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# how to write a c program to find the roots of the equation using bisection method 6
to write a c program to find the roots of the equation using bisection method
that too using array or pointers
Aug 29 '06 #1
8 22088 D_C
293 100+
1. You posted a similar thread to this, please refrain from doing that.
2. For which equation are we finding roots? What is the bijection method? A bijection is a 1-1 onto function, so "1-1 onto function method" really doesn't help. You should be more specific, then we might be able to help you.
Aug 29 '06 #2
Niheel
2,452 Expert Mod 2GB
Dup has been deleted.
Aug 29 '06 #3
aruna
6 i just wrote a c program to find the roots of equation using bisection method
this my program

Expand|Select|Wrap|Line Numbers
1. /* to find the roots using bisection method*/
2. # include <stdio.h>
3. # include <conio.h>
4. # include<stdlib.h>
5. # include <math.h>
6. # define EPS .00001
7. double F(double x);
8. double root(double fVal1,double fVal2);
9.
10. void main()
11. {
12.
13.
14.     double fVal1,fVal2;
15.     clrscr();
16.     printf("enter the values ");
17.     scanf("%f %f",&fVal1,&fVal2);
18.     printf("the roots of the euation is %1.9f\n",root(fVal1,fVal2));
19.     getch();
20. }
21.     double root(double fVal1,double fVal2)
22.     {
23.         int nCtr=0;
24.         double fTest1,fTest2,fTest3,fVal,fNum;
25.
26.         fTest1=F(fVal1);
27.         fTest2=F(fVal2);
28.         if((fTest1*fTest2)>0)
29.         {
30.             printf("both are of equal sign");
31.             exit(0);
32.         }
33.         else
34.
35.         {
36.             do
37.             {
38.                 nCtr++;
39.                 fVal1=(fVal1+fVal2)/2;
40.                 fTest3=F(fVal);
41.             if(fTest1*fTest3>0)
42.
43.             {
44.                 fTest1=fTest3;
45.                 fVal1=fVal;
46.             }
47.             else
48.             {
49.                  fTest2=fTest3;
50.                  fVal2=fVal;
51.
52.             }
53.             fNum=fabs (fVal2-fVal1);
54.
55.
56.             } while(fNum>EPS);
57.         }
58.             return fabs(fVal2)>fabs(fVal1)?fVal2:fVal1;
59.
60.
61.
62.
63.
64.            }
65.            double F(double x)
66.            {
67.             return (x*x*x)+(3*x)-5;
68.
69.            }
70.
Sep 3 '06 #4
Banfa
9,065 Expert Mod 8TB

double fVal1,fVal2;

in main to

float fVal1,fVal2;

in order to get the correct values into the program.

A more serious error is that in

double root(double fVal1,double fVal2)
{
...
}

you declare and use a value fVal but you never initialise it to anything.
Sep 3 '06 #5
aruna
6 i have changed my program still iam not getting it
# include <stdio.h>
# include <conio.h>
# include<stdlib.h>
# include <math.h>
# define EPS .00001
float F(float x);
float root(float fVal1,float fVal2);

void main()
{

float fVal1,fVal2,;
clrscr();
printf("enter the values ");
scanf("%f %f",&fVal1,&fVal2);
printf("the roots of the euation is %1.9f\n",root(fVal1,fVal2));
getch();
}
float root(float fVal1,float fVal2)
{
int nCtr=0;
float fTest1,fTest2,fTest3,fVal,fNum;

fTest1=F(fVal1);
fTest2=F(fVal2);
if((fTest1*fTest2)>0)
{
printf("both are of equal sign");
exit(0);
}
else

{
do
{
nCtr++;
fVal=(fVal1+fVal2)/2;
fTest3=F(fVal);
if(fTest1*fTest3>0)

{
fTest1=fTest3;
fVal1=fVal;
}
else
{
fTest2=fTest3;
fVal2=fVal;

}
fNum=fabs (fVal2-fVal1);

} while(fNum>EPS);
}
return fabs(fVal2)>fabs(fVal1)?fVal2:fVal1;

}
float (float x)
{
return (x*x)-5;

}
Sep 4 '06 #6
aruna
6 i have just wrote a program to find the roots of the equation using bisection method
but there is some mistake iam not getting the output.comment iam getting as abnormal termination.so please help me
here is my program.

Expand|Select|Wrap|Line Numbers
1. /* to find the roots using bisection method*/
2. # include <stdio.h>
3. # include <conio.h>
4. # include<stdlib.h>
5. # include <math.h>
6. # define EPS .00001
7. float F(float x);
8. float root(float fVal1,float fVal2);
9.
10. void main()
11. {
12.
13.
14.     float fVal1,fVal2,;
15.     clrscr();
16.     printf("enter the values ");
17.     scanf("%f %f",&fVal1,&fVal2);
18.     printf("the roots of the euation is %1.9f\n",root(fVal1,fVal2));
19.     getch();
20. }
21.     float root(float fVal1,float fVal2)
22.     {
23.         int nCtr=0;
24.         float fTest1,fTest2,fTest3,fVal,fNum;
25.
26.         fTest1=F(fVal1);
27.         fTest2=F(fVal2);
28.         if((fTest1*fTest2)>0)
29.         {
30.             printf("both are of equal sign");
31.             exit(0);
32.         }
33.         else
34.
35.         {
36.             do
37.             {
38.                 nCtr++;
39.                 fVal=(fVal1+fVal2)/2;
40.                 fTest3=F(fVal);
41.                 if(fTest1*fTest3>0)
42.
43.                 {
44.                     fTest1=fTest3;
45.                     fVal1=fVal;
46.                 }
47.                 else
48.                 {
49.                      fTest2=fTest3;
50.                      fVal2=fVal;
51.
52.                 }
53.                 fNum=fabs (fVal2-fVal1);
54.
55.
56.             } while(fNum>EPS);
57.         }
58.         return fabs(fVal2)>fabs(fVal1)?fVal2:fVal1;
59.
60.
61.
62.
63.
64.            }
65.            float (float x)
66.            {
67.             return (x*x)-5;
68.
69.            }
70.
Sep 7 '06 #7
Banfa
9,065 Expert Mod 8TB

Please DO you the [code] and [/code] to delimit you code.

The posted code does not compile, please post compiling code. I would suggest that rather than typeing your code into the message box you copy and paste it from the source file once you have got the code compiling so we are all working from the same base line.
Sep 7 '06 #8
priya mahajan
9 [

/*The Bisection Method to solve x*x*x+3*x-5***********/
#include<stdio.h>
#include<conio.h>
#include<math.h>

float f(float x);
main()
{

float a,b,y,r1,r2,r3;
do{
printf("\nenter approximate interval");
scanf("%f%f",&a,&b);
r1=f(a);
r2=f(b);
}while(r1*r2>0);
printf("a\t\tb x=(a+b)/2 f(x)\n\n");
do{
r1=f(a);
r2=f(b);
y= (a+b)/2.0;
r3=f(y);
printf( "%.4f %.4f %.4f %.4f",a,b,y,r3);
if(r1*r3<0)
b=y;
else
a=y;
printf("\n");
}while(fabs(r3)>0.0001);
printf("\n The approx sol is %.4f",y);
getch();
return 0;
}

float f(float x )
{
float f1;
f1=(x*x*x)+(3*x)-5;
return(f1);
}

its simple but it works!!.
Oct 5 '06 #9

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