Copy of 38 Programsmm

8/8/2019 Copy of 38 Programsmm

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Program no.1

Write a program in C to find sum and average of square offirst 100 natural numbers.

/* PROGRAM TO CALCULATE SUM AND AVERAGE OF 100 NATURALNUMBERS*/

#include

#include

#include

void main( )

{

int i, n=100,sum=0, p;

float avg;

for (i=1;i


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i=i+1;

prod=prod*i;

if(i


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OUTPUT:

Principal amount p =5000

Rate of interest r =10

Time period t =12

Simple Interest = 6000.000000

Compound Interest = 10692.14188

Program no 4

/* FIBBONACCI SERIES */

#include#include

#include

void main()

{

clrscr();

int i;

double x1,x2,x;

x1=1;

x2=1;

printf("%12.0f\t%12.0f",x1,x2);

for(i=3;i


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Program no.5

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

#include

#include

#include

void main()

{

int i,j,count=0;

clrscr();

printf("\nprime nos. between 1 and 100\n");

for(i=1;i


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31 37 41 43 47

53 59 61 67 71

73 79 83 89 97

Program no.6Write a program in C to generate prime nos. between 1 and

100 exceptthose divisible by 5.

/* GENERATE PRIME NOS. BETWEEN 1 AND 100 EXCEPT THOSEDIVISIBLE BY 5*/

#include

#include

void main()

{

clrscr();

int n1=1,n2=100,j,i,temp,k=0,flag=0;

for (i=n1;i


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flag=0;

break;

}

}

if (flag==1)

{

printf ("\t %d",temp);

}

}

}

OUTPUT:

2 3 7 11 13 17 19 23 29

31 37 41 43 47 53 59 61 67

71 73 79 83 89 97

Program no. 7

/* to sort a list of 5 numbers in ascending order*/

#includeD#include

#includevoid main()

{

int i,j,t;

int n[5]; clrscr();

//to enter the data

printf("\n enter the 5 numbers to be sorted\n");

for(i=0;i


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printf("output:");

for(i=0;i


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else

{

for(j=1;j


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enter any number == 2319

sum of digit given number == 15

Program no. 10/*CALCULATION OF TAXABLE INCOME*/

#include

#include

#include

void main()

{

float income,tax;clrscr();

printf("\nenter taxable income:");

scanf("%f",&income);

if(income


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amount of tax on Rs132000.000000 equals to

Rs32600.000program no. 11

/* CALCULATION OF SINE SERIES */

#include

#include

#include

int fact (int a);

void main()

{

double x,sum1=0.0,sum2=0.0,sum=0.0;

int y,i;

clrscr();

printf("\nenter the value for x:");

scanf("%lf",&x);

for(i=1;i


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enter the value for x:1

sum of sine series sin(1.000000)==0.841471

Program no 12

/*CALCULATION OF COSINE SERIES*/

#include

#include

#include

long fact (int a);

void main()

{

double x,sum1=0.0,sum2=0.0,sum=0.0;

int y,i;

clrscr();printf("\nenter the value for x:");

scanf("%lf",&x);

for(i=0;i


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Program no.13

Write a program in C to find out the average marks ofstudents and print the marks of the topper.

/* PROGRAM TO FIND OUT AVERAGE MARKS OF STUDENTS */

#include

#include

#include

void main()

{

int stu[20][20],i,j,m,n,roll=0;

float avg[20],sum=0.0,max=0.0;

clrscr();

printf("\n Enter the number of students =");

scanf("%d",&n);printf ("\n Enter the number of subjects =");

scanf ("%d",&m);

for(i=1;i


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average =%f",roll,max);

getch();

}

OUTPUT:

Enter the number of students =3

Enter the number of subjects =5

Enter marks for 1 student

Subject 1=69

Subject 2=58

Subject 3=45

Subject 4=10

Subject 5=35

Enter marks for 2 student

Subject 1=47

Subject 2=25

Subject 3=16

Subject 4=97

Subject 5=46

Enter marks for 3 studentSubject 1=30

Subject 2=90

Subject 3=76

Subject 4=58

Subject 5=47

The average of the 1 student = 42.400002



The topper of the class is student 3 with average =

59.200001

Program no.14

Write a program in C to reverse the digits of a number andfind the sum of its digits.

/* PROGRAM TO REVERSE THE DIGITS OF A NUMBER AND FIND THE


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SUM OF ITS DIGITS */

#include

#include

void main()

{

int n1,n2=0,rem,sum=0;

printf ("\n Enter the number to be reversed=");

scanf ("%d",&n1);

while (n1>0)

{r

rem=n1%10;

n1=n1/10;

n2=n2*10+rem;

}

printf ("\n The reversed number is %d",n2);

while (n2>0)

{rem=n2%10;

n2=n2/10;

sum=sum+rem;

}

printf ("\n The sum of digits of reversed number is

%d",sum);

}

OUTPUT:

Enter the number to be reversed =4567

The reversed number is 7654

The sum of digits of reversed number is 22

Programme no. 15Write a program in C to find the sum, mean standard deviation

andvariance of any numbers.

/*......STANDARD DEVIATION AND VARIATION ........*/

#include

#include

#include

void main()

{

clrscr();

int n;


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float x[20],sum;

float am,var,sd;

int i;

printf("\n enter the no. of terms=");

scanf("%d",&n);

printf("\n enter %d values \n",n);

for(i=0;i


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{

int num[10];

int i,x=2,j,y=0,n;

printf ("\n Enter the number of elements of binary

number=");

scanf ("%d",&n);

j=n-1;

for (i=0;i=0)

{

for (i=0;i


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int i=0,j,n;

int rem[10];

printf ("\n Enter the number=");

scanf ("%d",&n);

do

{

rem[i]=n%2;

i=i+1;

n=n/2;

}

while (n>0);

for (j=i-1;j>=0;j--)

{

printf ("The binary number is %d",rem[j]);

}

}

OUTPUT:

Enter the number= 123

The binary number is 1111011

program no.18Write 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 (j=0;j


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6 8

10 12

Program no. 19

/*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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}

output:

enter no. of row and col of matrix a:3 3

enter no. of row and col of matrix b:3 2

enter matrix a

0 1 2

1 2 3

2 3 4

enter matrix b

1 -2

-1 0

2 -1

the product ofmatrix is:

3 -25 -5

7 -8

program no.20

/*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(x1)

f(x2) f(x3)");


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l15: x3=(x1+x2)/2;

p=fun(x1);

q=fun(x2);

r=fun(x3);

i=i++;

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

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

if((p*r)>0)

x1=x3;

else

x2=x3;

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


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the root lies b/w 0.500000 & 1.000000

n x1 x2 x3 f(x1) f(x2)

f(x3)

1 0.500000 1.000000 0.750000 -0.175639

1.718282 0.587750

2 0.500000 0.750000 0.625000 -0.175639

0.587750 0.167654

3 0.500000 0.625000 0.562500 -0.175639

0.167654 -0.012782

4 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 -0.002035

8 0.566406 0.570312 0.568359 -0.0020350.008780 0.003364

9 0.566406 0.568359 0.567383 -0.002035

0.003364 0.000662

10 0.566406 0.567383 0.566895 -0.002035

0.000662 -0.000687

root of the equ is 0.566895

programme no 21

/*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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a10: else p=f(x0);q=df(x0);

x1=x0-(p/q);

i++;

printf("\n %d\t%f\t%f\t%f\t%f\n",i,x0,x1,p,q);

if(fabs((x1-x0)/x1)


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Programme no.22 /*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();

b printf("\n enter the first approximation :");

scanf("%f",&x1);printf("\n enter the second approximation :");

scanf("%f",&x2);

if(f(x1)*f(x2)=0.0001)

{

c++;

if(f(x1)*f(m)


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The 1,ilteration is 0.605959



The answer is repeated at 3 ilteration is 0.607100

Program no.23

/*NEWTON GREGORY FORWARD INTERPOLATION*/

#include

#include

#includevoid 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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y=my[i]+y1+y2+y3+y4;

printf("\nwhen x=%6.4f,y=%6.8f",x,y);

getch();

}

OUTPUT:

enter no. of terms:5

enter values x y:3 13

5 23

11 899

27 17315

34 35606

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

when x=7.0000,y=899.00000000

Program no.24

/*NEWTON GREGORY BACKWARD INTERPOLATION*/#include

#include


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#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 values x\ty:\n");

for(i=0;i


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20 41

40 103

60 168

80 218

100 235

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

when x=70.0000,y=196.00000000

program no.25

/*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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Enter No. of Points:6

Enter the given set of values:

x y

4 18

5 100

7 294

10 900

11 1210

13 2028

Enter the value of x at which f(x)is required:8

when x= 8.00 then y=445.62

Programme no.26Write 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

#includevoid main()

{

float ax[20], ay[20], diff[30],temp=1;

int n,j,m,z=0,A=0,k=0;

clrscr();

coutn;

for (int i=0;i


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// creating difference table

for (i=0;i1)

{

m=n-1;

A=0;

for(j=0;j


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OUTPUT:

Enter the number of points=5

Enter (x1 ,y1 ) 5 150

Enter (x2 ,y2 ) 7 392

Enter (x3 ,y3 ) 11 1452

Enter (x4 ,y4 ) 13 2366

Enter (x5 ,y5 ) 17 5202

Enter the value of x=9

The difference table is as follows:

121

265

457

709

24

3242

1

1

0

The value of y for x = 9 is: 810

Program no.27

/******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);


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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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Programe no 28

/******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;


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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 the number of item : 4

Enter the value in the form of x

Enter the value of x1 20




Enter the value in the form of y

Enter the value of y1 24




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

:- 25

When x = 25.0000 , y = 33.31251144

Program no.29

/* 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 ;


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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;

}

OUTPUT :-Enter number of parts : 6

lower and upper limits : 4 5.2

x y

4.00000 1.38629

4.24000 1.44456

4.48000 1.499624.72000 1.55181

4.96000 1.60141

5.20000 1.64866

result = 1.827570

program no.30

/* 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;


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clrscr();


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;

}


Enter lower and upper limits :0.2 1.4

x y

0.20000 3.02951

0.40000 2.79753

0.60000 2.89759

0.80000 3.16604

1.00000 3.559751.20000 4.06983

1.40000 4.70418

result = 4.052133program no.31

/*SIMPSION 3/8 RULE */#include


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#include

#include

void main()

{ float fun(float);

float h , k1=0.0 , k2=0.0 ;

float x[20] , y[20];

int n , i;

clrscr();


scanf("%d" , &n);


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];

else

k2 = 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;


return g;

}

OUTPUT : -

Enter number of part parts : 6

Enter lower and upper limits : 0.2 1.4

x y

0.20000 3.02951

0.40000 2.79753

0.60000 2.89759


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0.80000 3.16604

1.00000 3.55975

1.20000 4.06983

1.40000 4.70418

result = 4.05299

program no.32

/* 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

k2 = k2 + 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;


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}


Enter lower and upper limits : 4 5.2

x y

4.00000 1.38629

4.20000 1.43508

4.40000 1.48160

4.60000 1.52606

4.80000 1.56862

5.00000 1.60944

5.20000 1.64866

result = 1.814274

program no.33

/* WEEDEL'S 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);


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 % 6 == 0)

k1 = k1 + 2 * y[i];

else if(i % 3 == 0)

k1 = k1 + 6 * y[i];

else if(i % 2 == 0)


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k1 = k1 + y[i];

else

k4 = k4 + 5 * y[i];

}

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

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

y[0] = ((3 * h)/10) * (y[0] + y[n] + k1 + k2 + k3 + k4);

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

getch();

}

float fun(float x)

{ float g;


return g;

}


Enter lower and upper limits : 0.2 1.4x y

0.20000 3.02951

0.40000 2.79753

0.60000 2.89759

0.80000 3.16604

1.00000 3.55975

1.20000 4.06983

1.40000 4.70418

result = 4.051446

program no.34

/* GAUSS ELIMINATION METHOD */#include

#include

#include

#define n 3

void main()

{


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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 3

2 -8 1 -5

1 -2 9 8

matrix:-

3.000000 1.000000 -1.000000

3.000000

2.000000 -8.000000 1.000000

-5.000000

1.000000 -2.000000 9.000000

8.000000

Solution is :-

x[1]= 1.0000

x[2]= 1.0000

x[3]= 1.0000

program no 35/* 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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{

temp = matrix[i][j] / matrix[j][j];

for(k = 0 ; k


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printf ("\n Enter the number of unknowns=");

scanf ("%d",&n);

for(i=1;i


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Enter allowed error, max iteration=0.001 ,4

Iteration x[1]= 0.8500 -1.0275

x[2] = 1.01091 1.0025

x[3] = -0.9998 0.99981

1.0000 - 1.0000 1.0000

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

Program no.37

/* CURVE FITTING - STRAIGHT LINE */

# include

# include

# include

void main()

{

int i=0,ob;

float

x[10],y[10],xy[10],x2[10],sum1=0,sum2=0,sum3=0,sum4=0;clrscr();

double a,b;

printf("\n Enter the no. of observations :");

scanf(%d,&ob);

printf("\n Enter the values of x :\n");

for (i=0;i


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b=(sum2-ob*a)/sum1;

printf("\n\n Equation of the STRAIGHT LINE of the form

y=a+b*x is );

printf("\n\n\t\t\t y=[%f] + [%f]x.a,b);

getch();

}

OUTPUT:Enter the no. of observations : 5

Enter the values of x :

Enter the value of x1 : 1Enter the value of x2 : 2

Enter the value of x3 : 3



Enter the values of y :

Enter the value of y1: 14





Equation of the STRAIGHT LINE of the form y=a+b*x is :

y=0 + 13.6 x

Program no.38

/* RUNGA - KUTTA METHOD */

#include

#include

#include

void main()

{

float f(float x , float y);

float x0 = 0.1 , y0 = 1 , xn = 2.1 , h =0.2 , k1 , k2 ,

k3 , k4 ;

int i , n ;


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clrscr();

printf("\ndy/dx = (x^3 +y^2)/10 ");

printf("\ngiven y0=1 && x0= 0.1 && h=0.2 (in the range x0

< x < 2.1)\n");

n = (xn - x0) / h;

for(i = 0 ; i


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0.500000 y = 1.194833

The solution of differential equation is when x =

0.700000 y = 1.297048


0.900000 y = 1.436797


1.100000 y = 1.633150


1.300000 y = 1.914011


1.500000 y = 2.322503


1.700000 y = 2.931453


1.900000 y = 3.880822


2.100000 y = 5.495997

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