# 3.1 Polynomial interpolation ```Polynomial interpolation:
Objective: given n distinct points ( x1 , y1 ), ( x2 , y 2 ), , ( xn , y n ) , find a polynomial
a 0  a1 x  a 2 x 2    a n 1 x n 1
of degree n-1 or less,
10
interpolates the data.
8
(x4,y4)
y
6
(x3,y3)
2
4
(x1,y1)
0
(x2,y2)
-3
-2
-1
0
1
2
3
x
Special case:
y  a0  a1 x .
Given ( x1 , y1 ) and ( x2 , y2 ) , find the line
4
5
(x2,y2)
(x1,y1)
3
y
6
7
[solution:]
0.0
0.2
0.4
0.6
x
1
0.8
1.0
, that
The unknown coefficients a0 and a1 can be found by solving the system of two
linear equations
a 0  a1 x1  y1
a 0  a1 x 2  y 2
.
General case:
a0 , a1 ,, an1
can be found by solving for the following system of linear
equations:
a 0  a1 x1  a 2 x12    a n 1 x1n 1  y1
a 0  a1 x 2  a 2 x 22    a n 1 x 2n 1  y 2

a 0  a1 x n  a 2 x n2    a n 1 x nn 1  y n
1

1
 


1
x1

x2



xn

x1n 1   a 0   y1 

  
x 2n 1   a1   y 2 

      

  
x nn 1  a n 1   y n 
Thus, the augmented matrix used in Gauss-Jordan reduction is
1

1



1
x1
x2

xn
x12

x1n 1
x 22




x 2n 1

x n2
x nn 1
y1 

y2 
 .

yn 

Example:
Determine the equation of the polynomial of degree 2 whose graph passes through the
points (1,6), (2,3) and (3,2).
[solution:]
2
Suppose the polynomial of degree 2 is
y  a 0  a1 x  a 2 x 2
. Then, the
corresponding system of linear equations is
a 0  a1  a 2  6
a 0  2a1  4a 2  3 .
a 0  3a1  9a 2  2
Thus, the augmented matrix is
1 1 1 6 
1 2 2 2 3 inreduced
row echelon form





1 3 3 2 2 
1 0 0 11 
0 1 0  6


0 0 1 1 
 a 0  11, a1  6, a 2  1  y  x 2  6 x  11
3
```