If you are asked to find a polynomial function of degree n-1
whose graph passes through specified points in a collection of data represented
by n points in the xy-plane, then the procedure is called Polynomial Curve
Fitting. To solve for the n coefficients of p(x), substitute each of the n
points into the polynomial function and obtain n linear equations in n
variables a0, a1, a2, ... , an-1
a0 + a1x1 + a2x12
+ . . . an-1x1n-1 = y1
a0 + a1x2 + a2x22
+ . . . an-1x2n-1 = y2
a0 + a1xn + a2xn2
+ . . . an-1xnn-1 = yn
Example:
Determine the polynomial p(x) = a0+a1x+a2x2
whose graph passes through the points ( 2 ,8) (4, 0)
(6,12).
Solution: Substituting x = 2, 4, and 6 into p(x) and
equating the results to the respective y-values produces the system of linear
equations in the variables a0, a1, a2 shown
below.
p(2) = a0 + a1(2) + a2(2)2 = a0
+ 2a1 + 4a2 = 8
p(4) = a0 + a1(4) + a2(4)2 = a0
+ 4a1 + 16a2 = 0
p(6) = a0
+ a1(6) + a2(6)2
= a0 + 6a1 + 36a2 = 12
[ 1 2
4 8 ]
[ 1 4
16 0 ]
-1(R1) + R2 ---> R2
[ 1 6 36
12 ]
[ 1 2
4 8 ]
[ 0 2
12 -8 ]
1/2(R2) ---> R2
[ 1 6
36 12 ]
[ 1 2
4 8 ]
[ 0 1
6 -4 ]
-1(R1) + R3 ---> R3
[ 1 6
36 12 ]
[ 1 2
4 8 ]
[ 0 1
6 -4 ]
-4(R2) + R3 ---> R3
[ 1 4
32 4 ]
[ 1 2
4 8 ]
[ 0 1
6 -4 ]
1/8(R3) ---> R3
[ 0 0
8 20 ]
[ 1 2
4 8 ]
[ 0 1
6 -4 ]
-2(R2) + R1 ---> R1
[ 0 0
1 5/2 ]
[ 1 0
-8 16 ]
[ 0 1
6 -4 ]
8(R3) + R1 ---> R1
[ 0 0
1 5/2 ]
[ 1 0
0 36 ]
[ 0 1
6 -4 ]
-6(R3) + R2 ---> R2
[ 0 0
1 5/2 ]
[ 1 0
0 36 ]
---> a0
[ 0 1
0 -19 ] ---> a1
[ 0 0
1 5/2 ] ---> a2
Therefore,
p(x) = 5/2x2 –
19x + 36
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