First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 3. INTERPOLATION 95
for i =0, 1, ..., n, or in matrix form
⎡
⎤
1 0 0 ... 0 ⎡ ⎤ ⎡ ⎤
a 1 (x 1 − x 0 ) 0 0
0 y 0
1 (x 2 − x 0 ) (x 2 − x 0 )(x 2 − x 1 ) 0
a 1
y ⎢ ⎥ =
1
⎢ ⎥
⎢
⎣.
.
.
.
⎥ ⎣ . ⎦ ⎣ . ⎦
⎦
∏
1 (x n − x 0 ) (x n − x 0 )(x n − x 1 ) ... n−1
i=0 (x a n y n
n − x i ) } {{ } } {{ }
} {{ } a y
A
for [a 0 , ..., a n ] T . Note that the coefficient matrix A is lower-triangular, and a can be solved
by forward substitution, which is shown in the next example, in O(n 2 ) operations.
Example 49. Find the interpolating polynomial using Newton’s basis for the data:
(−1, −6), (1, 0), (2, 6).
Solution. We have p 2 (x) =a 0 + a 1 π 1 (x)+a 2 π 2 (x) =a 0 + a 1 (x +1)+a 2 (x +1)(x − 1). Find
a 0 ,a 1 ,a 2 from
p 2 (−1) = −6 ⇒ a 0 + a 1 (−1+1)+a 2 (−1+1)(−1 − 1) = a 0 = −6
p 2 (1) = 0 ⇒ a 0 + a 1 (1+1)+a 2 (1 + 1)(1 − 1) = a 0 +2a 1 =0
p 2 (2) = 6 ⇒ a 0 + a 1 (2+1)+a 2 (2 + 1)(2 − 1) = a 0 +3a 1 +3a 2 =6
or, in matrix form
Forward substitution is:
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
1 0 0 a 0 −6
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣1 2 0⎦
⎣a 1 ⎦ = ⎣ 0 ⎦ .
1 3 3 a 2 6
a 0 = −6
a 0 +2a 1 =0⇒−6+2a 1 =0⇒ a 1 =3
a 0 +3a 1 +3a 2 =6⇒−6+9+3a 2 =6⇒ a 2 =1.
Therefore a =[−6, 3, 1] T and
p 2 (x) =−6+3(x +1)+(x +1)(x − 1).
Factoring out and simplifying gives p 2 (x) =−4+3x + x 2 , which is the polynomial discussed
in Example 48.