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regula falsi c program

aashish-cd answered on March 25, 2022 Popularity 5/10 Helpfulness 3/10

answer regula falsi c program

related regula falsi method in c

regula falsi c program

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/* Program: Finding real roots of nonlinear
   equation using Regula Falsi or False Position Method
   Author: CodeSansar
   Date: November 18, 2018 */
/* Header Files */
#include<stdio.h>
#include<conio.h>
#include<math.h>
/* Defining equation to be solved.
   Change this equation to solve another problem. */
#define   f(x)   x*log10(x) - 1.2

int main()
{
	
	 float x0, x1, x2, f0, f1, f2, e;
	 int step = 1;
	 clrscr();
	 /* Inputs */
	 up:
	 printf("\nEnter two initial guesses:\n");
	 scanf("%f%f", &x0, &x1);
	 printf("Enter tolerable error:\n");
	 scanf("%f", &e);
	 /* Calculating Functional Values */
	 f0 = f(x0);
	 f1 = f(x1);
	 /* Checking whether given guesses brackets the root or not. */
	 if( f0*f1 > 0.0)
	 {
		  printf("Incorrect Initial Guesses.\n");
		  goto up;
	 }
	 /* Implementing Regula Falsi or False Position Method */
	 printf("\nStep\t\tx0\t\tx1\t\tx2\t\tf(x2)\n");
	 do
	 {
		  x2 = x0 - (x0-x1) * f0/(f0-f1);
		  f2 = f(x2);
		  printf("%d\t\t%f\t%f\t%f\t%f\n",step, x0, x1, x2, f2);
		
		  if(f0*f2 < 0)
		  {
			   x1 = x2;
			   f1 = f2;
		  }
		  else
		  {
			   x0 = x2;
			   f0 = f2;
		  }
		  step = step + 1;
	
	 }while(fabs(f2)>e);

	 printf("\nRoot is: %f", x2);
	 getch();
	 return 0;
}

Popularity 5/10 Helpfulness 3/10 Language whatever

Source: www.codesansar.com

Tags: c position regula whatever

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Contributed on Mar 25 2022

aashish-cd

0 Answers Avg Quality 2/10

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regula falsi method in c

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/*========================================================================!
!Program to find solutions to equations of the form f(x)=0 using          !
!Regula Falsi method (Method of False Position)                           !
!LANGUAGE :: C (compiler: GNU GCC)                                        !
!========================================================================*/

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

double f(double y){
    return  y*exp(y)-1;
}
double newApprox(double yL,double yH){
    return yL-f(yL)*(yH-yL)/(f(yH)-f(yL));
}

int main()
{
    double xL,xH,diff,tolerance;
    double xMid,prevXMid;
    printf("Enter the two numbers xL and xH between which root is to be found\n");
    scanf("%lf %lf",&xL,&xH);
    printf("\nEnter the tolerance\n");
    scanf("%lf",&tolerance);
    if (f(xL)==0){
        printf("\nThe root is : %lf",xL);
        return 0;
    }
    else if (f(xH)==0){
        printf("\nThe root is : %lf",xH);
        return 0;
    }
    else if (f(xL)*f(xH)>0){
        printf("\nf(xH) and f(xL) must be of opposite sign");
        return 0;
    }
    xMid = newApprox(xL,xH);
    printf("\nxL= %lf \t xH= %lf \t xMid= %lf \t f(xMid)= %lf",xL,xH,xMid,f(xMid));
    do{
        prevXMid = xMid;
        if (f(xMid)*f(xH) < 0){
                xL = xMid;
        }
        else if ( f(xMid)*f(xL) <0 ){
                xH = xMid;
        }
        xMid = newApprox(xL,xH);
        diff = fabs((xMid-prevXMid)/xMid);
        printf("\nxL= %lf \t xH= %lf \t xMid= %lf \t f(xMid)= %lf",xL,xH,xMid,f(xMid));
    }while(diff > tolerance);

    printf("\n\nThe solution is %lf\n",xMid);
    return 0;
}

/*---------------------  OUTPUT  -------------------------------------------
Enter the two numbers xL and xH between which root is to be found
0.1
0.9

Enter the tolerance
0.0000001

xL= 0.100000     xH= 0.900000    xMid= 0.438347          f(xMid)= -0.320500
xL= 0.438347     xH= 0.900000    xMid= 0.534792          f(xMid)= -0.087062
xL= 0.534792     xH= 0.900000    xMid= 0.559236          f(xMid)= -0.021707
xL= 0.559236     xH= 0.900000    xMid= 0.565224          f(xMid)= -0.005294
xL= 0.565224     xH= 0.900000    xMid= 0.566678          f(xMid)= -0.001284
xL= 0.566678     xH= 0.900000    xMid= 0.567031          f(xMid)= -0.000311
xL= 0.567031     xH= 0.900000    xMid= 0.567116          f(xMid)= -0.000075
xL= 0.567116     xH= 0.900000    xMid= 0.567137          f(xMid)= -0.000018
xL= 0.567137     xH= 0.900000    xMid= 0.567142          f(xMid)= -0.000004
xL= 0.567142     xH= 0.900000    xMid= 0.567143          f(xMid)= -0.000001
xL= 0.567143     xH= 0.900000    xMid= 0.567143          f(xMid)= -0.000000
xL= 0.567143     xH= 0.900000    xMid= 0.567143          f(xMid)= -0.000000
xL= 0.567143     xH= 0.900000    xMid= 0.567143          f(xMid)= -0.000000

The solution is 0.567143
---------------------------------------------------------------------------*/

Popularity 4/10 Helpfulness 1/10 Language whatever

Source: www.numericalmethods.in

Tags: regula c whatever

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Contributed on Sep 03 2022

Anisujjaman Shohan

0 Answers Avg Quality 2/10

regula falsi c program

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