Math 2 Spring 2011
Problem 7.1
For this problem, consider the space
L
2
[
−
1
,
1] with inner product and norm
(
f, g
) =
i
1

1
f
(
x
)
g
(
x
)
dx
and
b
f
b
2
= (
f, f
)
Let
P
i
(
x
), for
i
= 0
,
1
, . . .
be the Legendre polynomials of degree
i
and let
n
+ 1
−
st have
the form
P
n
+1
(
x
) =
ρ
n
(
x
−
x
0
)(
x
−
x
1
)
···
(
x
−
x
n
)
i.e.,
x
i
for 0
≤
i
≤
n
are the roots of
P
n
+1
(
x
).
Let the Lagrange interpolation functions that use the
x
i
be
ℓ
i
(
x
) for 0
≤
i
≤
n
. So, for
example,
L
n
(
x
) =
ℓ
0
(
x
)
f
(
x
0
) +
···
+
ℓ
n
(
x
)
f
(
x
n
)
is the Lagrange form of the interpolation polynomial of
f
(
x
) de±ned by the roots.
Let
P
n
be the space of polynomials of degree less than or equal to
n
. We can write the
least squares approximation of
f
(
x
) in terms of the
P
i
(
x
) using the generalized Fourier series
as
f
n
(
x
) =
α
0
P
0
(
x
) +
α
1
P
1
(
x
) +
···
+
α
n
P
n
(
x
) where
α
i
=
(
f, P
i
)
(
P
i
, P
i
)
7.1.a
Clearly, (
ℓ
i
, ℓ
i
)
n
= 0. Show that (
ℓ
i
, ℓ
j
) = 0 when
i
n
=
j
. Therefore, the functions
ℓ
0
(
x
)
, . . . , ℓ
n
(
x
)
are an orthogonal basis for
P
n
.
Solution:
Recall that
P
n
+1
(
x
)
⊥
P
n
so (
P
n
+1
, p
) = 0 for any
p
(
x
)
∈
P
n
. Also, we have for a constant
scale factor
γ
i
ℓ
i
(
x
) =
γ
i
P
n
+1
(
x
)
(
x
−
x
i
)
We therefore have
(
ℓ
i
, ℓ
j
) =
γ
i
γ
j
i
1

1
p
P
n
+1
(
x
)
(
x
−
x
i
)
Pp
P
n
+1
(
x
)
(
x
−
x
j
)
P
dx
=
γ
i
γ
j
i
1

1
P
n
+1
(
x
)
p
P
n
+1
(
x
)
(
x
−
x
i
)(
x
−
x
j
)
P
dx
=
γ
i
γ
j
(
P
n
+1
, p
n

1
) = 0
where
p
n

1
(
x
)
∈
P
n
since it is a polynomial of degree
n
−
1.
1
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 Spring '11
 gallivan
 Numerical Analysis, Taylor Series, p1, Chebyshev Economization

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