BISECTION METHOD
#include<stdio.h>
#include<math.h>
float fun (float x)
{
return (x*x*x - 4*x - 9);
}
void bisection (float *x, float a, float b, int *itr)
/* this function performs and prints the result of one iteration */
{
*x=(a+b)/2;
++(*itr);
printf("Iteration no. %3d X = %7.5f\n", *itr, *x);
}
Int main ()
{
int itr = 0, maxmitr;
float x, a, b, allerr, x1;
printf(“This code is written by Amisha \n”);
printf("\n Enter the values of a, b, allowed error and maximum iterations:\n");
scanf("%f %f %f %d", &a, &b, &allerr, &maxmitr);
bisection (&x, a, b, &itr);
do
{
if (fun(a)*fun(x) < 0)
b=x;
else
a=x;
bisection (&x1, a, b, &itr);
if (fabs(x1-x) < allerr)
{
printf("After %d iterations, root = %6.4f\n", itr, x1);
return 0;
}
x=x1;
}
while (itr < maxmitr);
printf("The solution does not converge or iterations are not sufficient");
return 1;
}
This code is written by Amisha
Enter the values of a, b, allowed error and maximum iterations:
3
2
0.004
20
Iteration no. 1 X = 2.50000
Iteration no. 2 X = 2.75000
Iteration no. 3 X = 2.62500
Iteration no. 4 X = 2.68750
Iteration no. 5 X = 2.71875
Iteration no. 6 X = 2.70312
Iteration no. 7 X = 2.71094
Iteration no. 8 X = 2.70703
After 8 iterations, root = 2.7070
=== Code Execution Successful ===
FALSE POSITION METHOD
#include<stdio.h>
#include<math.h>
float f(float x)
{
return cos(x) - x*exp(x);
}
void regula (float *x, float x0, float x1, float fx0, float fx1, int *itr)
{
*x = x0 - ((x1 - x0) / (fx1 - fx0))*fx0;
++(*itr);
printf("Iteration no. %3d X = %7.5f \n", *itr, *x);
}
Int main ()
{
int itr = 0, maxmitr;
float x0,x1,x2,x3,allerr;
printf(“This code is written by Amisha\n”);
printf("\nEnter the values of x0, x1, allowed error and maximum iterations:\n");
scanf("%f %f %f %d", &x0, &x1, &allerr, &maxmitr);
regula (&x2, x0, x1, f(x0), f(x1), &itr);
do
{
if (f(x0)*f(x2) < 0)
x1=x2;
else
x0=x2;
regula (&x3, x0, x1, f(x0), f(x1), &itr);
if (fabs(x3-x2) < allerr)
{
printf("After %d iterations, root = %6.4f\n", itr, x3);
return 0;
}
x2=x3;
}
while (itr<maxmitr);
printf("Solution does not converge or iterations not sufficient:\n");
return 1;
}
This code is written by Amisha
Enter the values of x0, x1, allowed error and maximum iterations:
0
1
0.0005
20
Iteration no. 1 X = 0.31467
Iteration no. 2 X = 0.44673
Iteration no. 3 X = 0.49402
Iteration no. 4 X = 0.50995
Iteration no. 5 X = 0.51520
Iteration no. 6 X = 0.51692
Iteration no. 7 X = 0.51748
Iteration no. 8 X = 0.51767
After 8 iterations, root = 0.5177
=== Code Execution Successful ===
SECANT METHOD
#include<stdio.h>
#include<math.h>
#define ESP 0.0001
#define F(x) x*x - 4*x - 10
void main()
{
float x1,x2,x3,f1,f2,t;
printf(“This code is written by Amisha\n”);
printf("\nEnter the value of x1: ");
scanf("%f",&x1);
printf("\nEnter the value of x2: ");
scanf("%f",&x2);
printf("\n________________________________________\n");
printf("\n x1\t x2\t x3\t f(x1)\t f(x2)");
printf("\n______________________________________________\n");
do
{
f1=F(x1);
f2=F(x2);
x3=x2-((f2*(x2-x1))/(f2-f1));
printf("\n%f %f %f %f %f",x1,x2,x3,f1,f2);
x1=x2;
x2=x3;
if(f2<0)
{
t=fabs(f2);
}
else
{
t=f2;
}
}
while(t>ESP);
printf("\n______________________________________________\n");
printf("\n\nApp.root = %f",x3);
getch();
}
This code is written by Amisha
Enter the value of x1: 4
Enter the value of x2: 5
________________________________________
x1 x2 x3 f(x1) f(x2)
______________________________________________
4.000000 5.000000 6.000000 -10.000000 -5.000000
5.000000 6.000000 5.714286 -5.000000 2.000000
6.000000 5.714286 5.740741 2.000000 -0.204082
5.714286 5.740741 5.741661 -0.204082 -0.006859
5.740741 5.741661 5.741657 -0.006859 0.000025
___________________________________________
App.root = 5.741657
=== Code Execution Successful ===
NEWTON RAPHSON METHOD
#include<stdio.h>
#include<math.h>
float f(float x)
{
return x*log10(x) - 1.2;
}
float df (float x)
{
return log10(x) + 0.43429;
}
void main()
{
int itr, maxmitr;
float h, x0, x1, allerr;
printf(“This code is written by Amisha\n”);
printf("\nEnter x0, allowed error and maximum iterations\n");
scanf("%f %f %d", &x0, &allerr, &maxmitr);
for (itr=1; itr<=maxmitr; itr++)
{
h=f(x0)/df(x0);
x1=x0-h;
printf(" At Iteration no. %3d, x = %9.6f\n", itr, x1);
if (fabs(h) < allerr)
{
printf("After %3d iterations, root = %8.6f\n", itr, x1);
return 0;
}
x0=x1;
}
printf(" The required solution does not converge or iterations are insufficient\n");
return 1;
}
This code is written by Amisha
Enter x0, allowed error and maximum iterations
3
0.005
20
At Iteration no. 1, x = 2.746148
At Iteration no. 2, x = 2.740649
At Iteration no. 3, x = 2.740646
After 3 iterations, root = 2.740646
=== Code Execution Successful ===
BIRGE VIETA METHOD
#include<stdio.h>
#include<math.h>
float p[6], ply[6],q[6];
float synth(int m, float r){
int i;
q[0] = p[0];
for(i=1;i<=m;i++){
q[i] = (q[i-1]*r)+p[i];
}
printf("\n");
for(i=0;i<m;i++){
printf("\t%f",q[i]);
}
printf("\t%f",q[m]);
return(q[m]);
}
void main(){
int m,i,flag=0;
float r, x,x1, fx, fdx;
printf("BIRGE-VIETA METHOD\n");
printf("This is written by Amisha\n");
printf("\nEnter the highest degree of the equation (max 5): ");
scanf("%d",&m);
for(i=0;i<=m;i++){
printf("\n Coefficient x[%d] = ",m-i);
scanf("%f",&p[i]);
ply[i] = p[i];
}
printf("\nEnter the initial value x0 : ");
scanf("%f",&r);
x = r;
do{
printf("\n%f\n",x);
fx = synth(m,x);
for(i=0;i<=m;i++){
p[i]=q[i];
}
fdx = synth(m-1,x);
x1 = x - (fx/fdx);
if(fabs(x1-x) <= 0.0009){
flag = 1;
}
x = x1;
for(i=0;i<=5;i++){
p[i]=ply[i];
}
}while(flag!=1);
printf("\nApproximate root = %f", x1);
}
BIRGE-VIETA METHOD
This is written by Amisha
Enter the highest degree of the equation (max 5): 3
Coefficient x[3] = 1
Coefficient x[2] = 1
Coefficient x[1] = -3
Coefficient x[0] = -3
Enter the initial value x0 : 2
2.000000
1.000000 3.000000 3.000000 3.000000
1.000000 5.000000 13.000000
1.769231
1.000000 2.769231 1.899408 0.360492
1.000000 4.538462 9.928994
1.732924
1.000000 2.732924 1.735949 0.008266
1.000000 4.465847 9.474922
Approximate root = 1.732051
=== Code Execution Successful ===
GAUSS ELIMINATION METHOD
#include<stdio.h>
int main()
{
int i,j,k,n;
float A[20][20],c,x[10],sum=0.0;
printf(“This code is written by Amisha”);
printf("\nEnter the order of matrix: ");
scanf("%d",&n);
printf("\nEnter the elements of augmented matrix row-wise:\n\n");
for(i=1; i<=n; i++)
{
for(j=1; j<=(n+1); j++)
{
printf("A[%d][%d] : ", i,j);
scanf("%f",&A[i][j]);
}
}
for(j=1; j<=n; j++) {
for(i=1; i<=n; i++)
{
if(i>j)
{
c=A[i][j]/A[j][j];
for(k=1; k<=n+1; k++)
{
A[i][k]=A[i][k]-c*A[j][k];
}
}
}
}
x[n]=A[n][n+1]/A[n][n];
for(i=n-1; i>=1; i--)
{
sum=0;
for(j=i+1; j<=n; j++)
{
sum=sum+A[i][j]*x[j];
}
x[i]=(A[i][n+1]-sum)/A[i][i];
}
printf("\nThe solution is: \n");
for(i=1; i<=n; i++)
{
printf("\nx%d=%f\t",i,x[i]); }
return(0);
}
This code is written by Amisha
Enter the order of matrix: 3
Enter the elements of augmented matrix row-wise:
A[1][1] : 10
A[1][2] : -7
A[1][3] : 3
A[1][4] : 5
A[2][1] : -6
A[2][2] : 8
A[2][3] : 4
A[2][4] : 7
A[3][1] : 2
A[3][2] : 6
A[3][3] : 9
A[3][4] : -1
The solution is:
x1=-7.809084
x2=-8.690902
x3=7.418177
=== Code Execution Successful ===
GAUSS JORDAN METHOD
#include<stdio.h>
int main()
{
int i,j,k,n;
float A[20][20],c,x[10];
printf(“This code is written by Amisha\n”);
printf("\nEnter the size of matrix: ");
scanf("%d",&n);
printf("\nEnter the elements of augmented matrix row-wise:\n");
for(i=1; i<=n; i++)
{
for(j=1; j<=(n+1); j++)
{
printf(" A[%d][%d]:", i,j);
scanf("%f",&A[i][j]);
}
}
/* Now finding the elements of diagonal matrix */
for(j=1; j<=n; j++)
{
for(i=1; i<=n; i++)
{
if(i!=j)
{
c=A[i][j]/A[j][j];
for(k=1; k<=n+1; k++)
{
A[i][k]=A[i][k]-c*A[j][k];
}
}
}
}
printf("\nThe solution is:\n");
for(i=1; i<=n; i++)
{
x[i]=A[i][n+1]/A[i][i];
printf("\n x%d=%f\n",i,x[i]);
}
return(0);
}
This code is written by Amisha
Enter the size of matrix: 3
Enter the elements of augmented matrix row-wise:
A[1][1]:10-
A[1][2]:7
A[1][3]:5
A[1][4]:9
A[2][1]:5
A[2][2]:6
A[2][3]:7
A[2][4]:8
A[3][1]:9
A[3][2]:7
A[3][3]:5
A[3][4]:4
The solution is:
x1=4.999996
x2=-10.631570
x3=6.684205
=== Code Execution Successful ===
GAUSS SIEDEL METHOD
#include<stdio.h>
#include<math.h>
#define X 2
int main()
{
float x[X][X+1],a[X], ae, max,t,s,e;
int i,j,r,mxit;
for(i=0;i<X;i++) a[i]=0;
printf(“This code is written by Amsha \n”);
puts(" Enter the elements of augmented matrix row wise\n");
for(i=0;i<X;i++)
{
for(j=0;j<X+1;j++)
{
scanf("%f",&x[i][j]);
}
}
printf(" Enter the allowed error and maximum number of iteration: ");
scanf("%f%d",&ae,&mxit);
printf("Iteration\tx[1]\tx[2]\n");
for(r=1;r<=mxit;r++)
{
max=0;
for(i=0;i<X;i++)
{
s=0;
for(j=0;j<X;j++)
if(j!=i) s+=x[i][j]*a[j];
t=(x[i][X]-s)/x[i][i];
e=fabs(a[i]-t);
a[i]=t;
}
printf(" %5d\t",r);
for(i=0;i<X;i++)
printf(" %9.4f\t",a[i]);
printf("\n");
if(max<ae)
{
printf(" Converses in %3d iteration\n", r);
for(i=0;i<X;i++)
printf("a[%3d]=%7.4f\n", i+1,a[i]);
return 0;
}
}
}
This code is written by Amisha
Enter the elements of augmented matrix row wise
3
45
6
7
8
9
Enter the allowed error and maximum number of iteration:
0.001
4
Iteration x[1] x[2]
1 2.0000 -0.6250
Converses in 1 iteration
a[ 1]= 2.0000
a[ 2]=-0.6250
=== Code Execution Successful ===
NEWTON FORWARD
INTERPOLATION
#include<stdio.h>
#include<math.h>
int main(){
float h,f,p,d,s;
int i,j,n;
printf("this code is written by Amisha \n");
printf("Enter the value of n (number of terms you want to enter): ");
scanf("%d",&n);
float x[n],y[n];
printf("Enter the elements of x\n");
for(i=1;i<=n;i++){
scanf("%f",&x[i]);
}
printf("Enter the elements of y\n");
for(i=1;i<=n;i++){
scanf("%f",&y[i]);
}
h=x[2]-x[1];
printf("Please enter the value of x for which you want to print y: ");
scanf("%f",&f);
p=1;
d=y[1];
s=(f-x[1])/h;
for(int i=1;i<=n-1;i++){
for(int j=1;j<=(n-i);j++){
y[j]=y[j+1]-y[j];
}
p=p*(s-i+1)/i;
d=d+p*y[1];
}
printf("For the value of x(%f) the value of y is %0.4f",f,d);
}
This program is written by Amisha
Enter the value of n (number of terms you want to enter): 4
Enter the elements of x
4
3
4
5
Enter the elements of y
6
7
8
7
6
6
Please enter the value of x for which you want to print y: 3
For the value of x(3.000000) the value of y is 7.0000
=== Code Execution Successful ===
SIMPSON’S 3/8 METHOD
#include<stdio.h>
#include<math.h>
#define f(x) 1/(1+x*x)
int main()
{
float lower, upper, intgrl=0.0, stepSize, k;
int i, subInterval;
printf("This code is written by Amisha\n");
printf("Enter lower limit integration limit: ");
scanf("%f", &lower);
printf("Enter upper integration limit: ");
scanf("%f", &upper);
printf("Enter sub intervals: ");
scanf("%d", &subInterval);
stepSize = (upper - lower)/subInterval;
intgrl = f(lower) + f(upper);
for(i=1; i<= subInterval-1; i++)
{
k = lower + i*stepSize;
if(i%3 == 0)
{
intgrl = intgrl + 2 * f(k);
}
else
{
intgrl= intgrl + 3 * f(k);
}
}
intgrl = intgrl * stepSize*3/8;
printf("The integration is: %.3f", intgrl);
return 0;
}
This code is written by Amisha
Enter lower limit integration limit: 0
Enter upper integration limit: 1
Enter sub intervals: 12
The integration is: 0.785
=== Code Execution Successful ===
RUNGE KUTTA METHOD (4TH
ORDER)
#include<stdio.h>
#include<math.h>
float f(float x,float y);
int main()
{
float x0,y0,m1,m2,m3,m4,m,y,x,h,xn;
printf(“this code is written by Amisha\n”);
printf("Enter x0,y0,xn,h:");
scanf("%f %f %f %f",&x0,&y0,&xn,&h);
x=x0;
y=y0;
printf("\n\nX\t\tY\n");
while(x<xn)
{
m1=f(x0,y0);
m2=f((x0+h/2.0),(y0+m1*h/2.0));
m3=f((x0+h/2.0),(y0+m2*h/2.0));
m4=f((x0+h),(y0+m3*h));
m=((m1+2*m2+2*m3+m4)/6);
y=y+m*h;
x=x+h;
printf("%f\t%f\n",x,y);
}
}
float f(float x,float y)
{
float m;
m=(x-y)/(x+y);
return m;
}
This code is written by Amisha
Enter x0,y0,xn,h:0
2
2
0.5
X Y
0.500000 1.621356
1.000000 1.242713
1.500000 0.864069
2.000000 0.485426
=== Code Execution Successful ===
MODIFIED EULAR METHOD
#include<stdio.h>
#include<math.h>
#include<string.h>
float fun(float,float);
int main()
{
int i,j,c;
float x[100],y[100],h,m[100],m1,m2,a,s[100],w;
printf("C program for Modified Euler Method \n");
printf("This code is written by Amisha\n");
printf("Enter the initial value of x:");
scanf("%f",&x[0]);
printf("\nEnter the value of increment h:");
scanf("%f",&h);
printf("\nEnter the final value of x:");
scanf("%f",&a);
printf("\nEnter the initial value of the variable y :");
scanf("%f",&y[0]);
s[0]=y[0];
for(i=1;x[i-1]<a;i++)
{
w=100.0;
x[i]= x[i-1]+h;
m[i]=fun(x[i-1],y[i-1]);
c=0;
while(w>0.0001)
{
m1=fun(x[i],s[c]);
m2=(m[i]+m1)/2;
s[c+1]=y[i-1]+m2*h;
w=s[c]-s[c+1];
w=fabs(w);
c=c+1;
}
y[i]=s[c];
}
printf("\nThe respective values of x and y are\n x \t y\n");
for(j=0;j<i;j++)
{
printf("%f\t%f",x[j],y[j]);
printf("\n");
}
}
float fun(float a,float b)
{
float c;
c=a*a+b;
return(c);
}
C program for Modified Euler Method
This code is written by Amisha
Enter the initial value of x:0
Enter the value of increment h:0.05
Enter the final value of x:0.1
Enter the initial value of the variable y :1
The respective values of x and y are
x y
0.000000 1.000000
0.050000 1.051345
0.100000 1.105579
=== Code Execution Successful ===
S NO DATE PARTICULARS SIGNATURES
1 Bisection
method
2 False position
method
3 Secant method
4 Newton
Raphson
method
5 Birge vieta
method
6 Gauss
elimination
7 Gauss Jordan
method
8 Gauss siedal
method
9 Newton forward
interpolation
10 Simpson’s 3/8
method
11 RK method
12 Modified eular
method
