C語言 正割法 Program to find root of an equations using secant method

2-7 正割法 Program to find root of an equations using secant method

源自於

https://www.geeksforgeeks.org/program-to-find-root-of-an-equations-using-secant-method/

Examples :

Input : equation = x3 + x - 1 
        x1 = 0, x2 = 1, E = 0.0001
Output : Root of the given equation = 0.682326
         No. of iteration=5

Algorithm

Initialize: x1, x2, E, n         // E = convergence indicator
calculate f(x1),f(x2)

if(f(x1) * f(x2) = E); //repeat the loop until the convergence
    print 'x0' //value of the root
    print 'n' //number of iteration
}
else
    print "can not found a root in the given interval"

程式:

// C Program to find root of an equations using secant method 
#include <stdio.h>

// function takes value of x and returns f(x) 
float f(float x) 
{ 
    // we are taking equation as x^3+x-1 
    float f = pow(x, 3) + x - 1; 
    return f; 
} 
  
void secant(float x1, float x2, float E) 
{ 
    float n = 0, xm, x0, c; 
    if (f(x1) * f(x2) < 0)
    { 
        do { 
            // calculate the intermediate value 
            x0 = (x1 * f(x2) - x2 * f(x1)) / (f(x2) - f(x1)); 
  
            // check if x0 is root of equation or not 
            c = f(x1) * f(x0); 
  
            // update the value of interval 
            x1 = x2; 
            x2 = x0; 
  
            // update number of iteration 
            n++; 
  
            // if x0 is the root of equation then break the loop 
            if (c == 0) 
                break; 
            xm = (x1 * f(x2) - x2 * f(x1)) / (f(x2) - f(x1));
            printf("xm=%0.5lf\n", xm); 
        
            
        } while (fabs(xm - x0) >= E); // repeat the loop 
                                // until the convergence 
  
        printf("Root of the given equation=%0.5lf", x0); 
        printf("\nNo. of iterations = %0.0lf", n); 
    
        
    } else
        
        printf("Can not find a root in the given inteval"); 
} 
  
// Driver code 
int main() 
{ 
    // initializing the values 
    float x1 = 0, x2 = 1, E = 0.0001; 
    secant(x1, x2, E); 
    return 0; 
} 

輸出結果

xm=0.69005                                                                                                                                     
xm=0.68202                                                                                                                                     
xm=0.68233                                                                                                                                     
xm=0.68233                                                                                                                                     
Root of the given equation=0.68233                                                                                                             
No. of iterations = 5