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Program no.1Write a program in C to find sum and average of squareoffirst 100 natural numbers./* PROGRAM TO CALCULATE SUM AND AVERAGE OF 100NATURAL NUMBERS*/#include#include#includevoid main( ){ int i, n=100,sum=0, p;

float avg;for (i=1;i

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program NO.2/*PROGRAM TO FIND PRODUCT OF 1ST 'N' NATURAL

NUMBERS*/#include#include#includevoid 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(i

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Program no. 3

Write a program in C to calculate simple and compoundinterest./* PROGRAM TO CALCULATE SIMPLE AND COMPOUNDINTEREST */#include#include#includevoid main(){

float p,r,t,si,ci,a,v;clrscr();printf ("\n Principal amount p=");scanf ("%f",&p);printf("\n rate of interest r= ");scanf("%f",&r);printf("\n Time period t=");scanf("%f",&t);si=(p*r*t)/100;v=1+(r/100);a=p*pow(v,t);ci=a-p;printf("\n Simple Interest = %f",si);printf("\n Compound Interest = %f",ci);getch();

}

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OUTPUT:Principal amount p =5000Rate of interest r =10Time period t =12Simple Interest = 6000.000000Compound Interest = 10692.14188

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Program no 4/* FIBBONACCI SERIES */

#include#include#includevoid 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*/


int i,j,count=0;clrscr();printf("\nprime nos. between 1 and 100\n");for(i=1;i

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prime nos. between 1 and 100

2 3 5 7 1113 17 19 23 2931 37 41 43 4753 59 61 67 7173 79 83 89 97

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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 EXCEPTTHOSE DIVISIBLE BY 5*/#include#includevoid main(){

clrscr();int n1=1,n2=100,j,i,temp,k=0,flag=0;for (i=n1;i

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}if (flag==1)

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

}}OUTPUT:

2 3 7 11 13 17 19 23 2931 37 41 43 47 53 59 61 6771 73 79 83 89 97

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Program no. 7

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


int i,j,t;int n[5]; clrscr();//to enter the dataprintf("\n enter the 5 numbers to be sorted\n");for(i=0;i

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enter the 5 numbers to be sorted231421189output:914182123

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Program no 8

/*CALCULATION OF THE VALUE OF nCr*/#include#include#includevoid main(){

int n,r,y;clrscr();printf("\nCALCULATION OF nCr");printf("\nenter the value of n ,r");scanf("%d%d",&n,&r);y=fact(n)/(fact(r)*fact(n-r));printf("\n%dC%d = %d",n,r,y);getch();

}int fact (int a)

{int f=1,j;if((a==0)&&(a==1))return(1);

else{

for(j=1;j

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8C3 = 56

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Program no. 9/*SUM OF DIGITS OF A GIVEN NUMBER*/


int num,p,sum=0;clrscr();printf("\nenter any number == ");scanf("%d",&num);while(num!=0)

{p=num%10;

sum=sum+p;num=num/10;

}printf("\nsum of digit of given number == ");printf("%d",sum);getch();}

OUTPUT:

enter any number == 2319sum of digit given number == 15

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program no. 10

/* CALCULATION OF SINE SERIES */#include#include#includeint 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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}

OUTPUT:

enter the value for x:1sum of sine series sin(1.000000)==0.841471

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Program no 11/*CALCULATION OF COSINE SERIES*/

#include#include#includelong 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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sum of cosine series cos(2.000000)==-1.350759

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Program no.12Write a program in C to find out the average marks of

students and print the marks of the topper.

/* PROGRAM TO FIND OUT AVERAGE MARKS OF STUDENTS */#include#include#includevoid 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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{max=avg[i];roll=i;

}}

printf("\n The topper of the class is student %d with average=%f",roll,max);

getch();}

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OUTPUT:Enter the number of students =3Enter the number of subjects =5Enter marks for 1 studentSubject 1=69Subject 2=58Subject 3=45Subject 4=10Subject 5=35Enter marks for 2 studentSubject 1=47Subject 2=25Subject 3=16Subject 4=97Subject 5=46Enter marks for 3 studentSubject 1=30Subject 2=90Subject 3=76Subject 4=58Subject 5=47The average of the 1 student = 42.400002The average of the 2 student = 45.200001The average of the 3 student = 59.200001The topper of the class is student 3 with average = 59.200001

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Program no.13Write 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 ANDFIND THE SUM OF ITS DIGITS */#include#includevoid main(){int n1,n2=0,rem,sum=0;printf ("\n Enter the number to be reversed=");scanf ("%d",&n1);while (n1>0)

{rrem=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 =4567The reversed number is 7654The sum of digits of reversed number is 22

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Programme no. 14Write a program in C to find the sum, mean standard deviation

andvariance of any numbers.

/*......STANDARD DEVIATION AND VARIATION ........*/#include#include#includevoid main(){

clrscr();int n;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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Enter 5 values498612Arithmetic MEAN, STD. DEVIATION and VARIANCE =7.8000007.360000 2.712932

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Programme no. 15Write a program in C to convert binary number to decimal number./* CONVERT BINARY NUMBER TO DECIMAL NUMBER */

#include#include#includevoid main(){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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Programme no.16

Write a program in C to convert decimal number to binary number./* CONVERT DECIMAL NUMBER TO BINARY NUMBER */

#include#includevoid main(){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= 123The binary number is 1111011

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program no.17Write a program in C to add two rectangular matrices./* ADDTION OF MATRIX */#include#includevoid 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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Program no. 18

/*MULTIPLICATION OF MATRIX*/#include#includevoid 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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}31

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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 2enter matrix a0 1 21 2 32 3 4enter matrix b1 -2

-1 02 -1

the product ofmatrix is:3 -25 -57 -8

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program no.19/*BISECTION METHOD*/

#include#include#includevoid 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;}

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

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

if((p*r)>0)x1=x3;

elsex2=x3;

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

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/*output:equation:x*(exp(x)-1)enter the app value of x1,x2:0.5 1the root lies b/w 0.500000 & 1.000000n 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 0.1676543 0.500000 0.625000 0.562500 -0.175639 0.167654-0.0127824 0.562500 0.625000 0.593750 -0.012782 0.167654 0.0751425 0.562500 0.593750 0.578125 -0.012782 0.075142 0.0306196 0.562500 0.578125 0.570312 -0.012782 0.030619 0.0087807 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 0.000662-0.000687root of the equ is 0.566895

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programme no 20/*NEWTON RALPHSON*/

#include#include#includevoid 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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/*output:The equation is X^3+1.2X^2-5X-7.2enter the initial value of x0:2i x0 x1 f(x0) df(x0)2 2.000000 2.372881 -4.400000 11.8000003 2.372881 2.313010 1.052938 17.5866154 2.313010 2.311227 0.029603 16.6012655 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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Programme no.21/*REGULA-FALSE METHOD */

#include#include#includefloat 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(x2)=0.0001){c++;if(f(x1)*f(m)

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OUTPUT:Enter the first approximation : 0Enter the second approximation : 1The 1,ilteration is 0.605959The 2,ilteration is 0.607057The 3,ilteration is 0.607100The answer is repeated at 3 ilteration is 0.607100

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Program no.22

/*NEWTON GREGORY FORWARD INTERPOLATION*/


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:5enter values x y:3 135 2311 89927 1731534 35606enter value of x at which y is to be calculated:7when x=7.0000,y=899.00000000programme no.24

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Program no.23/*NEWTON GREGORY BACKWARD INTERPOLATION*/


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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printf("\nwhen x=%6.4f,y=%6.8f",x,sum);getch();}OUTPUT:enter no. of terms:5enter values x y:20 4140 10360 16880 218100 235enter value of x at which y is to be calculated:70when x=70.0000,y=196.00000000

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program no.24/*LAGRANGE METHOD OF INTERPOLATION*/

#include#include#include#define max 50void 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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7 29410 90011 121013 2028Enter the value of x at which f(x)is required:8when x= 8.00 then y=445.62

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Programme no.25Write a program in C/C++ which can calculate the value of a function ata 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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265457709243242110The value of y for x = 9 is: 810

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Program no.26

/******BESSEL'S METHOD OFINTERPOLATION******/#include#include#includevoid 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++){ pr intf("\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++){ pr intf("\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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printf("\nWhen x = %6.4f , y = %6.8f" , x , y);getch();

}

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OUTPUT :-Enter number of item : 4Enter 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 32Enter the value in the form of yEnter the value of y1 24Enter the value of y2 32Enter the value of y3 35Enter the value of y4 40Enter the value of x for which you want the value of y :-

25When x = 25.0000 , y = 32.945313

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Programe no 27/******STIRLING METHOD OF

INTERPOLATION******/#include#include#includevoid 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]); }

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printf("\nEnter the value of x for which you want thevalue 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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Enter the value of x1 20Enter the value of x2 24Enter the value of x3 28Enter the value of x4 32Enter the value in the form of yEnter the value of y1 24Enter the value of y2 32Enter the value of y3 35Enter the value of y4 40Enter the value of x for which you want the value of y :-

25When x = 25.0000 , y = 33.31251144

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Program no.28/* TRAPEZOIDAL RULE */


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){

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float g;g = log(x);return g;

}

OUTPUT :-Enter number of parts : 6lower and upper limits : 4 5.2x y4.00000 1.386294.24000 1.444564.48000 1.499624.72000 1.551814.96000 1.601415.20000 1.64866result = 1.827570

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program no.29

/* SIMPSION 1/3 RULE */#include#include#includevoid main(){ float fun(float);

float h , k1=0.0 , k2=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 + 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);

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return g;}

OUTPUT :-Enter number of parts : 6Enter lower and upper limits :0.2 1.4x y0.20000 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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program no.30

/*SIMPSION 3/8 RULE */#include#include#includevoid main(){ float fun(float);

float h , k1=0.0 , k2=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 % 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;

g = sin(x) - log(x) + exp(x);return g;

}

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program no.31

/* BOOLS RULE */#include#include#includevoid 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);

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return g;}OUTPUT :-

Enter number of parts : 6Enter lower and upper limits : 4 5.2x y4.00000 1.386294.20000 1.435084.40000 1.481604.60000 1.526064.80000 1.568625.00000 1.609445.20000 1.64866result = 1.814274

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program no.32/* WEEDEL'S RULE */


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 % 6 == 0)k1 = k1 + 2 * y[i];else if(i % 3 == 0)k1 = k1 + 6 * y[i];else if(i % 2 == 0)k1 = k1 + y[i];elsek4 = 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);

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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 : 6

Enter lower and upper limits : 0.2 1.4x y0.20000 3.02951

0.40000 2.797530.60000 2.897590.80000 3.166041.00000 3.559751.20000 4.069831.40000 4.70418result = 4.051446

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program no.33/* 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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{s = 0;for(j = i + 1 ; j < n ; j++)s += matrix[i][j] * x[j];x[i] = (matrix[i][n] - s) / matrix[i][i];}

/ /now printing the resultprintf("\nSolution is :-\n");for(i = 0 ; i < n ; i++)printf("\nx[%d]=%7.4f" , i+1 ,x[i]);

getch();}

OUTPUT :-Enter the elements of the augment matrix row

wise :3 1 -1 32 -8 1 -51 -2 9 8matrix:-3.000000 1.000000 -1.000000 3.0000002.000000 -8.000000 1.000000

-5.0000001.000000 -2.000000 9.000000 8.000000Solution is :-x[1]= 1.0000x[2]= 1.0000

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x[3]= 1.0000

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program no 34/* GAUSS JORDAN METHOD */

#include#include#include#define n 3void 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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printf("\nx[%d]=%7.4f" , i+1 ,matrix[i][n]/matrix[i][i]);getch();

}OUTPUT :-

Enter the elements of the augment matrix row wise :-3 1 -1 32 -8 1 -51 -2 9 8The digonal matrix is :--3.000000 0.000000 0.000000 3.0000000.000000 -8.666667 0.000000 -8.6666670.000000 0.000000 8.884615 8.884615Solution is :-x[1]= 1.0000x[2]= 1.0000x[3]= 1.0000

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Programme no.35/* GAUSS SEIDAL METHOD */

#include#include#include#includevoid main(){

float a[10][10],x[10],aerr, maxerr, t, s, err;int i,j,itr,maxitr,n;clrscr();printf ("\n Enter the number of unknowns=");scanf ("%d",&n);for(i=1;i

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x[i]=t;}

printf ("%d",itr);for (i=1;i

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Program no.36/* CURVE FITTING - STRAIGHT LINE */

# include# include# includevoid 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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sum4+=x2[i];}

a=(sum2*sum4-sum3*sum1)/(ob*sum4-sum1*sum1);b=(sum2-ob*a)/sum1;printf("\n\n Equation of the STRAIGHT LINE of the formy=a+b*x is );printf("\n\n\t\t\t y=[%f] + [%f]x.a,b);getch();

}

OUTPUT:Enter the no. of observations : 5Enter the values of x :Enter the value of x1 : 1Enter the value of x2 : 2Enter the value of x3 : 3Enter the value of x4 : 4Enter the value of x5 : 5Enter the values of y :Enter the value of y1: 14Enter the value of y2: 27Enter the value of y3: 40Enter the value of y4: 55Enter the value of y5: 68

Equation of the STRAIGHT LINE of the form y=a+b*x is :y=0 + 13.6 x

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Program no.37/* RUNGA - KUTTA METHOD */


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 ;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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/* Function sub program */float f(float x , float y){

float g;g = (x*x*x + y*y) / 2 ;return g ;

}

OUTPUT :-dy/dx = (x^3 +y^2)/10given y0 = 1 && x0 = 0.1 && h = 0.2 (in the range x0

< x < 2.1)The solution of differential equation is when x =

0.100000 y = 1.053870The solution of differential equation is when x =

0.300000 y = 1.116833The solution of differential equation is when x =0.500000 y = 1.194833

The solution of differential equation is when x =0.700000 y = 1.297048



The solution of differential equation is when x =1.300000 y = 1.914011The solution of differential equation is when x =



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2.100000 y = 5.495997

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