C語言 習題1-4 已知下列座標 求牛頓向前差除表 Pn(1.35)=?

C語言 習題1-4 已知下列座標
x= [1.2 , 1.25 , 1.30 , 1.40 ,  1.45 , 1.50]
f= [[0.1823 ,  0.2231   , 0.2624  ,0.3365  , 0.3716 , 0.4055 ],
求牛頓向前差除表 Pn(1.35)=?

程式

// CPP program for implementing 
// Newton divided difference formula 
#include <stdio.h>

// Function to find the product term 
double proterm(int i, float value, float x[]) 
{ 
    double pro = 1; 
    for (int j = 0; j < i; j++) { 
        pro = pro * (value - x[j]); 
    } 
    return pro; 
} 
  
// Function for calculating 
// divided difference table 
void dividedDiffTable(float x[], float y[][10], int n) 
{ 
    for (int i = 1; i < n; i++) { 
        for (int j = 0; j < n - i; j++) { 
            y[j][i] = (y[j][i - 1] - y[j + 1] 
                         [i - 1]) / (x[j] - x[i + j]); 
        } 
    } 
} 
  
// Function for applying Newton's 
// divided difference formula 
double applyFormula(float value, float x[], 
                   float y[][10], int n) 
{ 
    double sum = y[0][0]; 
  
    for (int i = 1; i < n; i++) { 
      sum = sum + (proterm(i, value, x) * y[0][i]); 
    } 
    return sum; 
} 
  
// Function for displaying  
// divided difference table 
void printDiffTable(float y[][10],int n) 
{ 
    for (int i = 0; i < n; i++) { 
        for (int j = 0; j < n - i; j++) { 
            printf("    %0.4lf\t", y[i][j]); 
        } 
        printf("\n"); 
    } 
} 
  
// Driver Function 
int main() 
{ 
    // number of inputs given 
    int n = 6; 
    float value, sum, y[10][10]; 
    float x[] = { 1.2 , 1.25 , 1.30 , 1.40 ,  1.45 , 1.50 }; 

    // y[][] is used for divided difference 
    // table where y[][0] is used for input 
    // 0.1823 ,  0.2231   , 0.2624  ,0.3365  , 0.3716 , 0.4055
    y[0][0] = 0.1823; 
    y[1][0] = 0.2231; 
    y[2][0] = 0.2624; 
    y[3][0] = 0.3365;
    y[4][0] = 0.3716;
    y[5][0] = 0.4055;
    
  
    // calculating divided difference table 
    dividedDiffTable(x, y, n); 
  
    // displaying divided difference table 
    printDiffTable(y,n); 
  
    // value to be interpolated 
    value = 1.35; 
    double result = applyFormula(value, x, y, n);
    // printing the value 
    printf("\nValue at %.2lf  is %.5lf ",value , result); 
    return 0; 
}

輸出畫面

    0.1823     0.8160     -0.3000     -0.0000     0.8002     -3.9997 
    0.2231     0.7860     -0.3000     0.2000     -0.3998 
    0.2624     0.7410     -0.2600     0.1001 
    0.3365     0.7020     -0.2400 
    0.3716     0.6780 
    0.4055 

Value at 1.35  is 0.30016