nacp programs (1)

8/2/2019 NACP Programs (1)

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Numerical Analysis And ComputProgramming

Programs

Submitted By:

Mohd.Yawar Nihal Siddiqui

10-CSS-40B.Tech, Computer Engineering

h Semester


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Table of Contents

PROGRAM TO CALCULATE SUM AND AVERAGE OF 100 NATURAL NUMBERS ..........

PROGRAM TO FIND PRODUCT OF 1ST 'N' NATURAL NUMBERS ..............................

FIBBONACCI SERIES ........................................................................................

GENERATING PRIME NO. BETWEEN 1 AND 100 ....................................................

GENERATE PRIME NOS. BETWEEN 1 AND 100 EXCEPT THOSE DIVISIBLE BY 5 ........

TO SORT A LIST OF 5 NUMBERS IN ASCENDING ORDER .......................................

CALCULATION OF THE VALUE OF nCr .................................................................

SUM OF DIGITS OF A GIVEN NUMBER .................................................................

CALCULATION OF TAXABLE INCOME ...................................................................

CALCULATION OF SINE SERIES .........................................................................

CALCULATION OF COSINE SERIES ......................................................................

PROGRAM TO FIND OUT AVERAGE MARKS OF STUDENTS ...................................

PROGRAM TO REVERSE THE DIGITS OF A NUMBER AND FIND THE SUM OF ITS DIGIT

STANDARD DEVIATION AND VARIATION ............................................................

CONVERT BINARY NUMBER TO DECIMAL NUMBER ..............................................

CONVERT DECIMAL NUMBER TO BINARY NUMBER ..............................................

ADDITION OF MATRIX ......................................................................................

MULTIPLICATION OF MATRIX ............................................................................

BISECTION METHOD .........................................................................................

NEWTON RALPHSON ........................................................................................

REGULA-FALSE METHOD ..................................................................................

NEWTON GREGORY FORWARD INTERPOLATION ..................................................

NEWTON GREGORY BACKWARD INTERPOLATION ................................................

LAGRANGE METHOD OF INTERPOLATION ...........................................................


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NEWTON DIVIDED DIFFERENCE METHOD ...........................................................

BESSEL'S METHOD OF INTERPOLATION ..............................................................

STIRLING METHOD OF INTERPOLATION ..............................................................

TRAPEZOIDAL RULE ........................................................................................

SIMPSION 1/3 RULE .........................................................................................

SIMPSION 3/8 RULE ..........................................................................................

BOOLS RULE ...................................................................................................

WEDDLE'S RULE ..............................................................................................

GAUSS ELIMINATION METHOD .........................................................................

GAUSS JORDAN METHOD .................................................................................

GAUSS SEIDAL METHOD ..................................................................................

CURVE FITTING - STRAIGHT LINE ......................................................................

RUNGA - KUTTA METHOD .................................................................................

ROMBERG METHOD ...........................................................

NUMERICAL DOUBLE INTEGRATION..............................................


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1. Write a program in C to find sum and average of square of first 100

natural numbers.

/* PROGRAM TO CALCULATE SUM AND AVERAGE OF 100 NATURAL NUMBERS*/

# include#include void

main( )

{

int i,

n=100,sum=0, p;

float avg;

for

(i=1;i


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2. Write a program to find to find the product of first n natural numbers.

/*PROGRAM TO FIND PRODUCT OF 1ST 'N' NATURAL NUMBERS*/

#include

#include

void

main(){

int i,n;

long

int

prod;

clrscr(

);

printf("\nEnter the value of

n:\t"); scanf("%d",&n);i=1,prod=1;

a5:

i=i+1;

prod=

prod*

i;

if(i0);

for (j=i-

1;j>=0;j--)

{

printf ("The binary number is

%d",rem[j]);

}

}

OUTPUT:

Enter the number= 123The binary number is 1111011


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18. Write a program in C to add two rectangular matrices.

/* ADDTION OF MATRIX */

#include#include

void

main()

{

int mat1[10][10],mat2[10]

[10],mat3[10][10]; int i,j,m,n;

clrscr();

printf ("\n The number of rows

are="); scanf ("%d",&m);

printf ("\n The number of columns

are="); scanf ("%d",&n);

for

(i=0;i


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}}

for

(i=0;i


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19. Write a program to perform multiplication in matrices.

/*MULTIPLICATION OF MATRIX*/

#include

#include

void

main()

{int m,n,p,q,i,j,k;

int a[10][10],b[10]

[10],c[10][10]; clrscr();

printf("\nenter no. of row and col of

matrix a:"); scanf("%d%d",&m,&n);

printf("\nenter no. of row and col of

matrix b:"); scanf("%d%d",&p,&q);

if(n!=p)

{ printf("\nmatrix can't be

multiplied\n");goto end;

}

printf("\nenter matrix

a\n"); for(i=0;i


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c

h

(

)

;

}


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Output:

Enter no. of row and col of matrix a:3 3Enter no. of row and col of matrix b:3 2

Enter matrix a0 1 2

1 2 32 3 4

Enter matrix b1 -2-1 02 -1

The product of matrix is:3 -25 -57 -8


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20. Bisection Method

/*BISECTION METHOD*/

#include

#include

#include

void

main()

{ float fun(float m);

float

x1,x2,x3,p,q,r

; int i=0;clrscr();

l10: printf("\nequation:x*(exp(x)-1) ");

printf("\nenter the app value of

x1,x2:"); scanf("%f %f",&x1,&x2);

if(fun(x1)*fun(x2)>0){ printf("\n wrong values entered...enter

again:\n"); goto l10;}

else

printf("\n the root lies b/w %f & %f",x1,x2);

printf("\n n x1 x2

x3

f(x3)");

l15:

x3=(x1+x2)/2;

p=fun(x1);

q=fun(

x2);

r=fun(

x3);

i=i++;

printf("\n%d %f %f %f %f

%f

if((p*r

)>0)

x1=x3;

else

x2=x3;

if((fabs((x2-x1)/x2))


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f(x1) f(x2)

%f",i,x1,x2,x3,p,q,r);n x1 x2 x3 f(x1) f(x2) f(x3)

1 0.500000 1.000000 0.750000 -0.175639 1.718282 0.5877502 0.500000 0.750000 0.625000 -0.175639 0.587750 0.1676543 0.500000 0.625000 0.562500 -0.175639 0.167654 -0.0127824 0.562500 0.625000 0.593750 -0.012782 0.167654 0.075142

5 0.562500 0.593750 0.578125 -0.012782 0.075142 0.030619

6 0.562500 0.578125 0.570312 -0.012782 0.030619 0.008780

7 0.562500 0.570312 0.566406 -0.012782 0.008780 -0.0020358 0.566406 0.570312 0.568359 -0.002035 0.008780 0.0033649 0.566406 0.568359 0.567383 -0.002035 0.003364 0.00066210 0.566406 0.567383 0.566895 -0.002035 0.0006-0.000687

Root of the equation is 0.566895


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21.Newton Ralphson Method.

/*NEWTON RALPHSON*/

#include#include

#include

void

main()

{float

f(float a);

float

df(float a);

int i=1;

float x0,x1,p,q;

float error =0.0001,delta

=0.001; clrscr();

printf("\n\nThe equation is X^3+1.2X^2-

5X-7.2"); printf("\nenter the initial

value of x0:");

scanf("%f",&x0);

printf("\n i x0 x1 f(x0) df(x0)\n ");

if (fabs (df(x0))


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return(g1);

}

Output:

The equation isX^3+1.2X^2-5X-7.2Enter the initial valueof x0:2

i x0 x1 f(x0) df(x0)

2 2.000000 2.372881 -4.400000 11.800000

3 2.372881 2.313010 1.052938 17.586615

4 2.313010 2.311227 0.029603 16.601265

5 2.311227 2.311225 0.000026 16.572248

The root of equation X^3+1.2X^2-5X-7.2 is 2.311227


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22.Regula falsi Method (Method of false position).

/*REGULA-FALSE METHOD */

#include

#include#include

float f(float x)

{return(3*x-

cos(x)-1);

}void main()

{ float f(float x);

double

x1,x2,m;

int c=0;

clrscr();printf("\n enter the first

approximation :"); scanf("%f",&x1);

printf("\n enter the second

approximation :"); scanf("%f",&x2);

if(f(x1)*f(x

2)=0.0001)

{

c++;

if(f(x1)*f(

m)


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approximation :"); getch();}

OUTPUT:

Enter the first approximation :0

Enter the secondapproximation: 1The 1st iteration is 0.605959The 2nd iteration is 0.607057The 3rd iteration is 0.607100The answer is repeated at 3 iteration is 0.607100


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23. Newton Gregory forward Interpolation method.

/*NEWTON GREGORY FORWARD INTERPOLATION*/

#include

#include

#include

void

main()

{

int n,i,j,k;

float mx[10],my[10],x,y=0,h,p,diff[20]

[20],y1,y2,y3,y4; clrscr();

printf("\nenter no. of

terms:"); scanf("%d",&n);

printf("\nenter values

x\ty:\n"); for(i=0;i


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OUTPUT:

Enter no. of terms:5

Enter values x y:

3 135 2311 89927 1731534 35606

Enter value of x at which y is to be calculated:7

When

x=7.0000,y=899.00000000


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24. Newton Gregory backward Interpolation method.

/*NEWTON GREGORY BACKWARD INTERPOLATION*/

#include

#include

#include

void

main(){

int n,i,j,k;

float mx[10],my[10],x,x0=0,y0,sum=0,fun=1,h,p,diff[20][20],y1,y2,y3,y4;

clrscr();

printf("\nenter no. of

terms:"); scanf("%d",&n);

printf("\nenter valuesx\ty:\n"); for(i=0;i


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1)))/k;

sum=sum+fun*diff[i]

[k]; }

printf("\nwhen x=%6.4f,y=

%6.8f",x,sum); getch();

}

OUTPUT:

Enter no. of terms:5

Enter values x y:20 4140 103

60 16880 218100 235

Enter value of x at which y is to be calculated:70

When x=70.0000,y=196.00000000


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25. Lagranges method of Interpolation.

/*LAGRANGE METHOD OF INTERPOLATION*/

#include

#include

#include

#define

max 50

void

main()

{

float

ax[max],ay[max],nr,dr,x,y=0

; int i,j,n;

clrscr();

printf("\nEnter No. of

Points:"); scanf("%d",&n);printf("\nEnter the given set of

values:\nx\ty\n"); for(i=0;i


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10 90011 121013 2028

Enter the value of x at which f(x)is required:8 when x= 8.00 theny=445.62


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26. Write a program in C/C++ which can calculate the value of afunction at a point using Newton Divided Difference method.

/* NEWTON DIVIDED DIFFERENCE METHOD */

#include

#include

#include

void

main(){

float ax[20], ay[20],

diff[30],temp=1; int

n,j,m,z=0,A=0,k=0;clrscr();

cout>n;

for (int

i=0;i


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1

;

A

=

0

;

for(j=0;j


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A=0;

for

(z=0;z


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9

2432

42

11

0The value of y for x = 9 is: 810


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27. Write a program in C for Bessels Method of Interpolation.

/*BESSEL'S METHOD OF INTERPOLATION*/

#include

#include

#include

void

main()

{ int n , i , j ;

float ax[10] , ay[10] , x , y=0 , h , p , diff[20][20] , y1 , y2 , y3

, y4 ; clrscr();

printf("\nEnter the noumber of

item : "); scanf("%d" , &n);

printf("\nEnter the value in the form of

x\n"); for(i = 0 ; i < n ; i++)

{ printf("\nEnter the value of x%d\t" , i+1);

scanf("%f" , &ax[i]); }printf("\nEnter the value in the form of y\n");

for(i = 0 ; i < n ; i++)

{ printf("\nEnter the value of y%d\t" , i+1);

scanf("%f" , &ay[i]); }

printf("\nEnter the value of x for which you want the value of y

:- "); scanf("%f" , &x);

h = ax[1] - ax[0];

for(i = 0 ; i < n-1 ; i++)

diff[i][1] = ay[i+1] -

ay[i]; for(j = 2 ; j


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OUTPUT

Enter number of item: 4

Enter the value in the form of x

Enter the value of x1 20Enter the value of x2 24Enter the value of x3 28Enter the value of x4 32

Enter the value in the form of y

Enter the value of y1 24Enter the value of y2 32Enter the value of y3 35Enter the value of y4 40

Enter the value of x for which you want the value of y:25

When x = 25.0000 , y = 32.945313


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28. Write a program for Stirlings Method of Interpolation.

/*STIRLING METHOD OF INTERPOLATION*/

#include#include

#include

void

main()

{ int n , i , j ;

float ax[10] , ay[10] , x , y=0 , h , p , diff[20][20] ,

y1 , y2 , y3 ,y4;

clrscr();

printf("\nEnter the noumber of

item : "); scanf("%d" , &n);

printf("\nEnter the value in the form ofx\n"); for(i = 0 ; i < n ; i++)

{printf("\nEnter the value of x

%d\t" , i+1); scanf("%f" , &ax[i]); }

printf("\nEnter the value in the form of

y\n"); for(i = 0 ; i < n ; i++)

{printf("\nEnter the value of y

%d\t" , i+1); scanf("%f" , &ay[i]); }

printf("\nEnter the value of x for which you want the value of y

:- ");

scanf("%f"

, &x);

h = ax[1] -

ax[0];

for(i = 0 ; i < n-1 ; i++)diff[i][1] = ay[i+1] -

ay[i]; for(j = 2 ; j


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OUTPUT

Enter the number of item : 4

Enter the value in the form of xEnter the value of x1 20Enter the value of x2 24Enter the value of x3 28Enter the value of x4 32

Enter the value in the form of y

Enter the value of y124Enter the value of y2 32

Enter the value of y3 35Enter the value of y4 40

Enter the value of x for which you want the value of y:25

When x = 25.0000, y = 33.31251144


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29. Write a program for trapezoidal rule for Numerical Integration.

/* TRAPEZOIDAL RULE */

#include

#include

#include

void main()

{

float

fun(float);

float h ,

k1=0.0 ;

float x[20] ,

y[20];

int n , i;

clrscr();

printf("\nEnter number of

parts : "); scanf("%d" , &n);

printf("\nEnter lower and upper

limits : "); scanf("%f %f" , &x[0] ,

&x[n]);

y[0] = fun(x[0]);

h = (x[n] - x[0])/n ;

printf("\nx y");

printf("\n %8.5f %8.5f " , x[0] ,y[0]);

for(i=1 ; i < n ; i++)

{ x[i] = x[0] + i * h ;y[i] = fun(x[i]);

printf("\n %8.5f %8.5f " , x[i] , y[i]);

k1 = k1 + 2 * y[i];

}

y[n] = fun(x[n]);

printf("\n %8.5f %8.5f " , x[n] ,

y[n]);

y[0] = (h / 2.0 ) * (y[0] + y[n]

+ k1 );

printf("\nresult = %f \n" ,

y[0]);

getch();

}

float

fun(float

x)

{

float g;

g = log(x);

return g;

}


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OUTPUT:

Enter number of parts: 6Lower and upper limits: 4 5.2

x y4.00000 1.386294.24000 1.444564.48000 1.499624.72000 1.551814.96000 1.601415.20000 1.64866

Result = 1.827570


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30. Write a program for Simpsons One Third Rule.

/* SIMPSION 1/3 RULE */

#include#include

#include

void

main()

{ float fun(float);

float h , k1=0.0 ,

k2=0.0 ; float x[20]

, y[20];

int n , i;

clrscr();


parts : "); scanf("%d" , &n);printf("\nEnter lower and upper

limits :"); scanf("%f %f" , &x[0] ,

&x[n]);

y[0] = fun(x[0]);

h = (x[n] - x[0])/n ;

printf("\nx y");

printf("\n%8.5f %8.5f" , x[0] ,y[0]);

for(i=1 ; i < n ; i++)

{ x[i] = x[0] + i * h ;y[i] = fun(x[i]);

printf("\n %8.5f %8.5f " , x[i] , y[i]);

if(i % 2 == 0)

k1 = k1 + 2 * y[i];else

k2 = k2 + 4 * y[i];

}

y[n] = fun(x[n]);

printf("\n %8.5f %8.5f " , x[n] , y[n]);

y[0] = (h / 3.0 ) * (y[0] + y[n] + k1

+ k2 ); printf("\nresult =%f \n" ,

y[0]);getch();

}

float fun(float x)

{ float g;

g = sin(x) - log(x) +

exp(x); return g;}

OUTPUT :Enter number of parts : 6Enter lower and upper limits :0.2 1.4x y


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0.20000 3.029510.40000 2.797530.60000 2.897590.80000 3.166041.00000 3.559751.20000 4.069831.40000 4.70418Result = 4.052133


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31. Simpsons 3/8 rule.

/*SIMPSION 3/8 RULE */

#include

#include

#include

void

main()

{ float fun(float);

float h , k1=0.0 ,

k2=0.0 ; float x[20]

, y[20];

int n , i;

clrscr();


parts : "); scanf("%d" , &n);printf("\nEnter lower and upper

limits : "); scanf("%f %f" , &x[0] ,

&x[n]);

y[0] = fun(x[0]);

h = (x[n] - x[0])/n ;

printf("\nx y");

printf("\n%8.5f %8.5f" , x[0] ,y[0]);

for(i=1 ; i < n ; i++)

{ x[i] = x[0] + i * h ;

y[i] = fun(x[i]);

printf("\n %8.5f %8.5f " , x[i] , y[i]);

if(i % 3 == 0)

k1 = k1 + 2 * y[i];

elsek2 = k2 + 3 * y[i];

}

y[n] = fun(x[n]);

printf("\n %8.5f %8.5f " , x[n] , y[n]);

y[0] = ((3 *h) / 8.0 ) * (y[0] + y[n] + k1

+ k2 ); printf("\nresult =%f \n" , y[0]);getch();

}

float fun(float x)

{ float g;

g = sin(x) - log(x) +

exp(x); return g;

}


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OUTPUT :

Enter number of part parts : 6Enter lower and upper limits : 0.2 1.4

x y

0.20000 3.029510.40000 2.797530.60000 2.897590.80000 3.166041.00000 3.559751.20000 4.069831.40000 4.70418Result = 4.052991


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32.Booles Rule for Numerical Integration.

/* BOOLS RULE */

#include

#include

#include

void

main()

{ float fun(float);

float h , k1=0.0 , k2=0.0 , k3=0.0 ,

k4=0.0; float x[20] , y[20];int n , i;

clrscr();printf("\nEnter number of

parts"); scanf("%d" , &n);

printf("\nEnter lower and upper

limits :"); scanf("%f %f" , &x[0] ,

&x[n]);y[0] = fun(x[0]);

h = (x[n] - x[0]) / n;

printf("\nx y");

printf("\n%8.5f %8.5f" , x[0] ,y[0]);for(i=1 ; i < n ; i++)

{ x[i] = x[0] + i * h ;

y[i] = fun(x[i]);

printf("\n %8.5f %8.5f " , x[i] , y[i]);

if(i % 2 == 0)

k1 = k1 + 12 * y[i];

else

k1 = k1 + 32 * y[i];

}y[n] = fun(x[n]);

printf("\n %8.5f %8.5f " , x[n] , y[n]);

y[0] = ((2 * h)/45) * (7 * y[0] + 7 * y[n] + k1 + k2 + k3

+ k4); printf("\nresult =%f \n" , y[0]);

getch();

}

float fun(float x)

{ float g;g = log(x);

return g;

}

OUTPUT :Enter number of parts : 6Enter lower and upper limits : 4 5.2

x y4.00000 1.38629


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OUTPUT :Enter number of parts : 6

Enter lower and upper limits : 0.2 1.4x y0.20000 3.029510.40000 2.797530.60000 2.897590.80000 3.166041.00000 3.55975

1.20000 4.069831.40000 4.70418Result = 4.051446


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34. Write a program for Gauss Elimination Method in C.

/* GAUSS ELIMINATION METHOD */

#include

#include

#include

#define n

3void main()

{ float temp , s , matrix[n][n+1] , x[n];

int i , j , k;

clrscr();

printf("\nEnter the elements of the augment matrix row wise

:\n");for(i = 0 ; i < n ; i++)

for(j=0 ; j


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OUTPUT:

Enter the elements of the augment matrix row wise :3 1 -1 32 -8 1 -51 -2 9 8

Matrix :3.000000 1.000000 -1.000000 3.0000002.000000 -8.000000 1.000000 -5.0000001.000000 -2.000000 9.000000 8.000000

Solution is :

x[1]= 1.0000x[2]= 1.0000x[3]= 1.0000


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35. Write a program for Gauss Jordon Method in C.

/* GAUSS JORDAN METHOD */

#include

#include

#include

#define n

3

void main()

{ float temp , matrix[n][n+1];

int i , j , k;

clrscr();

printf("\nEnter the elements of the augment matrix row wise :-

\n");

for(i = 0 ; i < n ; i++)

for(j=0 ; j


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1 -2 9 8The digonal matrix is :--3.000000 0.000000

0.000000 -8.6666670.000000

0.000000 Solution is :-x[1]= 1.0000x[2]= 1.0000x[3]= 1.0000

0.000000 3.0000000.000000 -8.6666678.884615 8.884615


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36. Write a program for Gauss Seidal Method in C.

/* GAUSS SEIDAL METHOD */

#include

#include#include

#include

void main()

{

float a[10][10],x[10],aerr, maxerr, t,

s, err; int i,j,itr,maxitr,n;

printf ("\n Enter the number of

unknowns="); scanf ("%d",&n);

for(i=1;i

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