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Computational Methods
Vladimir [email protected]
Office hours : Th 11:30AM-12:30PM or by appointment
V. Yushutin (UMD) A(C)MSC460 1 / 11
Chapter 4 Sec. 4.7
Gaussian quadratureSection 4.7
V. Yushutin (UMD) A(C)MSC460 2 / 11
Chapter 4 Sec. 4.7
Gaussian quadrature
So far we’ve seen quadratures of the type∫ b
af (x)dx =
n∑j=0
cj f (xj)
where xj are equally spaced between a and b. These (n + 1)-point closedNewton-Cotes quadratures are exact for polynomials of degree n or less.We say that Newton-Cotes quadratures have degree of precision n.
Chebyshev nodesUsage of non-equidistant nodes improves the interpolation.
Gauss-Legendre nodesUsage of non-equidistant nodes improves the numerical integration.
V. Yushutin (UMD) A(C)MSC460 3 / 11
Chapter 4 Sec. 4.7
Gaussian quadrature
So far we’ve seen quadratures of the type∫ b
af (x)dx =
n∑j=0
cj f (xj)
where xj are equally spaced between a and b. These (n + 1)-point closedNewton-Cotes quadratures are exact for polynomials of degree n or less.We say that Newton-Cotes quadratures have degree of precision n.
Chebyshev nodesUsage of non-equidistant nodes improves the interpolation.
Gauss-Legendre nodesUsage of non-equidistant nodes improves the numerical integration.
V. Yushutin (UMD) A(C)MSC460 3 / 11
Chapter 4 Sec. 4.7
Gaussian quadrature
So far we’ve seen quadratures of the type∫ b
af (x)dx =
n∑j=0
cj f (xj)
where xj are equally spaced between a and b. These (n + 1)-point closedNewton-Cotes quadratures are exact for polynomials of degree n or less.We say that Newton-Cotes quadratures have degree of precision n.
Chebyshev nodesUsage of non-equidistant nodes improves the interpolation.
Gauss-Legendre nodesUsage of non-equidistant nodes improves the numerical integration.
V. Yushutin (UMD) A(C)MSC460 3 / 11
Chapter 4 Sec. 4.7
Gauss-Legendre quadrature on [-1,1]
A typical two-point quadrature has the following form, x1, x2 ∈ [−1, 1]:∫ 1
−1f (x)dx = c1f (x1) + c2f (x2)
Different quadratures have different values of 4 parameters: c1, c2, x1, x2.
For example, a two-point Newton-Cotes quadrature corresponds toc1 = 1/2, c2 = 1/2, x1 = −1, x2 = 1; and it’s exact for polynomials ofdegree 1 or less.
An typical polynomial of third order also has 4 parameters:
f (x) = a0 + a1x + a2x2 + a3x3
Now we require that a quadrature c1f (x1) + c2f (x2) computes the integralof any polynomial of third order exactly. It’s a system for c1, c2, x1, x2.
V. Yushutin (UMD) A(C)MSC460 4 / 11
Chapter 4 Sec. 4.7
Gauss-Legendre quadrature on [-1,1]
A typical two-point quadrature has the following form, x1, x2 ∈ [−1, 1]:∫ 1
−1f (x)dx = c1f (x1) + c2f (x2)
Different quadratures have different values of 4 parameters: c1, c2, x1, x2.For example, a two-point Newton-Cotes quadrature corresponds toc1 = 1/2, c2 = 1/2, x1 = −1, x2 = 1; and it’s exact for polynomials ofdegree 1 or less.
An typical polynomial of third order also has 4 parameters:
f (x) = a0 + a1x + a2x2 + a3x3
Now we require that a quadrature c1f (x1) + c2f (x2) computes the integralof any polynomial of third order exactly. It’s a system for c1, c2, x1, x2.
V. Yushutin (UMD) A(C)MSC460 4 / 11
Chapter 4 Sec. 4.7
Gauss-Legendre quadrature on [-1,1]
A typical two-point quadrature has the following form, x1, x2 ∈ [−1, 1]:∫ 1
−1f (x)dx = c1f (x1) + c2f (x2)
Different quadratures have different values of 4 parameters: c1, c2, x1, x2.For example, a two-point Newton-Cotes quadrature corresponds toc1 = 1/2, c2 = 1/2, x1 = −1, x2 = 1; and it’s exact for polynomials ofdegree 1 or less.
An typical polynomial of third order also has 4 parameters:
f (x) = a0 + a1x + a2x2 + a3x3
Now we require that a quadrature c1f (x1) + c2f (x2) computes the integralof any polynomial of third order exactly. It’s a system for c1, c2, x1, x2.
V. Yushutin (UMD) A(C)MSC460 4 / 11
Chapter 4 Sec. 4.7
Gauss-Legendre quadrature on [-1,1]
A typical two-point quadrature has the following form, x1, x2 ∈ [−1, 1]:∫ 1
−1f (x)dx = c1f (x1) + c2f (x2)
Different quadratures have different values of 4 parameters: c1, c2, x1, x2.For example, a two-point Newton-Cotes quadrature corresponds toc1 = 1/2, c2 = 1/2, x1 = −1, x2 = 1; and it’s exact for polynomials ofdegree 1 or less.
An typical polynomial of third order also has 4 parameters:
f (x) = a0 + a1x + a2x2 + a3x3
Now we require that a quadrature c1f (x1) + c2f (x2) computes the integralof any polynomial of third order exactly. It’s a system for c1, c2, x1, x2.
V. Yushutin (UMD) A(C)MSC460 4 / 11
Chapter 4 Sec. 4.7
Gauss-Legendre quadratureThe nonlinear system
c1 + c2 = 2c1x1 + c2x2 = 0c1x2
1 + c2x22 = 2/3
c1x31 + c2x3
2 = 0has a unique solution:
Two-point Gauss-Legendre quadraturec1 = c2 = 1, x1 = − 1√
3 , x2 = 1√3
TheoremFor any n ∈ N there exists a Gauss-Legendre quadrature which is exact forall polynomials of degree 2n + 1 or less.
To prove the theorem we will use another polynomial basis (Legendre) toconstruct a quadrature with the degree of precision of 2n + 1.
V. Yushutin (UMD) A(C)MSC460 5 / 11
Chapter 4 Sec. 4.7
Gauss-Legendre quadratureThe nonlinear system
c1 + c2 = 2c1x1 + c2x2 = 0c1x2
1 + c2x22 = 2/3
c1x31 + c2x3
2 = 0has a unique solution:
Two-point Gauss-Legendre quadraturec1 = c2 = 1, x1 = − 1√
3 , x2 = 1√3
TheoremFor any n ∈ N there exists a Gauss-Legendre quadrature which is exact forall polynomials of degree 2n + 1 or less.
To prove the theorem we will use another polynomial basis (Legendre) toconstruct a quadrature with the degree of precision of 2n + 1.
V. Yushutin (UMD) A(C)MSC460 5 / 11
Chapter 4 Sec. 4.7
continuedProof: Start with the monomial basis of Πn: 1, x , x2, ..., xn. Introduce aninner product (f , g) =
∫ 1−1 f (x)g(x)dx . Apply the Gram-Schmidt process
to get an orthogonal basis l0(x), l1(x), ..., ln(x) of the same Πn. Note thatdegree of polynomial lk(x) is exactly k. Do one extra iteration ofGram-Schmidt to get ln+1(x) - orthogonal to all previous Πk(x), k = [0, n].
Consider n + 1 distinct real roots xj of ln+1(x). Any polynomial pn(x)from Πn is defined by n + 1 values pn(xj) at these nodes and coincideswith the Lagrange interpolant of itself:
pn(x) =n∑
j=0pn(xj)
n∏i=0,i 6=j
(x − xi )(xj − xi )
constructed quadraturen∑
j=0cjpn(xj) =
n∑j=0
pn(xj)
∫ 1
−1
n∏i=0,i 6=j
(x − xi )(xj − xi )
dx
=∫ 1
−1pn(x)dx
V. Yushutin (UMD) A(C)MSC460 6 / 11
Chapter 4 Sec. 4.7
continued
Clearly, the suggested quadrature is exact for Πn. In fact, it has a largerdegree of precision. Consider a polynomial f ∈ Π2n+1 and the followingdecomposition for some q, r ∈ Πn:
f2n+1(x) = qn(x)ln+1(x) + rn(x)∫ 1
−1f (x)dx =
∫ 1
−1q(x)In+1(x)dx +
∫ 1
−1r(x)dx =
∫ 1
−1r(x)dx
since In+1(x) is orthogonal to all Πn. Finally, because r(x) is of degree nand due to the fact that In+1(xj) = 0:∫ 1
−1r(x)dx =
n∑j=0
cj r(xj) =n∑
j=0cj f (xj)
The quadrature∫ 1−1 f (x)dx =
∑nj=0 cj f (xj) is exact for polynomials of
degree 2n + 1 or less.
Legendre nodes x0, ..., xn are roots of theLegendre polynomial of degree n + 1
V. Yushutin (UMD) A(C)MSC460 7 / 11
Chapter 4 Sec. 4.7
continued
Clearly, the suggested quadrature is exact for Πn. In fact, it has a largerdegree of precision. Consider a polynomial f ∈ Π2n+1 and the followingdecomposition for some q, r ∈ Πn:
f2n+1(x) = qn(x)ln+1(x) + rn(x)∫ 1
−1f (x)dx =
∫ 1
−1q(x)In+1(x)dx +
∫ 1
−1r(x)dx =
∫ 1
−1r(x)dx
since In+1(x) is orthogonal to all Πn. Finally, because r(x) is of degree nand due to the fact that In+1(xj) = 0:∫ 1
−1r(x)dx =
n∑j=0
cj r(xj) =n∑
j=0cj f (xj)
The quadrature∫ 1−1 f (x)dx =
∑nj=0 cj f (xj) is exact for polynomials of
degree 2n + 1 or less. Legendre nodes x0, ..., xn are roots of theLegendre polynomial of degree n + 1
V. Yushutin (UMD) A(C)MSC460 7 / 11
Chapter 4 Sec. 4.7
Legendre polynomials
n + 1 ln+1(x)0 11 x2 (3x2 − 1)/23 (5x3 − 3x)/24 (35x4 − 30x2 + 3)/85 (63x5 − 70x3 + 15x)/8
V. Yushutin (UMD) A(C)MSC460 8 / 11
Chapter 4 Sec. 4.7
Gauss quadratures
In fact, the previous theorem would work if we choose any other system oforthogonal polynomials! Just choose weighted inner product (f , g)w ...
Gauss-Legengre for∫ 1−1 w(x)f (x)dx , w(x) = 1
Gauss-Chebyshev for∫ 1−1 w(x)f (x)dx , w(x) =
√1− x2
Gauss-Jacobi for∫ 1−1 w(x)f (x)dx , w(x) = (1− x)α(1 + x)β
Gauss-Laguerre for∫∞
0 w(x)f (x)dx , w(x) = e−x
V. Yushutin (UMD) A(C)MSC460 9 / 11
Chapter 4 Sec. 4.7
Gauss quadratures
In fact, the previous theorem would work if we choose any other system oforthogonal polynomials! Just choose weighted inner product (f , g)w ...
Gauss-Legengre for∫ 1−1 w(x)f (x)dx , w(x) = 1
Gauss-Chebyshev for∫ 1−1 w(x)f (x)dx , w(x) =
√1− x2
Gauss-Jacobi for∫ 1−1 w(x)f (x)dx , w(x) = (1− x)α(1 + x)β
Gauss-Laguerre for∫∞
0 w(x)f (x)dx , w(x) = e−x
V. Yushutin (UMD) A(C)MSC460 9 / 11
Chapter 4 Sec. 4.7
Gauss-Legendre quadrature on arbitrary domain
Apply a linear change of the variable x = (a + b)/2 + s(b − a)/2 toconvert: ∫ b
af (x)dx = (b − a)
2
∫ 1
−1f((a + b)
2 + s (b − a)2
)ds
ExampleConstruct an efficient quadrature for splines of degree less or equal 3
Error estimate for Gauss-Legendre quadratures∫ b
af (x)dx −
n∑j=0
cj f (xj) = (b − a)2n+3
2n + 3 f 2n+2(ξ) ((n + 1)!)4
(2(n + 1)!)3
V. Yushutin (UMD) A(C)MSC460 10 / 11
Chapter 4 Sec. 4.7
Gauss-Legendre quadrature on arbitrary domain
Apply a linear change of the variable x = (a + b)/2 + s(b − a)/2 toconvert: ∫ b
af (x)dx = (b − a)
2
∫ 1
−1f((a + b)
2 + s (b − a)2
)ds
ExampleConstruct an efficient quadrature for splines of degree less or equal 3
Error estimate for Gauss-Legendre quadratures∫ b
af (x)dx −
n∑j=0
cj f (xj) = (b − a)2n+3
2n + 3 f 2n+2(ξ) ((n + 1)!)4
(2(n + 1)!)3
V. Yushutin (UMD) A(C)MSC460 10 / 11
Chapter 4 Sec. 4.7
Round-off error analysisInstead of having exact values of f (xi ), we compute a Gauss-Legendrequadrature with f (xi ) + εi , where εi are related to the machine precision εand:
∣∣∣∣∣∫ b
af (x)dx −
n∑i=0
ci (f (xi ) + εi )∣∣∣∣∣ ≤
∣∣∣∣∣n∑
i=0ciεi
∣∣∣∣∣+ R ≤ maxi|εi |( n∑
i=0|ci |)
+ R
But all the weights are positive, which can be shown with a help of Fi (x)of degree 2n:
ci =n∑
j=0cjFi (xj) =
∫ b
aFi (x)dx > 0 , Fi (x) =
n∏j=0,j 6=i
(x − xj)2
(xi − xj)2
StabilityGauss-Legendre quadratures have positive weights and are stablenumerical procedures
V. Yushutin (UMD) A(C)MSC460 11 / 11
Chapter 4 Sec. 4.7
Round-off error analysisInstead of having exact values of f (xi ), we compute a Gauss-Legendrequadrature with f (xi ) + εi , where εi are related to the machine precision εand:∣∣∣∣∣∫ b
af (x)dx −
n∑i=0
ci (f (xi ) + εi )∣∣∣∣∣ ≤
∣∣∣∣∣n∑
i=0ciεi
∣∣∣∣∣+ R ≤ maxi|εi |( n∑
i=0|ci |)
+ R
But all the weights are positive, which can be shown with a help of Fi (x)of degree 2n:
ci =n∑
j=0cjFi (xj) =
∫ b
aFi (x)dx > 0 , Fi (x) =
n∏j=0,j 6=i
(x − xj)2
(xi − xj)2
StabilityGauss-Legendre quadratures have positive weights and are stablenumerical procedures
V. Yushutin (UMD) A(C)MSC460 11 / 11
Chapter 4 Sec. 4.7
Round-off error analysisInstead of having exact values of f (xi ), we compute a Gauss-Legendrequadrature with f (xi ) + εi , where εi are related to the machine precision εand:∣∣∣∣∣∫ b
af (x)dx −
n∑i=0
ci (f (xi ) + εi )∣∣∣∣∣ ≤
∣∣∣∣∣n∑
i=0ciεi
∣∣∣∣∣+ R ≤ maxi|εi |( n∑
i=0|ci |)
+ R
But all the weights are positive, which can be shown with a help of Fi (x)of degree 2n:
ci =n∑
j=0cjFi (xj) =
∫ b
aFi (x)dx > 0 , Fi (x) =
n∏j=0,j 6=i
(x − xj)2
(xi − xj)2
StabilityGauss-Legendre quadratures have positive weights and are stablenumerical procedures
V. Yushutin (UMD) A(C)MSC460 11 / 11
