c - C中梯形数值积分的实现

Question

我正在尝试使用以下公式使用梯形近似来实现数值积分:

我的问题是我不知道如何正确地实现它。为了测试，我写了一个文件，其中 22050 double 值都等于 2，例如:

....................
    value =2.0;
    for ( index = 0 ; index < 22050;index++){           
            fwrite(&value,sizeof(double),1,inp2);
        }

为了让问题简单化，假设我想要每 100 个样本的整数值:

 X Area                integral value 
0-100               should be  200  
100-200             should be  200
.....                ...........
22000-22050         should be  100

为此，我编写了一个应该执行此操作的程序，但得到的结果是 4387950 for 100 samples 这是我的代码:

..............................
    // opening the files 

double* inputData= NULL;
unsigned int N = 100;
double h= 0.0;
unsigned int index= 0;
FILE* inputFile=NULL;
double value =0.0;
int i =0,j=0;


    inputFile = fopen("sinusD","rb");
    outputFile=fopen("Trapez","wb+");
    if( inputFile==NULL || outputFile==NULL){
        printf("Couldn't open the files \n");
        return -1;
    }
    inputData = (double*) malloc(sizeof(double)*N);

    h=22050/2;
    while((i = fread(inputData,sizeof(double),N,inputFile))==N){
        value += inputData[0] +inputData[N];
        for(index=1;index<N;index++){
            value+=(2*inputData[index]);
        }
        value *=h;
        fprintf(outputFile,"%lf ",value);
        value =0;

    }
    if(i!=0){
        value = 0;
        i=-i;
        printf("i value %i\n", i);
        fseek(inputFile,i*sizeof(double),SEEK_END);
        fread(inputData,sizeof(double),i,inputFile);
            for(index=0;index<-i;index++){
            printf("index %d\n",index);
            value += inputData[0] +inputData[i];
            value+=(2*inputData[index]);
        }
        value *=h;
        fprintf(outputFile,"%lf ",value);
        value =0;
    }
    fclose(inputFile);
    fclose(outputFile);
    free(inputData);
    return 0;}

知道怎么做吗？

更新

while((i = fread(inputData,sizeof(double),N,inputFile))==N){
    value = (inputData[0] + inputData[N])/2.0;
    for(index=1;index<N;index++){
        value+=inputData[index];
    }
    value *=h;
    fprintf(outputFile,"%lf ",value);
    printf(" value %lf\n",value);
    value =0;

}

我得到 199.000 作为每个段的结果。

Answer 1

为什么您不从简单的事情开始。假设您有以下数据 {1,2,3,4,5,6,7,8,9,10} 并假设 h = 1。这个很简单，

#include <stdio.h>

#define SIZE 10

int main()
{
    double a[SIZE] = {1,2,3,4,5,6,7,8,9,10}, sum = 0.0, trapz;
    int h = 1;
    int i = 0;

    for ( i; i < SIZE; ++i){
        if ( i == 0 || i == SIZE-1 ) // for the first and last elements
            sum += a[i]/2;
        else
            sum += a[i]; // the rest of data
    }   
    trapz = sum*h; // the result

    printf("Result: %f \n", trapz);
    return 0;
}

这是结果

Result: 49.500000

用 Matlab 仔细检查你的工作:

Y = [1 2 3 4 5 6 7 8 9 10];
Q = trapz(Y)

Q =

   49.5000

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ $$$$$$$$$$$$$$$$$$$$$$$$$$

编辑:对于您在评论中的问题:这是 matlab 代码:

X = 0:pi/100:pi; % --> h = pi/100
Y = sin(X); % get values as many as the size of X
Q = trapz(X,Y);
Q =
    1.9998

现在要在 C 中实现相同的场景，请执行以下操作

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

#define SIZE 101

#define PI 3.14159265358

int main()
{
    double X[SIZE], Y[SIZE], incr = 0.0, h = PI/100.0, sum = 0.0, trapz;
    int i = 0, k = 0, j = 0;

    // Generate samples
    for ( i; i < SIZE; ++i)
    {
        X[i] = incr;
        incr += h;
    }

    // Generate the function Y = sin(X)
    for ( k; k < SIZE; ++k)
    {
        Y[k] = sin(X[k]);
    }

    // Compute the integral of sin(X) using Trapezoidal numerical integration method
    for ( j; j < SIZE; ++j){
        if ( j == 0 || j == SIZE-1 ) // for the first and last elements
            sum += Y[j]/2;
        else
            sum += Y[j]; // the rest of data
    }   
    trapz = sum * h; // compute the integral

    printf("Result: %f \n", trapz);

    return 0;
}

结果是

Result: 1.999836