Numeric Method

7/28/2019 Numeric Method

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Assignment on Numeric MethodsPRASANNA KUMAR BARIK

Roll_No=-:970126

12/15/2009

SCA , KIIT UNIVERSITY


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Index:

Sl.no Programs Page no.1 Bisection method 3

2 Regula Falsi 4

3 Newton Raphson 5

4 Secant Method 6

5 Solve problem using all methods 7-12

6 Gauss Elimination 13

7 Gauss Jordan 14-158 gauss Seidel 16

9 Gauss Inverse 17-19

10 Lagrange Interpolating 20


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01.Solve the equation: x^3 - 5x + 1 using Bisection

method.

#include#include

#includevoid main(){int i;float xl,xr,xu,fxr;clrscr();

xl=2;xu=3;xr=(xl+xu)/2;printf("Bisection method\n");

printf("xl xu xr f(xr)\n");

printf("------------------------------------------------\n");for(i=0;i0)xl=xr;elsexu=xr;xr=(xl+xu)/2;}printf("The root is %f ",xr);getch();}

OUTPUT


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02.Solve the equation: x^3 - 5x + 1 using Regula Falsi

method.

#include

#include#includevoid main(){int a;float xl,xr,xu,fxr,fxl,fxu;clrscr();

xl=2;xu=3;printf("REGULA FALSI method\n");printf("interval xl xu xr

f(xr)\n");printf("_____________________________________________________\n\n"); for(a=0;a0)xl=xr;elsexu=xr;

}printf("\nThe root is %f",xr);getch();}OUTPUT:


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03.Solve the equation: : x^3 - 5x + 1 using Newton

Raphson method.

#include#include

#include

void main(){

float x=0,fx,fx1,xi;int a;clrscr();for(a=0;a


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04.solve the equation: x^3 - 5x + 1 = 0 using secant

method.

#include#include

#includevoid main(){

int a;double xn,xn1,fxn,fxn1,xn2;clrscr();printf("\n SECANT METHOD\n");printf("\n Enter the value of x0:");scanf("%lf",&xn);printf("\n Enter the value of x1:");scanf("%lf",&xn1);

printf("interval xn xn+1 f(xn)f'(xn)\n");printf("-------------------------------------------------------------------------\n");

for (a=0;a


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05.solve the equation f(x)=cos x - xe^x and f(x)=x^ex -

1 using Bisection, Regula Falsi, Newton Raphson's and

Secant method.

/* Bisection Method */

#include

#include

#include

void main()

{

float xl=0,xu=1,xr,e=2.718281828;

int i,n;

float fx;

clrscr();

printf("how many iteration do u want?..");

scanf("%d",&n);

for(i=0;i0)

xl=xr;

elsexu=xr;

printf("\n%f",xr);

}

getch();

}

OUTPUT


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/*Regular Falsi */

#include

#include

#include

void main(){

int i;

float xl,xr,xu,fxr,fxl,fxu,e=2.718281828;

clrscr();

xl=2;

xu=3;

printf("Regular falsi method\n");

printf("interval xl xu xr

f(xr)\n");

printf("___________________________________________________

__________\n\n");

for(i=0;i0)

xl=xr;

else

xu=xr;

}

printf("\nThe root is %f",xr);

getch();

}


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OUTPUT:-

/* Newton Raphson's */

#include

#include

#include

void main()

{int i;

float xn,xn1,fxn,fxn1,e=2.718281828;;

clrscr();

printf("\n NEWTON RAPHSON METHOD\n");

printf("\n Enter the value of x0:");

scanf("%f",&xn);

printf("\n interval xn xn+1 f(xn)

f'(xn)\n");

printf("--------------------------------------------------------

-----------------\n");

for(i=0;i


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fxn=cos(xn)-((xn)*(pow(e,xn)));

fxn1=(-sin(xn))-((xn)*(pow(e,xn)))-(pow(e,xn));

xn1=xn-(fxn/fxn1);

printf("%d %f %f %f%f\n",i,xn,xn1,fxn,fxn1);

xn=xn1;

}

printf("\n The root is %f",xn);

getch();}

OUTPUT:


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/*Secant Method*/

#include

#include

#include

void main()

{

int i;

float xn,xn1,fxn,fxn1,e=2.718281828;

clrscr();

printf("\n NEWTON RAPHSON METHOD\n");

printf("\n Enter the value of x0:");

scanf("%f",&xn);

printf("\n interval xn xn+1 f(xn)

f'(xn)\n");

printf("--------------------------------------------------------

-----------------\n");

for(i=0;i


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getch();

}

OUTPUT


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06. Solve the equations

x1 + x2 + x3 = 6

3x1 + 3x2 + 4x3 = 20

2x1 + x2 + 3x3 = 13

Using: Gauss Elimination Method

#include#includevoid main(){int x3,x2,x1,m,n,a[3][4];

clrscr();for(m=0;m


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7)Gauss Jordan Method

#include#includevoid main()

{int m,n,k,x1,p,l;int ar[10][10];clrscr();printf("\n enter the number of unknowns to be found: ");scanf("%d",&l);printf("\n enter the value of equation : \n");for(m=0;m


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}}}}gotoxy(45,20);

printf("matrix after GAUSS JORDAN");for(m=0;m


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8)Gauss Seidel Method

#include

#include

#includevoid main()

{

float x1,x2=0,x3=0;

int a,n;

clrscr();

printf("\n GAUSS SEIDEL METHOD \n");

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

scanf("%d",&n);

for(a=0;a


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9) Inverse using Gauss Jordan Method:

1 1 1

3 3 42 1 3

#include

#include

#include

void main()

{

int i,j,k,l,n,m,t;

float a[8][8],p,g,d,x;

clrscr();printf("enter the orderof the matrix\n");

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

printf("enter the matrix with identity matrix:\n");

for(i=0;i


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t=a[p][j];

a[p][j]=a[i][j];

a[i][j]=t;

}

}

g=a[i][i];for(j=i;j


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OUTPUT:


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10.Find the value of f(2) using Langrange Interpolating:

f(0)=1

f(1)=3

f(3)=55

#include#includevoid main()

{

float x0=0.0,x1=1.0,x=2.0,x2=3.0,fx0=1.0,fx1=3.0,fx2=55.0;

float fx,st1,st2,st3;

clrscr();

st1=(((x-x1)*(x-x2))/((x0-x1)*(x0-x2)))*(fx0);

st2=(((x-x0)*(x-x2))/((x1-x0)*(x1-x2)))*(fx1);

st3=(((x-x0)*(x-x1))/((x2-x0)*(x2-x1)))*(fx2);

fx=st1+st2+st3;

printf("result of lagrange ,f(2)=%f",fx);

getch();

}

OUTPUT:

Numeric Method

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