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example of under-determined
polynomial interpolation∗
rmilson†
2013-03-21 13:59:24
Consider the following interpolation problem:
Given x1 , y1 , x2 , y2 ∈ R with x1 6= x2 to determine all cubic polynomials
p(x) = ax3 + bx2 + cx + d, x, a, b, c, d ∈ R
such that
p(x1 ) = y1 ,
p(x2 ) = y2 .
This is a linear problem. Let P3 denote the vector space of cubic polynomials.
The underlying linear mapping is the multi-evaluation mapping
E : P3 → R 2 ,
given by
p 7→
p(x1 )
,
p(x2 )
p ∈ P3
The interpolation problem in question is represented by the equation
y1
E(p) =
y2
where p ∈ P3 is the unknown. One can recast the problem into the traditional form by taking standard bases of P3 and R2 and then seeking all possible
a, b, c, d ∈ R such that
 
a
3
2
(x1 ) (x1 ) x1 1 
b
y1


=
3
2
y2
(x2 ) (x2 ) x2 1  c 
d
∗ hExampleOfUnderdeterminedPolynomialInterpolationi
created: h2013-03-21i
by:
hrmilsoni version: h32840i Privacy setting: h1i hExamplei h15A06i
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However, it is best to treat this problem at an abstract level, rather than mucking about with row reduction. The Lagrange interpolation formula gives us a
particular solution, namely the linear polynomial
p0 (x) =
x − x1
x − x2
y1 +
y2 ,
x2 − x1
x1 − x2
x∈R
The general solution of our interpolation problem is therefore given as p0 + q,
where q ∈ P3 is a solution of the homogeneous problem
E(q) = 0.
A basis of solutions for the latter is, evidently,
q1 (x) = (x − x1 )(x − x2 ),
q2 (x) = xq1 (x),
x∈R
The general solution to our interpolation problem is therefore
p(x) =
x − x2
x − x1
y1 +
y2 + (ax + b)(x − x1 )(x − x2 ),
x2 − x1
x1 − x2
x ∈ R,
with a, b ∈ R arbitrary. The general under-determined interpolation problem is
treated in an entirely analogous manner.
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