Numeric :Gauss Seidel

 

Gauss Seidel

 

#include<stdio.h>

int main()

{

    double a[50],b[50],c[50],X1,X2,X3,eax1,eax2,eax3,p,q,r;

    int i,n=4;

   // freopen("input.txt","r",stdin);

//printf("Enter length :");

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

//............................

printf("Enter 1st equation :");

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

{

scanf("%lf",&a[i]);

}

//..........................

printf("Enter 2nd equation :");

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

{

scanf("%lf",&b[i]);

}

//...........................

printf("Enter 3rd equation :");

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

{

scanf("%lf",&c[i]);

}

printf("ENter the value of X1 :");

scanf("%lf",&X1);

printf("ENter the value of X2 :");

scanf("%lf",&X2);

printf("ENter the value of X3 :");

scanf("%lf",&X3);

//...........................

X1=p;

X2=q;

X3=r;

X1=((a[3]-(a[2]*X2)-(a[1]*X3))/a[0]);

printf("X1 is %lf\n",X1);

X2=((b[3]-(b[2]*X3)-(b[0]*X1))/b[1]);

printf("X2 is %lf\n",X2);

X3=((c[3]-(c[1]*X2)-(c[0]*X1))/c[2]);

printf("X2 is %lf\n",X3);

//...........................

eax1=(((X1-p)/X1)*100);

eax2=(((X2-q)/X2)*100);

eax3=(((X3-r)/X3)*100);

printf("eax1=%.1lf\n",eax1);

printf("eax2=%.1lf\n",eax2);

printf("eax3=%.1lf\n",eax3);

return 0;

}

Numerical :Curve Fitting

 

Numerical :Curve Fitting

sample input

 

T	700	720	740	760
V	0.0977	0.12184	0.1406	0.15509

 

#include<stdio.h>

int main()

{

    float y[50],x[50],sumx=0,X,xx,sumxx=0,sumy=0,sumxy=0,xsum,ysum,a1,a0;

    int n;

printf("Enter the Range :");

scanf("%d",&n);

//....................................

printf("ENTER x =\n");

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

{

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

    sumx=sumx+x[i];

    sumxx=sumxx+(x[i])*(x[i]);

}

xsum=(sumx/n);

//....................................

printf("ENTER y =\n");

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

{

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

    sumy=sumy+y[i];

}

ysum=(sumy/n);

//.....................................

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

{

    sumxy=sumxy+x[i]*y[i];

}

//.....................................

a1=((n*sumxy)-(sumx*sumy))/((n*sumxx)-((sumx)*(sumx)));

printf("a1=%f",a1);

a0=((ysum)-(a1*xsum));

printf("\n\na0=%f",a0);

printf("\n NOw ENTER the value of X :");

printf("\nY=a0+a1X=");

scanf("%f",&X);

xx=a0+a1*X;

printf("\nANS= %f",xx);

return 0;

}

NUmeric :LINEAR INTERPOLATION–QUADRATIC -LAGRANCE

NUmeric :LINEAR INTERPOLATION–QUADRATIC -LAGRANCE

sample input

T	700	720	740	760
V	0.0977	0.12184	0.1406	0.15509

 

//LINEAR INTERPOLATION

//QUADRATIC

//LAGRANCE

#include<stdio.h>

int main()

{

    double y[50],x[50],X,f1,f11,b2,f2,sum=0,G;

    int n,op;

printf("Enter the Range :");

scanf("%d",&n);

//....................................

printf("ENTER x =\n");

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

{

    scanf("%lf",&x[i]);

}

//....................................

printf("ENTER y =\n");

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

{

    scanf("%lf",&y[i]);

    //((y[2]-y[1])/(x[2]-x[1]));

}

printf("ENTER the value of X =\n");

scanf("%lf",&X);

//.....................................

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

    {    G=1;

        for(int j=0;j<n;j++)

        {

            if(i==j)

            {

                continue;

            }

            G=G*((X-x[j])/(x[i]-x[j]));

        }

        sum=sum+G*y[i];

    }

//.....................................

printf("\n\nEnter Your Option\n__________________________\n");

printf("1.lINEAR INTERPOLATION\n\n2.QUADRATIC\n\n3.LAGRANCE\n__________________________\n");

while(scanf("%d",&op)==1)

{

switch(op)

{

case 1:

f1=y[0]+(((y[1]-y[0])*(X-x[0]))/(x[1]-x[0]));

printf("LINEAR INTERPOLATION RESULT is = %lf\n\n",f1);

    break;

case 2:

    f11=y[0]+(((y[1]-y[0])*(X-x[0]))/(x[1]-x[0]));

    b2=((((y[2]-y[1])/(x[2]-x[1]))-((y[1]-y[0])/(x[1]-x[0])))/(x[2]-x[0]));

    f2=(f11+(b2*((X-x[0])*(X-x[1]))));

    printf("\nQUADRATIC RESULT is = %lf\n\n",f2);

    break;

case 3:

    printf("Lagrance RESULT is = %lf \n",sum);

    break;

default:

    printf("ERORR");

    break;

}

}

//.....................................

return 0;

}

Numerical Methods for Engineers (5th and 6th Edition)-Steven C Chapra & Raymond P.Canale (PDF BOOK) with (SOlution)

Numerical Methods for Engineers (5th and 6th Edition)-Steven C Chapra & Raymond P.Canale (PDF BOOK) with (SOlution)

                                 Download link->Numerical Methods for Engineers (6th Edition)

                                 Download link->Numerical Methods for Engineers (5th Edition)

                                               SOLUTION LINK (Numerica-3-4-5-6-9-11-13)

NUmeric :Secant Method code

 

Sample equation for Secant Method

where X-1=3

and x0=4 ;

F(x)=2x³-11.7x²+17.7x¹-5

//scant method

#include<stdio.h>

#include<math.h>

#include<iostream>

using namespace std;

int main()

{

    float xi,Xii,newxi,Es,a[40],old=0,Ea;

    int n,K;

    cout<< "HIghest power in the equation : " ;

    cin>>n;

    //.........................................

    printf("Enter Xi =");

    scanf("%f",&xi);

    printf("Enter Xii =");

    scanf("%f",&Xii);

    printf("Enter Es =");

    scanf("%f",&Es);

    //..........................................

    printf("\n");

    printf("enter coefficient :\n");

    for (int i=n; i>=0; i--)

    {

        printf("X^%d =",i);

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

    }

    //...........................................

    while(1)

    {

        float fxi=0,fXii=0;

//

       for (int i=n; i>=0; i--)

        {

             fxi=fxi+a[i]*pow(xi,i);

             fXii=fXii+a[i]*pow(Xii,i);

        }

        newxi=xi-((fxi*(Xii-xi))/(fXii-fxi));

//   printf("\n\n\nSTEP no: %d\n",K+1);

        printf("\nXi   =  %.2f",xi);

        printf("\nfxi  =  %.2f",fxi);

//

        printf("\nXii   =  %.2f",Xii);

        printf("\nfXii  =  %.2f",fXii);

//

        printf("\nnew Xi   =  %.2f",newxi);

//

        Ea=((newxi-old)/newxi)*100;

        old =newxi;

        //cout <<" now =" <<old;

        if(Ea<0)

        {

            Ea=Ea*(-1);

        }

        printf("\nEa= %f\n",Ea);

        if(Ea<Es)

        {

            break;

        }

        Xii=xi;

        xi=newxi;

        K++;

    }

    return 0;

}