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Math 128A, Spring 2007
Homework 3 Solution
4.1.12 (a)
L
0
(
x
) +
L
1
(
x
) +
L
2
(
x
) +
L
3
(
x
) is the unique polynomial of degree
≤
3
interpolating the data (
x
0
,
1)
,
(
x
1
,
1)
,
(
x
2
,
1)
,
(
x
3
,
1). Since the polynomial
1 also has degree
≤
3 and interpolates the data, we conclude
L
0
(
x
) +
L
1
(
x
) +
L
2
(
x
) +
L
3
(
x
) = 1
.
(b) In general,
L
0
(
x
) +
···
L
n
(
x
) and 1 are both degree
≤
n
interpolants
of (
x
0
,
1)
, . . . ,
(
x
n
,
1), and therefore must be equal.
4.1.13 For
j
≤
3, note that
x
j
0
L
0
(
x
)+
x
j
1
L
1
(
x
)+
x
j
2
L
2
(
x
)+
x
j
3
L
3
(
x
) and
x
j
are both
polynomials of degree
≤
3 interpolating (
x
0
, x
j
0
)
,
(
x
1
, x
j
1
)
,
(
x
2
, x
j
2
)
,
(
x
3
, x
j
3
).
Since such interpolants are unique, the two polynomials must be equal.
For
j >
3, note that each
L
i
(
x
) has degree
≤
3, so a linear combination
of them can’t have the same degree as
x
j
.
4.1.15 Note: this is the complicated solution.
In analogy to the Lagrange basis
{
L
i
(
x
)
}
, we want to ﬁnd a basis
{
H
i
(
x
)
}
for the (vector space of) polynomials of degree
≤
3 so that the coordinates
of
P
(
x
) with respect to
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This note was uploaded on 05/07/2008 for the course MATH 128A taught by Professor Rieffel during the Spring '08 term at University of California, Berkeley.
 Spring '08
 Rieffel
 Math

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