Home | | Numerical Methods | | Statistics and Numerical Methods | Solution of Equations and Eigenvalue Problems

# Solution of Equations and Eigenvalue Problems

1 Numerical solution of Non-Linear Equations Method of false position Newton Raphson Method Iteration Method 2 System of Linear Equations Gauss Elimination Method Gauss Jordan Method Gauss Jacobi Method Gauss Seidel Method 3 Matrix Inversion Inversion by Gauss Jordan Method 4 Eigen Value of a Matrix Von Mise`s power method

SOLUTION OF EQUATIONS AND EIGENVALUE PROBLEMS

1 Numerical solution of Non-Linear Equations

Method of false position

Newton Raphson Method

Iteration Method

2 System of Linear Equations

Gauss Elimination Method

Gauss Jordan Method

Gauss Jacobi Method

Gauss Seidel Method

3 Matrix Inversion

Inversion by Gauss Jordan Method

4 Eigen Value of a Matrix

Von Mise`s power method

1 Numerical solution of Non-Linear equations

Introduction

The problem of solving the equation is of great importance in science and Engineering.

In this section, we deal with the various methods which give a solution for the equation

Solution of Algebraic and transcendental equations

The equation of the form f(x) =0 are called algebraic equations if f(x) is purely a polynomial in x.For example: are algebraic equations .

If f(x) also contains trigonometric, logarithmic, exponential function etc. then the equation is known as transcendental equation.

Methods for solving the equation

The following result helps us to locate the interval in which the roots of

Method of false position.

Iteration method

Newton-Raphson method

Method of False position (Or) Regula-Falsi method (Or) Linear interpolation method

In bisection method the interval is always divided into half. If a function changes sign over an interval, the function value at the mid-point is evaluated. In bisection method the interval from a to b into equal intervals ,no account is taken of the magnitude of .An alternative method that exploits this graphical insights is to join by a straight line.The intersection of this line with the X-axis represents an improved estimate of the root.The replacement of the curve by a straight line gives a “ the origin of the name ,method of false position ,or in Latin ,Regula falsi .It is also called the linear interpolation method.

Problems

1. Find a real root of that lies between 2 and 3 by the method of false position and correct

to three decimal places.

Sol:

Let

The root lies between 2.5 and 3.

The approximations are given by 2. Obtain the real root of,correct to four decimal places using the method of false position.

Sol:

Given

Taking logarithmic on both sides,

x log10 x 2

f (x) x log10 x 2

f (3.5) 0.0958 0 and f (3.6) 0.00027 0

The roots lies between 3.5 and 3.6

The approximation are given by The required root is 3.5973

Exercise:

1. Determine the real root of correct to four decimal places by Regula-Falsi method.

Ans: 1.0499

2. Find the positive real root of correct to four decimals by the method of False position .

Ans: 1.8955

3.Solve the equation by Regula-Falsi method,correct to 4 decimal places.

Ans: 2.7984

Newton’smethod(or) Newton-Raphson method (Or) Method of tangents

This method starts with an initial approximation to the root of an equation, a better and closer approximation to the root can be found by using an iterative process.

Derivation of Newton-Raphson formula

Let   be the root of  f (x) 0 and x0   be an approximation to   .If hx0

Then   by   the   Tayor’s   series Note:

The error at any stage is proportional to the square of the error in the previous stage.

The order of convergence of the Newton-Raphson method is at least 2 or the convergence of N.R method is Quadratic.

Problems

1.Using Newton’s method ,find the correct root five decimal between places.

Sol:

Given  First approximation:=0.71429

Second approximation:  0.73109

Third approximation =0.73205

Fourth approximation: =0.73205

The root is 0.73205, correct to five decimal places.

By Newton-Raphson formula, we have approximation

The initial approximation is x0 = 0.5

2. Find a real root of x=1/2+sinx near 1.5, correct to 3 decimal places by newton-Rapson method. Sol: By Newton-Raphson formula, we have approximation First approximation: =1.497

Second approximation: =1.497.

The required root is 1.497

Exercise:

1. Find the real root of e x 3x ,that lies between 1 and 2 by New places.

Ans:1.5121

2.Use Newton-Raphson method to solve the equation Ans: 0.03226

Iteration method (Or) Method of successive approximations(Or)Fixed point method

For solving the equation f (x) = 0 by iteration method, we start with an approximation value of the root.The equation  f (x) Theorem(Fixed point theorem) The order of convergence

Theorem Note

The order of convergence in general is linear (i.e) =1

Problems  Hence the root is 0.7549

Exercise:

1. Find the cube root of 15, correct to four decimal places, by iteration method

Ans: 2.4662

2 System of linear equation

Introduction

Many problems in Engineering and science needs the solution of a system of simultaneous linear equations .The solution of a system of simultaneous linear equations is obtained by the following two types of methods

Direct methods (Gauss elimination and Gauss Jordan method)

Indirect methods or iterative methods (Gauss Jacobi and Gauss Seidel method)

(a) Direct methods are those in which

The computation can be completed in a finite number of steps resulting in the exact solution

The amount of computation involved can be specified in advance.

The method is independent of the accuracy desired.

(b) Iterative methods (self correcting methods) are which

Begin with an approximate solution and

Obtain an improved solution with each step of iteration

But would require an infinite number of steps to obtain an exact solution without round-off errors

The accuracy of the solution depends on the number of iterations performed.

Simultaneous linear equations

The system of equations in n unknowns x1 , x2 , x3 ,..... xn  is given by The system of equation can be solved by using determinants (Cramer’s rule) or by means of matrices.

These involve tedious calculations. These are other methods to solve such equations .In this chapter we will discuss four methods viz.

(i)                Gauss –Elimination method

(ii)             Gauss –Jordan method

(iii)           Gauss-Jacobi method

(iv)           Gauss seidel method

Ø Gauss-Elimination method

This is an Elimination method and it reduces the given system of equation to an equivalent upper triangular system which can be solved by Back substitution.

Consider the system of equations Gauss-algorithm is explained below:

Step 1. Elimination of x1 from the second and third equations .If a11 != 0 the first equation is used to eliminate x1 from the second and third equation. After elimination, the reduced system is Step 2:

Elimination of x2 from the third equation. If a’22 != 0 We eliminate x2 from third equation and the reduced upper triangular system is Step 3:

From third equation x3  is known. Using x3 in the second equation x2  is obtained. using both x2 And x3  in the first equation, the value of x1 is obtained.

Thus the elimination method, we start with the augmented matrix (A/B) of the given system and transform it to (U/K) by eliminatory row operations. Finally the solution is obtained by back substitution process. 2. Gauss –Jordan method

This method is a modification of Gauss-Elimination method. Here the elimination of unknowns is performed not only in the equations below but also in the equations above. The co-efficient matrix A of the system AX=B is reduced into a diagonal or a unit matrix and the solution is obtained directly without back substitution process. Examples  method (2) Gauss –Jordan method Ans: x=1; y=3; z=5

Iterative method

These methods are used to solve a special of linear equations in which each equation must possess one large coefficient and the large coefficient must be attached to a different unknown in that equation.Further in each equation, the absolute value of the large coefficient of the unknown is greater than the sum of the absolute values of the other coefficients of the other unknowns. Such type of simultaneous linear equations can be solved by the following iterative methods.

Gauss-Jacobi method

Gauss seidel method i.e., the co-efficient matrix is diagonally dominant.

Solving the given system for x,y,z (whose diagonals are the largest values),we have Gauss-Jacobi method

If the rth iterates are  x(r ) , y (r ) , z (r ) ,then the iteration scheme for this method is The iteration is stopped when the values x, y, z start repeating with the desired degree of accuracy.

Gauss-Seidel method

This method is only a refinement of Gauss-Jacobi method .In this method ,once a new value for a unknown is found ,it is used immediately for computing the new values of the unknowns.

If the rth iterates are, then the iteration scheme for this method is Hence finding the values of the unknowns, we use the latest available values on the R.H.S The process of iteration is continued until the convergence is obtained with desired accuracy.

Conditions for convergence

Gauss-seidel method will converge if in each equation of the given system ,the absolute values o the largest coefficient is greater than the absolute values of all the remaining coefficients This is the sufficient condition for convergence of both Gauss-Jacobi and Gauss-seidel iteration methods.

Rate of convergence

The rate of convergence of gauss-seidel method is roughly two times that of Gauss-Jacobi method. Further the convergences in Gauss-Seidel method is very fast in gauss-Jacobi .Since the current values of the unknowns are used immediately in each stage of iteration for getting the values of the unknowns.

Problems

1.Solve by Gauss-Jacobi method, the following system   3 Matrix inversion

Introduction

A square matrix whose determinant value is not zero is called a non-singular matrix. Every non-singular square matrix has an inverse matrix. In this chapter we shall find the inverse of the non-singular square matrix A of order three. If X is the inverse of A,

Then

Inversion by Gauss-Jordan method

By Gauss Jordan method, the inverse matrix X is obtained by the following steps:

Step 1:        First consider the augmented matrix

Step 2:        Reduce the matrix A in to the identity matrix I by employing row transformations.

The row transformations used in step 2 transform I to A-1

Finally write the inverse matrix A-1.so the principle involved for finding A-1 is as shown below

Note

The answer can be checked using the result

Problems

1.Find the inverse of by Gauss-Jordan method. 4 Eigen value of a Matrix

Introduction

For every square matrix A, there is a scalar λ and a non-zero column vector X such that AX= λ X .Then the scalar λ  is called an Eigen value of A and X, the corresponding Eigen vector. We have studied earlier the computation of Eigen values and the Eigen vectors by means of analytical method. In this chapter, we will discuss an iterative method to determine the largest Eigen value and the corresponding Eigen vector.

Power method (Von Mise’s   power   method)

Power method is used to determine numerically largest Eigen value and the corresponding Eigen vector of a matrix A

Working Procedure Problems

1.Use the power method to find the dominant value and the corresponding Eigen vector of the matrix

Sol:

Let

--- - - ----

Hence the dominant Eigen value is 15.97 and the corresponding Eigen vector is 2.

Determine the dominant Eigen value and the corresponding Eigen vector of  using power method.

Exercise:

1. Find the dominant Eigen value and the corresponding Eigen vector of A Find also the other two Eigen values.

2. Find the dominant Eigen value of the corresponding Eigen vector of By power method. Hence find the other Eigen value also. Tutorial 2 1.Solve the following system by Gauss-elimination method 3.Determine the dominant Eigen value and the corresponding Eigen vector of using power method. Ans:[0.062,1,0.062]T

4.Find the smallest Eigen value and the corresponding Eigen vector of using power method. Ans:[0.707,1,0.707]T

5.Find the dominant Eigen value and the corresponding Eigen vector of by power method. Hence find the other Eigen value also . Ans:λ=2.381

Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
Mathematics (maths) - Solution of Equations and Eigenvalue Problems : Solution of Equations and Eigenvalue Problems |

Related Topics