Pointers Coming soon..........
Definition:
Syntax:
Uses:
Types of pointer:
Dynamic Memory Allocation:
Examples:
Pros and Cons:
Definition:
Syntax:
Uses:
Types of pointer:
Dynamic Memory Allocation:
Examples:
Pros and Cons:
Monday, July 30, 2012
lvalue and rvalue in c
In c
lvalue are reference to memory their value persist even after execution of statement. lvalue are mainly identifier, variables which can be on both sides of '=' assignment operator.
whereas rvalue are the value(result) of the expression,it's value is persist just for a statement execution, rvalue (value, r optional) can only be on right side of '=' assignment operator
lvalue are reference to memory their value persist even after execution of statement. lvalue are mainly identifier, variables which can be on both sides of '=' assignment operator.
whereas rvalue are the value(result) of the expression,it's value is persist just for a statement execution, rvalue (value, r optional) can only be on right side of '=' assignment operator
for example
1. int x=6;
here x is lvalue and 6 is rvalue.
2. int x=x+y+z;
x- lvalue and (x+y+z) is rvalue.
3. x+5=y+z
its and error because left operand must be lvalue but given is rvalue.
4. Also contact variable are rvalue so following expression is illegal
int const x=6;
x=7;
Sunday, July 29, 2012
How to make Bootable Pendrive
Insert Pen-drive, Open command prompt and type following command one after another.
1. diskpart
2. list disk
3. select disk X
4. clean
5. create partition primary
6. format fs=fat32 quick
7. active
8. assign
9. exit
Now copy your operating system to pendrive.
(Avoid burning CD/DVD for every os...... use bootable pendrive, Save Environment)
Friday, July 20, 2012
C Program for Simphson 1/3 rd rule
/*Simphson's 1/3rd Rule */
#include<stdio.h>
#include<conio.h>
#include<alloc.h>
double value(double);
double **e,n;
void main()
{
double p,i,j,a,b,m,*x,*f,h,odd,even,I;
clrscr();
printf("Enter Degree of the Equation");
scanf("%lf",&n);
p=n;
e=(double**)malloc(((n+1)*2)*sizeof(double));
printf("Enter Values of coeff. in decreasing order");
for(i=0;i<=n;i++)
{
scanf("%lf",&e[i][0]);
e[i][1]=p;
p--;
}
printf("Enter values of a,b,n");
scanf("%lf %lf %lf",&a,&b,&m);
x=(double*)malloc((m+1)*sizeof(double));
f=(double*)malloc((m+1)*sizeof(double));
h=(b-a)/m;
x[0]=a;
for(i=1;i<=m;i++)
{
x[i]=x[i-1]+h;
printf(" %lf",x[i]);
}
for(i=0;i<=m;i++)
{
f[i]=value(x[i]);
printf(" %lf",f[i]);
}
odd=0;
for(i=1;i<m;i+=2)
{
odd+=f[i];
}
even=0;
for(i=2;i<m-1;i+=2)
{
even+=f[i];
}
I=((b-a)/3*n)*(f[0]+4*odd+2*even+f[m+1]);
printf("Result=%lf",I);
getch();
}
double value(double a)
{
double s=0,i;
for(i=0;i<=n;i++)
{
s=s+e[i][0]*pow(a,e[i][1]);
}
return s;
}
C Program for Langrange Interpolating Polynomial
/* L.I.P */
#include<stdio.h>
#include<conio.h>
#include<alloc.h>
float **x;
void main()
{
float n,i,j,*l,*f,R,a;
clrscr();
puts("-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-");
puts("Lagrange Interpolating Polynomial");
puts("-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-");
printf("How many Values you are Entering");
scanf("%f",&n);
puts("-------------------------------------------------------");
x=(float**)malloc(((n+1)*2)*4);
l=(float*)malloc(n*4);
printf("Enter Value of xi and f(xi) Simultaneously");
for(i=0;i<n;i++)
{
scanf("%f %f",&x[i][0],&x[i][1]);
}
puts("-------------------------------------------------------");
printf("Enter value of x");
scanf("%f",&a);
puts("-------------------------------------------------------");
for(i=0;i<n;i++)
{
l[i]=1;
for(j=0;j<n;j++)
{
if(i!=j)
l[i]*=(a-x[j][0])/(x[i][0]-x[j][0]);
}
}
R=0;
for(i=0;i<n;i++)
{
R+=l[i]*x[i][1];
}
printf("Result=%f",R);
puts("\n-------------------------------------------------------");
getch();
}
Output:-
-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-
Lagrange Interpolating Polynomial
-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-
How many Values you are Entering
3
-------------------------------------------------------
Enter Value of xi and f(xi) Simultaneously
1 0
4 1.386294
6 1.791760
-------------------------------------------------------
Enter value of x
2
-------------------------------------------------------
Result=0.565844
-------------------------------------------------------
C Program for Quadratic Interpolation
/*Quadratic Interpolation*/
#include<stdio.h>
#include<conio.h>
void main()
{
float x,b0,b1,b2,x0,x1,x2,f,f0,f1,f2;
clrscr();
puts("*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*");
puts("Quadratic Interpolation");
puts("*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*");
printf("Enter value of x0,f(x0)");
scanf("%f %f",&x0,&f0);
puts("----------------------------------------");
printf("Enter value of x1,f(x1)");
scanf("%f %f",&x1,&f1);
puts("----------------------------------------");
printf("Enter value of x2,f(x2)");
scanf("%f %f",&x2,&f2);
puts("----------------------------------------");
printf("Enter value of x");
scanf("%f",&x);
puts("----------------------------------------");
b0=f0;
b1=(f1-f0)/(x1-x0);
b2=(((f2-f1)/(x2-x1))-((f1-f0)/(x1-x0)))/(x2-x0);
f=b0+b1*(x-x0)+b2*(x-x0)*(x-x1);
printf("b0=%f\nb1=%f\nb2=%f\nf(x)=%f",b0,b1,b2,f);
puts("\n----------------------------------------");
getch();
}
Output:-
*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*
Quadratic Interpolation
*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*
Enter value of x0,f(x0)
1 0
----------------------------------------
Enter value of x1,f(x1)
4 1.5863
----------------------------------------
Enter value of x2,f(x2)
6 1.7918
----------------------------------------
Enter value of x
2
----------------------------------------
b0=0.000000
b1=0.528767
b2=-0.085203
f(x)=0.699173
----------------------------------------
C Program for Linear Interpolation
/*Linear Interpolation*/
#include<stdio.h>
#include<conio.h>
void main()
{
float x1,x0,x,f0,f1,f;
clrscr();
puts("-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-");
puts("Program for Linear Interpolation");
puts("-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*"-);
printf("Enter the values of x0 and f(x0)\n");
scanf("%f %f",&x0,&f0);
puts("----------------------------------------");
printf("Enter the values of x1 and f(x1)");
scanf("%f %f",&x1,&f1);
puts("----------------------------------------");
printf("Enter the values of x ");
scanf("%f",&x);
puts("----------------------------------------");
f=f0+((f1-f0)*(x-x0)/(x1-x0));
printf("%f",f);
puts("\n--------------------------------------");
getch();
}
Output:-
-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-
Program for Linear Interpolation
-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-
Enter the values of x0 and f(x0)
1 0
----------------------------------------
Enter the values of x1 and f(x1)
6 1.791759
----------------------------------------
Enter the values of x
2
----------------------------------------
0.358352
--------------------------------------
C Program for Polynomial Regression
Program:-15
/* Polynmial Regresion */
#include<stdio.h>
#include<conio.h>
#include<alloc.h>
void main()
{
float *x,*y,n,i,x1=0,x12=0,y1=0,x1y1=0,a1,a0,e;
printf("How many values You are Entering");
scanf("%f",&n);
x=(float*)malloc(n*4);
y=(float*)malloc(n*4);
puts("Enter coressponding Elements X & Y");
for(i=0;i<n;i++)
{
scanf("%f %f",&x[i],&y[i]);
x1+=x[i];
y1+=y[i];
x1y1+=(x[i]*y[i]);
x12+=(x[i]*x[i]);
}
a1=(n*x1y1-x1*y1)/(n*x12-x1*x1);
a0=y1/n-a1*x1/n;
e=y[0]-a0-a1*x[0];
printf("ao=%f\na1=%f\ne%f",a0,a1,e);
}
Output of Program:-
How many values You are Entering
6
Enter coressponding Elements X & Y
0 2.1
1 7.7
2 13.6
3 27.2
4 40.4
5 61.1
Matrix
6.000000 15.000000 55.000000 152.100006
15.000000 55.000000 225.000000 583.599976
55.000000 225.000000 979.000000 2480.800049
Required Equation
Y=4.632166+2.271759X+1.869648X2
C Program for Multiple Regression
/*Multiple Regression */
#include<stdio.h>
#include<conio.h>
#include<alloc.h>
#include<math.h>
void main()
{
float f,k,sum=0,a[3][4],*x1,*x2,*y,n,i,j,p[]={0,0,0};
float x_1=0,x_2=0,x12=0,x22=0,x1x2=0,y1=0,x1y1=0,x2y1=0,a1,a0,e;
clrscr();
printf("Multiple Regression");
printf("\nHow many values You are Entering");
scanf("%f",&n);
x1=(float*)malloc(n*4);
x2=(float*)malloc(n*4);
y=(float*)malloc(n*4);
puts("Enter coressponding Elements X1,X2 & Y");
for(i=0;i<n;i++)
{
scanf("%f %f %f",&x1[i],&x2[i],&y[i]);
x_1+=x1[i];
x_2+=x2[i];
x1x2+=x1[i]*x2[i];
x12+=x1[i]*x1[i];
x22+=x2[i]*x2[i];
y1=y1+y[i];
x1y1=x1y1+x1[i]*y[i];
x2y1=x2y1+x2[i]*y[i];
}
a[0][0]=n;
a[0][1]=a[1][0]=x_1;
a[0][2]=a[2][0]=x_2;
a[1][2]=a[2][1]=x1x2;
a[1][1]=x12;
a[2][2]=x22;
a[0][3]=y1;
a[1][3]=x1y1;
a[2][3]=x2y1;
puts("\nMatrix");
for(i=0;i<3;i++)
{
for(j=0;j<4;j++)
{
printf("\t%6.6f",a[i][j]);
}
printf("\n");
}
/* Gauss Elimination */
for(k=0;k<2;k++)
{
for(i=k+1;i<3;i++)
{
for(i=k+1;i<3;i++)
{
f=a[i][k]/a[k][k];
for(j=0;j<4;j++)
}
}
for(i=2;i>=0;i--)
{
sum=a[i][3];
for(j=0;j<3;j++)
{
sum=sum-a[i][j]*p[j];
}
p[i]=sum/a[i][i];
}
e=y[0]-p[0]-p[1]*x1[0]-p[2]*x2[0];
p[0]+=e;
puts("Required Equation");
printf("Y=%f+%fX1+%fX2",p[0],p[1],p[2]);
}
Output of Program:-
Multiple Regression
How many values You are Entering
6
Enter coressponding Elements X1,X2 & Y
0 0 5
2 1 10
2.5 2 9
1 3 0
4 6 3
7 2 27
Matrix
6.000000 16.500000 14.000000 54.000000
16.500000 76.250000 48.000000 243.500000
14.000000 48.000000 54.000000 100.000000
Required Equation
Y=5.000000+4.000000X1+-3.000000X2
C Program for Gauss Sedial Method
/* Gauss Sedial Method */
#include<stdio.h>
#include<conio.h>
void main()
{
float a[3][4],i,j,k,sum,x[]={0,0,0};
clrscr();
printf("Enter coeff. in format\nax1+bx2+c3=d");
for(i=0;i<3;i++)
{
for(j=0;j<4;j++)
{
scanf("%f",&a[i][j]);
}
}
for(k=0;k<2;k++)
{
for(i=0;i<3;i++)
{
sum=a[i][3];
for(j=0;j<3;j++)
{
if(i!=j)
sum=sum-a[i][j]*x[j];
}
x[i]=sum/a[i][i];
}
puts("\n-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-");
printf("\nValue After %d Iteration\n",k+1);
for(i=0;i<3;i++)
{
printf("\tx%d=%f",(int)i+1,x[i]);
}
puts("\n-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-");
}
getch();
}
Output Of Program:-
Enter coeff. in format
ax1+bx2+c3=d
3 -0.1 -0.2 7.85
0.1 7 -0.3 -19.3
0.3 -0.2 10 71.4
-*-*-*-*-*-*-*-*-*-*-*-*-*-**-*-*-*-*-*-*-*-*-*-*-
Value After 0 Iteration
x1=2.616667 x2=-2.794524 x3=7.005609
-*-*-*-*-*-*-*-*-*-*-*-*-*-**-*-*-*-*-*-*-*-*-*-*-
-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*--*-*-*-*-*-*-*-*-*-
Value After 0 Iteration
x1=2.990556 x2=-2.499624 x3=7.000291
-*-*-*-*-*-*-*-*-*-*-**-*-*-*-*-*-*-*-*-*-*-*-*-*-
C Program for Linear Regression
Program:-14
/*Linear Regression*/
#include<stdio.h>
#include<conio.h>
#include<alloc.h>
void main()
{
float *x,*y,n,i,x1=0,x12=0,y1=0,x1y1=0,a1,a0,e;
puts("-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*");
printf("How many values You are Entering");
scanf("%f",&n);
puts("-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*");
x=(float*)malloc(n*4);
y=(float*)malloc(n*4);
puts("-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*");
puts("Enter coressponding Elements X & Y");
for(i=0;i<n;i++)
{
scanf("%f %f",&x[i],&y[i]);
x1+=x[i];
y1+=y[i];
x1y1+=(x[i]*y[i]);
x12+=(x[i]*x[i]);
}
a1=(n*x1y1-x1*y1)/(n*x12-x1*x1);
a0=y1/n-a1*x1/n;
e=y[0]-a0-a1*x[0];
a0+=e;
puts("-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*");
printf("\nY=%f+%fX",a0,a1);
puts("-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*");
}
Output of Program:-
-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-
How many values You are Entering7
-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-
Enter coressponding Elements X & Y
1 0.5
2 2.5
3 2
4 4
5 3.5
6 6
7 5.8
-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-
Y=-0.371429+0.871429X
-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-
C Program for Navive Gauss Elimination
/* Naive Gauss Elimination */
#include<stdio.h>
#include<conio.h>
void main()
{
float a[3][4],i,j,k,f,sum,x[]={0,0,0};
clrscr();
printf("Enter coeff. in format\nax1+bx2+c3=d");
for(i=0;i<3;i++)
{
for(j=0;j<4;j++)
{
scanf("%f",&a[i][j]);
}
}
for(k=0;k<2;k++)
{
for(i=k+1;i<3;i++)
{
f=a[i][k]/a[k][k];
for(j=0;j<4;j++)
{
a[i][j]=a[i][j]-a[k][j]*f;
}
}
}
printf("\nForward Elimination\n");
for(i=0;i<3;i++)
{
for(j=0;j<4;j++)
{
printf("\t %10.6f",a[i][j]);
}
printf("\n");
}
printf("Backward substituation\n");
for(i=2;i>=0;i--)
{
sum=a[i][3];
for(j=0;j<3;j++)
{
sum=sum-a[i][j]*x[j];
}
x[i]=sum/a[i][i];
}
for(i=0;i<3;i++)
{
printf("\tx%d=%f",i,x[i]);
}
getch();
}
Output Of Program:-
Enter coeff. in format
ax1+bx2+c3=d
3 -0.1 -0.2 7.85
0.1 7 -0.3 -19.3
0.3 -0.2 10 71.6
Forward Elimination
3.000000 -0.100000 -0.200000 7.850000
-0.000000 7.003334 -0.293333 -19.561666
0.000000 0.000000 10.012042 70.284286
Backward substituation
x1=3.001360 x2=-2.499163 x3=7.019975
C Program For Secant method 2
/*Secant Method*/
#include<stdio.h>
#include<conio.h>
#include<math.h>
#include<alloc.h>
#include<process.h>
#define f(x) sin(x)+cos(1+x*x)-1
double deri(double);
void main()
{
double i,p,it=50;
double v,xo,x1,ea,x_1;
char ch;
clrscr();
puts("-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-");
puts("------- Finding Root Using Secant Method --------");
puts("-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-");
puts("-------------------------------------------------");
puts("Enter Value of X-1,X0\n");
scanf("%lf %lf",&x_1,&xo);
puts("------------------------------------------------");
puts("\nEnter no. of Iteration\n");
scanf("%lf",&it);
puts("------------------------------------------------");
i=0;
puts("Iteration \tXi\t\tEa");
while(1)
{
i++;
x1=xo-(f(xo)*(x_1-xo))/(f(x_1)-f(xo));
ea=(x1-xo)*100/x1;
if(ea<0) /*-- Eliminating error due to --*/
ea=ea*(-1); /*-- Negative sign of Ea --*/
if(f(x1)==0||i==it)
{
printf("\n%3d\t\t%5.6lf\t%5.4lf",(int)i,x1,ea);
puts("\n---------------------------------------------");
getch();
exit(0);
}
printf("\n%3d\t\t%5.6lf\t%5.4lf",(int)i,x1,ea);
x_1=xo;
xo=x1;
}
getch();
}
Output Of Program:-
-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*
------- Finding Root Using Secant Method --------
-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-
-------------------------------------------------
Enter Value of X-1,X0
1 3
------------------------------------------------
Enter no. of Iteration
10
------------------------------------------------
Iteration Xi Ea
1 3.509634 14.5210
2 4.057609 13.5049
3 3.964200 2.3563
4 3.399639 16.6065
5 3.489194 2.5667
6 9.558526 63.4965
7 12.883433 25.8076
8 13.638773 5.5382
9 14.151677 3.6243
10 16.520063 4.3364
--------------------------------------------
C Program for Secant Method
/*-*Secant Method*-*/
#include<stdio.h>
#include<conio.h>
#include<math.h>
#include<alloc.h>
#include<process.h>
float value(float);
float **e,n;
void main()
{
int i,p,it=50;
float v,xo,es,s=0,x1,ea,x_1;
char ch;
clrscr();
printf("Enter Degree of the Equation\n");
scanf("%f",&n);
puts("-------------------------------------------------");
e=(float**)malloc(((n+1)*2)*sizeof(float));
printf("Enter coeff.in decreasing order of Degree\n");
for(i=0,p=n;i<=n;i++,p--)
{
scanf("%f",&e[i][0]);
e[i][1]=p;
}
puts("-------------------------------------------------");
printf("Enter Value of X-1,X0\n");
scanf("%f %f",&x_1,&xo);
puts("------------------------------------------------");
printf("What would You Like to enter Es or Sig.fig or Iteration =>E/S/I");
ch=getche();
if(ch=='E'||ch=='e')
{
printf("\nEnter value of Es\n");
scanf("%d",&es);
}
else if(ch=='s'||ch=='S')
{
printf("\nEnter number of Sig. figure\n");
scanf("%d",s);
es=0.5*pow(10,(s-2));
}
else
{
printf("\nEnter no. of Iteration\n");
scanf("%d",&it);
}
puts("------------------------------------------------");
i=0;
printf("Iteration \tXi\t\tEa");
while(1)
{
i++;
x1=xo-(value(xo)*(x_1-xo))/(value(x_1)-value(xo));
ea=(x1-xo)*100/x1;
if(ea<0)
ea=ea*(-1);
if(ea<es||value(x1)==0||i==it)
{
printf("\n%3d\t\t%5.6f\t%5.4f",i,x1,ea);
puts("\n---------------------------------------------");
getch();
exit(0);
}
printf("\n%3d\t\t%5.6f\t%5.4f",i,x1,ea);
x_1=xo;
xo=x1;
}
}
float value(float x)
{
int i;
float s=0;
for(i=0;i<=n;i++)
{
s=s+e[i][0]*pow(x,e[i][1]);
}
return s;
}
Output of Program:-
Enter Degree of the Equation
3
-------------------------------------------------
Enter coeff.in decreasing order of Degree
1 -6 11 -6.1
-------------------------------------------------
Enter Value of X-1,X0
2.5 3.5
------------------------------------------------
What would You Like to enter Es or Sig.fig or Iteration =>E/S/Ii
Enter no. of Iteration
3
------------------------------------------------
Iteration Xi Ea
1 2.711111 29.0984
2 2.871090 5.5721
3 3.221925 10.8890
---------------------------------------------
C Program for Newton Raphson Method
/*Newton Raphson Method*/
#include<stdio.h>
#include<conio.h>
#include<math.h>
#include<alloc.h>
#include<process.h>
float value(float);
float **e,n;
float deri(float);
void main()
{
int i,p,it=50;
float v,xo,es=0,s,x1,ea;
char ch;
clrscr();
printf("Enter Degree of the Equation\n");
scanf("%f",&n);
puts("--------------------------------------------------");
e=(float**)malloc(((n+1)*2)*sizeof(float));
printf("Enter coeff.in decreasing order of Degree\n");
for(i=0,p=n;i<=n;i++,p--)
{
scanf("%f",&e[i][0]);
e[i][1]=p;
}
puts("-------------------------------------------------");
printf("Enter Value of Xo\n");
scanf("%f",&xo);
puts("-------------------------------------------------");
printf("What would You Like to enter Es or Sig.fig or Iteration =>E/S/I\n");
ch=getche();
puts("\n----------------------------------------------“);
if(ch=='E'||ch=='e')
{
printf("Enter value of Es\n");
scanf("%f",&es);
}
else if(ch=='s'||ch=='S')
{
printf("Enter number of Sig. figure\n");
scanf("%f",&s);
es=0.5*pow(10,(s-2));
}
else
{
printf("Enter no. of Iteration\n");
scanf("%d",&it);
}
i=0;
puts("-------------------------------------------------");
printf("Iteration\tX\t\tEa");
while(1)
{
i++;
x1=xo-value(xo)/deri(xo);
ea=(x1-xo)*100/x1;
if(ea<0)
ea=ea*(-1);
if(ea<es||value(x1)==0||i==it)
{
printf("\n%3d\t\t%6.5f\t\t%5.3f",i,x1,ea);
puts("\n---------------------------------------------");
getch();
exit(0);
}
printf("\n%3d\t\t%6.5f\t\t%5.3f",i,x1,ea);
xo=x1;
}
float value(float x)
{
int i;
float s=0;
for(i=0;i<=n;i++)
{
s=s+e[i][0]*pow(x,e[i][1]);
}
return s;
}
float deri(float x)
{
int i;
float s=0,t,h;
for(i=0;i<=n;i++)
{
t=e[i][0]*e[i][1];
if(e[i][1]>0)
h=e[i][1]-1;
else
h=0;
s=s+t*pow(x,h);
}
return s;
Output Of Program:-
First:-
Enter Degree of the Equation
2
--------------------------------------------------
Enter coeff.in decreasing order of Degree
-1 1.8 2.5
-------------------------------------------------
Enter Value of Xo
5
-------------------------------------------------
What would You Like to enter Es or Sig.fig or Iteration =>E/S/I
e
-------------------------------------------------
Enter value of Es
0.05
-------------------------------------------------
Iteration X Ea
1 3.35366 49.091
2 2.80133 19.717
3 2.72111 2.948
4 2.71934 0.065
5 2.71934 0.000
-----------------------------------------------
===============================
Second:-
Enter Degree of the Equation
3
--------------------------------------------------
Enter coeff.in decreasing order of Degree
2 -11.7 17.7 -5
-------------------------------------------------
Enter Value of Xo
3
-------------------------------------------------
What would You Like to enter Es or Sig.fig or Iteration =>E/S/I
i
-------------------------------------------------
Enter no. of Iteration
3
-------------------------------------------------
Iteration X Ea
1 5.13333 41.558
2 4.26975 20.226
3 3.79293 12.571
-----------------------------------------------
C Program for False Position Method 2
/*False Position Method*/
#include<stdio.h>
#include<conio.h>
#include<alloc.h>
#include<math.h>
#include<process.h>
#define f(x) log10(x*x*x)-0.7
double value(double);
void main()
{
double x,v,v1,v2,xr,xu,xl,xo=0,ea,et,tv;
double i,p,it=50;
clrscr();
puts("-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*");
puts("Finding Root By Using False Position Method");
puts("-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*");
puts("-------------------------------------------");
puts("Enter value of xl and xu ");
scanf("%lf %lf ",&xl,&xu);
puts("---------------------------------------------");
puts("Enter True Value and Iterations");
scanf("%lf %lf",&tv,&it);
puts("--------------------------------------------------");
i=0;
puts("Iteration\t Xr \t\t Ea\t\tEt");
while(1)
{
i++;
v1=f(xl);
v2=f(xu);
if(v1*v2>=0)
{
puts("Invalid Value of xl and xu");
getch();
exit(0);
}
xr=xu-(v2*(xl-xu)/(v1-v2));
ea=(xr-xo)*100/xr;
et=(tv-xr)*100/tv;
if(ea<0) /*--- Eliminating Error due to ---*/
ea=ea*-1; /*--- Negative sign of Ea ---*/
v=f(xr);
if(v==0||i==it)
{
printf("\n%3d\t\t%6.6lf\t\t%5.6lf\t\t%5.6lf",(int)i,xr,ea,et);
puts("\n-----------------------------------------------------------");
getch();
exit(0);
}
printf("\n%3d\t\t%6.6lf\t\t%5.6lf\t\t%5.6lf",(int)i,xr,ea,et);
if(v1*v<0)
xu=xr;
else
xl=xr;
xo=xr;
}
getch();
}
Output Of Program:-
-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-
Finding Root By Using False Position Method
-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-
---------------------------------------------------------------
Enter value of xl and xu
1
3
---------------------------------------------------------------
Enter True Value and Iterations
1.262803
10
---------------------------------------------------------------
Iteration Xr Ea Et
1 1.978088 100.000000 -56.642659
2 1.770376 11.732652 -40.194166
3 1.724625 2.652832 -36.571162
4 1.714334 0.600252 -35.756283
5 1.712009 0.135855 -35.572101
6 1.711482 0.030750 -35.530425
7 1.711363 0.006960 -35.520992
8 1.711336 0.001575 -35.518857
9 1.711330 0.000357 -35.518374
10 1.711329 0.000081 -35.518264
---------------------------------------------------------------
C Program for False Position Method
/*False Position Method*/
#include<stdio.h>
#include<conio.h>
#include<alloc.h>
#include<math.h>
#include<process.h>
float **e,n;
float value(float);
void main()
{
float x,v,v1,v2,xr,xu,xl,xo=0,ea,es;
int i,p,it=50;
char ch;
clrscr();
printf("Enter degree\n");
scanf("%f",&n);
e=(float**)malloc(((n+1)*2)*4);
puts("------------------------------------------");
printf("Enter coeff.in decreasing order of degree\n");
for(i=0,p=n;i<=n;i++,p--)
{
scanf("%f",&e[i][0]);
e[i][1]=p;
}
puts("------------------------------------------");
printf("Enter value of xl and xu\n");
scanf("%f %f",&xl,&xu);
puts("------------------------------------------");
puts("What would you like to Enter\nEs or Number of significant fig or Iteration .=>E/S/I");
ch=getche();
if(ch=='E'||ch=='e')
{
printf("\nEnter value of Es\n");
scanf("%f",&es);
}
else if(ch=='i'||ch=='I')
{printf("Enter no. of Iteration"); scanf("%d",&it);
}
else
{
printf("\nEnter no. of sig. fig\n");
scanf("%d",&p);
es=0.5*pow(10,(2-p));
}
puts("-------------------------------------------");
i=0;
printf("Iteration\tX\t\tEa");
while(1)
{
i++; /*-- Iteration number --*/
v1=value(xl);
v2=value(xu);
if(v1*v2>=0)
{
printf("Invalid Value of xl and xu");
getch();
exit(0);
}
xr=xu-(v2*(xl-xu)/(v1-v2));
ea=(xr-xo)*100/xr;
if(ea<0)
ea=ea*-1;
v=value(xr);
if(v==0||ea<es)
{
printf("\n%3d\t\t%6.5f\t\t%5.4f",i,xr,ea);
puts("\n-----------------------------------------------");
getch();
exit(0);
}
printf("\n%3d\t\t%6.5f\t\t%5.4f",i,xr,ea);
if(v1*v<0)
xu=xr;
else
xl=xr;
xo=xr; }
}
float value(float x)
{
int i;
float s=0;
for(i=0;i<=n;i++)
{
s=s+e[i][0]*pow(x,e[i][1]);
}
return s;
}
Output Of Program:-
First:-
Enter degree
2
------------------------------------------
Enter coeff.in decreasing order of degree
-0.4 2.2 4.7
------------------------------------------
Enter value of xl and xu
5 10
------------------------------------------
What would you like to Enter
Es or Number of significant fig.or Iteration=>E/S/I
e
Enter value of Es
1
-------------------------------------------
Iteration X Ea
1 6.50000 100.0000
2 6.97727 6.8404
3 7.10297 1.7696
4 7.13435 0.4399
-----------------------------------------------
Second:-
Enter degree
2
------------------------------------------
Enter coeff.in decreasing order of degree
-0.5 2.5 4.5
------------------------------------------
Enter value of xl and xu
5 10
------------------------------------------
What would you like to Enter
Es or Number of significant fig.or Iteration=>E/S/I
I
Enter no. of Iteration
4
-------------------------------------------
Iteration X Ea
1 5.90000 100.0000
2 6.23853 5.4265
3 6.35184 1.7838
4 6.38825 0.5700
-----------------------------------------------
C Program for Bisection Method 2
/*Bisection Method*/
#include<stdio.h>
#include<conio.h>
#include<alloc.h>
#include<math.h>
#include<process.h>
#define f(x) log10(x*x*x)-0.7
double value(double);
void main()
{
double x,v,v1,v2,xr,xu,xl,xo=0,ea,et,tv;
double i,p,it=50;
clrscr();
puts("-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*");
puts("Finding Root By Using Bisection Method");
puts("-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*");
puts("--------------------------------------------------");
puts("Enter value of xl and xu ");
scanf("%lf %lf ",&xl,&xu);
puts("--------------------------------------------------");
puts("Enter True Value and Iterations");
scanf("%lf %lf",&tv,&it);
puts("--------------------------------------------------");
i=0;
puts("Iteration\t Xr \t\t Ea\t\tEt");
while(1)
{
i++;
v1=f(xl);
v2=f(xu);
if(v1*v2>=0)
{
puts("Invalid Value of xl and xu");
getch();
exit(0);
}
xr=(xl+xu)/2.0;
ea=(xr-xo)*100/xr;
et=(tv-xr)*100/tv;
if(ea<0)
ea=ea*-1;
v=f(xr);
if(v==0||i==it)
{
printf("\n%3d\t\t%6.5lf\t\t%5.6lf\t\t%5.6lf",(int)i,xr,ea,et);
puts("\n----------------------------------------------");
getch();
exit(0);
}
printf("\n%3d\t\t%6.5lf\t\t%5.6lf\t\t%5.6lf",(int)i,xr,ea,et);
if(v1*v<0)
xu=xr;
else
xl=xr;
xo=xr;
}
getch();
}
Output Of Program:-
-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*
Finding Root By Using Bisection Method
-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*
---------------------------------------------------------------Enter value of xl and xu
1
3
---------------------------------------------------------------
Enter True Value and Iterations
1.262803
10
---------------------------------------------------------------
Iteration Xr Ea Et
1 2.00000 100.000000 -58.377831
2 1.50000 33.333333 -18.783373
3 1.75000 14.285714 -38.580602
4 1.62500 7.692308 -28.681988
5 1.68750 3.703704 -33.631295
6 1.71875 1.818182 -36.105948
7 1.70312 0.917431 -34.868622
8 1.71094 0.456621 -35.487285
9 1.71484 0.227790 -35.796617
10 1.71289 0.114025 -35.641951
--------------------------------------------------------------
C Program for bisection method
/*Bisection Method*/
#include<stdio.h>
#include<conio.h>
#include<alloc.h>
#include<math.h>
#include<process.h>
float **e,n;
float value(float);
void main()
{
float x,v,v1,v2,xr,xu,xl,xo=0,ea,es;
int i,p,it=50;
char ch;
clrscr();
puts("-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*");
puts("Finding Root By Using Bisection Method");
puts("-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*");
printf("Enter degree\n");
scanf("%f",&n);
e=(float**)malloc(((n+1)*2)*4);
puts("------------------------------------------");
printf("Enter coeff.in decreasing order of degree\n");
for(i=0,p=n;i<=n;i++,p--)
{
scanf("%f",&e[i][0]);
e[i][1]=p;
}
puts("------------------------------------------");
printf("Enter value of xl and xu\n");
scanf("%f %f",&xl,&xu);
puts("------------------------------------------");
puts("What would you like to Enter\nEs or Number of sig fig.or Iteration=>E/S/I");
ch=getche();
if(ch=='E'||ch=='e')
{
printf("\nEnter value of Es\n");
scanf("%f",&es);
}
else if(ch=='i'||ch=='I')
{
printf("\nEnter no. of Iteration");
scanf("%d",&it);
}
else
{
printf("\nEnter no. of sig. fig\n");
scanf("%f",&p);
es=0.5*pow(10,(2-p));
}
puts("-------------------------------------------");
i=0;
puts("Iteration\t Xr \t\t Ea");
while(1)
{
i++;
v1=value(xl);
v2=value(xu);
if(v1*v2>=0)
{
printf("Invalid Value of xl and xu");
getch();
exit(0);
}
xr=(xl+xu)/2.0;
ea=(xr-xo)*100/xr;
if(ea<0)
ea=ea*-1;
v=value(xr);
if(v==0||ea<es||i==it)
{
printf("\n%3d\t\t%6.5f\t\t%5.3f",i,xr,ea);
puts("\n---------------------------------------------");
getch();
exit(0);
}
printf("\n%3d\t\t%6.5f\t\t%5.3f",i,xr,ea);
if(v1*v<0)
xu=xr;
else
xl=xr;
xo=xr;
}
}
float value(float x)
{
int i;
float s=0;
for(i=0;i<=n;i++)
{
s=s+e[i][0]*pow(x,e[i][1]);
}
return s;
}
Output Of Program:-
First:-
Enter degree
2
------------------------------------------
Enter coeff.in decreasing order of degree
-0.4 2.2 4.7
------------------------------------------
Enter value of xl and xu
5 10
------------------------------------------
What would you like to Enter
Es or Number of sig fig.or Iteration=>E/S/I
e
Enter value of Es
1
-------------------------------------------
Iteration Xr Ea
1 7.50000 100.000
2 6.25000 20.000
3 6.87500 9.091
4 7.18750 4.348
5 7.03125 2.222
6 7.10938 1.099
7 7.14844 0.546
-----------------------------------------------
Second:-
Enter degree
5
------------------------------------------
Enter coeff.in decreasing order of degree
0.65 -9 45.5 -88 82.3 -26
------------------------------------------
Enter value of xl and xu
0.5 1
------------------------------------------
What would you like to Enter
Es or Number of sig fig.or Iteration=>E/S/I
e
Enter value of Es
10
-------------------------------------------
Iteration Xr Ea
1 0.75000 100.000
2 0.62500 20.000
3 0.56250 11.111
4 0.59375 5.263
----------------------------------------------
C Program to find e^x
/* To find e^x */
#include<stdio.h>
#include<conio.h>
#include<math.h>
#define f0(x) pow(2.34176,x)
#define f1(x) pow(2.34176,x)
#define f2(x) pow(2.34176,x)
#define f3(x) pow(2.34176,x)
#define f4(x) pow(2.34176,x)
void main()
{
float t,x,h,v,i=0,ea=100;
clrscr();
puts("-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*");
puts("Enter value of x & h to find ln(x)");
scanf("%f %f",&x,&h);
puts("-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*");
puts("------------------------------------------------");
puts("Iter. ln(x+h) \tEa");
puts("------------------------------------------------");
v=f0(x);
printf("%2.0f\t%f\t%f",i++,v,ea);
t=v;
v=v+h*f1(x);
ea=(v-t)*100/v;
printf("\n%2.0f\t%f\t%f",i++,v,ea);
t=v;
v=v+h*h*f2(x)/2;
ea=(v-t)*100/v;
printf("\n%2.0f\t%f\t%f",i++,v,ea);
t=v;
v=v+h*h*h*f3(x)/(3*2);
ea=(v-t)*100/v;
printf("\n%2.0f\t%f\t%f",i++,v,ea);
t=v;
v=v+h*h*h*h*f4(x)/(4*3*2);
ea=(v-t)*100/v;
printf("\n%2.0f\t%f\t%f",i++,v,ea);
puts("\n------------------------------------------------");
getch();
}
Output of Program:-
First:-
Enter value of x & h to find e^(x+h)
2 1
-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*
------------------------------------------------
Iter. e^(x+h) Ea
------------------------------------------------
0 5.483840 100.000000
1 10.967680 50.000000
2 13.709599 19.999998
3 14.623572 6.249997
4 14.852066 1.538464
------------------------------------------------
Second:-
-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*
Enter value of x & h to find e^(x+h)
5 0.5
-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*
------------------------------------------------
Iter. e^(x+h) Ea
------------------------------------------------
0 70.422577 100.000000
1 105.633865 33.333332
2 114.436691 7.692311
3 115.903831 1.265826
4 116.087227 0.157981
------------------------------------------------
C program to find sin(x)
/* To find sin(x) */
#include<stdio.h>
#include<conio.h>
#include<math.h>
#define f0(x) sin(x)
#define f1(x) cos(x)
#define f2(x) -sin(x)
#define f3(x) -cos(x)
#define f4(x) sin(x)
void main()
{
float t,x,h,v,i=0,ea=100;
clrscr();
puts("Enter value of x & h to find sin(x)");
scanf("%f %f",&x,&h);
x=x*3.14231/180;
h=h*3.14231/180;
puts("------------------------------------------------");
puts("Iter. ln(x+h) \tEa");
puts("------------------------------------------------");
v=f0(x);
printf("%2.0f\t%f\t%f",i++,v,ea);
t=v;
v=v+h*f1(x);
ea=(v-t)*100/v;
printf("\n%2.0f\t%f\t%f",i++,v,ea);
t=v;
v=v+h*h*f2(x)/2;
ea=(v-t)*100/v;
printf("\n%2.0f\t%f\t%f",i++,v,ea);
t=v;
v=v+h*h*h*f3(x)/(3*2);
ea=(v-t)*100/v;
printf("\n%2.0f\t%f\t%f",i++,v,ea);
t=v;
v=v+h*h*h*h*f4(x)/(4*3*2);
ea=(v-t)*100/v;
printf("\n%2.0f\t%f\t%f",i++,v,ea);
puts("\n-----------------------------------------------");
getch();
}
Output of Program:-
Enter value of x & h to find sin(x)
44 1
------------------------------------------------
Iter. sin(x+h) Ea
------------------------------------------------
0 0.694658 100.000000
1 0.707213 1.775254
2 0.707107 -0.014962
3 0.707106 -0.000093
4 0.707106 0.000000
------------------------------------------------
C Program to find ln(x)
Program.no:-2.
/* To find ln(x) */
#include<stdio.h>
#include<conio.h>
#include<math.h>
#define f0(x) log(x)
#define f1(x) 1/x
#define f2(x) -1/(x*x)
#define f3(x) 2/(x*x*x)
void main()
{
float tv,t,x,h,v,i=0,et=100,ea=100;
clrscr();
puts("Enter value of x & h to find ln(x)");
scanf("%f %f",&x,&h);
tv=f0(x+h);
puts("-------------------------------------------------");
puts("Iter. ln(x+h) \tEa\t\tEt");
puts("-------------------------------------------------");
v=f0(x);
printf("%2.0f\t%f",i++,v);
t=v;
v=v+h*f1(x);
ea=(v-t)*100/v;et=(tv-v)*100/tv;
printf("\n%2.0f\t%f\t%f\t%f",i++,v,ea,et);
t=v;
v=v+h*h*f2(x)/2;
ea=(v-t)*100/v;et=(tv-v)*100/tv;
printf("\n%2.0f\t%f\t%f\t%f",i++,v,ea,et);
t=v;
v=v+h*h*h*f3(x)/(3*2);
ea=(v-t)*100/v;et=(tv-v)*100/tv;
printf("\n%2.0f\t%f\t%f\t%f",i++,v,ea,et);
puts("\n----------------------------------------------");
getch();
}
Output of Program:-
First:-
Enter value of x & h to find ln(x)
1 3
-------------------------------------------------------------
Iter. ln(x+h) Ea Et
-------------------------------------------------------------
0 0.000000
1 3.000000 100.000000 -116.404259
2 -1.500000 300.000000 208.202133
3 7.500000 120.000000 -441.010651
-------------------------------------------------------------
Second:-
Enter value of x & h to find ln(x)
3 1
--------------------------------------------------------------
Iter. ln(x+h) Ea Et
--------------------------------------------------------------
0 1.098612
1 1.431946 23.278353 -3.293046
2 1.376390 -4.036325 0.714442
3 1.388736 0.888986 -0.176110
---------------------------------------------------------------
C Program for Taylor Series
/*Taylor Series*/
/*ax^n+bx^(n-1)+..........*/
#include<stdio.h>
#include<conio.h>
#include<math.h>
#include<alloc.h>
float **m,x,n,**o,*f;
float value(float);
float sum(float);
float deri();
int fact(int);
void main()
{
int i,p;
float h,ea;
clrscr();
printf("Enter the Degree of the Equation");
scanf("%f",&n);
puts("--------------------------------------------------");
m=(float**)malloc((n+1)*2)*sizeof(float));
f=(float*)malloc(n*4);
printf("Enter the value of coefficient\n");
p=n;
for(i=0;i<=n;i++)
{
scanf("%f",&m[i][0]);
m[i][1]=p;
p--;
}
puts("--------------------------------------------------");
printf("Enter value of x\n");
scanf("%f",&x);
puts("--------------------------------------------------");
printf("Enter value of h\n");
scanf("%f",&h);
puts("-------------------------------------------------");
for(i=0;i<=n;i++)
{
if(i==0)
f[i]=value(x);
else
f[i]=deri();
printf("\nf%d(x)=%f",i,f[i]);
}
puts("\n-----------------------------------------------");
printf("\nResult=f(x+h)=%f",sum(h));
puts("\n-----------------------------------------------");
getch();
}
float sum(float h)
{
int i;
float s=0;
for(i=0;i<=n;i++)
{
s=s+pow(h,i)*f[i]/fact(i);
}
return s;
}
float value(float x)
{
int i;
float v=0;
for(i=0;i<=n;i++)
{
v=v+m[i][0]*pow(x,m[i][1]);
}
return v;
}
float deri()
{
float v=0;
int i,t;
for(i=0;i<=n;i++)
{
t=m[i][1];
m[i][1]--;
m[i][0]=m[i][0]*t;
if(m[i][1]<0)
break;
v=v+m[i][0]*pow(x,m[i][1]);
}
return v;
}
int fact(int i)
{
int j,fact=1;
for(j=i;j>0;j--)
{
fact=fact*j;
}
return fact;
}
Output of Program:-
First:-
Enter the Degree of the Equation2
------------------------------------------------------
Enter the value of coefficient
1 1 1
-------------------------------------------------------
Enter value of x
2
------------------------------------------------------
Enter value of h
1
------------------------------------------------------
f0(x)=7.000000
f1(x)=5.000000
f2(x)=2.000000
-------------------------------------------------------
Result=f(x+h)=13.000000
-------------------------------------------------------
=============================
Second:-
Enter the Degree of the Equation
3
------------------------------------------------------
Enter the value of coefficient
2.2 -5.3 6.2 2.5
-------------------------------------------------------
Enter value of x
2
------------------------------------------------------
Enter value of h
1
------------------------------------------------------
f0(x)=11.299999
f1(x)=11.400001
f2(x)=15.800001
f3(x)=13.200001
-------------------------------------------------------
Result=f(x+h)=32.800003
-------------------------------------------------------
/*ax^n+bx^(n-1)+..........*/
#include<stdio.h>
#include<conio.h>
#include<math.h>
#include<alloc.h>
float **m,x,n,**o,*f;
float value(float);
float sum(float);
float deri();
int fact(int);
void main()
{
int i,p;
float h,ea;
clrscr();
printf("Enter the Degree of the Equation");
scanf("%f",&n);
puts("--------------------------------------------------");
m=(float**)malloc((n+1)*2)*sizeof(float));
f=(float*)malloc(n*4);
printf("Enter the value of coefficient\n");
p=n;
for(i=0;i<=n;i++)
{
scanf("%f",&m[i][0]);
m[i][1]=p;
p--;
}
puts("--------------------------------------------------");
printf("Enter value of x\n");
scanf("%f",&x);
puts("--------------------------------------------------");
printf("Enter value of h\n");
scanf("%f",&h);
puts("-------------------------------------------------");
for(i=0;i<=n;i++)
{
if(i==0)
f[i]=value(x);
else
f[i]=deri();
printf("\nf%d(x)=%f",i,f[i]);
}
puts("\n-----------------------------------------------");
printf("\nResult=f(x+h)=%f",sum(h));
puts("\n-----------------------------------------------");
getch();
}
float sum(float h)
{
int i;
float s=0;
for(i=0;i<=n;i++)
{
s=s+pow(h,i)*f[i]/fact(i);
}
return s;
}
float value(float x)
{
int i;
float v=0;
for(i=0;i<=n;i++)
{
v=v+m[i][0]*pow(x,m[i][1]);
}
return v;
}
float deri()
{
float v=0;
int i,t;
for(i=0;i<=n;i++)
{
t=m[i][1];
m[i][1]--;
m[i][0]=m[i][0]*t;
if(m[i][1]<0)
break;
v=v+m[i][0]*pow(x,m[i][1]);
}
return v;
}
int fact(int i)
{
int j,fact=1;
for(j=i;j>0;j--)
{
fact=fact*j;
}
return fact;
}
Output of Program:-
First:-
Enter the Degree of the Equation2
------------------------------------------------------
Enter the value of coefficient
1 1 1
-------------------------------------------------------
Enter value of x
2
------------------------------------------------------
Enter value of h
1
------------------------------------------------------
f0(x)=7.000000
f1(x)=5.000000
f2(x)=2.000000
-------------------------------------------------------
Result=f(x+h)=13.000000
-------------------------------------------------------
=============================
Second:-
Enter the Degree of the Equation
3
------------------------------------------------------
Enter the value of coefficient
2.2 -5.3 6.2 2.5
-------------------------------------------------------
Enter value of x
2
------------------------------------------------------
Enter value of h
1
------------------------------------------------------
f0(x)=11.299999
f1(x)=11.400001
f2(x)=15.800001
f3(x)=13.200001
-------------------------------------------------------
Result=f(x+h)=32.800003
-------------------------------------------------------
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