The document discusses solving linear equations using numerical methods in MATLAB. It provides code implementations for Gauss-Seidel method, Jacobi method, and Cholesky method. It also includes routines for calculating the inverse of a 3x3 matrix, matrix norm, and forward substitution. Examples are given applying each method to different 3x3 matrices. Spectral radius is also calculated for some examples to analyze convergence properties.
Representation of of Stochastic Processes in Stochastic Processes in Real and Spectral Domains Real and Spectral Domains and and Monte Monte-Carlo sampling
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Representation of of Stochastic Processes in Stochastic Processes in Real and Spectral Domains Real and Spectral Domains and and Monte Monte-Carlo sampling
Calculus 10th edition anton solutions manualReece1334
Download at: https://goo.gl/e1svMM
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The Estimation for the Eigenvalue of Quaternion Doubly Stochastic Matrices us...ijtsrd
The purpose of this paper is to locate and estimate the eigen values of quaternion doubly stochastic matrices. We present several estimation theorems about the eigen values of quaternion doubly stochastic matrices. Mean while, we obtain the distribution theorem for the eigen values of tensor products of two quaternion doubly stochastic matrices. We will conclude the paper with the distribution for the eigen values of generalized quaternion doubly stochastic matrices. Dr. Gunasekaran K | Mrs. Seethadevi R ""The Estimation for the Eigenvalue of Quaternion Doubly Stochastic Matrices using Gerschgorin Balls"" Published in International Journal of Trend in Scientific Research and Development (ijtsrd), ISSN: 2456-6470, Volume-3 | Issue-3 , April 2019, URL: https://www.ijtsrd.com/papers/ijtsrd22860.pdf
Paper URL: https://www.ijtsrd.com/mathemetics/other/22860/the-estimation-for-the-eigenvalue-of-quaternion-doubly-stochastic-matrices-using-gerschgorin-balls/dr-gunasekaran-k
The Estimation for the Eigenvalue of Quaternion Doubly Stochastic Matrices us...ijtsrd
The purpose of this paper is to locate and estimate the eigen values of quaternion doubly stochastic matrices. We present several estimation theorems about the eigen values of quaternion doubly stochastic matrices. Mean while, we obtain the distribution theorem for the eigen values of tensor products of two quaternion doubly stochastic matrices. We will conclude the paper with the distribution for the eigen values of generalized quaternion doubly stochastic matrices. Dr. Gunasekaran K | Mrs. Seethadevi R ""The Estimation for the Eigenvalue of Quaternion Doubly Stochastic Matrices using Gerschgorin Balls"" Published in International Journal of Trend in Scientific Research and Development (ijtsrd), ISSN: 2456-6470, Volume-3 | Issue-3 , April 2019, URL: https://www.ijtsrd.com/papers/ijtsrd22860.pdf
Paper URL: https://www.ijtsrd.com/mathemetics/other/22860/the-estimation-for-the-eigenvalue-of-quaternion-doubly-stochastic-matrices-using-gerschgorin-balls/dr-gunasekaran-k
Computer Oriented Numerical Analysis
What is interpolation?
Many times, data is given only at discrete points such as .
So, how then does one find the value of y at any other value of x ?
Well, a continuous function f(x) may be used to represent the data values with f(x) passing through the points (Figure 1). Then one can find the value of y at any other value of x .
This is called interpolation
Newton’s Divided Difference Formula:
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Applied Numerical Methods Curve Fitting: Least Squares Regression, InterpolationBrian Erandio
Correction with the misspelled langrange.
and credits to the owners of the pictures (Fantasmagoria01, eugene-kukulka, vooga, and etc.) . I do not own all of the pictures used as background sorry to those who aren't tagged.
The presentation contains topics from Applied Numerical Methods with MATHLAB for Engineers and Scientist 6th and International Edition.
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This is the entrance exam paper for ISI MSQE Entrance Exam for the year 2010. Much more information on the ISI MSQE Entrance Exam and ISI MSQE Entrance preparation help available on http://crackdse.com
Positive and negative solutions of a boundary value problem for a fractional ...journal ijrtem
: In this work, we study a boundary value problem for a fractional
q, -difference equation. By
using the monotone iterative technique and lower-upper solution method, we get the existence of positive or
negative solutions under the nonlinear term is local continuity and local monotonicity. The results show that we
can construct two iterative sequences for approximating the solutions
1. EC 313: Numerical Methods Mayank Awasthi(2006033)
MATLAB Assignment – 2 B.Tech 3rd Year(ECE)
Pdpm IIITDM, JABALPUR
Ques1:
Gauss Siedel Method for Solving the Linear Equations:
% Implementing Gauss Siedel method
% a is matrix whose diagonal elements are non-zero else rearrange it
function x = gauss_siedel(a,b,x)
n=length(a);
l=zeros(n,n);
u=zeros(n,n);
d=zeros(n,n);
id=[1,0,0;0,1,0;0,0,1];
for i=1:n
for j=1:n
a(i,j)= a(i,j)/a(i,i); %making the diagonal elements 1.
% Breaking the matrix as a sum of ( L+U+I )
if i>j
l(i,j)=a(i,j);
end
if i<j
u(i,j)=a(i,j);
end
if i==j
d(i,j)=a(i,j);
end
end
end
%Implementing Norm of c where c = inverse(I+L)*U
norm2(inv2(id+l)*u);
if norm2(inv2(id+l)*u)>1
fprintf('Norm of c is greater than 1, Solution will diverge');
return;
end
for i=1:n
b(i,1)=b(i,1)/a(i,i);
end
% Calculating x using FORWARD DIFFERENCE CONCEPT
x=[0;0;0];
for i=1:100
x= forward_subs((l+d),(b-u*x)); m=a*x-b;
2. % Calculating(dis(A*x,b)<1e-6)to stop iteration at the given tolerance
temp=0;
for j=1:n
temp=temp+power(m(j,1),2);
end
temp=sqrt(temp);
if temp<0.0000001
break;
end
end
fprintf('nThe maximum no. of iteration required is %d',i);
end
MATLAB routine for Inverse of 3x3 matrix :
% Calculating Inverse of 3x3 matrix
function x = inv2(A)
I=[1,0,0;0,1,0;0,0,1];
n=length(A);
a21=A(2,1)/A(1,1);
for k = 1:n-1 % Elimination Phase
for i= k+1:n
if A(i,k) ~= 0
lambda = A(i,k)/A(k,k);
A(i,k:n) = A(i,k:n) - lambda*A(k,k:n);
I(i,k:n)= I(i,k:n) - lambda*I(k,k:n);
I(3,1)=I(2,1)*I(3,2)+I(3,1);
end
end
end
x1=back_subs(A,I(:,1)); % Backward Substitution
x2=back_subs(A,I(:,2));
x3=back_subs(A,I(:,3));
x=[x1,x2,x3];
end
MATLAB routine for Norm:
% Calculating Norm of matrix
function n0= norm2(c)
l=length(c);
n0=0;
for m=1:l
for n=1:l
n0 = n0 + c(m,n)*c(m,n);
end
end
n0=sqrt(n0);
end
3. MATLAB routine for Forward Substitution:
% Implementing Forward Substitution
function x=forward_subs(A,b)
n=length(A);
for i = 1:3
t=0;
for j = 1:(i-1)
t=t+A(i,j)*x(j);
end
x(i,1)=(b(i,1)-t)/A(i,i);
end
Part1: Part2:
>> a=[1 ,.2 .2;.2, 1, .2;.2, .2, 1] >> a=[1 ,.5 .5;.5, 1, .5;.5, .5, 1]
a= a=
1.0000 0.2000 0.2000 1.0000 0.5000 0.5000
0.2000 1.0000 0.2000 0.5000 1.0000 0.5000
0.2000 0.2000 1.0000 0.5000 0.5000 1.0000
>> b=[2;2;2] >> b=[2;2;2]
b= b=
2 2
2 2
2 2
>> x=[0;0;0] >> x=[0;0;0]
x= x=
0 0
0 0
0 0
>> gauss_siedel(a,b,x) >> gauss_siedel(a,b,x)
The maximum no. of iteration required is 8 The maximum no of iteration required is 17
ans = ans =
1.4286 1.0000
1.4286 1.0000
1.4286 1.0000
4. Part3:
>> a=[1 ,.9 .9;.9, 1, .9;.9, .9, 1]
a=
1.0000 0.9000 0.9000
0.9000 1.0000 0.9000
0.9000 0.9000 1.0000
>> b=[2;2;2]
b=
2
2
2
>> x=[0;0;0]
x=
0
0
0
>> gauss_siedel(a,b,x)
Norm of c is greater than 1, Solution will diverge
ans =
0
0
0
************************************************************************
Spectral radius of C:
Spectral radius of C is maximum eigenvalue of C*CT.
In this case C=(I+L)-1 * U
9. Ques2:
Jacobi Method for Solving the Linear Equations:
%Implementing Jacobi Method
function x = jacobi(A,b,x)
I=[1 0 0; 0 1 0; 0 0 1];
c=I-A;
%Checking Norm
if norm2(c) > 1
fprintf('nNorm is greater than one so it will diverge');
return;
end
for j=1:50
x=b+(I-A)*x;
n=A*x-b;
%checking the condition of tolerance to stop the iterations
temp=0;
for i = 1:3
temp= temp+ power(n(i,1),2);
end
temp=sqrt(temp);
if temp < 0.0001
break;
end
end
fprintf('nThe iteration required is %d',j);
end
MATLAB routine for Norm:
% Calculating Norm of matrix
function n0= norm2(c)
l=length(c);
n0=0;
for m=1:l
for n=1:l
n0 = n0 + c(m,n)*c(m,n);
end
end
n0=sqrt(n0);
end
10. Part1: Part2
>> a=[1 ,.2 .2;.2, 1, .2;.2, .2, 1] >> a=[1 ,.5 .5;.5, 1, .5;.5, .5, 1]
a= a=
1.0000 0.2000 0.2000 1.0000 0.5000 0.5000
0.2000 1.0000 0.2000 0.5000 1.0000 0.5000
0.2000 0.2000 1.0000 0.5000 0.5000 1.0000
>> b=[2;2;2] >> b=[2;2;2]
b= b=
2 2
2 2
2 2
>> x=[0;0;0] >> x=[0;0;0]
x= x=
0 0
0 0
0 0
>> jacobi(a,b,x) >> jacobi(a,b,x)
The iteration required is 12 Norm is greater than one so it will diverge
ans = ans =
1.4285 0
1.4285 0
1.4285 0
11. Part3 :
>> a=[1 ,.9 .9;.9, 1, .9;.9, .9, 1]
a=
1.0000 0.9000 0.9000
0.9000 1.0000 0.9000
0.9000 0.9000 1.0000
>> b=[2;2;2]
b=
2
2
2
>> x=[0;0;0]
x=
0
0
0
>> jacobi(a,b,x)
Norm is greater than one so it will diverge
ans =
0
0
0
From the Data we have calculated we find that Gauss_Siedel Method Diverge more
fastly than Jacobi Method.
12. Ques3:
Cholesky Method for Solving the Linear Equations:
% Implementing Cholesky Method
function x = cholesky(A,b,x)
n=length(A)
l(1,1)=sqrt(A(1,1));
for i=2:n
l(i,1)=A(i,1)/l(1,1);
end
for j=2:(n-1)
temp=0;
for k=1:j-1
temp=temp+l(j,k)*l(j,k);
end
l(j,j)= sqrt(A(j,j)-temp);
end
for j=2:(n-1)
for i=j+1:n
temp=0;
for k=1:j-1
temp=temp+l(i,k)*l(j,k);
end
l(i,j)=(A(i,j)-temp)/l(j,j);
end
end
temp1=l(n,1)*l(n,1)+l(n,2)*l(n,2);
l(n,n)=sqrt(A(n,n)-temp1);
y=forward_subs(l,b); %Using Forward Substitution Concept
x=back_subs(l',y);
end
15. Ques4:
Matlab Subroutine to calculate Condition Number:
% Calculating Condition Number
function [] = cond(a,b,b1);
x0=[0;0;0];
val1=max(abs(eig(a)));
val2=min(abs(eig(a)));
val=val1/val2 %definition of Condition number
x=gauss_siedel(a,b,x0);
x1=gauss_siedel(a,b1,x0);
errip=norm(x-x1)/norm(x);
errop=norm(b-b1)/norm(b);
fprintf('nerror in i/p is %d nand error in o/p is
%dn',errip,errop);
end
1).
>> a=[1 ,.2 .2;.2, 1, .2;.2, .2, 1]
a=
1.0000 0.2000 0.2000
0.2000 1.0000 0.2000
0.2000 0.2000 1.0000
>> b=[2;2;2]
b=
2
2
2
>> b1=[1.1;2.1;3.1]
b1 =
1.1000
2.1000
3.1000
16. >> cond(a,b,b1)
a=
1.0800 0.4400 0.4400
0.4400 1.0800 0.4400
0.4400 0.4400 1.0800
val =
3.0625 Condition Number.
The maximum no. of iteration required is 14
error in i/p is 1.309812e+000
and error in o/p is 4.112988e-001
2).
>> a=[1 ,.5 .5;.5, 1, .5;.5, .5, 1]
a=
1.0000 0.5000 0.5000
0.5000 1.0000 0.5000
0.5000 0.5000 1.0000
>> b=[2;2;2]
b=
2
2
2
>> b1=[1.1;2.1;3.1]
b1 =
1.1000
2.1000
3.1000
17. >> cond(a,b,b1)
a=
1.5000 1.2500 1.2500
1.2500 1.5000 1.2500
1.2500 1.2500 1.5000
val =
16.0000 Condition Number
Norm of c is greater than 1, Solution will diverge.
3).
>> a=[1 ,.9 .9;.9, 1, .9;.9, .9, 1]
a=
1.0000 0.9000 0.9000
0.9000 1.0000 0.9000
0.9000 0.9000 1.0000
>> b1=[1.1;2.1;3.1]
b1 =
1.1000
2.1000
3.1000
>> b1=[1.1;2.1;3.1]
b1 =
1.1000
2.1000
3.1000
>> b=[2;2;2]
b=
2
2
2
18. >> cond(a,b,b1)
a=
2.6200 2.6100 2.6100
2.6100 2.6200 2.6100
2.6100 2.6100 2.6200
val =
28.0000 Condition Number.
Norm of c is greater than 1, Solution will diverge.