All in One Maths Lab

7/30/2019 All in One Maths Lab

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C/C++

PROGRAMMING

FOR

NUMERICAL LAB

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CONTENTS

1. BISECTION METHOD

2. SECANT METHOD

3. EULER METHOD

4. TRAPEZOIDAL METHOD

5. REGULA FALSI METHOD

6. FIXED POINT METHOD

7. LAGRANGES INTERPOLATION

8. RUNGE-KUTTA METHOD

9. SIMPSONS 1/3 RULE

10. SIMPSONS 3/8 RULE

11. GAUSS-SEIDEL METHOD12. NEWTON-RAPHSON METHOD

13. NEWTONS DIVIDED DIFFERENCE METHOD

14. HUENS METHOD


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BISECTION METHOD

AIM: To find solution of x2

-4x+3 by BISECTION METHOD.

ALGORITHM:

1. Start2. Define the function f(x)3. Read initial values say x1 & x2.

4. if (((f(x1)*f(x2))


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cout


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SECANT METHOD

AIM:To find solution of equation x2-5x+4 by SECANT METHOD.

ALGORITHM:

1. Start

2. Function f(x) is declared.

3. Range in which the root lies is determined.

4. Then consecutive values are calculated:

Xn+1=Xn-(f(Xn)(Xn+1-Xn)/(f(Xn+1)-f(Xn))

Till((Xn+1-Xn)>0.000001))

5. Print the root.8. Stop

PROGRAM:

#include

main()

{

float a,b,c,f,g,s,h;

printf("enter two value :");

scanf("%f %f",&a,&s);f=(a*a)-(5*a)+4;

g=s*s-5*s+4;

b=a-((f*(a-s))/(f-g));

printf("\nITERATION SOLUTION");

printf(" \n\n 1 \t\t %f \n",b);

float e;

e=0.1;

int i=1;

while(e>0.001){

f=(a*a)-(5*a)+4;

h=(b*b)-(5*b)+4;

c=b-((h*(b-a))/(h-f));

if(h==f)

{

printf("\t 1 \t\t %f",b);

}

else

printf("\n %d \t\t %f \n",i+1,c);

if(c>b)e=(c-b);


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else

e=(b-c);

a=b;

b=c;

i++;

}printf("\n Solution after %d iterations is %f \n",i,c);

}

SAMPLE OUTPUT:

Enter two values: 5 6

ITERATION SOLUTION

1 4.333333

2 4.0769233 4.007519

4 4.000187

5 4.000000

Solution after 5 iterations is 4.000000

EULER METHOD

AIM: To find value of y at x=0.3 when given differential equation is y= (x+y).

ALGORITHM:

1.INPUT:

Initial values x0 ,y0, stepsize h , number of steps N

2.OUTPUT:

Approximation y(n+1) to the solution y(x(n+1)) at

x(n+1) = x0+ (n+1)h where n=0,1,2...N-1

3.For n=0,1,2...N-1 do

4.y=y+h*(x+y)

5.x=x+h

6.OUTPUT x(n+1) , y(n+1)

7.End

8.stop

PROGRAM:


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#include

float f(float x,float y)

{

float z;z=x+y;

return z;

}

int main()

{

float h,x,y;

int i,n;

printf("enter values of x , y , h,n \n ");

scanf(" %f %f %f %d", &x ,&y ,&h ,&n);

for (i=0;i


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x5 0.250000

y5 1.302563

x6 0.300000

y6 1.380191

TRAPEZOIDAL METHOD

AIM:

To find the solution of integration of x3

from 0 to 1.

ALGORITHM:

1.Input: enter limits a,b,n=1

2.compute:

3.h=(b-a)/n

4.while(x0.0001)


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{

h=(b-a)/n;

m=0; x=a+h;

while(xy)

e=z-y;

else

e=y-z;

z=y;

n++;}

printf(\n Solution of integration is : %f,y);

}

SAMPLE OUTPUT:

Enter the limits : 0 1

Iteration Solution

1 0.3125002 0.2777783 0.2656254 0.2600005 0.2569446 0.2551027 0.2539068 0.2530869 0.25250010 0.25206611 0.33506912 0.25147913 0.25127614 0.25111115 0.25097716 0.25086517 0.250772

Solution of integration is : 0.250772


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REGULA FALSI METHOD

AIM: To find solution of the function : x2-5x+4 by REGULA FALSI METHOD.

ALGORITHM:

1. Start.

2. Set values of x1 & x2.

3. A function f is defined where functional values are

calculated.

4. Repeat the steps until you obtain a sign change.

5. If((f(x1)*f(x2))0.000001

7. z=(x*f(y)-y*f(x))/(f(y)-f(x))

8. If (f(z)*f(y))0)

printf("invalid input");else if(f==0 && p==0)


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printf("solution is %f",a);

else if(f==0 && q==0)

printf("\n solution is %f",b);

else

{

while(r>0.000001 || r


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17 1.003042

18 1.001826

19 1.001096

20 1.000658

21 1.000395

22 1.00023723 1.000142

24 1.000085

25 1.000051

26 1.000031

27 1.000018

28 1.000011

29 1.000007

30 1.000004

31 1.000002

32 1.000001

33 1.00000134 1.000000

Solution after 34 iterations is 1.000000

FIXED POINT METHOD

AIM:To find the solution of equation : 1/(x2+1) by fixed point method.

ALGORITHM:

1.Read the first value

2.Calculate z(x) as (1/(x*x)+1)

3.while (g-z)>0.0001

4.g=(1/(z*z)+1)

5.z=g

6.End of while loop

PROGRAM:

#includemain()

{

int i=0;

float x,y,g,z,e=0.1;

printf("\n Enter the first value");

scanf("%f",&x);

z=(1/((x*x)+1));

while(e>0.0001)

{

g=1/((z*z)+1);printf("\n%d iteration %f",(i+1),g);


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if(g>z)

e=g-z;

else

e=z-g;

z=g;

i++;}

printf("\n Solution of equation is %f",g);

}

SAMPLE OUTPUT:

Enter the first value : 0.5

1 iteration 0.609756




5 iteration 0.6704726 iteration 0.689878










16 iteration 0.68240917 iteration 0.682276


Solution of equation is 0.682360

LAGRANGES INTERPOLATION

AIM-

To determine value at a given point using Lagranges interpoation when values are given as f(0)=1,

f(1)=3, f(3) = 13,f(8)=123

ALGORITHM-

1. Read the number of points(n),given points and its functional values.2. Declare d,p,sum3. For i=0 to n-1


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4. p=15. for j=0 to n-16. if (i!=j)7. d= (a-x[j])/(x[i]-x[j])8. p=p*d9. End of if statement10. End of inner for loop11.P1=p*f[i]12.Sum=sum+p113.End of for loop14. Print the result

PROGRAM:

#include

int main()

{

int n;

float a,x[10],f[10];

couta;

coutn;

for(int k=0;k


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}

SAMPLE OUTPUT:

Enter given point : 5Enter number of points : 4

Enter value of x0 : 0

Enter functional value : 1







Value at given point : 38.14286

RUNGE-KUTTA METHOD

AIM- To solve (dy/dx=(y2-x

2)/(y

2+x

2)) with y(0) =1 at 0.2 and 0.4

ALGORITHM-

1.Define given function f(x,y)

2.Read x0,y0,h,n.

3. For i=0 to (n-1)

4.K1=h*(f(x[i],y[i])

5.K2=h*(f(x[i]+(h/2), y[i]+(k1/2)))

6.K3=h*(f(x[i]+(h/2),y[i]+(k2/2))

7.K4=h*f(x[i]+h,y[i]+k3)

8.y[i+1]=y[i]+(k1+k2+k3+k4)/6

9.Print x[i],y[i]

10.Next i

11.End

PROGRAM-

9

#include

#include


{

return(((y*y)-(x*x))/((y*y)+(x*x)));

}


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int main()

{

float x[10],y[10],h,k1,k2,k3,k4;

int n,i;

printf("enter x.,y.,h");

scanf("%f%f%f",&x[0],&y[0],&h);printf("Enter value of n");

scanf("%d",&n);

printf(" \n x0 and y0 : %f %f ",x[0],y[0]);

for( i=0;i


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printf("enter x.,y.,h");

scanf("%f%f%f",&x[0],&y[0],&h);

printf("Enter value of n");

scanf("%d",&n);

printf("x0 and y0 :%f %f ",x[0],y[0]);

for( i=0;i


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#include

main()

{

int n,i;

i=1;float a,y,m,s,b,h,f,g,z,x,k,e;

printf("\n Input the limits");

scanf("%f %f",&a,&b);

f=a*a*a; g=b*b*b;

n=2;

h=(b-a)/n;

m=h*h*h;

z=h*(f+g+(4*m))/3;

n=n+2; e=0.1;

printf("\n ITERATION SOLUTION");

while(e>0.001){

h=(b-a)/n;

x=a+h;

s=0;int c=1;

while(xy)

e=z-y;

else

e=y-z;

z=y; n=n+2;

}

printf(\n Solution of integration is :%f,y);

}

SAMPLE OUTPUT:

Input the limits: 8.2 9.0

ITERATION SOLUTION

1 607.145630

2 574.745667

3 509.945984

4 548.825623


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5 542.345642

6 509.946136

7 509.945923

Solution of integration is : 509.945923

SIMPSONS 3/8 RULE

AIM: To find the solution of integration of sqrt(1-0.692sin2(m)) from 0 to 6.

ALGORITHM:

1.INPUT:read lmits-a,b

2.h=(b-a)/n

3.for i=1 to n

4.x[i]=a+ih

5.y[i]=f(x[i])

6.if(i%3=0)

S2+=y[i]

7.else s1+=y[i]

8.s=3h/8(f(a)+f(b)+3s1+2s2)

9.print s.10.end.

PROGRAM:

#include

#include

float f(float m);

main()

{

float a,b,n;float h,s1=0,s2=0,s=0;

b=M_PI_2;

printf("\n Enter the limits : ");

scanf("%f %f",&a,&n);

float x[6],y[6];

x[0]=a;x[n]=b;

h=(b-a)/n;

for(int i=1;i


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if(i%3==0)

s2+=y[i];

else

s1+=y[i];

}

s=(3*h/8)*((f(a)+f(b)+(3*s1)+(3*s2)));

printf("\n The value of integration is : %f",s);

}

float f(float m)

{

float p;

p=sqrt(1-(0.162*sin(m)*sin(m)));

return p;

}

SAMPLE OUTPUT:

Enter the limits : 0 6

The value of integration is : 1.599218

GAUSS SEIDEL METHOD

AIM:

To find the solution of system of equations by GAUSS-SEIDEL METHOD.

10w-2x-y-z=3

-2w+10x-y-z=15

-w-x+10y-2z=27

-w-x-2y+10z=-9

ALGORITHM:

1.INPUT: Input the equations s1 , s2 , s3 , s4..sn and no.of iterations

set values of the solutions to zero

2.OUTPUT: Approximate solutions x1 , x2 ,x3 ,xn will be out

3. For n = 0 , 1 , 2..N-1

4. assign values of x1 , x2 , x3.xn to k1 , k2 ,.kn

X1 = s1 ( x2 , x3 ,x4 ,..xn )


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X2= s2 ( x1 , x3 ,x4 .xn )

.

.

.

.

.

Xn = sn ( x1 ,x2,x3 ,.xn-1)

5.Print the approximate value of x1 , x2...xn

6.End

7.stop

PROGRAM:

#include

int main( ){

float x[10][10],k0,k1,k2,k3 ;

int i,j,k;

x[0][0]=0;x[1][0]=0;x[2][0]=0;x[3][0]=0;

printf (" iteration values are \n ") ;

for(i=0;i


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w x y z

0.300000 1.500000 2.700000 -0.900000

0.780000 1.740000 2.700000 -0.180000

0.900000 1.908000 2.916000 -0.108000

0.962400 1.960800 2.959200 -0.0360000.984480 1.984800 2.985120 -0.015840

0.993888 1.993824 2.993760 -0.006048

NEWTON RAPHSON METHODAIM:

To find the solution of the function x2-4x+3 by NEWTON-RAPHSON METHOD.

ALGORITHM:

1.The function f(x) is declared.

2.The derivative of the function is found out and stored in another variable(say f1(x)).

3.If f1(xn)=0,output fails & exit

4. Else xn+1=xn-(f(xn)/f1(xn))

5.The process continues till |xn+1-xn|0.000001)

{

c=b-((b*b)-(4*b)+3)/((2*a)-4);

if(c==2.0){printf("Solution is %f ",b);}

printf("\n %d \t %f",i+2,c);if(c>b)


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e=(c-b);

else

e=(b-c);

b=c;

i++;

}

printf("\n\nSolution of function is %f",c);

}

SAMPLE OUTPUT:

Enter the first value : 0.5

ITERATION SOLUTION1. 0.916667

2. 0.974537

3. 0.991728

4. 0.997266

5. 0.999091

6. 0.999697

7. 0.999899

8. 0.999966

9. 0.999989

10. 0.999996

11. 0.99999912. 1.000000

Solution of the function is 1.000000

NEWTONS DIVIDED

DIFFERENCE METHOD

AIM :

To find value of a function at a given point using NEWTONS DIVIDED DIFFERENCE METHOD Values given as f(5)=150, f(7)=392, f(11)=1452 , f(13)=2366, f(17)=5202

ALGORITHM:

1. Define function as given.


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2. Define f[x0 x1] as f(x1)-f(x0)/(x1-x0)

3. Define f[x0 x1 x2] as (f[x1 x2]-f[x0 x1])/(x2-x0)

4. Define f[x0 x1 x2 x3] as (f[x1 x2 x3]-f[x0 x1 x2])/(x3-x0)

5. Define f[x0 x1 x2 x3 x4] as (f[x1 x2 x3 x4]-f[x0 x1 x2 x3])/(x4-x0)

6. Define polynomial p(a) in terms of f[x0 x1],f[x0 x1 x2 ],f[x0 x1 x2 x3],f[x0 x1 x2 x3 x4]

7. In main, read the 5 values8. Print result.

PROGRAM:

#include

float x[20],f[20];

float func(int low,int high)

{

if(low==high-1)return (f[high]-f[low])/(x[high]-x[low]);

else

return (func(low+1,high)-func(low,high-1))/(x[high]-x[low]);

}

float calcpx(float y,int n)

{

float px;

px=f[0];

float fact=1;

int j;

for(int i=1;i


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cin>>n;

for(int i=0;i


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ALGORITM:

1.INPUT: Initial values x0 , y0 , step size h , number of steps N

2.OUTPUT: Approximation y(n+1) to the solution y(x(n+1)) at

x ( n + 1 ) = x0 + ( n + 1 ) h where n = 0 ,1 ,2 ,..N - 1

3.for n = 0 , 1 , 2 , ..N - 1 do

4.y = y+ h * ( x + y )

5.x = x + h

6. OUTPUT x ( n + 1 ) , y ( n + 1 )

7.End

8.stop

PROGRAM:

#include

#include


{

float z ;

z=x+y;

return z;

}

int main()

{

float x0,y0,y1,h,y11,e;

int i,n;

printf (" enter values of x0 , y0 , h , n \n ");

scanf ("%f%f%f%d",&x0,&y0,&h,&n);

printf(" x y \n ");

for (i=0;i0)

e=(y1-y11);

else

e=(y11-y1);

while (e>0.001)

{


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y11=y1 ;

y1=y0+((h/2)*(f(x0,y0)+f(x0,y11)));

printf(" %f \n ",y1);

if((y1-y11)>0)

e=(y1-y11);

else

e=(y11-y1);

}

y0=y1;

x0=x0+h;

}

return 0;

}

SAMPLE OUTPUT:

Enter values of x0 , y0 , h , n : 0 1 0.05 6

Iteration values of 0.050000 are

1.051250

1.051281


1.107722

1.107756


1.169654

1.169692


1.237326

1.237367

Iteration values of 0.250000 are1.311032

1.311077


1.391082

1.391131

****************************************

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