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This is a console application, a project which implements various numerical methods to solve mathematical problems.

Implemented Numerical Methods

1. Solution of Linear Equations
   • Jacobi iterative method
   • Gauss-Seidel iterative method
   • Gauss elimination
   • Gauss-Jordan elimination
   • LU factorization
2. Solution of Non-linear Equations
   • Bi-section method
   • False position method
   • Secant method
   • Newton-Raphson method
3. Solution of Differential Equations
   • Runge-Kutta method
4. Matrix Inversion

Jacobi Iterative Method

1. Arrange the given system of linear equations in diagonally dominant form.
2. set the tolerable error;
3. Convert the first equation in terms of first variable , second equation in the term of second variable and so on. for example, x1=1/a11(b1-a12*x2-a13*x3). x2=1/a12(b2-a11*x1-a13*x3). x3=1/a13(b3-a11*x1-a12*x2).
4. Set the initial guesses for x10,x20,x30 and so on;
5. Substitute the value of x10,x20,x30,.......in equation obtained from step 4 to calculate new values of x1,x2,x3 and so on;
6. Repeat step 5 until (|x10-x1|)>e and (|x20-x2|)>e and (|x30-x3|)>e and so on ;
7. Print the value of x1,x2,x3 and so on;

The Visual Representation of Jacobi Method:

Gauss-Seidel Iterative Method

1. Arrange the given system of linear equations in diagonally dominant form.
2. set the tolerable error;
3. Convert the first equation in terms of first variable , second equation in the term of second variable and so on. for example, x1=1/a11(b1-a12*x2-a13*x3). x2=1/a12(b2-a11*x1-a13*x3). x3=1/a13(b3-a11*x1-a12*x2).
4. Set the initial guesses for x10,x20,x30 and so on;
5. Substitute the value of x20,x30,.......in the first equation obtained from step 4 to calculate new value x1.Then use x1,x20,x30,...... in second equation to calculate new value of x2 and use x1,x2,x30,..... in third equation to calculate new value of x3 and so on;
6. Repeat step 5 until (|x10-x1|)>e and (|x10-x1|)>e and (|x10-x1|)>e;
7. Print the value of x1,x2,x3 and so on;

The Visual Representation of Guass-Scidel Iterative Method:

Gauss Elimination

Gauss Elimination

Purpose: The Gauss Elimination method is a systematic procedure to solve a system of linear equations. It converts the matrix into an upper triangular form, from which you can use back-substitution to find the solution.

How It Works

1. Forward Elimination:

• Convert the coefficient matrix to an upper triangular form by performing row operations.
• The goal is to have all zeros below the main diagonal of the matrix.
• To achieve this, use the following operations:
  • Subtract a multiple of one row from another to make the elements below the pivot (diagonal) equal to zero.

2. Back Substitution:

• Once the matrix is in upper triangular form, start from the last row and work your way up to solve for each variable.
• Use previously computed variable values to solve the current equation.

Example

Given System of Equations

x + y + z = 6 
2x + 3y + 7z = 18 
3x + 5y + 2z = 14

Convert to Augmented Matrix Form

 ____     ____
|             |
| 1  1  1 |6  |
| 2  3  7 |18 |
| 3  5  2 |14 |
|___       ___|

• Perform row operations to make it upper triangular.
• Use back-substitution to find the values of (x), (y), and (z).

Forward Elimination

1. For each column i in A:
   • Find pivot element in row i
   • For each row j below the pivot row:
     • Factor = A[j][i] / A[i][i]
     • For each column k in row j:
       • A[j][k] = A[j][k] - Factor * A[i][k]
     • b[j] = b[j] - Factor * b[i]

Back Substitution

2. For row i from last to first:
   • x[i] = (b[i] - sum(A[i][j] * x[j] for j from i+1 to end)) / A[i][i]

The Visual Representation of the Matrix is:

Gauss-Jordan Elimination

Purpose: Gauss-Jordan Elimination is an extension of Gauss Elimination. It simplifies the system further to the Reduced Row Echelon Form (RREF), from which the solution can be read directly without back-substitution.

How It Works

1. Forward Elimination:
   • Similar to Gauss Elimination, perform row operations to make the elements below the main diagonal zero.
2. Backward Elimination:
   • Normalize the pivot elements to 1 by dividing the entire row by the pivot value.
   • Use the pivot rows to make all entries above and below each pivot equal to zero.
   • The solutions x, y, and z can be directly read from the matrix.
3. The resulting matrix will be in Reduced Row Echelon Form, which looks like:

 ____     ____
|             |
| 1  0  0 |x  |
| 0  1  0 |y  |
| 0  0  1 |z  |
|___       ___|

Example

From the above augmented matrix:

1. Use additional row operations to make elements above the diagonal zero.
2. Normalize the diagonal elements to 1.
3. Directly extract the solutions from the final matrix.

Pseudocode:

1. For each column i in A:
   • Normalize row i by dividing by the pivot A[i][i]
   • For each row j (j != i):
     • Factor = A[j][i]
     • For each column k in row j:
       • A[j][k] = A[j][k] - Factor * A[i][k]
     • b[j] = b[j] - Factor * b[i]

The Visual Representation of the Matrix is:

LU Factorization

LU Factorization is a fundamental technique in linear algebra used to solve systems of linear equations. It expresses a given square matrix A as a product of two matrices, one lower L and one upper U i.e. A = LU. This simplifies many matrix operations, making it an essential tool in various engineering applications.

How it works

1. Given a set of linear equations of the form
        a11x1 + a12x2 + a13x3 + ... + a1nxn = b
     a21x1 + a22x2 + a23x3 + ... + a2nxn = b
     a31x1 + a32x2 + a33x3 + ... + a3nxn = b
     .....................................
     an1x1 + an2x2 + an3x3 + ... + annxn = b
   convert them into matrix form AX = B where A is the coefficient matrix,X is variable matrix and B is the matrix of constants on the right side of the equation.
2. Create two matrices naming L and U of the same size as A. Calculate the values of the elements of both matrices such that LU = A
3. Now equation can be written as LUX = B or LY = B where UX = Y
4. Solve Y from LY = B and then using Y solve X from UX = Y

Bi-section Method

Bi-section method is used to find root of an equation in a given interval i.e. value of x for which f(x) = 0.
The method is based on The Intermediate Value Theorem which states that if f(x) is a continuous function and there are two real numbers a and b such that f(a)*f(b) < 0, then it is guaranteed that it has at least one root between them.

How it works

Assumptions

1. f(x) is a continuous function in interval [a, b]
2. f(a) * f(b) < 0

Steps

1. Find middle point c = (a + b)/2 .
2. If f(c) == 0, then c is the root of the solution.
3. Else f(c) != 0
   1. If value f(a)*f(c) < 0 then root lies between a and c. So we recur for a and c
   2. Else If f(b)*f(c) < 0 then root lies between b and c. So we recur b and c.
   3. Else given function doesn’t follow one of assumptions.
4. Repeat the steps until difference between a and b is less then a value (very small value).

The Visual Representation For Bisection Method :

False Position Method

Like Bi-section method, False Position method is also an approximation method to find the roots of a given equation by repeatedly dividing the interval.
False position method is the same as Bi-section method with the only difference is that instead of finding the middle point we find the point that touches the x-axis.

Steps

1. Find middle point c = (a*func(b)-b*func(a))/(func(b)-func(a)) .
2. If f(c) == 0, then c is the root of the solution.
3. Else f(c) != 0
   1. If value f(a)*f(c) < 0 then root lies between a and c. So we recur for a and c
   2. Else If f(b)*f(c) < 0 then root lies between b and c. So we recur b and c.
   3. Else given function doesn’t follow one of assumptions.
4. Repeat the steps until difference between a and b is less then a value (very small value).

The Visual Representation For False Position Method :

Secant Method

1. Define the function as f(x).
2. Assume initial guesses x1,x2.
3. Calculate f(x1) and f(x2) and if f(x1)=f(x2) , then print "Mathematical Error" and terminate the program.
4. Determine x3=x2-(x2-x1)*f(x2)/f(x2)-f(x1).
5. Check if f(x3)=0,
   1. if true , then print the solution x3.
   2. else replace x1=x2 and x3=x2.
6. Repeat step 3 ,4,5 until f(x3)=0.

The Visual Representation For Secant Method:

Newton-Raphson Method

1. Define the function as f(x).
2. Assume initial guesses x0.
3. Calculate the derivatives of f(x) and define it as f '(x).
4. Calculate f(x0) and f '(x0) .
5. Check if f'(x0)=0 , then then print "Mathematical Error" and terminate the program.
6. Determine x1=x0-f(x0)/f '(x0).
7. Check if f(x1)=0,
   1. if true then print the solution x1.
   2. else replace x0=x1.
8. Repeat step 3 ,4,5 until f(x3)=0.

The Visual Representation For Newton-Raphson Method:

Runge-Kutta Method

Purpose: The Runge-Kutta method is a numerical technique used to solve ordinary differential equations (ODEs) of the form:

• dy/dx=f(x,y)

This method estimates the value of y at a given x using an iterative approach.

How It Works

The 4th-Order Runge-Kutta Method is the most commonly used variant due to its balance of accuracy and computational efficiency. Here’s how it works:

1. `Initialize`:
    -> Start with the initial condition: x0, y0.
    -> Define the step size hand the endpoint x_end.

2. Iterative Steps:

-> Compute four slopes (k-values) for each step:
-> k1 = h*f(x,y)
-> k2 = h*f( x + 0.5h, y + 0.5k1)
-> k3 = h*f(x + 0.5h,y + 0.5k2)
-> k4 = h*f(x + h,y + k3)

3. Update the next value:

-> y_next = y + 1/6(k1 + 2k2 + 2k3 + k4)
-> Increment x by h.
-> Repeat until the target x_end is reached.

Example

Given:

dy/dx = x + y, x0 = 0, y0 = 1,x_end = 2, h = 0.1

The program calculates intermediate values using the 4th-order Runge-Kutta method. It prints out the x and y values at each step.

Pseudocode:

1. Initialize x = x0, y = y0
2. While x < x_end:

   • k1 = h * f(x, y)

   • k2 = h * f(x + 0.5 * h, y + 0.5 * k1)

   • k3 = h * f(x + 0.5 * h, y + 0.5 * k2)

   • k4 = h * f(x + h, y + k3)

   • y = y + (1/6) * (k1 + 2k2 + 2k3 + k4)

   • x = x + h

The Visual Representation For Runge Kutta Method :

Matrix Inversion

Purpose

Matrix inversion is used to find the inverse of a matrix, which can then be applied to solve a system of linear equations. The inverse of a matrix A, denoted by A^-1 in such that: A * A^-1= I where I is the identity matrix.

How It Works

1. For a system of equations:
   • AX=B
   • A is the coefficient matrix.
   • X is the column vector of unknowns.
   • B is the column vector of constants.
2. If A is invertible, the solution is given by:
   • -X= A^-1*B
3. If the determinant of A is zero, the matrix is singular (non-invertible).

Example

If you have a coefficient matrix A:

 ____   ___
|          |
| 1  1  1  |
| 2  3  7  |
| 3  5  2  |
|___    ___|

The program computes A^-1 and displays it.

Pseudocode:

1. If det(A) == 0:
   • Return "Matrix is singular and cannot be inverted"
2. Else:
   • Use np.linalg.inv(A) to compute the inverse
   • Return A^{-1}

The Visual Representation of the Matrix is: