Category Archives: Mathematics

Mental Math Tricks

Posted on December 31, 2016 | Leave a comment

Will share that soon

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Posted in GRE, Mathematics

Area of Circle using function

Posted on November 5, 2015 | Leave a comment

for formula: http://www.mathsisfun.com/area.htm

code:

#include<stdio.h>
double pi=3.1416;
double britto(double r)
{
    double area=pi*r*r;
    return area;
 
}
int main()
{
double r=3;
double khetrofol=britto(r);
printf("%lf",khetrofol);
 
    return 0;
}

#include<stdio.h>

double pi=3.1416;

double britto(double r)

{

double area=pi*r*r;

return area;

}

int main()

{

double r=3;

double khetrofol=britto(r);

printf("%lf",khetrofol);

return 0;

}

We have to always remember that value that define is always pass from main function to customized function named by user so here r=3 is passing to function britto and then the britto formula pi*r*r is applying and returned the value and it is printed in the main function where the variable defined as khetrofol.

Here pi declared as global variable with it’s value.

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Posted in A ll Codes, ACM-ICPC, Function, Mathematics

Tagged circle, function

Find Maximum and Minimum and Summation value from array

Posted on November 5, 2015 | Leave a comment

Did this problem from subeen vai’s book.

For maximum value:

#include<stdio.h>
int find_max(int ara[],int n);
int main()
{
 
    int ara[]={-100,0,53,22,83,23,89,-132,201,3,85};
    int n=11;
    int max=find_max(ara,n);
    printf("%d\n",max);
    return 0;
}
 
int find_max(int ara[],int n)
{
    int max=ara[0]; //assume that first value of array according to index is maximum we will compare later with this
    int i;
    for(i=1;i<n;i++)
    {
        if(ara[i]>max) //here it is iterating the whole array
        {
            max=ara[i];
        }
    }
 
    return max;
}

#include<stdio.h>

int find_max(int ara[],int n);

int main()

{

int ara[]={-100,0,53,22,83,23,89,-132,201,3,85};

int n=11;

int max=find_max(ara,n);

printf("%d\n",max);

return 0;

}

int find_max(int ara[],int n)

{

int max=ara[0]; //assume that first value of array according to index is maximum we will compare later with this

int i;

for(i=1;i<n;i++)

{

if(ara[i]>max) //here it is iterating the whole array

{

max=ara[i];

}

return max;

}

For minimum value:

#include<stdio.h>
int find_min(int array[],int n);

int main()
{
    int array[]= {20,10,-123,980,60,50,70};
    int n=7;
    int min;
    min=find_min(array,n);
    printf("%d\n",min);
    return 0;
}

int find_min(int array[],int n)
{
    int min=array[0];
    int i;
    for(i=0; i<n; i++)
    {
        if(array[i]<min)
        {
            min=array[i];
        }
    }
    return min;
}

#include<stdio.h>

int find_min(int array[],int n);

int main()

{

int array[]= {20,10,-123,980,60,50,70};

int n=7;

int min;

min=find_min(array,n);

printf("%d\n",min);

return 0;

}

int find_min(int array[],int n)

{

int min=array[0];

int i;

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

{

if(array[i]<min)

{

min=array[i];

}

return min;

}

Summation:

#include<stdio.h>
int find_sum(int ary[],int valu);
int main()
{
    int ary[]={10,20,30};
    int n=3;
    int sum;
    sum=find_sum(ary,n);
    printf("%d\n",sum);
    return 0;
}
 
int find_sum(int ary[],int n)
{
    int i;
    int sum=0;
    for(i=0;i<n;i++)
    {
        sum=sum+ary[i];
    }
return sum;
}

#include<stdio.h>

int find_sum(int ary[],int valu);

int main()

{

int ary[]={10,20,30};

int n=3;

int sum;

sum=find_sum(ary,n);

printf("%d\n",sum);

return 0;

}

int find_sum(int ary[],int n)

{

int i;

int sum=0;

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

{

sum=sum+ary[i];

}

return sum;

}

Average:

Will give here

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Posted in A ll Codes, ACM-ICPC, Array, Mathematics

Math Motivation

Posted on November 4, 2015 | Leave a comment

http://www.prothom-alo.com/bangladesh/article/580627/%E0%A6%97%E0%A6%A3%E0%A6%BF%E0%A6%A4-%E0%A6%85%E0%A6%B2%E0%A6%BF%E0%A6%AE%E0%A7%8D%E0%A6%AA%E0%A6%BF%E0%A7%9F%E0%A6%BE%E0%A6%A1%E0%A7%87%E0%A6%B0-%E0%A6%A4%E0%A6%BE%E0%A6%B0%E0%A6%95%E0%A6%BE%E0%A6%B0%E0%A6%BE-%E0%A6%95%E0%A7%87-%E0%A6%95%E0%A7%8B%E0%A6%A5%E0%A6%BE%E0%A7%9F

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Posted in Mathematics

Tagged math olympiad

LCM – Least Common Multiple

Posted on November 1, 2015 | Leave a comment

Multiple of 3 is
3×1=3
3×2=6
3×3=9
3×4=12
3×5=15
3×6=18
3×7=21
3×8=24
3×9=27
3×10=30

Multiple of 4 is
4×1=4
4×2=8
4×3=12
4×4=16
4×5=20
4×6=24
4×7=28
4×8=32
4×9=36
4×10=40

Here in 3 and 4 mutilples common is 12
and it is only one value and so it is the lowest also then but if there were some more values than result could be something changed.

So here LCM=12
Here I implemented the code
applied this formula
$lcm (a, b) = \frac{a \cdot b}{\gcd (a, b)};$

code goes here :

#include<stdio.h>
 int main()
{
    int a,b,x,gcd,lcm;
    scanf("%d %d",&a,&b);

    if(a<b)//it is not possible for gcd to be greater than the lowest value
    {
        x=a;
    }
    else
    {
        x=b;
    }
    for( ;x>=1;x--){ //here i haven't initialized the value of x as x is first at outside
        if(a%x==0 && b%x==0){
            gcd=x;
            break; //here i have used break as if it is not here then when the division will happen it will run the loop till 1 so answer will be always 1
        }
    }

printf("GCD is %d\n",gcd);
lcm=a*b/gcd;
printf("LCM is %d\n",lcm);

    return 0;
}

#include<stdio.h>

int main()

{

int a,b,x,gcd,lcm;

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

if(a<b)//it is not possible for gcd to be greater than the lowest value

{

x=a;

}

else

{

x=b;

}

for( ;x>=1;x--){ //here i haven't initialized the value of x as x is first at outside

if(a%x==0 && b%x==0){

gcd=x;

break; //here i have used break as if it is not here then when the division will happen it will run the loop till 1 so answer will be always 1

}

printf("GCD is %d\n",gcd);

lcm=a*b/gcd;

printf("LCM is %d\n",lcm);

return 0;

}

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Posted in ACM-ICPC, LCM, Mathematics

Tagged LCM - Least Common Multiple

GCD Greatest Common Divisor(Factor)/Highest Common Factor

Posted on November 1, 2015 | Leave a comment

এখানে লজিকটা হচ্ছে…Here the logic is:
For 12:
1×12=12
2×6=12
3×4=12
So factor of 12=1,2,3,4,6,12
(We don’t need the negative values here)

For 30:
1×30=30
2×15=30
3×10=30
5×6=30
So factor of 30=1,2,3,5,6,10,15,30

Now which is the common factor among 12 and 30
So factor of 12=1,2,3,4,6,12
So factor of 30=1,2,3,5,6,10,15,30

common factor=1,2,3,6
Largest common factor here=6

So now if we divide 12/6 then we get 2
and for 30/6 we get 5

that is all divided perfectly so the logic should be like this

1.Find all the factors of each number
2. Circle the common factors
3. Choose the greatest of those
4. Divide the existing number with greatest common factor thern also get a perfect divide

Here is my code:

#include<stdio.h>
 int main()
{
    int a,b,x,gcd;
    scanf("%d %d",&a,&b);

    if(a<b)//it is not possible for gcd to be greater than the lowest value
    {
        x=a;
    }
    else
    {
        x=b;
    }
    for( ;x>=1;x--){ //here i haven't initialized the value of x as x is first at outside
        if(a%x==0 && b%x==0){
            gcd=x;
            break; //here i have used break as if it is not here then when the division will happen it will run the loop till 1 so answer will be always 1
        }
    }

printf("GCD is %d\n",gcd);

    return 0;
}

#include<stdio.h>

int main()

{

int a,b,x,gcd;

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

if(a<b)//it is not possible for gcd to be greater than the lowest value

{

x=a;

}

else

{

x=b;

}

for( ;x>=1;x--){ //here i haven't initialized the value of x as x is first at outside

if(a%x==0 && b%x==0){

gcd=x;

break; //here i have used break as if it is not here then when the division will happen it will run the loop till 1 so answer will be always 1

}

printf("GCD is %d\n",gcd);

return 0;

}

Here is a link that described many thinks about this clearly: http://www.mathsisfun.com/greatest-common-factor.html

Another method can also apply here:
link for understanding the modarithmetic euclidean algorithmhttps://www.khanacademy.org/computing/computer-science/cryptography/modarithmetic/a/the-euclidean-algorithm

Code:

#include<stdio.h>

int main()
{

    int a,b,t,x,gcd;
    scanf("%d %d",&a,&b);
    if(a==0)
        gcd=a;
    else if(b==0)
        gcd=b;
    else
    {
        while(b!=0)
        {

            t=b;
            b=a%b;
            a=t;
        }
        gcd=a;
    }
    printf("GCD is %d\n",gcd);

    return 0;
}

#include<stdio.h>

int main()

{

int a,b,t,x,gcd;

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

if(a==0)

gcd=a;

else if(b==0)

gcd=b;

else

{

while(b!=0)

{

t=b;

b=a%b;

a=t;

}

gcd=a;

}

printf("GCD is %d\n",gcd);

return 0;

}

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Posted in ACM-ICPC, GCD, Mathematics

Tagged GCD Greatest Common Divisor(Factor)/Highest Common Factor

URI Online Judge Solution| 1036 Bhaskara’s Formula

Posted on February 10, 2015 | Leave a comment

An equation ax² + bx + c = 0 where a,b and c are any number and called the coefficient of the equation
It is known as quadratic equation because any seconbd degree polynomial equation is known as quadratic equation.This equation has two roots x1 and x2

This is Bhaskara’s Formula or Bhaskaracharya’s Formula where discriminant D or Del = b² − 4ac
If D>0 then root of quadratic equation are real and unequal
If D=0 then root of quadratic equation are real and equal
If D<0 then root of quadratic equation are conjugate complex number
Source:http://encyclopedia2.thefreedictionary.com/Bhaskaracharya%2
7s+Formula

http://www.wolframalpha.com/input/?i=Bhaskara%20formula
So my code is here

#include<iostream>
#include<cstdio>
#include<cmath>
int main()
{
double a,b,c,r1,r2,d;
scanf("%lf %lf %lf",&a,&b,&c);
d=(pow(b,2)-(4*a*c));
r1=(-b+sqrt(d))/(2*a);
r2=(-b-sqrt(d))/(2*a);
if(a!=0 && d>0)
{
    printf("R1 = %.5lf\nR2 = %.5lf\n",r1,r2);
}
else{
    printf("Impossivel calcular\n");
}

return 0;
}

#include<iostream>

#include<cstdio>

#include<cmath>

int main()

{

double a,b,c,r1,r2,d;

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

d=(pow(b,2)-(4*a*c));

r1=(-b+sqrt(d))/(2*a);

r2=(-b-sqrt(d))/(2*a);

if(a!=0 && d>0)

{

printf("R1 = %.5lf\nR2 = %.5lf\n",r1,r2);

}

else{

printf("Impossivel calcular\n");

}

return 0;

}

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Posted in A ll Codes, Mathematics, Problem Solution, Programming, URI Online Judge

Tagged uri, urionlinejudge

C code using Forward Interpolation

Posted on December 9, 2014 | Leave a comment

Problem: The population of a town is given below as thousands
Year : 1891 1901 1911 1921 1931
Population : 46 66 81 93 101

Find the population of 1895 ?

Code:

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
main()
{
  float x[20],y[20],f,s,h,d,p;
  int j,i,n;
  printf("enter the value of n :");
  scanf("%d",&n);
  printf("enter the elements of x:");
  for(i=1;i<=n;i++)
   {
        scanf("\n%f",&x[i]);
         }
                   printf("enter the elements of y:");
               for(i=1;i<=n;i++)
               {
              scanf("\n%f",&y[i]);
                                }
  h=x[2]-x[1];
  printf("Enter the value of f(to findout value):");
  scanf("%f",&f);
s=(f-x[1])/h;
p=1;
d=y[1];
for(i=1;i<=(n-1);i++)
 {
                   for(j=1;j<=(n-i);j++)
                    {
                          y[j]=y[j+1]-y[j];

                    }
                    p=p*(s-i+1)/i;
                    d=d+p*y[1];
 }
printf("For the value of x=%6.5f THe value is %6.5f",f,d);
 getch();
}

#include<stdio.h>

#include<math.h>

#include<stdlib.h>

main()

{

float x[20],y[20],f,s,h,d,p;

int j,i,n;

printf("enter the value of n :");

scanf("%d",&n);

printf("enter the elements of x:");

for(i=1;i<=n;i++)

{

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

}

printf("enter the elements of y:");

for(i=1;i<=n;i++)

{

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

}

h=x[2]-x[1];

printf("Enter the value of f(to findout value):");

scanf("%f",&f);

s=(f-x[1])/h;

p=1;

d=y[1];

for(i=1;i<=(n-1);i++)

{

for(j=1;j<=(n-i);j++)

{

y[j]=y[j+1]-y[j];

}

p=p*(s-i+1)/i;

d=d+p*y[1];

}

printf("For the value of x=%6.5f THe value is %6.5f",f,d);

getch();

}

Output:

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Posted in A ll Codes, Mathematics, Numerical Method

Tagged newton forward interpolation

C code using lagrange method

Posted on December 8, 2014 | Leave a comment

Problem: The population of a town is given below as thousands
Year : 1891 1901 1911 1921 1931
Population : 46 66 81 93 101

Find the population of 1895 ?

Code:

#include<stdio.h>
#include<math.h>
int main()
{
  float x[10],y[10],temp=1,f[10],sum,p;
  int i,n,j,k=0,c;

  printf("\nhow many record you will be enter: ");
  scanf("%d",&n);
  for(i=0; i<n; i++)
  {
   printf("\n\nenter the value of x%d: ",i);
   scanf("%f",&x[i]);
   printf("\n\nenter the value of f(x%d): ",i);
   scanf("%f",&y[i]);
  }
  printf("\n\nEnter X for finding f(x): ");
  scanf("%f",&p);

  for(i=0;i<n;i++)
  {
    temp = 1;
    k = i;
    for(j=0;j<n;j++)
    {
      if(k==j)
      {
        continue;
      }
      else
      {
        temp = temp * ((p-x[j])/(x[k]-x[j]));
      }
    }
    f[i]=y[i]*temp;
  }

  for(i=0;i<n;i++)
  {
     sum = sum + f[i];
  }
  printf("\n\n f(%.1f) = %f ",p,sum);
  getch();
}

#include<stdio.h>

#include<math.h>

int main()

{

float x[10],y[10],temp=1,f[10],sum,p;

int i,n,j,k=0,c;

printf("\nhow many record you will be enter: ");

scanf("%d",&n);

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

{

printf("\n\nenter the value of x%d: ",i);

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

printf("\n\nenter the value of f(x%d): ",i);

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

}

printf("\n\nEnter X for finding f(x): ");

scanf("%f",&p);

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

{

temp = 1;

k = i;

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

{

if(k==j)

{

continue;

}

else

{

temp = temp * ((p-x[j])/(x[k]-x[j]));

}

f[i]=y[i]*temp;

}

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

{

sum = sum + f[i];

}

printf("\n\n f(%.1f) = %f ",p,sum);

getch();

}

Output:

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Posted in A ll Codes, Mathematics, Numerical Method

Tagged lagrange mehtod

C code using Backward Interpolation

Posted on December 8, 2014 | Leave a comment

Problem: The population of a town is given below as thousands
Year : 1891 1901 1911 1921 1931
Population : 46 66 81 93 101

Find the population of 1895 ?

Code:

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
main()
{
      float x[20],y[20],f,s,d,h,p;
      int j,i,k,n;
      printf("enter the value of the elements :");
      scanf("%d",&n);
      printf("enter the value of x:\n");
      for(i=1;i<=n;i++)
      {
                       scanf("%f",&x[i]);
                       }
             printf("enter the value of y:\n");
      for(i=1;i<=n;i++)
      {
                       scanf("%f",&y[i]);
                       }
                       h=x[2]-x[1];
           printf("enter the searching point f:");
scanf("%f",&f);
s=(f-x[n])/h;
d=y[n];
p=1;
for(i=n,k=1;i>=1,k<n;i--,k++)
{
                 for(j=n;j>=1;j--)
                 {
                                 y[j]=y[j]-y[j-1];
                                 }
                                 p=p*(s+k-1)/k;
                                 d=d+p*y[n];
}
printf("for f=%f ,ans is=%f",f,d);
getch();
}

#include<stdio.h>

#include<math.h>

#include<stdlib.h>

main()

{

float x[20],y[20],f,s,d,h,p;

int j,i,k,n;

printf("enter the value of the elements :");

scanf("%d",&n);

printf("enter the value of x:\n");

for(i=1;i<=n;i++)

{

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

}

printf("enter the value of y:\n");

for(i=1;i<=n;i++)

{

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

}

h=x[2]-x[1];

printf("enter the searching point f:");

scanf("%f",&f);

s=(f-x[n])/h;

d=y[n];

p=1;

for(i=n,k=1;i>=1,k<n;i--,k++)

{

for(j=n;j>=1;j--)

{

y[j]=y[j]-y[j-1];

}

p=p*(s+k-1)/k;

d=d+p*y[n];

}

printf("for f=%f ,ans is=%f",f,d);

getch();

}

Output:

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Posted in A ll Codes, Mathematics, Numerical Method

Tagged newton backward interpolation

C code using Gauss Seidel Method

Posted on December 8, 2014 | Leave a comment

Problem: Solve the following systems using gauss seidel method
5×1-x2-x3-x4=-4
-x1+10×2-x3-x4=12
-x1-x2+5×3-x4=8
-x1-x2-x3+10×4=34

Code:

#include<stdio.h>
#include<conio.h>
#include<math.h>
#define acc 0.0001
#define X1(x2,x3,x4) ((x2 + x3 + x4 -4)/5)
#define X2(x1,x3,x4) ((x1 + x3 + x4 +12)/10)
#define X3(x1,x2,x4) ((x1 + x2 + x4 +8)/5)
#define X4(x1,x2,x3) ((x1 + x2 + x3 +34)/10)

void main()
{
  double x1=0,x2=0,x3=0,x4=0,y1,y2,y3,y4;
  int i=0;
  system("cls");
  printf("\n______________________________________________________________\n");
  printf("\n   x1\t\t   x2\t\t   x3\t\t    x4\n");
  printf("\n______________________________________________________________\n");
  printf("\n%f\t%f\t%f\t%f",x1,x2,x3,x4);
  do
  {
   y1=X1(x2,x3,x4);
   y2=X2(x1,x3,x4);
   y3=X3(x1,x2,x4);
   y4=X4(x1,x2,x3);
   if(fabs(y1-x1)<acc && fabs(y2-x2)<acc && fabs(y3-x3)<acc &&fabs(y4-x4) )
   {
     printf("\n_____________________________________________________________\n");
     printf("\n\nx1 = %.3lf",y1);
     printf("\n\nx2 = %.3lf",y2);
     printf("\n\nx3 = %.3lf",y3);
     printf("\n\nx4= %.3lf",y4);
     i = 1;
   }
   else
   {
     x1 = y1;
     x2 = y2;
     x3 = y3;
     x4 = y4;
     printf("\n%f\t%f\t%f\t%f",x1,x2,x3,x4);
   }
  }while(i != 1);
getch();
}

#include<stdio.h>

#include<conio.h>

#include<math.h>

#define acc 0.0001

#define X1(x2,x3,x4) ((x2 + x3 + x4 -4)/5)

#define X2(x1,x3,x4) ((x1 + x3 + x4 +12)/10)

#define X3(x1,x2,x4) ((x1 + x2 + x4 +8)/5)

#define X4(x1,x2,x3) ((x1 + x2 + x3 +34)/10)

void main()

{

double x1=0,x2=0,x3=0,x4=0,y1,y2,y3,y4;

int i=0;

system("cls");

printf("\n______________________________________________________________\n");

printf("\n x1\t\t x2\t\t x3\t\t x4\n");

printf("\n______________________________________________________________\n");

printf("\n%f\t%f\t%f\t%f",x1,x2,x3,x4);

{

y1=X1(x2,x3,x4);

y2=X2(x1,x3,x4);

y3=X3(x1,x2,x4);

y4=X4(x1,x2,x3);

if(fabs(y1-x1)<acc && fabs(y2-x2)<acc && fabs(y3-x3)<acc &&fabs(y4-x4) )

{

printf("\n_____________________________________________________________\n");

printf("\n\nx1 = %.3lf",y1);

printf("\n\nx2 = %.3lf",y2);

printf("\n\nx3 = %.3lf",y3);

printf("\n\nx4= %.3lf",y4);

i = 1;

}

else

{

x1 = y1;

x2 = y2;

x3 = y3;

x4 = y4;

printf("\n%f\t%f\t%f\t%f",x1,x2,x3,x4);

}

}while(i != 1);

getch();

}

Output:

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Posted in A ll Codes, Mathematics, Numerical Method

Tagged gauss seidel method

C code using Euler Method

Posted on December 8, 2014 | Leave a comment

Problem: Here we have to find dy/dx=x+y where y(0)=1 at the point x=0.05 and x=0.10 taking h=0.05

Algorithm:

Start
Define function
Get the values of x0, y0, h and xn
*Here x0 and y0 are the initial conditions
h is the interval
xn is the required value
n = (xn – x0)/h + 1
Start loop from i=1 to n
y = y0 + h*f(x0,y0)
x = x + h
Print values of y0 and x0
Check if x < xn
If yes, assign x0 = x and y0 = y
If no, goto 9.
End loop i
Stop

Code:

#include<stdio.h>
float fun(float x,float y)
{
    float f;
    f=x+y;
    return f;
}
main()
{
    float a,b,x,y,h,t,k;
    printf("\nEnter x0,y0,h,xn: ");
    scanf("%f%f%f%f",&a,&b,&h,&t);
    x=a;
    y=b;
    printf("\n  x\t  y\n");
    while(x<=t)
    {
        k=h*fun(x,y);
        y=y+k;
        x=x+h;
        printf("%0.3f\t %0.3f\n",x,y);
    }
}

#include<stdio.h>

float fun(float x,float y)

{

float f;

f=x+y;

return f;

}

main()

{

float a,b,x,y,h,t,k;

printf("\nEnter x0,y0,h,xn: ");

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

x=a;

y=b;

printf("\n x\t y\n");

while(x<=t)

{

k=h*fun(x,y);

y=y+k;

x=x+h;

printf("%0.3f\t %0.3f\n",x,y);

}

Output:

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Posted in A ll Codes, Algorithm, Mathematics, Numerical Method

C code using Runge-Kutta 4th order method

Posted on December 8, 2014 | Leave a comment

Problem: Here we have to find y(0,2) and y(0,4), Given dy/dx=1+y^2 where y=0 when x=0

Algorithm:

Step 1: input x0,y0,h,last point n

Step 2:m1=f(xi,yi)

Step 3:m2=f(xi+h/2,yi+m1h/2)

Step 4:m3=f(xi+h/2,yi+m2h/2)

Step 5:m4=f(xi+h,yi+m3h)

Step 6:yi+1=yi+(m1+2m2+2m3+m4/6)h

Step 5:Display output

Code:

#include<stdio.h>
#include <math.h>
#include<conio.h>
#define F(x,y)  1 + (y)*(y)
void main()
{
    double y0,x0,y1,n,h,f,k1,k2,k3,k4;
    system("cls");
    printf("\nEnter the value of x0: ");
    scanf("%lf",&x0);
    printf("\nEnter the value of y0: ");
    scanf("%lf",&y0);
    printf("\nEnter the value of h: ");
    scanf("%lf",&h);
    printf("\nEnter the value of last point: ");
    scanf("%lf",&n);
    for(; x0<n; x0=x0+h)
    {
        f=F(x0,y0);
        k1 = h * f;
        f = F(x0+h/2,y0+k1/2);
        k2 = h * f;
        f = F(x0+h/2,y0+k2/2);
        k3 = h * f;
        f = F(x0+h/2,y0+k2/2);
        k4 = h * f;
        y1 = y0 + ( k1 + 2*k2 + 2*k3 + k4)/6;
        printf("\n\n  k1 = %.4lf  ",k1);
        printf("\n\n  k2 = %.4lf ",k2);
        printf("\n\n  k3 = %.4lf ",k3);
        printf("\n\n  k4 = %.4lf ",k4);
        printf("\n\n  y(%.4lf) = %.3lf ",x0+h,y1);
        y0=y1;
    }
    getch();
}

#include<stdio.h>

#include <math.h>

#include<conio.h>

#define F(x,y) 1 + (y)*(y)

void main()

{

double y0,x0,y1,n,h,f,k1,k2,k3,k4;

system("cls");

printf("\nEnter the value of x0: ");

scanf("%lf",&x0);

printf("\nEnter the value of y0: ");

scanf("%lf",&y0);

printf("\nEnter the value of h: ");

scanf("%lf",&h);

printf("\nEnter the value of last point: ");

scanf("%lf",&n);

for(; x0<n; x0=x0+h)

{

f=F(x0,y0);

k1 = h * f;

f = F(x0+h/2,y0+k1/2);

k2 = h * f;

f = F(x0+h/2,y0+k2/2);

k3 = h * f;

f = F(x0+h/2,y0+k2/2);

k4 = h * f;

y1 = y0 + ( k1 + 2*k2 + 2*k3 + k4)/6;

printf("\n\n k1 = %.4lf ",k1);

printf("\n\n k2 = %.4lf ",k2);

printf("\n\n k3 = %.4lf ",k3);

printf("\n\n k4 = %.4lf ",k4);

printf("\n\n y(%.4lf) = %.3lf ",x0+h,y1);

y0=y1;

}

getch();

}

Output:

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Posted in A ll Codes, Mathematics, Numerical Method

Tagged runge-kutta using c code

C code for newton-raphson method

Posted on November 27, 2014 | Leave a comment

Problem: Here we have to find root for the polynomial x^3-8*x-4 upto 6D(decimal places)

Solution in C:

#include<stdio.h>
#include<math.h>
#define f(x) pow(a,3)-8*a-4;
#define fd(x) 3*pow(a,2)-8;
int main()
{
    double a,b,c,d,h,k,x,y;
    int i,j,m,n;
    printf("Enter the value of xn: ");
    scanf("%lf",&a);
    printf("Enter itteration number: ");
    scanf("%d",&n);
    printf("  xn           f(x)         f'(x)       hn=-f(x)/f'(xn)   xn+1=xn+h\n");
    printf("-----------------------------------------------------------------------------------------\n");
    for(i=1;i<=n;i++)
    {
    x=f(a);
    y=fd(x);
    h=-(x/y);
    k=h+a;
    printf(" %.7lf     %.7lf     %.7lf     %.7lf     %.7lf\n",a,x,y,h,k);
a=k;
    }
     printf("\nThe approximation to the root is %.6lf which is upto 6D",k);
    return 0;

}

#include<stdio.h>

#include<math.h>

#define f(x) pow(a,3)-8*a-4;

#define fd(x) 3*pow(a,2)-8;

int main()

{

double a,b,c,d,h,k,x,y;

int i,j,m,n;

printf("Enter the value of xn: ");

scanf("%lf",&a);

printf("Enter itteration number: ");

scanf("%d",&n);

printf(" xn f(x) f'(x) hn=-f(x)/f'(xn) xn+1=xn+h\n");

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

for(i=1;i<=n;i++)

{

x=f(a);

y=fd(x);

h=-(x/y);

k=h+a;

printf(" %.7lf %.7lf %.7lf %.7lf %.7lf\n",a,x,y,h,k);

a=k;

}

printf("\nThe approximation to the root is %.6lf which is upto 6D",k);

return 0;

}

Output:

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Posted in A ll Codes, Mathematics, Numerical Method

Tagged N-R method, newton-raphson

C code using Weddle’s rule

Posted on November 27, 2014 | Leave a comment

Problem: Here we have to find integration for the (1/1+x*x)dx with lower limit =0 to upper limit = 6

Algorithm:

Step 1: input a,b,number of interval n

Step 2:h=(b-a)/n

Step 3:If(n%6==0)

Then , sum=sum+((3*h/10)*(y(a)+y(a+2*h)+5*y(a+h)+6*y(a+3*h)+y(a+4*h)+5*y(a+5*h)+y(a+6*h)));
a=a+6*h
and Weddle’s rule is applicable then go to step 6

Step 4: else, Weddle’s rule is not applicable

Step 5:Display output

Code:

#include<stdio.h>
float y(float x){
    return 1/(1+x*x); //function of which integration is to be calculated
}
int main(){
    float a,b,h,sum;
    int i,n,m;

    printf("Enter a=x0(lower limit), b=xn(upper limit), number of subintervals: ");
    scanf("%f%f%d",&a,&b,&n);
    h = (b-a)/n;
    sum=0;

        if(n%6==0){
            sum=sum+((3*h/10)*(y(a)+y(a+2*h)+5*y(a+h)+6*y(a+3*h)+y(a+4*h)+5*y(a+5*h)+y(a+6*h)));
            a=a+6*h;

        printf("Value of integral is %f\n", sum);
        }
        else{
            printf("Sorry ! Weddle rule is not applicable");
        }
    }

#include<stdio.h>

float y(float x){

return 1/(1+x*x); //function of which integration is to be calculated

}

int main(){

float a,b,h,sum;

int i,n,m;

printf("Enter a=x0(lower limit), b=xn(upper limit), number of subintervals: ");

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

h = (b-a)/n;

sum=0;

if(n%6==0){

sum=sum+((3*h/10)*(y(a)+y(a+2*h)+5*y(a+h)+6*y(a+3*h)+y(a+4*h)+5*y(a+5*h)+y(a+6*h)));

a=a+6*h;

printf("Value of integral is %f\n", sum);

}

else{

printf("Sorry ! Weddle rule is not applicable");

}

Output:

Leave a comment

Posted in A ll Codes, Mathematics, Numerical Method

Category Archives: Mathematics

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C code using Forward Interpolation

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C code using Gauss Seidel Method

C code using Euler Method

C code using Runge-Kutta 4th order method

C code for newton-raphson method

C code using Weddle’s rule

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Category Archives: Mathematics

Caution and Tips:

Category

About Me & MyBlog

Recent Posts

Recent Comments

My Facebook

My Algorithmic Progress in OJ’s

My Github

My Linkedin

Tags

Count per Day

Archives

This site is powered by:

Bookmarks

My Coding Page

Top Posts & Pages

My Skills

My Poster/Presentations

My Videos

My Bookshelf

Quote From Life