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Unformatted text preview: ne) that interpolates the
data (x0 , y0 ) and (x1 , y1 ).
Answer. Let p(x) = a0 + a1 x. Then solve the simultaneous equations
p ( x0 ) = y 0
p ( x1 ) = y 1 =⇒ a0 + a1 x0 = y0
=⇒ a0 + a1 x1 = y1 to obtain a0 and a1 .
Alternatively, a quick informal approach is to equate the ratios
y1 − y0
p ( x) − y0
=
,
x − x0
x 1 − x0
and then rearrange to obtain
y1 − y0
( x − x0 ) .
p ( x) = y 0 +
x 1 − x0
Example. Find the interpolating polynomial through the points (−2, 4), (−1, 2),
(0, −3), (1, 5), (3, 4).
Answer. We have 5 data points. The interpolating polynomial has degree 4.
See Matlab polinterp1.m.
2 Polynomial interpolation in Lagrange form
n + 1 data points: (xj , yj ) for j = 0, . . . , n
Lagrange polynomials of degree n as basis functions:
n i ( x) x − xj
xi − xj =
j =0
j =i i ( xj ) = 1 for i = j , and i ( xj ) for i = 0, . . . , n = 0 if i = j Interpolating polynomial of degree n: p(x) =
No linear system to solve, but i ( x) n
i=0 y i i ( x) diﬃcult to evaluate If f ∈ C n+1 ([a, b]), i.e., n + 1 times continually diﬀerentiable in [a, b],
then the error in approximating f (x) by p(x) is
f (n+1) (ξ )
f ( x) − p ( x) =
(n + 1)! n ( x − xj )
j =1 for some unknown ξ ∈ [a, b] depending on x.
Mat...
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This note was uploaded on 05/26/2013 for the course MATHEMATIC 2089 taught by Professor Briancox during the Fall '13 term at National University of Singapore.
 Fall '13
 BrianCox
 Polynomials, Approximation, Least Squares

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