The document contains programs written in C programming language to solve problems using numerical methods like Bisection method, False position method, Trapezoidal rule, Simpson's rule, Runge-Kutta method, Euler's method, and Lagrange interpolation. The programs take user input, perform the respective calculations, and output the results. They provide examples of implementing numerical techniques to find roots of equations, compute integrals, and solve differential equations.

1. PROGRAM
/*program for legranges interpolation method*/
#include<stdio.h>
#include<conio.h>
void main()
{
float x[20],y[20],unknown,temp,result=0,n,i,j;
clrscr();
printf("nntLEGRANGES INTERPOLATION -FIRST ORDERn");
printf("nt---------------------------------------n");
printf("nEnter the limitn");
scanf("%f",&n);
printf("Enter the values for xn");
for(i=0;i<n;i++)
{
scanf("%f",&x[i]);
}
printf("Enter the values of yn");
for(i=0;i<n;i++)
{
scanf("%f",&y[i]);
}
printf("Enter the value whose f(x) to be foundn");
scanf("%f",&unknown);
for(i=0;i<n;i++)
{
temp=1;
for(j=0;j<n;j++)
{
if(i!=j)
{
temp*=((unknown-x[j])/(x[i]-x[j]));
}
}
result+=temp*y[i];
}
printf("f( %f )= ",unknown);
printf("%f",result);
getch();
}
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2. OUTPUT
LEGRANGES INTERPOLATION -FIRST ORDER
------------------------------------------------------------
Enter the limit
4
Enter the values for x
300
304
305
307
Enter the values of y
2.4771
2.4829
2.4843
2.4871
Enter the value whose f(x) to be found
301
F(301.000000) = 2.478597
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3. PROGRAM
/* program to claculate the integral using trezoidal rule*/
#define f(x) (1/(x+1))
#include<stdio.h>
#include<conio.h>
void main()
{
float h,a,b,result=0,temp=0;
int i,j,n;
float x[50],y[50];
clrscr();
printf("Enter the Lower & Upper limit a&bn");
scanf("%f %f",&a,&b);
printf("nEnter the number of intervalsn");
scanf("%d", &n);
h=(b-a)/n;
printf("nCOMPUTING INTEGRAL USING-TRAPEZOIDAL RULE n");
for(i=0;i<=n;i++)
{
x[i]=a+(i*h);
}
for(i=0;i<=n;i++)
{
y[i]=f(x[i]);
}
printf("xttty");
for(i=0;i<=n;i++)
{
printf("n%ftt%f",x[i],y[i]);
}
result=y[0]+y[n];
for(i=1;i<n;i++)
{
temp+=y[i];
}
result+=(temp*2);
result*=(h/2);
printf("nn%f",result);
getch();
}
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4. OUTPUT
Enter the Lower & Upper limit a&b
0
1
Enter the number of intervals
2
COMPUTING INTEGRAL USING-TRAPEZOIDAL RULE
x y
0.000000 1.000000
0.500000 0.666667
1.000000 0.500000
0.708333
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5. PROGRAM
/*program to solve differential equation using Eulers method*/
#include<stdio.h>
#include<conio.h>
#define f(x,y) ((y-x)/(y+x))
void main()
{
float x0,y0,h,unknown,result=0,temp=0;
printf("nntEULERS METHODn");
printf("nEnter the value for x0n");
scanf("%f",&x0);
printf("Enter the value for y0n");
scanf("%f",&y0);
printf("Enter the steplengthn");
scanf("%f",&h);
printf("Enter the value for unknownn");
scanf("%f",&unknown);
while(x0<=unknown)
{
temp=y0+h*f(x0,y0);
x0+=h;
result=y0;
y0=temp;
}
printf("Result= %f",result);
getch();
}
OUTPUT
EULERS METHOD
--------------
Enter the value for x0
1
Enter the value for y0
2
Enter the steplength
.5
Enter the value for unknown
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6. 2
Result= 2.257576
PROGRAM
/* program to compute integral using legranges interpolation 2nd order*/
#include<stdio.h>
#include<conio.h>
#define f(x) 1/(1+(x*x))
void main()
{
float u,l,b,a,h,result=0;
clrscr();
printf("ntLEGRANGES INTERPOLATION - 2nd ORDERn");
printf("nt-----------------------------------n");
printf("nEnter the upper limit:nt");
scanf("%f",&a);
printf("Enter the lower limit:nt");
scanf("%f",&b);
h=(a-b)/2;
result=f(-0.57735)+f(0.57735);
result*=h;
printf("nnRESULT = %f",result);
getch();
}
LEGRANGES INTERPOLATION - 2nd ORDER
-----------------------------------------------------------
Enter the upper limit:
1
Enter the lower limit:
-1
RESULT = 1.583338
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7. PROGRAM
/* program to solve the 3rd order differential equation using Runge Kutta 2nd order*/
#define f(x,y) x+y
#include<stdio.h>
#include<conio.h>
void main()
{
float x0,y0,y1,k1,k2,h,unknown;
clrscr();
printf("ntRUNGE KUTTA METHOD - 2nd ORDERn");
printf("nt------------------------------n");
printf("nEnter the values for the following:n");
printf("ntx0: ") ;
scanf("%f",&x0);
printf("nty0: ");
scanf("%f",&y0);
printf("nth: ");
scanf("%f",&h);
printf("ntUnknown: ");
scanf("%f",&unknown);
while(unknown!=x0)
{
k1=h*(f(x0,y0));
k2=h*(f(x0+h,y0+k1));
y1=y0+((k1+k2)/2);
x0+=h;
y0=y1;
}
printf("nUNKNOWN = %fnRESULT = %f",unknown,y1);
getch();
}
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8. OUTPUT
RUNGE KUTTA METHOD - 2nd ORDER
-----------------------------------------------------
Enter the values for the following:
x0: 0
y0: 1
h: .1
Unknown: .2
UNKNOWN = 0.200000
RESULT = 1.242050
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9. PROGRAM
/* program using Runge kutta 3rd order method*/
#define f(x,y) x+y
#include<stdio.h>
#include<conio.h>
void main()
{
float y1,x0,y0,k1,k2,k3,h,unknown;
clrscr();
printf("nntRUNGE KUTTA - 3rd ORDERn");
printf("nt-----------------------n");
printf("Enter the values for the folowingn");
printf("ntx0: ");
scanf("%f",&x0);
printf("nty0: ");
scanf("%f",&y0);
printf("nth: ");
scanf("%f",&h);
printf("ntUnknown: ");
scanf("%f",&unknown);
while(unknown!=x0)
{
k1=h*(f(x0,y0));
k2=h*(f(x0+h/2,y0+k1/2));
k3=h*(f(x0+h,y0+2*k2-k1));
y1=y0+(k1+4*k2+k3)/6;
x0+=h;
y0=y1;
}
printf("nn Result of %f = %f",unknown,y1);
getch();
}
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10. OUTPUT
RUNGE KUTTA METHOD - 3rd ORDER
--------------------------------------------------
Enter the values for the folowing
x0: 0
y0: 1
h: .1
Unknown: .2
Result of 0.200000 = 1.242787
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11. PROGRAM
/* program to compute integral using Simpsons rule*/
#define f(x) 1/(1+x)
#include<stdio.h>
#include<conio.h>
void main()
{
float a,b,h,result,temp1=0,temp2=0;
float x[20],y[20];
int i,j,n;
clrscr();
printf("nntSIMPSONS 1/3rd RULEn");
printf("nt---------------------n");
printf("Enter the Lower and Upper limitsn");
scanf("%f %f",&a,&b);
printf("nEnter the number of intervalsn");
scanf("%d",&n);
h=(b-a)/n;
for(i=0;i<=n;++i)
{
x[i]=a+(i*h);
}
for(i=0;i<=n;++i)
{
y[i]=f(x[i]);
}
printf("nnxttyn");
printf("n-------n");
for(i=0;i<=n;++i)
{
printf("n%ft%f",x[i],y[i]);
}
h=(b-a)/(2*n);
result=y[0]+y[n];
for(i=1;i<=n-1;i+=2)
{
temp1=temp1+y[i];
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12. }
result+=(temp1*4);
for(i=2;i<=n-2;i+=2)
{
temp2=temp2+y[i];
}
result+=temp2*2;
result*=h/3;
printf("nntResult = %f",result);
getch();
}
OUTPUT
SIMPSONS 1/3rd RULE
-----------------------------
Enter the Lower and Upper limits
0
1
Enter the number of intervals
2
x y
---------------------------
0.000000 1.000000
0.500000 0.666667
1.000000 0.500000
Result = 0.347222
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13. PROGRAM
/*program to find the integral at a point using modified Eulers method*/
#include<stdio.h>
#include<conio.h>
#define f(x,y) (y-(x*x))
void main()
{
float x0,x1,y1,y0,y2,y3,h,unknown,result;
clrscr();
printf("nntMODIFIED EULERS METHODn");
printf("nt----------------------n");
printf("nEnter the values for the following:n");
printf("ntx0: ");
scanf("%f",&x0);
printf("nty0: ");
scanf("%f",&y0);
printf("ntSteplength: ");
scanf("%f",&h);
printf("ntThe value whose f(x) to be found: ");
scanf("%f",&unknown);
while(x0<=unknown)
{
y1=y0+(h*f(x0,y0));
x1=x0+h;
while(1)
{
y2=y0+(h/2)*(f(x0,y0)+f(x1,y1));
y3=y0+(h/2)*(f(x0,y0)+f(x1,y2));
if(y2==y3)
{
break;
}
y1=y2;
}
x0+=h;
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14. result=y0;
y0=y1;
}
printf("nResult:ntf(%f) = %f",unknown,result);
getch();
}
OUTPUT
MODIFIED EULERS METHOD
--------------------------------------
Enter the values for the following:
x0: 1
y0: 2
Steplength: .5
The value whose f(x) to be found: 2
Result:
f(0.600000) = 2.180510
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15. PROGRAM
/*program to compute the integral value at a point using legranges interpolation- 3rd
order*/
#include<stdio.h>
#include<conio.h>
#define f(x) (1/(1+(x*x)))
void main()
{
int a,b;
float h,result,r1,r2;
clrscr();
printf("nntLEGRANGES INTERPOLATION - 3rd ORDERn");
printf("nt-----------------------------------nn");
printf("Enter the Upper limit & Lower limit:n");
scanf("%d %d",&b,&a);
h=(b-a)/2.0;
result=(f(-0.77459))+(f(0.77459));
r1=result*(5.0/9.0);
r2=(8.0/9.0)*f(0);
result=(r1+r2)*h;
printf("nResult:ntf(.6) = %f",result);
getch();
}
OUTPUT
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16. LEGRANGES INTERPOLATION - 3rd ORDER
----------------------------------------------------------
Enter the Upper limit & Lower limit:
1
-1
Result:
f(.6) = 1.583338
PROGRAM
/*program for Bisection method*/
#define f(x) (x*x*x-x-1)
#include<stdio.h>
#include<conio.h>
void main()
{
float i=0,j=1,f1,f2,x0;
clrscr();
while(1)
{
f1=f(i); /* initial guess*/
f2=f(j);
if((f1<0&&f2>0)||(f1>0&&f2<0))
break;
else
{
i++;
j=i+1;
}
}
printf("ntROOT OF THE (x*x*x-x-1) USING BISECTION METHOD");
printf("nt---------------------------------------------------nn");
while((j-i)>.001)
{
x0=(i+j)/2; /* finding the root of the equation*/
printf("t%f %f %f %fn",i,j,x0,f(x0));
if(f(x0)<0)
i=x0;
else
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17. j=x0;
}
printf("nntRoot of the equation=%f",x0);
getch();
}
OUTPUT
ROOT OF (x*x*x-x-1) USING BISECTION METHOD
--------------------------------------------------------------------
1.000000 2.000000 1.500000 0.875000
1.000000 1.500000 1.250000 -0.296875
1.250000 1.500000 1.375000 0.224609
1.250000 1.375000 1.312500 -0.051514
1.312500 1.375000 1.343750 0.082611
1.312500 1.343750 1.328125 0.014576
1.312500 1.328125 1.320312 -0.018711
1.320312 1.328125 1.324219 -0.002128
1.324219 1.328125 1.326172 0.006209
1.324219 1.326172 1.325195 0.002037
Root of the equation=1.325195
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18. PROGRAM
/*Program for find the root of the equation(x*x*x-x-1) by False position method*/
#define f(x) (x*x*x-x-1)
#include<stdio.h>
#include<conio.h>
void main()
{
float i=0,j=1,f1,f2,x0,fp,fq,x1,x2=0,prev;
clrscr();
while(1)
{
f1=f(i);
f2=f(j);
if((f1<0&&f2>0)||(f1>0&&f2<0)) /* Finding the intervals*/
break;
else
{
i++;
j=i+1;
}
printf("n The Intervals are:t%f,t%fn",i,j);
x0=i;
x1=j;
do
{
prev=x2;
fp=f(x0);
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19. fq=f(x1);
x2=x0-(fp*(x1-x0)/(fq-fp));
printf("Value of x0=%ftValue of x1=%ftValue of x2=%ftValue of
F(x2)=%fn",x0,x1,x2,f(x2));
if(f(x2)<0)
x0=x2;
else
x1=x2;
}while((x2-prev)>.001);
}
printf("nRoot of the equation=%f",x2);
getch();
}
OUTPUT
The Intervals are: 1.000000, 2.000000
Value of x0=1.000000
Value of x1=2.000000
Value of x2=1.166667
Value of F(x2)=-0.578704
Value of x0=1.166667
Value of x1=2.000000
Value of x2=1.253112
Value of F(x2)=-0.285363
Value of x0=1.253112
Value of x1=2.000000
Value of x2=1.293437
Value of F(x2)=-0.129542
Value of x0=1.293437
Value of x1=2.000000
Value of x2=1.311281
Value of F(x2)=-0.056589
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20. Value of x0=1.311281
Value of x1=2.000000
Value of x2=1.318988
Value of F(x2)=-0.024304
Value of x0=1.318988
Value of x1=2.000000
Value of x2=1.322283
Value of F(x2)=-0.010362
Value of x0=1.322283
Value of x1=2.000000
Value of x2=1.323684
Value of F(x2)=-0.004404
Value of x0=1.323684
Value of x1=2.000000
Value of x2=1.324279
Value of F(x2)=-0.001869
Root of the equation=1.324279
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21. PROGRAM
/*Program for find the root of the eqn(x*x*x-2x-5) by Newton Raffson Method
method*/
#define f(x) (x*x*x-2x-5)
#define q(x) (3*x*x-2)
#include<stdio.h>
#include<conio.h>
void main()
{
float i=0,j=1,f1,f2,fp,fq,x1,x2,prev,x0;
clrscr();
printf("nntNEWTON RAFFSON METHODn");
printf("nt---------------------n");
while(1)
{
f1=f(i);
f2=f(j);
if((f1<0&&f2>0)||(f1>0&&f2<0)) /* Finding the intervals*/
break;
else
{
i++;
j=i+1;
}
printf("n The Intervals are:t%f,t%fnn",i,j);
x0=i;
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22. x1=j;
}
x2=(x0+x1)/2;
printf("x2ttf(x2)ttf'(x2)nnn");
while(1)
{
prev=x2; /*Assigning the current value to the previous value*/
x2=prev-(f(prev)/q(prev));
if(x2==prev)
break;
else
printf("%ft%ft%fn",x2,f(x2),q(x2));
}
printf("nn THE ROOT IS =%f",x2);
getch();
}
OUTPUT
NEWTON RAFFSON METHOD
-------------------------------------------
The Intervals are: 1.000000, 2.000000
The Intervals are: 2.000000, 3.000000
x2 f(x2) f'(x2)
2.164179 0.807945 12.051013
2.097135 0.028881 11.193929
2.094555 0.000041 11.161484
2.094552 0.000001 11.161439
THE ROOT IS =2.094552
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23. PROGRAM:
/*program for gauss elimination method…*/
#include<conio.h>
#include<stdio.h>
void main()
{
int m,n,p,q,i,j;
float a[10][10],b[10][10],x,y,z,t; /*…Declaration…*/
clrscr();
printf("nGAUSS ELIMINATION METHODn");
printf("********************************n");
printf("nInput the raw size of first matrix:");
scanf("%d",&m);
printf("Input the column size of first matrix:");
scanf("%d",&n);
printf("nInput the %d elements to %d*%d matrix:nn",m*n,m,n);
for(i=0;i<m;i++) /*…For loop…*/
{
for(j=0;j<n;j++)
{
scanf("%f",&a[i][j]);
}
}
printf("nnInput the raw size of second matrix:");
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24. scanf("%d",&p);
printf("Input the column size of second matrix:");
scanf("%d",&q);
printf("nInput the %d elements to %d*%d matrix:nn",p*q,p,q);
for(i=0;i<p;i++) /*…For loop…*/
{
for(j=0;j<q;j++)
{
scanf("%f",&b[i][j]);
}
}
printf("ntMatrix A:nn");
for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
{
printf("%ft",a[i][j]); /*…Print matrix A…*/
}
printf("n");
}
printf("ntMatrix B:nn");
for(i=0;i<p;i++)
{
for(j=0;j<q;j++)
{
printf("%ft",b[i][j]); /*…Print matrix B…*/
}
printf("n");
}
t=a[0][0];
for(j=0;j<m;j++)
{
a[0][j]/=t;
}
b[0][0]/=t;
for(i=1;i<p;i++)
{
t=a[i][0];
for(j=0;j<m;j++)
{
a[i][j]-=(t*a[0][j]); /*…Calculations…*/
}
b[i][0]-=(t*b[0][0]);
}
t=a[1][1];
for(j=1;j<=m;j++)
{
a[1][j]/=t;
}
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25. b[1][0]/=t;
t=a[2][1];
for(j=1;j<m;j++)
{
a[2][j]-=(t*a[1][j]);
}
b[2][0]-=(t*b[1][0]);
printf("ntThe result is:n");
z=b[2][0]/a[2][2];
y=b[1][0]-(a[1][2]*z); /*…Printing the output…*/
x=b[0][0]-((a[0][1]*y)+(a[0][2]*z));
printf("ntx=%fnty=%fntz=%fn",x,y,z);
getch();
}
OUTPUT:
GAUSS ELIMINATION METHOD
*****************************
Input the raw size of first matrix: 3
Input the column size of first matrix: 3
Input the 9 elements to 3*3 matrix:
4 1 1
3 4 2
2 3 1
Input the raw size of second matrix: 3
Input the column size of second matrix: 1
Input the 3 elements to 3*1 matrix :
11
11
7
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26. Matrix A:
4.000000 1.000000 1.000000
3.000000 4.000000 2.000000
2.000000 3.000000 1.000000
Matrix B:
11.000000
11.000000
7.000000
The result is:
x=2.333333
y=0.333333
z=1.333333
PROGRAM
/* program using Newtons divided difference formula*/
#include<stdio.h>
#include<conio.h>
#include<math.h>
void main()
{
float d[100][100],result,temp,unknown;
int m,n,i,j,k;
clrscr();
printf("Enter the limitn");
scanf("%d %d",&m,&n);
printf("Enter the values of xn");
for(i=0;i<m;i++)
{
scanf("%f",&d[i][0]);
}
printf("Enter the values of yn");
for(i=0;i<n;i++)
{
St.Mary’scollege,Thrissur

27. scanf("%f",&d[i][1]);
}
printf("Enter the value where f(x) to be foundn");
scanf("%f",&unknown);
printf(" x t y");
for(i=0;i<=(n-2);i++)
{
printf(" ty%d",i);
printf(“---------------------------------------------------------------------“);
}
printf("n");
for(j=2;j<(n+1);j++)
{
k=j-1;
for(i=0;i<(n+1)-j;i++)
{
d[i][j]=(d[i+1][j-1]-d[i][j-1])/(d[k][0]-d[k-(j-1)][0]);
k++;
}
}
k=0;
for(i=0;i<n;i++)
{
printf("nn");
for(j=0;j<=n-k;j++)
{
printf(" %f ",d[i][j]);
}
k++;
}
result=d[0][1];
for(j=2;j<n+1;j++)
{
temp=1;
for(i=0;i<=j-2;i++)
{
temp=temp*(unknown-d[i][0]);
}
result+=temp*d[0][j];
}
printf("nnf(%f) = %f",unknown,result);
getch();
}
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28. OUTPUT
Enter the limit
6
6
Enter the values of x
4
5
7
10
11
13
Enter the values of y
48
100
294
900
1210
2028
Enter the value where f(x) to be found
8
x y y0 y1 y2 y3 y4
--------------------------------------------------------------------------------------
4.000 48.000 52.000 15.000 1.000 0.000 0.000
5.000 100.000 97.000 21.000 1.000 0.000
7.000 294.000 202.000 27.000 1.000
10.000 900.000 310.000 33.000
11.000 1210.000 409.000
13.000 2028.000
f(8.000000) = 448.000000
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29. PROGRAM
/*find the integral using taylor series*/
#include<stdio.h>
#include<conio.h>
#define f1(x,y) (x+(y*y))
#define f2(y,y1) (1+(2*y*y1))
#define f3(y,y1,y2) (2*(y*y2+y1*y1))
#define f4(y,y1,y2,y3) (2*((y*y3+y2*y1)+(2*y1*y2)))
void main()
{
float x0,y0,y1,y2,y3,y4,unknown,h,result=0;
clrscr();
printf("ntTAYLOR SERIESn");
printf("nEnter the initial valuesn");
scanf("%f %f",&x0,&y0);
printf("Enter the unknown value, whose integral to be foundnn");
scanf("%f",&unknown);
St.Mary’scollege,Thrissur

30. h=unknown-x0;
/*first*/
y1=f1(x0,y0);
y2=f2(y0,y1);
y3=f3(y0,y1,y2);
y4=f4(y0,y1,y2,y3);
result=y0+(h*y1)+((h*h)/2)*y2+((h*h*h)/6)*y3+((h*h*h*h)/24)*y4;
printf("nResult:ntf(%f)= %f",unknown,result);
getch();
}
OUTPUT
TAYLOR SERIES
Enter the initial values
0
1
Enter the unknown value, whose integral to be found
.2
Result:
f(0.200000)= 1.271067
St.Mary’scollege,Thrissur