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Sabado, Hulyo 12, 2014
Sabado, Hulyo 5, 2014
POLYNOMIAL CURVE FITTING
This is used when there are given points and you need to find the polynomial to see the whole graph..
1st find the system of the polynomial by substituting points to
y= a(sub 0)+a(sub1)x + a(sub2) x^2 +a(sub3) x^3+.......a(subn)x^n
the number of points minus one determine the highest degree you of x in
y= a(sub 0)+a(sub1)x + a(sub2) x^2 +a(sub3) x^3+.......
solve using any methos you want to find a(sub 0),a(sub1),a(sub2),a(sub3)...a(subn)
after you find those values... write it in the form of p(x)..
like this given
a(sub 0)=12
a(sub1)=2
a(sub2)=3
a(sub3)=5
p(x)= 12+2x + 3x^2 +5x^3 so this is the polynomial function
1st find the system of the polynomial by substituting points to
y= a(sub 0)+a(sub1)x + a(sub2) x^2 +a(sub3) x^3+.......a(subn)x^n
the number of points minus one determine the highest degree you of x in
y= a(sub 0)+a(sub1)x + a(sub2) x^2 +a(sub3) x^3+.......
solve using any methos you want to find a(sub 0),a(sub1),a(sub2),a(sub3)...a(subn)
after you find those values... write it in the form of p(x)..
like this given
a(sub 0)=12
a(sub1)=2
a(sub2)=3
a(sub3)=5
p(x)= 12+2x + 3x^2 +5x^3 so this is the polynomial function
Sabado, Hunyo 28, 2014
Gaussian Jordan Method in solving in Solving System of Equations
(read the previous posts so you'll not be lost)
Tip:
dodge making the rows with fraction values
you can eliminate them by not just only multiplying a scalar.. you can add a multiple of other equation to another equatiion
Steps:
Put system of Equation to matrix form
You need to put the matrix to echelon form using row operations
you must leave each row only one variable ( diff variables each row)
after that put to system and write coordinate form of the solution

(1,4,-2)
(read the previous posts so you'll not be lost)
Tip:
dodge making the rows with fraction values
you can eliminate them by not just only multiplying a scalar.. you can add a multiple of other equation to another equatiion
Steps:
Put system of Equation to matrix form
You need to put the matrix to echelon form using row operations
you must leave each row only one variable ( diff variables each row)
after that put to system and write coordinate form of the solution
(1,4,-2)
Biyernes, Hunyo 20, 2014
matrix
a rectangular array of number
its size is row x column or m x n
-------> this is a 2 x 3 matrix
matix with same number of columns and rows is called square matrix

2 types of matrix coefficient and augmented
coefficient matrix-- matrix which only the coefficient of the variables in the given system is written
x+9y-13z=2
20x+5y+-6z=9

augmented matrix-- matrix which the constant is written
x+9y-13z=2
20x+5y+-6z=9
a rectangular array of number
its size is row x column or m x n
matix with same number of columns and rows is called square matrix
2 types of matrix coefficient and augmented
coefficient matrix-- matrix which only the coefficient of the variables in the given system is written
x+9y-13z=2
20x+5y+-6z=9
augmented matrix-- matrix which the constant is written
x+9y-13z=2
20x+5y+-6z=9
Elementary Row Operations
used in system of equation to solve for solutions
1. Interchange two rows (or columns).
2. Multiply each element in a row (or column) by a non-zero number.
3. Multiply a row (or column) by a non-zero number and add the result to another row (or column).
EcHELON Form
form of system of equation in which the a variable in the first is not included in the second up to the last and the variables of the first variable in each equation is one.
to get echelon arrange the variables first and use elementary row operatiions to get the echelon form..
ex.of echelon form
x+3y+4z=22
y+3z=12
z=3
after that use back subtitution to find the solutions.. substitute the value of the variable on the third equation in the echelon form and substitute the corresponding values to the first to find the variable leading in the first equation.
so z=3 then y+3(3)= 12 then y= 3 then substitute to the first x+3(3)+4(3)=22 then x= 1
so the solution is in this form (x,y,z) then (1,3,3) is the solution..
Sabado, Hunyo 14, 2014
Inconsistent -----------------------> parallel LINES
this is when a1/a2 = b1/b2 and not equal to c1/c2
when the coefficient in the variables are the same but different constant
http://www.algebra.com/algebra/homework/coordinate/Types-of-systems-inconsistent-dependent-independent.lesson <---- for more info
This lesson concerns systems of two equations, such as:
2x + y = 10
3x + y = 13.
The equations can be viewed algebraically or graphically. Usually, the problem is to find a solution for x and y that satisfies both equations simultaneously. Graphically, this represents a point where the lines cross. There are 3 possible outcomes to this (shown here in blue, green, and red):
The two lines might not cross at all, as in

y = x
y = x + 10.
y = x
y = x + 10.
This means there are no solutions, and the system is called inconsistent.
If you try to solve this system algebraically, you'll end up with something that's not true, such as 0 = 10.
Whenever you end up with something that's not true, the system is inconsistent.
Sabado, Hunyo 7, 2014
consistent(dependent)---> coinsiding lines
If the equations on a give equation have they only have one
solution
a1/a2 is equal to b1/b2 is equal to c1/c2<<< construction of the
coefficient of the variables in the equations
there are infinite solutions... but you need to put parameters

The graphical solution is any point on the two identical straight lines.

to find the solution of the given system you need to find the value interms of x and y
and replace them with other variable.. like the solution of given system : (1-b,b) which y=t... it is most ad visable to change the variable of the rightmost in the equation...
If the equations on a give equation have they only have one
solution
a1/a2 is equal to b1/b2 is equal to c1/c2<<< construction of the
coefficient of the variables in the equations
there are infinite solutions... but you need to put parameters
The graphical solution is any point on the two identical straight lines.
to find the solution of the given system you need to find the value interms of x and y
and replace them with other variable.. like the solution of given system : (1-b,b) which y=t... it is most ad visable to change the variable of the rightmost in the equation...
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