
//计算sin
void sinx(double x) {
       int sign = 1;//正负号标识符
    int count=1;//用于计数
    double fraction,sinx=x;
            int i=1;
    do{
        count++;//计数
        i = i+2;
        sign = -sign;//正负号切换
        fraction = pow_n(x, i) / Fact(i);//计算每个分母
        fraction = fraction * sign;
        sinx=sinx + fraction;
    }while(fabs(fraction) >= 1e-5);
    printf("sin(x)=%.3f,count=%d\n", sinx, count);
}
//阶乘
double Fact(double n)
{
    double result = 1;
    for (int i = 1; i <= n; i++)
    {
        result = result * i;
    }
    return result;
}
////自定义pow函数x的n次幂
double pow_n(double x,double n)
{
    double m=1;
    if(n==0)
        return 1;
    else {
        for (int i = 1; i <= n; i++) {
            m = m * x;
        }
        return m;
    }
}