C语言的梯形Riemann和

Question

在下面的代码中，我尝试用梯形和矩形两种方式来近似(显然，梯形方式会更好)。

我试着在纸上做了一个算法，得到了如下结果:注意:N是矩形(或梯形)的数量，dx是使用a，b和N( dx = (b-a)/N )计算的。f(x) = x^2

矩形方法：

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<a href="http://www.codecogs.com/eqnedit.php?latex=\int_a^b&space;x^2&space;dx&space;\approx&space;\sum_{i=1}^N&space;f(a&space;&plus;&space;(i-1)dx)dx" target="_blank"><img src="http://latex.codecogs.com/png.latex?\int_a^b&space;x^2&space;dx&space;\approx&space;\sum_{i=1}^N&space;f(a&space;&plus;&space;(i-1)dx)dx" title="\int_a^b x^2 dx \approx \sum_{i=1}^N f(a + (i-1)dx)dx" /></a>

​

梯形法：

<a href="http://www.codecogs.com/eqnedit.php?latex=\int_a^b&space;x^2&space;dx&space;\approx&space;\sum_{i=1}^N&space;[f(a&space;&plus;&space;(i-1)dx)&space;&plus;&space;f(a&space;&plus;&space;i\cdot&space;dx)]dx" target="_blank"><img src="http://latex.codecogs.com/png.latex?\int_a^b&space;x^2&space;dx&space;\approx&space;\sum_{i=1}^N&space;[f(a&space;&plus;&space;(i-1)dx)&space;&plus;&space;f(a&space;&plus;&space;i\cdot&space;dx)]dx" title="\int_a^b x^2 dx \approx \sum_{i=1}^N [f(a + (i-1)dx) + f(a + i\cdot dx)]dx" /></a>

​

code (在下面的代码中，f(x)=x^2，F(x)是它的逆导数(x^3/3)：

int main() {
    int no_of_rects;
    double  a, b;

    printf("Number of subdivisions = ");
    scanf("%d", &no_of_rects);

    printf("a = ");
    scanf("%lf", &a);

    printf("b = ");
    scanf("%lf", &b);

    double dx = (b-a)/no_of_rects;

    double rectangular_riemann_sum = 0;

    int i;
    for (i=1;i<=no_of_rects;i++) {
        rectangular_riemann_sum +=  (f(a + (i-1)*dx)*dx);
    }

    double trapezoidal_riemann_sum = 0;

    int j;
    for (j=1;j<=no_of_rects;j++) {
        trapezoidal_riemann_sum += (1/2)*(dx)*(f(a + (j-1)*dx) + f(a + j*dx));
        printf("trapezoidal_riemann_sum: %lf\n", trapezoidal_riemann_sum);
    }

    double exact_integral = F(b) - F(a);
    double rect_error = exact_integral - rectangular_riemann_sum;
    double trap_error = exact_integral - trapezoidal_riemann_sum;

    printf("\n\nExact Integral: %lf", exact_integral);
    printf("\nRectangular Riemann Sum: %lf", rectangular_riemann_sum);
    printf("\nTrapezoidal Riemann Sum: %lf", trapezoidal_riemann_sum);
    printf("\n\nRectangular Error: %lf", rect_error);
    printf("\nTrapezoidal Error: %lf\n", trap_error);

    return 0;
}

其中：

double f(double x) {
    return x*x;
}

double F(double x) {
    return x*x*x/3;
}

我已经包含了数学和stdio头文件。实际情况是矩形riemann和是可以的，但是梯形riemann和由于某种原因总是为0。

有什么问题吗？是不是我的公式里有什么？还是我的代码？(顺便说一下，我是C语言的新手)

提前谢谢。

Answer 1

在此语句中：

trapezoidal_riemann_sum += (1/2)*(dx)*(f(a + (j-1)*dx) + f(a + j*dx));

1/2 ==为零，所以整个语句为零。至少将分子或分母更改为双精度形式，以获得双精度值。也就是说，1/2.0、1.0/2或1.0/2.0都可以工作。