Jacobi method and lagrange interpolation

Mohamad Tafrikan (moh.tafrikan@gmail.com) Page 1
GAUSS AND JACOBY ELIMINATION METHOD
1) Solve the linear equation system follow using Gauss Method
{
Solution:
{
( , ( + ( ,
The linear equation system (1) is , where:
Since . We will use the Gauss formula:
and . The linear equation system (1) will be ,
we will compute values of .
( *
( *
( *
( *
( *
( *
Now presented by:

{
The linear equation system (2) is , where:
Since . We will use the Gauss formula:
and . The linear equation system (2) will be , we
will compute values of .
( *
( *
Now presented by:
{
We get , and then , we get: and .
using this formula to get :
{
( ∑ )
( ∑ ) ( )
( ∑ ) ( ( )*
So, solution for (1) is { }

2) Solve the linear equation system follow using Gauss Elimination Method
{
Solution:
{
By change of row:
[ ] ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ [ ]
[ ] [ ]
The linear equation system (1) will be , we will compute values of
by , where .
( *
( *
( *
( *
( *
( *
* + ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ * +
We get a new linear equation system , that is
* +

We will use the Gauss formula: and .
The linear equation system (2) will be , we will compute values of .
( *
( *
[ ]
Now presented by:
{
We get , , .
By change of column:
[ ] [ ]
The linear equation system (3) will be , we will compute values of
by , where .
( *
( *
( *
( *
( *
( *
* + ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ [ ]

We get a new linear equation system , that is
[ ]
We will use the Gauss formula: and .
The linear equation system (2) will be , we will compute values of .
( *
( *
[ ]
Now presented by:
{
We get , , .
BY DECOMPOSITION
[ ] [ ] [ ]
Where each entry of is determined by:
{
∑
∑
We get each entry of and ,
{
∑
(
∑
+

If we have a system we can solve its system by Decomposition Assume
, defined:
[ ] [ ] [ ]
We get:
{
∑
And form of defined by
[ ] [ ] [ ]
So, we get the solutions of is
{
( ∑ )
Example 1: Solve this system
{
Solution:
(3)3 (2) 2 (5) 5 (6) 6
(-1) (4) (3) (5)
(1) (-1) (3) (1)
And then we get:
[ ], * + * +
And we get
[ ] [ ] [ ]
So we get solution
Exercise 2. No 3.

Solve this a linear equation system by Decomposition!
* + * + * +
Solution:
(1)1 (2) 2 (3) 3 (4) 4 (2)2
(1) (4) (9) (16) (10)
(1) (8) (27) (64) (44)
(1) (16) (81) (256) (190)
And we get
* + * + * +
So we get solution
Solution by MATLAB:
>> A=[1 2 3 4;1 4 9 16;1 8 27 64;1 16 81 256]
A =
1 2 3 4
1 4 9 16
1 8 27 64
1 16 81 256
>> b=[2;10;44;190]
b =
2
10
44
190
>> x=(Ab)'
x = -1.0000 1.0000 -1.0000 1.0000
JACOBI METHOD
Suppose a system n-equation,
{ (1)

If , Eq. (1) can we write:
{ (2)
Where , and
( *
[ ] [ ] (3)
So, solution of Eq. (1) is
(4)
Alghoritm for Jacobi Method:
1. ( ) with initial value ( )
2.
3.
( ∑ )
4. ‖ ‖
5. For
Example 1:
{
By using (4), we get:
{
With initial value , we get:

And the next we will calculate by using MS Excel, we obtain:
Table 1
0 0 0 0
1 7.2 8.3 8.4
2 9.71 10.7 11.5
3 10.57 11.571 12.482
4 10.8535 11.8534 12.8282
5 10.95098 11.95099 12.94138
6 10.983375 11.983374 12.980394
7 10.9944162 11.9944163 12.9933498
8 10.99811159 11.99811158 12.9977665
9 10.99936446 11.99936446 12.99924463
Assume , since ‖ ‖ ‖ ‖ , then,
process stopped. So, we get final solution:
, or .
GAUSS-SEIDEL METHOD
Suppose a system n-equation,
{ (5)
If , Eq. (1) can we write:
{
(6)
Where , and

[ ] [ ]
Solution of Eq.(5) is
(7)
Alghoritm for Gauss-Seidel Method:
1. ( ) with initial condition ( )
2.
3.

( ∑ )

( ∑ ∑ )

( ∑ )
4. ‖ ‖ , then
5. For
Example 2.
{
By using (6), with initial value , we get:
( )
( )
( )
Tabel 2
0 0 0 0
1 7.2 9.02 11.644
2 10.4308 11.67188 12.820536

3 10.9312952 11.95723672 12.97770638
4 10.99126495 11.99466777 12.99718654
5 10.99890409 11.99932772 12.99964636
6 10.99986204 11.99991548 12.9999555
Assume , since ‖ ‖ ‖ ‖
,then, process stopped. So, we get final solution:
, or
Exercise page 89 number (1)
Use Jacobi method and Gauss-Seidel method to solve this system:
[ ] [ ] [ ]
(8)
With initial condition:
 By using Jacobi method
By using (4), we get:
{
With initial condition , we get:
Tabel 1
0 0 0 0 0 0 0

1 1.2 -3 2 -1.42 0.8 0.8
2 1.206 -2.8636 3.2794 -1.646 0.44086 0.35914
3 1.147936 -2.5552898 3.1008986 -2.0192692 0.87358 0.43272
4 1.09169802 -2.62921629
3.54091715
2
-
1.977922984
0.86700482
2 0.006575178
5
1.00131316
5 -2.58155443
3.54377891
4
-
2.061826638
1.01538009
7 0.148375275
6
0.96179561
3 -2.601802339
3.70257364
1
-
2.036178359
1.04463921
6 0.029259119
7
0.91266332
7 -2.595286119
3.72566523
2
-
2.060214779
1.10538006
8 0.060740852
8
0.89030797
2 -2.602097962
3.79128381
5
-
2.049953094
1.12595881
4 0.020578746
9
0.86655543
6 -2.601469385
3.80838238
3
-
2.057964588
1.15297405
7 0.027015243
1
0
0.85467021
6 -2.604128265
3.83723832
1
-
2.054246735
1.16461871
4 0.011644657
1
1
0.84339169
7 -2.604307402
3.84744106
3
-
2.057105832
1.17706750
4 0.012448789
1
2
0.83724727
6 -2.605423563 3.86053428
-
2.055810531
1.18325222
1 0.006184718
Assume , since ‖ ‖ ‖ ‖ ,
then, process stopped. So, we get final solution:
,
or .
 By using Gauss-Seidel method
By using (6), with initial value , we get:
( )
( )
( )
( )
( )

Tabel 1
0 0 0 0 0 0 0
1 1.2 -2.8636 2 -1.42 0.8 0.8
2 1.19236 -2.55979 3.260304 -1.66919 0.46814 0.33186
3 1.117419 -2.6365 3.091333 -2.06055 0.930169 0.462029324
4 1.091481 -2.57466 3.618334 -1.94957 0.853079 0.077090124
5 0.977438 -2.60858 3.514119 -2.09152 1.053798 0.200718593
6 0.96397 -2.59109 3.752752 -2.01513 1.033902 0.019895339
7 0.899536 -2.60569 3.706912 -2.07598 1.124904 0.091001323
8 0.89202 -2.59942 3.816874 -2.03865 1.120177 0.004726487
Assume , since ‖ ‖ ‖ ‖ , then,
process stopped. So, we get final solution:
, or
.
LAGRANGE INTERPOLATION
Suppose the data pair in [ ]available, the Lagrange interpolation is
the n-order polynomial interpolation, such that

Thus, the Lagrange interpolation results at the observation point are equal to the observed
value of itself. At point x which is not the point of observation Lagrange value is calculated
through the following formula:
∑
where ∏ ( ) is called coefficient of Lagrange.
Note that
{
So, for we get:
With the error is determined by the formula:
∏
So, the form of Lagrange interpolation for order-2 formula is as follows:
with a error determined as follows:
| |
Example 1.1
Given and a table of data pair :
10 11 12 13 14

2.302 6 2.397 9 2.484 9 2.564 9 2.639 1
By using Lagrange interpolation, find value of !
Solution:

First, we must compute , with


Then,
So,
With error:
, we get:
| |
Then

First, we must compute , with



Then,

So,
With error:
, we get:
| |
Then
Lagrange Interpolation Algorithm:
Declaration
Int ;
Real
Deskription
1. Input order polynomial (n)
2. Input pair for
3. Input point try xk;
4. Initial value ;
5. Load for i=0 to i=n
a. P(i)=1;
b. for j=0 to j=n

i. if then
c.
6. print (‘for x= ’, xk, ‘then y= ‘,L);
7. Stop
Program’s MATLAB Lagrange method
clear; help lagrang;
n = input('Order polinomial n = ');
x = zeros(n+1, 1);
y = zeros(n+1, 1);
z = zeros(n+1, 1);
for i = 1: n+1
fprintf('data ke-%2dn', i-1);
x(i) = input('x : ');
y(i) = input('y : ');
end
disp(' x y');
for i = 1: n+1
fprintf('%8.4f %10.6fn', x(i),y(i));
end
again=1;
while (lagi==1)
xk = input('input point xk: ');
L = 0;
for i = 1:n+1
P(i)=1;
for j= 1:n+1
if (i~=j)
P(i) = P(i)*(xk-x(j))/(x(i)-x(j));
end
end
L = L + P(i)*y(i);
end
fprintf('Value y= %10.8f on x= %8.6f n', L,xk);
disp(‘Press 1 to reload, Press 0 to exit');
again=input(' ');
end
Example 1.2
Estimate the error of √ at the point x=116 in the interval [100,144] using Lagrange
orde-2, calculated at , and .
Solution:

√ then
√
,
√
and
√
[ ]
√
|√ | ( *
Example 1.3
Draw the Lagrange chart for the following table, then estimate the value of y at point
x 1 3 5 7 9 11
y -3 4 5 -8 -3 0
>> myLAGRANG
input vektor sb-x: [1 3 5 7 9 11]
input vektor sb-y: [-3 4 5 -8 -3 0]
input point xk: 4.5
Value yk= 7.14111328 at xk= 4.500000

Example 1.4 : Interpolate at point x = 1.5 in the following observation table:
x -2 -1 1 2
y -6 0 0 6
>> myLAGRANG
Input vector sb-x: [-2 -1 1 2]
Input vector sb-y: [-6 0 0 6]
Input point xk: 1.5
Value yk= 1.87500000 at xk= 1.500000

NEWTON INTERPOLATION
Suppose is the -th Lagrange polynomial whose value corresponds to the value of
at points then model of divided difference of can written:
For coefficents that corresponds.
The coefficient or constant can be calculated in the following way:
 for , then
 for then such that
[ ]
using notation of divided difference:
 Divided Difference (Zeroth): [ ]
 Divided Difference (First): [ ]
{ [ ] [ ]}
 Divided Difference (Second):
[ ]
{ [ ] [ ]}

Divided Difference is more easily understood through the following table:
[ ] [ ] [ ] [ ] [ ]
[ ]
[ ]
{ [ ] [ ]}
[ ]
{ [ ] [ ]}
[ ]
[ ] [ ]
[ ]
[ ] [ ]
[ ]
[ ]
{ [ ] [ ]}
[ ]
{ [ ] [ ]}
[ ]
[ ] [ ]
[ ]
[ ]
{ [ ] [ ]}
[ ]
{ [ ] [ ]}
[ ]
[ ]
{ [ ] [ ]}
[ ]
-
Example 2.1
Given the following table. Look for the polynomial approach y value on x=2.
X -1 0 3 6 7
Y 3 -6 39 822 1611
Solution:
Consider the following Dived Difference table:
[ ]
[ ] [ ] [ ] [ ]
-
From the table, we get:
[ ] [ ] [ ] [ ] and
[ ] ; such that:
( ) ( ) ( ) ( )
So, for x=2, we obtain

Program’s MATLAB
clear; help NForward;
N = input(‘pait data : ');
T = zeros(N, N+1);
disp('input data: ');
for i = 1: N
fprintf('data th-%2dn', i);
x(i) = input('x : ');
y(i) = input('y : ');
T(i,1) = x(i);
T(i,2) = y(i);
end
for kol = 3:N+1
br = (N+2-kol);
for brs = 1:br
T(brs,kol)=T(brs+1, kol-1)-T(brs, kol-1);
end
end
disp(' x y delta1 delta2 ..........');
for i = 1: N
for j=1:N+1
fprintf('%8.4f ',T(i,j));
end
fprintf('n');
end
%interpolasi
lagi=1;
while (lagi==1)
h=x(2)-x(1);
xk = input(‘input point interpolation: ');
t = (xk - x(1))/h;
yk = T(1,2) + t*T(1,3);
for kol = (N+1):-1:4
fak = kol-2;
t1 = 1;
for j = (kol-3) : -1 : 1
t1 = t1*(t-j);
fak = fak*j;
end
t1 = t*t1;
yk = yk + t1*T(1,kol)/fak;
end
fprintf('At x= %8.6f value of y = %12.8f n',xk,yk);
disp(‘Press 1 to reload, Press 0 to exit');
lagi=input(' ');
end

Jacobi method and lagrange interpolation

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