Numerical Method Model Question

Comparison between Gauss Elimination method and Gauss Jordan method

• In Gauss- elimination method, the augmented matrix is reduced to Echelon form using row operations and then back substitution is applied to get values of unknowns, occurring in a system of linear equations.
• In Gauss-Jordan method, the augmented matrix is reduced to reduced Echelon form using elementary row operations to obtain values of unknowns of a system of linear equations.
• For small system it more convenient to use Gauss Jordan method.
• Gauss Jordan requires more computational work than Gauss Elimination.

Given system of equation,

    2x+3y+4z=5

    3x+4y+5z=6

    4x+5y+6z=7

The augmented matrix of given system is;

Applying, 

Applying, 

Now, using backward substitution;

3z = 0

i.e. z = 0

Finding the value of y using the value of z

Again, finding the value of x using the value of y and z

Hence the value of x, y, z are -2, 3, 0 respectively.

Algorithm for Gauss elimination method

1. Start

2.  Read the order of matrix ‘n’ and take the coefficients of linear equation.

3.  Do for k=1 to n-1

     Do for i=k+1 to n

     Do for j=k+1 to n+1

     a[i][j]=a[i][j]-a[i][k]/a[k][k]*a[k][j]

     end for j

     end for i

     end for k

4. Compute x[n]=a[n][n+1]/a[n][n]

5. Do for k=n-1 to 1

       Sum=0

    Do for j=k+1 to n

     Sum=Sum+a[k][j]*x[j]

    end for j

      x[k]=(a[k][n+1]-Sum)/a[k][k]

     end for k

6. Display the result x[k]

7. Stop

Algorithm For Lagrange’s interpolation

1.      Read the number of data ‘n’.

2.      Read the value at which value is needed (say y).

3.      Read available data points x and f(x).

4.      for i=0 to n

for j=0 to n

if(i!=j)

L[i]=L[i]*(y-x[j])/(x[i]-x[j])

end if

end for

end for

5.      for i=0 to n

sum=sum+L[i]*f[i]

end for

6.      Print interpolation value ‘sum’ at y.

Stop

Program

#include<stdio.h>

#include<conio.h>

int main()

{

  float x[10], f[10], y, sum=0.0, l;

 int n, i, j;

 printf("\\nInput number of data:");

 scanf("%d", &n);

 printf("\\nInput data points x(i) & f(i):\\n"); for(i=0;i<n;i++)

{

   printf("x[%d]=",i);

   scanf("%f", &x[i]);

   printf("f[%d]=",i);

   scanf("%f", &f[i]);

}

printf("\\nFunctional value:");

scanf("%f", &y);

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

{

   l=1;

   for(j=0;j<n;j++)

  {

    if(j!=i)

    {

      l=l*(y-x[j])/(x[i]-x[j]);

    }

 }

sum=sum+l*f[i];

}

printf("\\nValue at %f=%f", y, sum);

getch();

return 0;

}

Here, the integration interval is divided into n number of segments of equal width, where n is even number. Then the step size is

We have from simpson’s 1/3 rule,

Therefore,

This equation is called composite Simpson’s 1/3 rule.

It improves the accuracy of integration formulae by locating integration points in special locations. 

#include<stdio.h>

#include<conio.h>

#include<math.h>

#include<stdlib.h>

float f ( float x)

{

   float y;

   y= pow(x, 2)+x -2;

   return y;

}

void main()

{

   float x1, x2, x0, error=0.0001;

   int i=0;

   printf("\\nEnter two initial guess:");

   scanf("%f%f", &x1, &x2);

   if (f(x1 )*f(x2 )>0)

   {

      printf("\\nWrong Input!!");

      exit(0);

   }

else

{

   do

      {

         x0=(x1+x2)/2;

         if(f(x0 )*f(x1 )>0)

                   x1=x0;

         else

                   x2=x0;

        i++;

      }while(fabs (f(x0))>error);

   }

printf("\\nRoot=%f", x0);

printf("\\nNumber of iteration=%d",i);

getch();

}

Given,

Let the initial guess be 0.5.

Now, let us calculate the root using tabular form.

Here, the 4^th and 5^th iteration has same value of x₂ up to 2 decimal place, so that root of given equation is 1.42.

The difference table is

Here,

Now according to Newton’s backward difference formula;

We have,

Therefore the linear equation is

y = a + bx

i.e. y = 1.6 + 1.2x

Given,

     f(x)=ln x

let h=0.1

We have central divided difference formula for second order is

= -1.009

Both the Jacobi and Gauss-Seidel methods are iterative methods for solving the linear system Ax = b.

In the Jacobi method the updated vector x is used for the computations only after all the variables (i.e. all components of the vector x) have been updated. On the other hand in the Gauss-Seidel method, the updated variables are used in the computations as soon as they are updated.

Thus in the Jacobi method, during the computations for a particular iteration, the “known” values are all from the previous iteration. However in the Gauss-Seidel method, the “known” values are a mix of variable values from the previous iteration (whose values have not yet been evaluated in the current iteration), as well as variable values that have already been updated in the current iteration.

Even though the Gauss-Seidel’s method uses the improved values as soon as they are computed, this does not ensure that the Gauss-Seidel’s method would converge faster than Jacobi iterations.

Consider a system of linear equation;

Then, these equation can be written as;

From Gauss Jacobi approximation; we have, (with initial approximation x₁ = a, x₂ = b & x₃ = c)

And, from Gauss Seidel approximation;

Boundary Value Problem

Consider the following linear second order differential equation,

Suppose we are interested in solving this differential equation between the values x = a & z = b. Hence a & b are two values such that a<b. Let us divide the interval [a, b] into n equal subintervals of length h each.

Let x₀, x₁,.....,x_n be the pivotal points and a & b are called the boundary points. Solving the differential equation means finding the values of y₀, y₁, ....., y_n. Suppose y₀ and y_n are given. That is the solution values at the boundary points are given. Then the differential equation is called a Boundary Value Problem. So, the following is the general form of a boundary value problem.

Difference between Initial value and Boundary value problem

An initial value problem has all of the conditions specified at the same value of the independent variable in the equation (and that value is at the lower boundary of the domain, thus the term “initial” value). On the other hand, a boundary value problem has conditions specified at the extremes of the independent variable. For example, if the independent variable is time over the domain [0,1], an initial value problem would specify a value of y(t) and y‘(t) at time t = 0, while a boundary value problem would specify values for y(t) at both t = 0 and t = 1.

Here,

1^st iteration:

2^nd iteration:

Here,

Given Poisson’s eqn. is

h = 1

Let’s divide the domain into grids of 3 x 3 with f = 0 at the boundary as below

Now, from the difference equation for the Poisson’s equation,

Solving these equations,

Using eq.(ii) in (iii)

Using eq.(ii) in (iv)

& we have eq. (i)

Solving equation (a), (b) & (i) we get,

Using these values in eq. (ii)