numlabreport

8/2/2019 numlabreport

1/41

1.Write a c program to evaluate value of e correct upto 3 decimal places.

#include

#include

main()

{

int count=1;

float sum1=1,k=1,i=1,sum2=1,term=1;

printf(" Term no. || term || sum || difference\n"); /* format to print the output*/

printf(" %d || %f || %f ||%f ",count,term,sum1,sum2-sum1);

while(1)

{ k=k*i;

term=1/k;

sum1=sum2;

sum2=sum2+term;

if((sum2-sum1)


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---------------------------------------

3 || 0.500000 || 2.000000 ||0.500000

---------------------------------------

4 || 0.166667 || 2.500000 ||0.166667

---------------------------------------

5 || 0.041667 || 2.666667 ||0.041667

---------------------------------------

6 || 0.008333 || 2.708333 ||0.008333

---------------------------------------

7 // 0.001389 || 2.716667 ||0.001389

Value of e is:: 2.718

2..Write a program to implement bisection method for finding root of an non linear

equation.

Equation: 3xcosx = 1 /* equation from which value of x is to be found*/

#include

#include

#include

#define f(x) (3*x-cos(x)-1)

main()

{ float a,b,m,e,error=0,x,error1;

double n;

int d=1,c=0,i=0;

while(d==1)

{ printf("Enter value of a and b::"); /* getting initial interval as input*/

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

if(f(a)*f(b)


3/41

else if(f(a)==0)

{

printf("Interval limit %f is the result",a);

exit(0);

}

else if(f(b)==0)

{

printf("Interval limit %f is the result",b);

exit(1) }

else

printf("No root exists in this interval\n");

}

printf("Enter degree of precision::"); /*takes degree of precision as input*/

scanf("%f",&e);

e=5*pow(0.1,(e+1));

printf(" i\t a\t\t b\t\t m\t abserror\t\t order of convergence \n\n"); /* format to print the output*/

while(fabs(a-b)>e)

{ if(i==0)

{ m=(a+b)/2;

printf(" %d\t %f\t %f\t %f\t ------\t\t -----\n",i,a,b,m) ;

}

else { x=m;

m=(a+b)/2;

error1=error;

error=fabs(x-m);

printf(" %d\t %f\t %f\t %f\t %f\t",i,a,b,m,error) ;

if(i==1)

{


4/41

printf("------\n");

}

else

{

n=log(error)/log(error1); /* calculates order of convergence*/

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

}

}

if(f(m)==0)

{ c=1;

break;

}

else

{ if(f(a)*f(m)


5/41

Enter value of a and b::0 1

Enter degree of precision::4

i a b m abserror order of convergence

0 0.000000 1.000000 0.500000 ------ -----

1 0.500000 1.000000 0.750000 0.250000 ------

2 0.500000 0.750000 0.625000 0.125000 1.500000

3 0.500000 0.625000 0.562500 0.062500 1.333333

4 0.562500 0.625000 0.593750 0.031250 1.250000

5 0.593750 0.625000 0.609375 0.015625 1.200000

6 0.593750 0.609375 0.601563 0.007813 1.166667

7 0.601563 0.609375 0.605469 0.003906 1.142857

8 0.605469 0.609375 0.607422 0.001953 1.125000

9 0.605469 0.607422 0.606445 0.000977 1.111111

10 0.606445 0.607422 0.606934 0.000488 1.100000

11 0.606934 0.607422 0.607178 0.000244 1.090909

12 0.606934 0.607178 0.607056 0.000122 1.083333

13 0.607056 0.607178 0.607117 0.000061 1.076923

14 0.607056 0.607117 0.607086 0.000031 1.071429

one of the root is:: 0.607101

3. Write a program to implement Regular falsi method for finding root of an non linear equation.

Equation: 3xcosx = 1 /* equation from which value of x is to be found*/

#include

#include

#include

#define f(x) (3*x-cos(x)-1) /*macro defining the function*/

main()

{ float a,b,m1=0,m2=0,error1,error=0,e;


6/41

double n;

int d=1,i=0;

while(d==1)

{ printf("Enter value of a and b::"); /* getting initial interval as input*/

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

if(f(a)*f(b)


7/41


scanf("%f",&e);

e=5*pow(0.1,(e+1));

printf(" i\t |b-a|\t\t m\t\t abserror\t order of convergence \n\n");

do

{

if(i==0)

{ m2=(a*f(b)-b*f(a))/(f(b)-f(a));

printf(" %d\t %f\t %f\t ------\t\t -----\n",i,fabs(b-a),m2) ;

}

else

{ m1=m2;

m2=(a*f(b)-b*f(a))/(f(b)-f(a)); /* calculation of approximation to root*/

error1=error;

error=fabs(m2-m1);

printf(" %d\t %f\t %f\t %f\t",i,fabs(b-a),m2,error) ;

if(i==1)

{ printf("------\n");

}

else

{ n=log(error)/log(error1); /* calculation of order of convergence*/

printf("%lf\n",fabs(n));

}

}

If ( f(m2)==0)

break;

else

{ if ( f(a)*f(m2)


8/41

b= m2;

else

a= m2;

}

i++;

}while(fabs(m2-m1)>e);

printf("one of the root is:: %f\n",m2); /*prints output as result*/

}

OUTPUT

Enter value of a and b::0 1


i |b-a| m abserror order of convergence

0 1.000000 0.578085 ------ ----

1 0.421915 0.605959 0.027873 ------

2 0.394041 0.607057 0.001099 1.903251

3 0.392943 0.607100 0.000043 1.476279

one of the root is:: 0.607100

4.. Write a program to implement Repeated substitution method for finding root of an non linear

equation.

Equation:: Log10(x)-2*x+7=0 /* equation from which value of x is to be found*/

#include

#include

#include

float f(float x) /* expressing as x = g(x)*/

{ return((log10(x)+7)/2);

}

float der(float x) /* finds derivative of g(x)*/


9/41

{

return (1/(2*x*log(10)));

}

main()

{ double x1,x2,error=0,error1;

int i=0,e;

double n;

printf("Enter initial approximation to root::"); /* inputs initial root from user*/

scanf("%lf",&x1);

if(x11) /*checks for condition of convergence*/

{

printf("condition of convergence is not satisfied.");

exit(0);

}

x2=f(x1);

while(fabs(x2-x1)>e)

{ if((fabs(der(x2)))>1) /*checks for condition of convergence*/

{

printf("condition of convergence is not satisfied.");


10/41

exit(0);

}

error1=error;

error=fabs(x2-x1);

printf(" %d\t %lf\t %lf\t %lf\t",i,x1,x2,error);

if(i==0)

{ printf("--------\n");

}

else

{ n=log(error)/log(error1);


}

x1=x2;

x2=f(x1);

i++;

}

printf(" root is %lf\n",x2); /*prints output as result*/

}

OUTPUT

Enter initial approximation to root::1


i xi f(xi) abserror order of convergence

0 1.000000 3.500000 2.500000 --------

1 3.500000 3.772034 0.272034 1.420759

2 3.772034 3.788288 0.016254 3.164333

3 3.788288 3.789222 0.000934 1.693543

4 3.789222 3.789275 0.000053 1.410122

5 3.789275 3.789278 0.000003 1.289369


11/41

6 3.789278 3.789278 0.000000 1.202215

root is 3.789278

5. Write a program to implement Newton - Raphson method for finding root of an non linear

equation.

a) Equation:x e^(--x) = 0 /* equation from which value of x is to be found*/

#include

#include

#include

float f(float x) /* finds value of f(x)*/

{ float k;

k= x*exp(-x);

return k;

}

float der(float x) /* finds derivative of f (x) */

{

float k;

k=exp(-x)*(1-x);

return k; }

float f2(float x) /* finds second order derivative of f(x) */

{

float k;

k=exp(-x)*(x-2);

return k;

}

main()

{ double x1,x2,error=0,error1,e;

int i=0;


12/41

double n;


scanf("%lf",&x1);


scanf("%lf",&e);

e=5*pow(0.1,(e+1));

printf(" i\t Xi\t\t f(Xi)\t\tf'(Xi)\t\tabserror\t order of convergence \n");

if(der(x1)==0) /* if der(x1)==0,process is stopped and no root exists*/

{

printf("condition of convergence is not satisfied. \n value of f'(x) is zero at at x=%lf",x1);

exit(0);

}

if(!(fabs((f(x1)*f2(x1))/pow(der(x1),2))e)

{

if(der(x1)==0)

{

printf("condition of convergence is not satisfied. \n value of f'(x) is zero at at x=%lf",x1);

exit(0);

}


13/41

if(!(fabs((f(x2)*f2(x2))/pow(der(x2),2))


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4 14.380524 0.000008 -0.000008 1.074736 0.922029

5 15.455260 0.000003 -0.000003 1.069179 0.928081

6 16.524439 0.000001 -0.000001 1.064415 0.933232

7 17.588853 0.000000 -0.000000 1.060281 0.937677

8 18.649135 0.000000 -0.000000 1.056660 0.941549

9 19.705795 0.000000 -0.000000 1.053459 0.944956

10 20.759254 0.000000 -0.000000 1.050609 0.947979

11 21.809863 0.000000 -0.000000 1.048054 0.950680

12 22.857917 0.000000 -0.000000 1.045750 0.953108

13 23.903667 0.000000 -0.000000 1.043661 0.955303

14 24.947329 0.000000 -0.000000 1.041758 0.957299

15 25.989087 0.000000 -0.000000 1.040017 0.959116

16 27.029104 0.000000 -0.000000 1.038419 0.960790

17 28.067523 0.000000 -0.000000 1.036945 0.962323

18 29.104468 0.000000 -0.000000 1.035582 0.963741

19 30.140049 0.000000 -0.000000 1.034317 0.965055

20 31.174366 0.000000 -0.000000 1.033141 0.966275

21 32.207507 0.000000 -0.000000 1.032044 0.967413

22 33.239550 0.000000 -0.000000 1.031018 0.968472

23 34.270568 0.000000 -0.000000 1.030057 0.969464

24 35.300625 0.000000 -0.000000 1.029154 0.970398

25 36.329779 0.000000 -0.000000 1.028305 0.971270

26 37.358084 0.000000 -0.000000 1.027504 0.972101

27 38.385588 0.000000 -0.000000 1.026748 0.972875

28 39.412336 0.000000 -0.000000 1.026033 0.973611

29 40.438369 0.000000 -0.000000 1.025356 0.974308

30 41.463725 0.000000 -0.000000 1.024713 0.974965

31 42.488439 0.000000 -0.000000 1.024103 0.975596

32 43.512542 0.000000 -0.000000 1.023522 0.976186


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33 44.536065 0.000000 -0.000000 1.022969 0.976756

34 45.559034 0.000000 -0.000000 1.022442 0.977295

35 46.581476 0.000000 -0.000000 1.021939 0.977813

36 47.603415 0.000000 -0.000000 1.021458 0.978299

37 48.624873 0.000000 -0.000000 1.020997 0.978775

38 49.645870 0.000000 -0.000000 1.020557 0.979224

39 50.666427 0.000000 -0.000000 1.020134 0.979656

40 51.686561 0.000000 -0.000000 1.019729 0.980066

41 52.706290 0.000000 -0.000000 1.019340 0.980464

42 53.725630 0.000000 -0.000000 1.018966 0.980849

43 54.744596 0.000000 -0.000000 1.018606 0.981214

44 55.763203 0.000000 -0.000000 1.018261 0.981572

45 56.781463 0.000000 -0.000000 1.017927 0.981904

46 57.799390 0.000000 -0.000000 1.017606 0.982236

47 58.816996 0.000000 -0.000000 1.017296 0.982548

48 59.834292 0.000000 -0.000000 1.016997 0.982853

49 60.851289 0.000000 -0.000000 1.016708 0.983147

50 61.867997 0.000000 -0.000000 1.016429 0.983432

51 62.884426 0.000000 -0.000000 1.016159 0.983701

52 63.900585 0.000000 -0.000000 1.015898 0.983977

53 64.916483 0.000000 -0.000000 1.015645 0.984232

54 65.932129 0.000000 -0.000000 1.015401 0.984474

55 66.947529 0.000000 -0.000000 1.015164 0.984716

56 67.962693 0.000000 -0.000000 1.014934 0.984959

57 68.977627 0.000000 -0.000000 1.014711 0.985173

58 69.992337 0.000000 -0.000000 1.014494 0.985400

59 71.006832 0.000000 -0.000000 1.014284 0.985610

60 72.021116 0.000000 -0.000000 1.014080 0.985817

61 73.035197 0.000000 -0.000000 1.013882 0.986022


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62 74.049079 0.000000 -0.000000 1.013689 0.986211

63 75.062768 0.000000 -0.000000 1.013502 0.986406

64 76.076270 0.000000 -0.000000 1.013320 0.986591

65 77.089590 0.000000 -0.000000 1.013142 0.986767

66 78.102732 0.000000 -0.000000 1.012970 0.986948

67 79.115702 0.000000 -0.000000 1.012801 0.987106

68 80.128504 0.000000 -0.000000 1.012638 0.987287

69 81.141141 0.000000 -0.000000 1.012478 0.987441

70 82.153619 0.000000 -0.000000 1.012322 0.987602

71 83.165942 0.000000 -0.000000 1.012171 0.987755

72 84.178112 0.000000 -0.000000 1.012022 0.987899

73 85.190135 0.000000 -0.000000 1.011878 0.988054

74 86.202012 0.000000 -0.000000 1.011737 0.988186

75 87.213749 0.000000 -0.000000 1.011599 0.988333

76 88.225348 0.000000 -0.000000 1.011464 0.988465

77 89.236813 0.000000 -0.000000 1.011333 0.988605

78 90.248146 0.000000 -0.000000 1.011205 0.988733

79 91.259351 0.000000 -0.000000 1.011079 0.988857

80 92.270430 0.000000 -0.000000 1.010956 0.988982

81 93.281386 0.000000 -0.000000 1.010836 0.989088

82 94.292222 0.000000 -0.000000 1.010719 0.989228

83 95.302941 0.000000 -0.000000 1.010603 0.989270

84 96.313544 0.000000 -0.000000 1.010488 0.989231

85 97.324033 0.000000 -0.000000 1.010391 0.990795

86 98.334424 0.000000 -0.000000 1.010244 0.985923

87 99.344669 0.000000 -0.000000 1.010143 0.990171

88 100.354812 0.000000 -0.000000 1.009730 0.959434

Condition of convergence is not satisfied.

value of |g'(x)| is greater than 1 at x=102.374821


17/41

b) Equation :: x^3x3=0 /* equation from which value of x is to be found*/

#include

#include

#include


{ float k;

k= pow(x,3)-x-3;

return k;

}


{

float k;

k=(3*x*x-1);

return k;

}


{

float k;

k= (6*x);

return k;

}

main()


int i=0;


18/41

double n;


scanf("%lf",&x1);


scanf("%lf",&e);

e=5*pow(0.1,(e+1));



{

printf("condition of convergence is not satisfied. \n value of f'(x) is zero at x=%lf",x1);

exit(0);

}


{ if(der(x1)==0)

{


exit(0);

}


19/41



20/41

Condition of convergence is not satisfied.

value of |g'(x)| is greater than 1 at x=-1.147176

c)

Equation: tan(inverse)x =0 /* equation from which value of x is to be found*/

#include

#include

#include

double f(double x) /* finds value of f(x)*/

{ double k;

k=(double)atan(x);

return k;

}

double der(double x) /* finds derivative of f (x) */

{

double k;

k=1+(x*x);

k=1/k;

return k;

}

double f2(double x) /* finds second order derivative of f(x) */

{

double k;

k= pow(2,(1+(x*x)));

k=(-2)*(x)/k;

return k;

}

main()

{ double x1,x2,error=0,error1,t1,e;


21/41

int i=0;

double n;


scanf("%lf",&x1);


scanf("%lf",&e);

e=5*pow(0.1,(e+1));



{


exit(0);

}


{ if(der(x1)==0)

{


exit(0);

}


22/41



23/41

#include

#include

#include


{ float k;

k=(x*x*x)-(3*x)+2;

return k;

}


{

float k;

k=(3*x*x)-3;

return k;

}


{

float k;

k= (6*x);

return k;

}

main()


int i=0;

double n;


scanf("%lf",&x1);


scanf("%lf",&e);


24/41

e=5*pow(0.1,(e+1));



{


exit(0);

}

if(!(fabs((f(x1)*f2(x1))/pow(der(x1),2)) e )

{ if(der(x1)==0)

{


exit(0);

}



25/41

if(i==0)

printf("--------\n");

else

{ printf(" %d\t %lf\t %lf\t %lf\t%lf ",i,x1,f(x1),der(x1),error);

n=log(error)/log(error1); /* calculation of order of convergence*/


}

x1=x2;

x2=x1-(f(x1)/der(x1));

i++;

}

printf("Root is %lf\n",x2);

}

OUTPUT

Enter initial approximation to root::1.2


i Xi f(Xi) f'(Xi) abserror order of convergence

0 1.200000 0.128000 1.320000 0.096970 --------

1 1.103030 0.032939 0.650028 0.050674 1.278135

2 1.052356 0.008367 0.322362 0.025956 1.224328

3 1.026401 0.002109 0.160496 0.013143 1.186366

4 1.013258 0.000530 0.080074 0.006614 1.158513

5 1.006643 0.000133 0.039993 0.003318 1.137463

6 1.003325 0.000033 0.019985 0.001662 1.121138

7 1.001664 0.000008 0.009990 0.000832 1.108179

8 1.000832 0.000002 0.004995 0.000416 1.097663

9 1.000416 0.000001 0.002497 0.000208 1.089029

10 1.000208 0.000000 0.001248 0.000104 1.081755

11 1.000104 0.000000 0.000624 0.000052 1.075515


26/41

Root is 1.000026

6. Write a program to implement GaussElimination method for finding roots of simultaneous

equations.

Equation::

x1 + x2 x3 + x4 =2;

2 x1 + x2 - x3 3 x4 =1;

3 x1 -- x2 x3 + x4 =2;

5 x1 + x2 +3 x32 x4 =7;

#include

#include

#include

#define rows 4

#define cols 5

void printmatrix( float [][cols] );

void pivot(float(*)[cols],int,int);

main()

{ float matrix[rows][cols]={ {1,1,-1,1,2}, /* defining coefficient matrix */

{2,1,1,-3,1},

{3,-1,-1,1,2},

{5,1,3,-2,7},

};

int k=0,i,j,p;

float m,max,result[rows],temp,t;

printf("initial coefficient matrix is:: \n");

printmatrix(matrix);

while(k


27/41

for(i=k+1;imax)

{ max=matrix[i][k];

p=i;

}

}

if(p!=k)

pivot(matrix,p,k);

for(i=k+1;i


28/41

{

temp=matrix[i][cols-1];

for(j=rows-1;j>i;j--)

temp-=matrix[i][j]*result[j];

temp=temp/matrix[i][i];

result[i]=temp;

}

for(i=0;i


29/41

}

return;

}

OUTPUT

initial coefficient matrix is::

1.00 1.00 -1.00 1.00 2.00

2.00 1.00 1.00 -3.00 1.00

3.00 -1.00 -1.00 1.00 2.00

5.00 1.00 3.00 -2.00 7.00

5.00 1.00 3.00 -2.00 7.00

0.00 0.60 -0.20 -2.20 -1.80

3.00 -1.00 -1.00 1.00 2.00

1.00 1.00 -1.00 1.00 2.00

5.00 1.00 3.00 -2.00 7.00

0.00 0.60 -0.20 -2.20 -1.80

0.00 -1.60 -2.80 2.20 -2.20

1.00 1.00 -1.00 1.00 2.00

5.00 1.00 3.00 -2.00 7.00

0.00 0.60 -0.20 -2.20 -1.80

0.00 -1.60 -2.80 2.20 -2.20

0.00 0.80 -1.60 1.40 0.60


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5.00 1.00 3.00 -2.00 7.00

0.00 0.80 -1.60 1.40 0.60

0.00 0.00 -6.00 5.00 -1.00

0.00 0.60 -0.20 -2.20 -1.80

5.00 1.00 3.00 -2.00 7.00

0.00 0.80 -1.60 1.40 0.60

0.00 0.00 -6.00 5.00 -1.00

0.00 0.00 1.00 -3.25 -2.25

5.00 1.00 3.00 -2.00 7.00

0.00 0.80 -1.60 1.40 0.60

0.00 0.00 1 .00 -3.25 -2.25

0.00 0.00 0.00 -14.50 -14.50

Final matrix after all arrangements ::::

5.00 1.00 3.00 -2.00 7.00

0.00 0.80 -1.60 1.40 0.60

0.00 0.00 1.00 -3.25 -2.25

0.00 0.00 0.00 -14.50 -14.50

Solution is ::

X1----1.00

X2----1.00

X3----1.00


31/41

X4----1.00

7. Write a program to implement GaussSeidel Iterative method for finding roots of simultaneous

equations.

x1 +6 x2 =2;

x1 -- 2 x2 -- 6x3 =14;

9 x1 4 x2 + x3 = --17;

#include

#include

#include

#define rows 3

#define cols 4

void printmatrix(float[][cols]);

void pivot(float(*)[cols],int,int);

main()

{

float matrix[rows][cols]={ {1,6,0,4}, /*defining coefficient matrix*/

{1,-2,-6,14},

{9,4,1,-17}

};

int k=0,i,j,p;

float m,max,Xo[rows],Xn[rows],temp,t,sum,e;

printf("initial coefficient matrix is:: \n");


while(k


32/41

{ if(matrix[i][k]>max)

{ max=matrix[i][k];

p=i;

}

}

if(p!=k)

pivot(matrix,p,k);

k++;

}

printf("matrix after pivoting is::\n");


for(i=0;i


33/41

while(1)

{ count++;

for(i=0;i


34/41

for(i=0;i


35/41


36/41

for(i=0;i


37/41

Enter the (x,y) values:

Enter (X0,Y0): 1 1

Enter (X1,Y1): 2 1

Enter (X2,Y2): 3 2

Enter the value to interpolate: 2.5

Corresponding Value of y is:: 1.375000

9.Write a program to find value of integration by Trapezoidal method.

F(x) = 1/(1+x)

#include

#include

#include

double f(double x) /* finds value of f(x) */

{ double k;

k=1+x;

k=1/k;

return k;

}

main()

{

int i=0,j,c=1;

double h,a,b,sum,I0,I1,e,k,x,error;

while(c==1)

{printf("Enter the interval values of the integral::"); /* inputs the lower and upper limit of integral */

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

if(b


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}

printf("Enter degree of precision::"); /* inputs degree of precision*/

scanf("%lf",&e);

e=5*pow(0.1,(e+1));

h=b-a;

I0=h*(f(b)+f(a))/2;

printf("ITERATION\t INTERVAL\t INTEGRAL\t ERROR \n "); /* format for printing */

printf("%d\t\t%lf\t %lf\t -------- \n",i,h,I0);

I1=I0;

do

{ i++;

h=h/2; /* divides the interval into half */

I0=I1;

I1=f(a);

x=a+h;

while(xe);

printf("Value of the integral is:: %lf\n",I1);

}

OUTPUT

Enter the interval values of the integral::0 1


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ITERATION INTERVAL INTEGRAL ERROR

0 1.000000 0.750000 --------

1 0.500000 0.708333 0.041667

2 0.250000 0.697024 0.011310

3 0.125000 0.694122 0.002902

4 0.062500 0.693391 0.000731

5 0.031250 0.693208 0.000183

6 0.015625 0.693162 0.000046

Value of the integral is:: 0.693162

10.Write a program to solve a differential equation by Eulers modified method.

dy/dx = 2x^2+2y

y(exact)=1.5e^2xx^2x0.5

#include

#include

#include

#define f(x,y) ((2*x*x)+(2*y)) /* macro for dy/dx */

#define exact(x) (1.5*exp(2*x)-x*x-x-0.5) /* macro for exact value of y*/

main()

{ int c=1;

float a,b,y0,x0,h,e,Yold,Ynew,k,error,x,y;

printf("Enter initial value of y::");

scanf("%f",&y0); /*inputs initial value of y*/

printf("Enter value of interval in between values of x::");

scanf("%f",&h); /*inputs interval in between values of x*/

while(c==1)

{

printf("Enter range for the value of x::\n");


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printf("lower bound::");

scanf("%f",&a);

printf("Upper bound::");

scanf("%f",&b);

if(b


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y0=Ynew;

x0=x;

x=x+h;

}

}

OUTPUT

Enter initial value of y::1

Enter value of interval in between values of x::0.1

Enter range for the value of x::

lower bound::0

Upper bound::1


X Y Yactual ERROR

0.000000 1.000000 1.000000 0.000000

0.100000 1.223331 1.222104 0.001004

0.200000 1.500735 1.497737 0.002002

0.300000 1.848672 1.843178 0.002981

0.400000 2.287261 2.278311 0.003928

0.500000 2.841096 2.827423 0.004836

0.600000 3.540227 3.520175 0.005696

0.700000 4.421388 4.392800 0.006508

0.800000 5.529473 5.489550 0.007273

0.900000 6.919354 6.864473 0.007995

1.000000 8.658097 8.583587 0.008681

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