Solution of equations and eigenvalue problems

 To find an approximate real root of given
equation.
 ITERATION FORMULA OF NEWTON-
RAPHSON METHOD
 𝒙𝒊+𝟏 = 𝒙𝒊 −
𝒇 𝒙 𝒊
𝒇′ 𝒙 𝒊
, 𝒊 = 𝟎, 𝟏, 𝟐 … .

 THE CONDITION FOR CONVERGENCE OF
NEWTON-RAPHSON METHOD FOR 𝒇(𝒙) = 𝟎
 The condition is |𝑓(𝑥). 𝑓′(𝑥)| < 𝑓′ 𝑥
2
in a
neighbourhood of the root
 ORDER OF CONVERGENCE OF NEWTON-
RAPHSON METHOD
 The order of convergence is 2.

 To find the root of the equation 𝑓(𝑥) = 0
 Step 1. Find the numbers a and b in such a way that
𝑓(𝑎) and 𝑓(𝑏) are in opposite signs. (From this we
conclude that there is one real root between a and b)
 Step 2. Choose the initial approximate root as 𝑥0
 Step 3. Using the Newton Raphson’s formula 𝒙𝒊+𝟏 =
𝒙𝒊 −
𝒇 𝒙𝒊
𝒇′ 𝒙 𝒊
, 𝒊 = 𝟎, 𝟏, 𝟐 … ., find the sequence
𝑥0, 𝑥1, … 𝑥 𝑛, . . and the point of convergence of the
sequence is the root of the given equation.

 1. It can be used for finding root of both
algebraic and transcendental equations.
 2. The convergence of Newton’s method is
faster and so it is preferred compared to
other methods.
 3. It is simple and easy to deal with and it is
used to improve the results obtained by
other methods.

 To find the root of the equation 𝑓(𝑥) = 0
 Step 1. Find the numbers a and b in such a way that
𝑓(𝑎) and 𝑓(𝑏) are in opposite signs. (From this we
conclude that there is one real root between a and b)
 Step 2. Write the given equation on the form 𝑥 =
∅(𝑥)
 Step 3. Choose the initial approximate root as 𝑥0
 Step 4. Replace 𝑥 by 𝑥0 in step 2, and take 𝑥1 =
∅(𝑥0)
 Step 5. Further , find 𝑥2 = ∅(𝑥1)
 Step 6. Continuing this way, we will get a sequence
𝑥0, 𝑥1, … 𝑥 𝑛, . . and the point of convergence of the
sequence is the root of the given equation

 ORDER OF CONVERGENCE OF FIXED POINT
ITERATION METHOD
 The order of convergence is 1.
 THE CONDITION FOR CONVERGENCE OF
FIXED POINT ITERATION METHOD FOR
𝒇(𝒙) = 𝟎
 The condition is |∅′ 𝑥 | < 1, for all values of
𝑥

 DIRECT METHOD
 1. Gauss-elimination method
 2. Gauss-Jordan method
 INDIRECT METHOD (ITERATION METHOD)
 1. Gauss-Jacobi method
 2. Gauss- Seidel method

 This is a direct method. In this method the
given ‘n’ system of simultaneous linear
equations
 𝑎𝑖1 𝑥1 + 𝑎𝑖2 𝑥2 + ⋯ + 𝑎𝑖𝑛 𝑥 𝑛 = 𝑏𝑖 , 𝑖 = 1,2, … 𝑛
 can be written in the form AX = B
 Where 𝐴 = 𝑎𝑖𝑗 𝑛×𝑛
𝑋 = 𝑥𝑖 𝑛×1 𝐵 =
𝑏𝑖 𝑛×1 matrices
 And in the augmented matrix (A, B), the
coefficient matrix A can be reduced to upper
triangular matrix, which can be solved by
using back substitution.

 This is a direct method. In this method the
given ‘n’ system of simultaneous linear
equations
 𝑎𝑖1 𝑥1 + 𝑎𝑖2 𝑥2 + ⋯ + 𝑎𝑖𝑛 𝑥 𝑛 = 𝑏𝑖 , 𝑖 = 1,2, … 𝑛
 can be written in the form AX = B
 Where 𝐴 = 𝑎𝑖𝑗 𝑛×𝑛
𝑋 = 𝑥𝑖 𝑛×1 𝐵 =
𝑏𝑖 𝑛×1 matrices
 And in the augmented matrix (A, B), the
coefficient matrix A can be reduced to a
diagonal matrix, which can be solved by
using back substitution.

 This is an iterative method.
 Suppose the given linear equations are
 𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 𝑧 = 𝑑1 , 𝑎2 𝑥 + 𝑏2 𝑦 + 𝑐2 𝑧 = 𝑑2 , 𝑎3 𝑥 + 𝑏3 𝑦 + 𝑐3 𝑧 = 𝑑3
 Let the coefficient matrix 𝐴 =
𝑎1 𝑏1 𝑐1
𝑎2 𝑏2 𝑐2
𝑎3 𝑏3 𝑐3
 To apply Gauss – Jacobi, the coefficient matrix A should be diagonally dominant.
 (i.e) 𝑎1 > 𝑏1 + 𝑐1
 𝑏2 > 𝑎2 + 𝑐2
 𝑐3 > 𝑎3 + |𝑏3|
 To solve the given system, we have
 𝑥 =
1
𝑎1
(𝑑1 − 𝑏1 𝑦 − 𝑐1 𝑧)
 𝑦 =
1
𝑏2
(𝑑2 − 𝑎2 𝑥 − 𝑐2 𝑧)
 𝑧 =
1
𝑐3
(𝑑3 − 𝑎3 𝑥 − 𝑏3 𝑦)
 If 𝑥(0), 𝑦(0), 𝑧(0) are the initial values of x, y, z respectively
 First iteration values are
 𝑥 1 =
1
𝑎1
(𝑑1 − 𝑏1 𝑦 0 − 𝑐1 𝑧 0 )
 𝑦 1 =
1
𝑏2
(𝑑2 − 𝑎2 𝑥 0 − 𝑐2 𝑧 0 )
 𝑧 1 =
1
𝑐3
(𝑑3 − 𝑎3 𝑥 0 − 𝑏3 𝑦 0 )
 Second iteration values are
 𝑥 2 =
1
𝑎1
(𝑑1 − 𝑏1 𝑦 1 − 𝑐1 𝑧 1 )
 𝑦 2 =
1
𝑏2
(𝑑2 − 𝑎2 𝑥 1 − 𝑐2 𝑧 1 )
 𝑧 2 =
1
𝑐3
(𝑑3 − 𝑎3 𝑥 1 − 𝑏3 𝑦 1 )
 Proceeding like this and stop the iteration when the values of x, y, z are start repeating with required degree of
accuracy.

 This is an iterative method.
 Suppose the given linear equations are
 𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 𝑧 = 𝑑1 , 𝑎2 𝑥 + 𝑏2 𝑦 + 𝑐2 𝑧 = 𝑑2 , 𝑎3 𝑥 + 𝑏3 𝑦 + 𝑐3 𝑧 = 𝑑3
 Let the coefficient matrix 𝐴 =
𝑎1 𝑏1 𝑐1
𝑎2 𝑏2 𝑐2
𝑎3 𝑏3 𝑐3
 To apply Gauss – Seidel, the coefficient matrix A should be diagonally dominant.
 (i.e) 𝑎1 > 𝑏1 + 𝑐1
 𝑏2 > 𝑎2 + 𝑐2
 𝑐3 > 𝑎3 + |𝑏3|
 To solve the given system, we have
 𝑥 =
1
𝑎1
(𝑑1 − 𝑏1 𝑦 − 𝑐1 𝑧)
 𝑦 =
1
𝑏2
(𝑑2 − 𝑎2 𝑥 − 𝑐2 𝑧)
 𝑧 =
1
𝑐3
(𝑑3 − 𝑎3 𝑥 − 𝑏3 𝑦)
 If 𝑥(0), 𝑦(0), 𝑧(0) are the initial values of x, y, z respectively
 First iteration values are
 𝑥 1 =
1
𝑎1
(𝑑1 − 𝑏1 𝑦 0 − 𝑐1 𝑧 0 )
 𝑦 1 =
1
𝑏2
(𝑑2 − 𝑎2 𝑥 1 − 𝑐2 𝑧 0 )
 𝑧 1 =
1
𝑐3
(𝑑3 − 𝑎3 𝑥 1 − 𝑏3 𝑦 1 )
 Second iteration values are
 𝑥 2 =
1
𝑎1
(𝑑1 − 𝑏1 𝑦 1 − 𝑐1 𝑧 1 )
 𝑦 2 =
1
𝑏2
𝑑2 − 𝑎2 𝑥 2 − 𝑐2 𝑧 1
 𝑧 2 =
1
𝑐3
(𝑑3 − 𝑎3 𝑥 2 − 𝑏3 𝑦 2 )
 Proceeding like this and stop the iteration when the values of x, y, z are start repeating with required degree of
accuracy.

 The condition is the absolute value of largest
coefficient is greater than the sum of the
absolute value of all the remaining
coefficients.
 (i.e) the coefficient matrix is diagonally
dominant
 (i.e) 𝑎𝑖𝑖 > 𝑗=1,𝑖≠𝑗
𝑛
|𝑎𝑖𝑗| ∀ 𝑖 = 1,2 … 𝑛
 The rate of convergence in Gauss –Seidel
method is very fast than in Gauss-Jacobi

S.No Gauss Jorden Gauss Jacobi
1 Direct method Iterative method
2 This method produce exact
solution
after a finite number of
steps
This method gives a sequence
of approximate solutions, which
ultimately approach the actual
solution.
3 Applicable if the coefficient
matrix
is non-singular
Applicable if the coefficient
matrix is diagonally dominant.

S.No Gauss elimination Gauss Jacobi
1 Direct method Iterative method
2 This method produce exact
solution after a finite number
of steps
This method gives a sequence
of approximate solutions,
which ultimately approach the
actual solution.
3 Applicable if the coefficient
matrix
is non-singular
Applicable if the coefficient
matrix is diagonally dominant.

 Procedure
 Step 1. Write the augmented matrix (𝐴, 𝐼) ,
where A is the given matrix
 Step 2. Reduce the matrix A in (𝐴, 𝐼) to the
identity matrix by using row transformation.
 Step 3. From step 2, you will get (𝐼 𝐴−1
)

 If A is any square matrix, then there exist a
scalar 𝜆 and a non-zero column vector X such
that AX = 𝜆 X, the scalar 𝜆 is called Eigen value
of A and the corresponding X is called Eigen
vector.
 By the property of Eigen values and Eigen
vectors , if 𝜆 is an Eigen value of A and X is the
corresponding Eigen vector, then
1
𝜆
is an Eigen
value of 𝐴−1
with the same Eigen vector X
 Also the smallest Eigen value of A is
1
𝜆
and the
corresponding Eigen vector is X
 And sum of the Eigen values of A= sum of the
principal diagonal elements of A

 Procedure - (for a square matrix A of order 3 x 3)
 1. Let 𝑋0 be the initial which is usually chosen as a vector
with all components equal to 1. 𝑖. 𝑒 𝑋0 =
1
1
1
(i.e.,
normalized)
 2. Find the product 𝐴𝑋0 and express it in the form 𝐴𝑋0 =
𝜆1 𝑋1, where 𝑋1 is normalized by taking out the largest
component𝜆1.
 3. Find 𝐴𝑋1 and express it in the form𝐴𝑋1 = 𝜆2 𝑋2, where
𝑋2 is normalized by taking out the largest component 𝜆2
and continue the process.
 4. Thus we have a sequence of equations
 𝐴𝑋0 = 𝜆1 𝑋1, 𝐴𝑋1 = 𝜆2 𝑋2, 𝐴𝑋2 = 𝜆3 𝑋3 … . ·
 5. We stop at the stage where 𝑋 𝑟−1, 𝑋 𝑟 are almost same.
 Then 𝜆 𝑟 is the largest Eigen value and 𝑋 𝑟 is the
corresponding Eigen vector.

 1. Symmetric Matrix
 A square matrix 𝑎𝑖𝑗 𝑛×𝑛
is said to be
symmetric matrix if 𝑎𝑖𝑗 = 𝑎𝑗𝑖 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖 ≠ 𝑗
 2. Orthogonal Matrix
 A square matrix 𝑎𝑖𝑗 𝑛×𝑛
is said to be
Orthogonal matrix if 𝐴𝐴 𝑇
= 𝐼
 (Or) 𝐴 𝑇 = 𝐴−1

 3. For an Orthogonal matrix A, 𝑑𝑒𝑡 𝐴 = 𝐴 =
± 1
 4. The diagonal elements of a diagonal
matrix D are its Eigen values

 Working Rule (Jacobi Method)
 If 𝐴 =
𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33
is the given symmetric matrix (𝑖𝑒. 𝑎𝑖𝑗 = 𝑎𝑗𝑖)
 Step 1. Choose the largest off diagonal element (say 𝑎13)
 Step 2. Then take the rotation matrix 𝑆1 =
𝑐𝑜𝑠𝜃 0 −𝑠𝑖𝑛𝜃
0 1 0
𝑠𝑖𝑛𝜃 0 𝑐𝑜𝑠𝜃
 Step 3. Define 𝑡𝑎𝑛2𝜃 =
2𝑎13
𝑎11−𝑎33
 𝑖. 𝑒 𝜃 =
1
2
tan−1 2𝑎13
𝑎11−𝑎33
𝑖𝑓 𝑎11 ≠ 𝑎33
 (or) 𝜃 =
𝜋
4
𝑖𝑓 𝑎11 = 𝑎33 & 𝑎12 > 0
 (or) 𝜃 = −
𝜋
4
𝑖𝑓 𝑎11 = 𝑎33 & 𝑎12 < 0
 Step 4. Substitute 𝜃 value in 𝑆1
 Step 5. Find 𝐴1 = 𝑆1
−1
𝐴𝑆1 = 𝑆1
𝑇
𝐴 𝑆1
 Step 6. Again choose the largest off-diagonal element of 𝐴1
 Step 7. Proceed again from Step 2 to step 5 until you get a diagonal matrix 𝐴 𝑛
 The diagonal elements in 𝐴 𝑛 are the Eigen values of A
 Step 8. For Eigen vector , to find the matrix 𝑆 = 𝑆1 𝑆2 . . = 𝑠𝑖𝑗
 The corresponding vectors are 𝑠𝑖1 , 𝑠𝑖2 , 𝑠𝑖3

Solution of equations and eigenvalue problems

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to Solution of equations and eigenvalue problems

Similar to Solution of equations and eigenvalue problems (20)

More from Santhanam Krishnan

More from Santhanam Krishnan (20)

Recently uploaded

Recently uploaded (20)

Solution of equations and eigenvalue problems