Numerical Methods File

7/31/2019 Numerical Methods File

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[Type the document titl

e]INDEX

KIIT COLLEGE OF ENGINEERING


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S.no

Topic Date Signature

1. Bisection Method

2. Newton Raphson Method

3. Gauss Elimination Method

4. Gauss Jordan Method

5. Gauss Seidal Method

6. Trapezoidal Method

7. Simpsons Rule

8. Runge Kutta Method

9. Eulars Method

10. Lagranges Interpolation

/* Program 1:-Bisection

AMIT KUMAR


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Method*/

#include#include#includeFigure 1float f(float x){return(x*x*x-4*x-9);}void bisect(float *x,float a,float b,int*itr){*x=(a+b)/2;++(*itr);printf("itration no. %3d X=%7.5f\n",*itr, *x);}main(){int itr=0,matrix;clrscr();float x,a,b,aree,x1;printf("enter the value of a,b","allowed error, maximum itration\n");scanf("%f%f%f%d", & a,& b,& aree,& matrix);bisect(&x,a,b,& itr);do{if (f(a)*f(x)


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return 0;}

x=x1;}while (itr


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/* Output*/

enter the value of a,b320.000120itration no. 1 X=2.50000after 1 iteration, root= %6.4

AMIT KUMAR


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/*Program 2:- NewtonRaphson Method*/

#include#include#include

float f(float x){return x*log10(x)-1.2;}

float df(float x)

{return log10(x)+.43429;}

main(){int itr,maxitr;float h,x0,x1,aerr;clrscr();printf("enter x0,allowed error, maximum iterations\n");scanf("%f %f %d",&x0,&aerr,&maxitr);

for (itr=1;itr


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/*output*/

Enter x0,allowed error, maximum iterations2.7 0.000001 18

Iteration no. 1, x= 2.740799Iteration no. 2, x= 2.740646Iteration no. 3, x= 2.740646After 3 iterations root = 2.7406

AMIT KUMAR


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/* Program 3:- GaussElimination Method*/

#include#include#include#define N 3main(){float a[N][N+1],t;int i,j,k;clrscr();printf("enter the elements of the augmented matrix rowise\n");for(i=0;i


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/*Output*/

Enter the elements of augmented matrix row wise10 -7 3 5 6-6 8 -1 -4 53 1 4 11 25 -9 -2 4 7The upper triangular matrix is:-10.000 -7.0000 3.0000 5.0000 6.00000.0000 3.8000 0.8000 -1.0000 8.6000-0.0000 -0.0000 2.4474 10.3158 -6.81580.0000 -0.0000 -0.0000 9.9247 9.9247

The solution is: -X [1] = 5.0000X [2] = 4.0000X [3] = -7.0000X [4] = 1.0000

AMIT KUMAR


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/*Program 4:- Gauss JordonMethod*/

#include#include#include#define N 3main(){float a[N][N+1],t;int i,j,k;clrscr();printf("enter the elements of the augmented matrix rowise\n");for(i=0;i


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printf("the solution is:-\n");for(i=0;i


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/*Output*/

Enter the elements of augmented matrix row wise-10 -7 3 5 6-6 8 -1 -4 53 1 4 11 25 -9 -2 4 7The diagonal matrix is:-10.0000 -0.0000 -0.0000 -0.0000 50.00000.0000 3.8000 -0.0000 0.0000 15.2000-0.0000 0.0000 2.4474 0.0000 -17.13160.0000 -0.0000 0.0000 9.9247 9.9247

The solution is: -X [1] = 5.0000X [2] = 4.0000X [3] = -7.0000X [4] = 1.0000

AMIT KUMAR


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/*Program 5:- Gauss-SeidalMethod*/

#include#include#include#define N 3

main(){float a[N][N+1],x[N],aerr,maxerr,t,s,err;int i,j,itr,maxitr;clrscr();

printf("ENTER the elements of the augmented matrix rowwise\n");for(i=0;i


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for(i=0;i


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/*Output*/

Enter the elements of augmented matrix row wise20 1 -2 173 20 -1 -182 -3 20 25Enter the allowed error, maximum iteration.0001 10Iteration X(1) X(2) X(3)1 0.8500 -1.02375 1.01092 1.0025 -0.9998 0.99983 1.0000 -1.0000 1.0000

4 1.0000 -1.0000 1.0000Converges in 4 iterationsX [1] = 1.0000X [2] = -1.0000X [3] = 1.0000

AMIT KUMAR


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/*Program 6:-TrapezoidalMethod*/


float y(float x){return 1/(1+x*x);}void main(){float x0, xn, h, s;int i,n;printf("enter the lower limit, upper limit, no. of intervals\n");scanf("%f%f%d",&x0,&xn,&n);h=(xn-x0)/n;s=y(x0)+y(xn);for(i=1; i


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/*output*/

enter the lower limit, upper limit, no. of intervals1 2 2the value of definite integrals is 0.32885enter the lower limit, upper limit, no. of intervals

1 3 2the value of definite integrals is 0.50000

enter the lower limit, upper limit, no. of intervals1 5 9the value of definite integrals is 0.59599

AMIT KUMAR


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/*Program 7:-Simpsonsmethod*/


float y(float x){

return 1/(1+x*x);}void main(){

float x0, xn, h, s;int i,n;clrscr();printf("enter the lower limit, upper limit, no. of intervals\n");scanf("%f%f%d",&x0,&xn,&n);h=(xn-x0)/n;s=y(x0)+y(xn)+y(x0+h);

for(i=3;i


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/*Output*/

enter the lower limit, upper limit, no. of intervals232the value of definite integrals is 0.07299

AMIT KUMAR


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/*Program 8:-Runge kuttaMethod*/

#include#include#includefloat f(float x,float y)

{return x+y*y;}main(){float x0,y0,x,x1,y1,h;clrscr();printf("Enter the values of x0,y0,x,h \n");scanf("%f%f%f%f",&x0,&y0,&x,&h);x1=x0;y1=y0;while(1){if(x==x1) break;

y1+=h*f(x1,y1);x1+=h;printf("whenx=%6.4f,y=%6.4f\n",x1,y1);getch();}}

AMIT KUMAR


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/*output*/

Enter the values of x0,y0,x,h0 1 3 .1whenx=0.1000,y=1.1000whenx=0.2000,y=1.2310whenx=0.3000,y=1.4025

whenx=0.4000,y=1.6292whenx=0.5000,y=1.9347whenx=0.6000,y=2.3590whenx=0.7000,y=2.9755whenx=0.8000,y=3.9308whenx=0.9000,y=5.5560whenx=1.0000,y=8.7329whenx=1.1000,y=16.4591whenx=1.2000,y=43.6594whenx=1.3000,y=234.3935whenx=1.4000,y=5728.5537

whenx=1.5000,y=3287361.5000whenx=1.6000,y=1080677892096.0000whenx=1.7000,y=116786471838643992000000.0000Floating point error: Overflow.

Abnormal program termination

AMIT KUMAR


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/*Program 9:-Euler Method*/


float f(float x, float y)

{return x+y*y;}main(){float x0,y0,x1,y1,xn,h;clrscr();printf ("Enter the values of x0,y0,xn,& h\n");scanf ("%f%f%f%f", &x0,&y0,&xn,&h);x1=x0; y1=y0;while (1)

{if (x1==xn) break;y1+=h*f(x1,y1);x1+=h;printf ("when x=%7.4f, y=%8.5f\n",x1,y1);getch();}}

AMIT KUMAR


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/*output*/

Enter the values of x0,y0,xn,& h1 2 3 4when x= 5.0000, y=22.00000when x= 9.0000, y=1978.00000

when x=13.0000, y=15651950.00000when x=17.0000, y=979934168219648.00000when x=21.0000, y=3841084041118427740000000000000.00000Floating point error: Overflow.

Abnormal program termination

AMIT KUMAR


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/*Program10:-LagrangeMethod*/


#define MAX 100main(){float ax[MAX+1],ay[MAX+1],nr,dr,x,y=0;int i,j,n;printf("enter the values of n\n");scanf("%d",&n);printf("enter the values\n");for(i=0;i


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/*Output*/

enter the values of n2enter the values4 5 64 5 8enter the values of x for which y is wanted2when x= 2.0000,y=-22.00000

AMIT KUMAR

Numerical Methods File

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