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NUMERICAL METHODS

School Of Engineering Of Oils

Diego F.

Industrialist university Of Santander

I Semester of 2010

Gauss-Seidel

It is a method iterativo that turns out to be a

quite efficient method.

a11x1+a12x2 + = b1

a21x1+a22x2 + = b2

Of the equation 1 we clear x1, of the equation 2

let's clear x2, , of the equation n let's clear

xn. This gives us the following set of equations:

The set of equations forms the formulae

iterativas. To begin the process iterativo,

it gives the value of zero to the variables x2., xn;

this gives

the first value for x1.


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NUMERICAL METHODS


Diego F.


I Semester of 2010

2replaces the value of x1 in the equation, and

variables keep on having the value of zero. This

givesthe following value for x2:

These values of x1 and x2, we replaced them in

equation 3, while x3..., xn they keep on havingvalue of zero; and this way successively up to

coming to

last equation. With this first step of the process

iterativo the first list of values is obtained for

the unknowns:

The process is repeated replacing these

last information instead of zeros as to the

beginning, and

it obtains the second list of values for each oneof the unknowns:


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NUMERICAL METHODS


Diego F.


I Semester of 2010

Now it is possible to calculate the approximate

errors

relative, with regard to each of the unknowns:

The process is even repeated

that the error is minor than one

value prearranged

Example of the method of

Gauss-Seidel

To bring the solution of the system near until the

error is a minor of 1 % first We Clear the

unknowns:


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NUMERICAL METHODS


Diego F.


I Semester of 2010

We clear the unknowns:

The process begins iterativo, replacing the valuesof x2=x3=0 in the first equation, to calculate the

first value of x1=2,66667

It is replaced x1=2,66667 and x3=0 in the second

equation, to obtain x2=-2.82381

-2.82381is replaced x1=2,66667 and x2= in the third

one

equation, to obtain x3= 7.1051

Since we cannot calculate any approximate error

yet, we repeat the process but now with the last

information obtained for the unknowns.


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NUMERICAL METHODS


Diego F.


I Semester of 2010

Replacing x2=-2.82381 and x3= 7.1051 in the

equation 1 obtains x1= 3.6626.

Replacing x1= 3.6626 and x3= 7.1051 in

equation 2 obtains x2=-3.24404.

Finally, replacing x1= 3.6626 and

x2=-3.24404 in the equation 3 is obtained

x3= 7.06106.

This way, we have the second list of values of

approach to the solution of the system:

Now if it is possible to calculate the absolute

errors for each of the unknowns:

Since the target has not been achieved, we must

repeat the same process with the last obtained

values of each of the unknowns. One notice that


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NUMERICAL METHODS


Diego F.


I Semester of 2010

although the approximate error of x3 expires

already in spite of being a minor to 1 %, this

must be fulfilled for the threeapproximate errors

Therefore repeating the same process

it obtains:

And in this case the approximate errors are:

Now the target has been fulfilled for each of the

approximate errors. Therefore the approximatesolution is the obtained one

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