Thus,
Z
b
a
f
(
x
)
w
(
x
)
dx
=
n
X
j
=0
A
j
f
(
x
j
)
.
(8.25)
This proves that the Gaussian quadrature has degree of precision
k
= 2
n
+1.
Now suppose that the interpolatory quadrature (8.18) has maximal degree
of precision 2
n
+ 1. Take
f
(
x
) =
P
(
x
)(
x

x
0
)(
x

x
1
)
· · ·
(
x

x
n
) where
P
is
a polynomial of degree
≤
n
. Then,
f
is a polynomial of degree
≤
2
n
+ 1 and
Z
b
a
f
(
x
)
w
(
x
)
dx
=
Z
b
a
P
(
x
)(
x

x
0
)
· · ·
(
x

x
n
)
w
(
x
)
dx
=
n
X
j
=0
A
j
f
(
x
j
) = 0
.
Therefore, the polynomial (
x

x
0
)(
x

x
1
)
· · ·
(
x

x
n
) of degree
n
+ 1 is
orthogonal to all polynomials of degree
≤
n
. Thus, it is a multiple of
ψ
n
+1
.
Example 24.
Consider the interval[1,1]and the weight functionw≡1.The corresponding orthogonal the Legendre Polynomials1, x, x213, x3
3
5
x,
· · ·
. Take
n
= 1
. The roots of
ψ
2
are
x
0
=

q
1
3
and
x
1
=
q
1
3
. There
fore, the corresponding Gaussian quadrature is
Z
1

1
f
(
x
)
dx
≈
A
0
f
r

1
3
!
+
A
1
f
r
1
3
!
(8.26)
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CHAPTER 8.
NUMERICAL INTEGRATION
where
A
0
=
Z
1

1
l
0
(
x
)
dx,
(8.27)
A
1
=
Z
1

1
l
1
(
x
)
dx.
(8.28)
We can evaluate evaluate these integrals directly or use the
method of un
determined coefficients
to find
A
0
and
A
1
. The latter is generally easier
and we illustrate it now. Using that the quadrature has to be exact for
1
and
x
we have
2 =
Z
1

1
1
dx
=
A
0
+
A
1
,
(8.29)
0 =
Z
1

1
xdx
=

A
0
r
1
3
+
A
1
r
1
3
.
(8.30)
Solving this
2
×
2
linear system we get
A
0
=
A
1
= 1
.
So the Gaussian
quadrature for
n
= 1
in
[

1
,
1]
is
Q
1
[
f] =fT[
r
1
3
!
+
f
r
1
3
!
(8.31)
Let us compare this quadrature to the elementary Trapezoidal Rule.
Take
f
(
x
) =
x
2
. The Trapezoidal Rule,
f
]
, gives
T
[
x
2
] =
2
2
[
f
(

1) +
f
(1)] = 2
(8.32)
whereas the Gaussian quadrature
Q
1
[
f
]
yields the exact result:
Q
1
[
x
2
] =

r
1
3
!
2
+
r
1
3
!
2
=
2
3
.
(8.33)
Example 25.
Let us take again the interval
[

1
,
1]
but now
w
(
x
) =
1
√
1

x
2
.
As we know (see 2.4 ),
ψ
n
+1
(
x
) =
T
n
+1
(
x
)
, i.e.
the
(
n
+ 1)
st Chebyshev
Polynomial. Its zeros are
x
j
= cos[
2
j
+1
2(
n
+1)
π
]
for
j
= 0
, . . . , n
. For
n
= 1
we
have
x
0
= cos
π
4
=
r
1
2
,
(8.34)
x
1
= cos
5
π
4
=

r
1
2
.
(8.35)
8.3.
GAUSSIAN QUADRATURES
119
We can use again the method of undetermined coefficients to find
A
0
and
A
1
:
π
=
Z
1

1
1
1
√
1

x
2
dx
=
A
0
+
A
1
,
(8.36)
0 =
Z
1

1
x
1
√
1

x
2
dx
=

A
0
r
1
2
+
A
1
r
1
2
,
(8.37)
which give
A
0
=
A
1
=
π
2
. Thus, the corresponding Gaussian quadrature to
approximate
R
1

1
f
(
x
)
1
√
1

x
2
dx
is
Q
1
[
f
] =
π
2
"
f

r
1
2
!
+
f
r
1
2
!#
.
(8.38)
8.3.1
Convergence of Gaussian Quadratures
Let
f
∈
C
[
a, b
] and consider the interpolation quadrature (8.18).
Can we
guarantee that the error converges to zero as
n
→ ∞
, i.e.,
Z
b
a
f
(
x
)
w
(
x
)
dx

n
X
j
=0
A
j
f
(
x
j
)
→
0
,
as
n
→ ∞
?
The answer is no. As we know, convergence of the interpolating polynomial
to
f
depends on the smoothness of
f
and the distribution of the nodes.
However, if the interpolatory quadrature is a Gaussian one the answer is yes.
This follows from the following special properties of the quadrature weights
A
0
, A
1
, . . . , A
n
in the Gaussian quadrature.