尝试使用 C 中的泰勒级数计算 SINE 时输出不正确

Question

我们的教授给我们这个作业，要求我们使用泰勒级数计算给定数字的正弦，直到连续两项的差异小于 10^-6，但由于我是计算机科学的新手，所以我不会真的很确定如何从这里继续下去。我相信阶乘计算正确。

我期望 sum 的输出与内置库的 sin(x) 输出相匹配

e.x.对于输入：1.0，sum 的输出应该是 0.841471，与在预制函数中输入 1.0 的结果匹配，但其输出是 0.158529。

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

int main()
{
    int val = 0,neg;
    long double sum = 0.0, x, prev = 0.0, curr, factorial = 1.0;
    printf("Type X:\n");
    scanf("%Lf",&x);
    curr = x;
    while (fabs(curr- prev) >= 0.000001){
        val++;
 
        prev = curr;
        factorial *= (2*val) * (2*val+1);

        curr = (pow(x, 2*val+1))/factorial;
 
        if (val % 2 == 1)
            neg = -1;
        else
            neg = 1;
        sum += neg*curr;     
    }
    printf("taylor series sinx:%Lf\n", sum);
    printf("math library sinx:%lf\n", sin(x));
    return 0;
}

Answer 1

您设置了

curr = x;

但也没有设置

sum = x;

。

当您打印操作时，这是显而易见的：

/* SO Staging Ground 7916-4500 */
/* SO 7916-4500 */
#include <stdio.h>
#include <math.h>

int main(void)
{
    int val = 0, neg;
    long double sum = 0.0, x, prev = 0.0, curr, factorial = 1.0;
    printf("Type X:\n");
    scanf("%Lf", &x);
    curr = x;
    while (fabsl(curr - prev) >= 0.000001)
    {
        val++;

        prev = curr;
        factorial *= (2 * val) * (2 * val + 1);

        curr = (powl(x, 2 * val + 1)) / factorial;

        if (val % 2 == 1)
            neg = -1;
        else
            neg = 1;
        sum += neg * curr;
        printf("val = %2d: neg = %+2d; factorial = %12.0Lf; term = %15.9Le; "
               "old = %15.9Le; new = %15.9Le\n",
               val, neg, factorial, curr, prev, sum);
    }
    printf("Taylor series sinx: %Lf\n", sum);
    printf("math library sinx:  %Lf\n", sinl(x));
    return 0;
}

产生：

Type X:
1
val =  1: neg = -1; factorial =            6; term = 1.666666667e-01; old = 1.000000000e+00; new = -1.666666667e-01
val =  2: neg = +1; factorial =          120; term = 8.333333333e-03; old = 1.666666667e-01; new = -1.583333333e-01
val =  3: neg = -1; factorial =         5040; term = 1.984126984e-04; old = 8.333333333e-03; new = -1.585317460e-01
val =  4: neg = +1; factorial =       362880; term = 2.755731922e-06; old = 1.984126984e-04; new = -1.585289903e-01
val =  5: neg = -1; factorial =     39916800; term = 2.505210839e-08; old = 2.755731922e-06; new = -1.585290154e-01
val =  6: neg = +1; factorial =   6227020800; term = 1.605904384e-10; old = 2.505210839e-08; new = -1.585290152e-01
Taylor series sinx: -0.158529
math library sinx:  0.841471

第一次计算的输出显示错误。 解决方法是将

sum

（以及

curr

）设置为

x

：

/* SO Staging Ground 7916-4500 */
/* SO 7916-4500 */
#include <stdio.h>
#include <math.h>

int main(void)
{
    int val = 0, neg;
    long double sum = 0.0, x, prev = 0.0, curr, factorial = 1.0;
    printf("Type X:\n");
    scanf("%Lf", &x);
    sum = curr = x;
    while (fabsl(curr - prev) >= 0.000001)
    {
        val++;

        prev = curr;
        factorial *= (2 * val) * (2 * val + 1);

        curr = (powl(x, 2 * val + 1)) / factorial;

        if (val % 2 == 1)
            neg = -1;
        else
            neg = 1;
        sum += neg * curr;
        printf("val = %2d: neg = %+2d; factorial = %12.0Lf; term = %15.9Le; "
               "old = %15.9Le; new = %15.9Le\n",
               val, neg, factorial, curr, prev, sum);
    }
    printf("Taylor series sinx: %Lf\n", sum);
    printf("math library sinx:  %Lf\n", sinl(x));
    return 0;
}

输出：

Type X:
1
val =  1: neg = -1; factorial =            6; term = 1.666666667e-01; old = 1.000000000e+00; new = 8.333333333e-01
val =  2: neg = +1; factorial =          120; term = 8.333333333e-03; old = 1.666666667e-01; new = 8.416666667e-01
val =  3: neg = -1; factorial =         5040; term = 1.984126984e-04; old = 8.333333333e-03; new = 8.414682540e-01
val =  4: neg = +1; factorial =       362880; term = 2.755731922e-06; old = 1.984126984e-04; new = 8.414710097e-01
val =  5: neg = -1; factorial =     39916800; term = 2.505210839e-08; old = 2.755731922e-06; new = 8.414709846e-01
val =  6: neg = +1; factorial =   6227020800; term = 1.605904384e-10; old = 2.505210839e-08; new = 8.414709848e-01
Taylor series sinx: 0.841471
math library sinx:  0.841471