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Gauss quadrature Elena Celledoni Department of Mathematical Sciences, NTNU October 23rd 2012 Elena Celledoni NA
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Page 1: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

Gauss quadrature

Elena Celledoni

Department of Mathematical Sciences, NTNU

October 23rd 2012

Elena Celledoni NA

Page 2: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

Gauss quadrature

Consider [a, b] ⊂ R and

I (f ) :=

∫ b

a

w(x) f (x) dx ,

and w(x) is a weight function. Given x0 < x1 · · · < xn, xi ∈ [a, b],

consider the Hermite interpolation polynomial for f on these nodes:

p2n+1(x) =n∑

i=0

Hi (x)f (xi ) +n∑

i=0

Ki (x)f ′(xi ),

Hi (x) = (Li (x))2(1− 2L′i (xi )(x − xi )),

Ki (x) = (Li (x))2(x − xi ),

Li (x) :=n∏

j=0,j 6=i

x − xjxi − xj

L0(x) ≡ 1,putting p2n+1 in place of f in the integral gives an approximation of theintegral:∫ b

a

w(x) f (x) dx ≈n∑

i=0

Wi f (xi )+n∑

i=0

Vi f′(xi ),

Wi :=∫ b

aw(x)Hi (x) dx ,

Vi :=∫ b

aw(x)Ki (x) dx .

Elena Celledoni NA

Page 3: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

Gauss quadrature

Consider [a, b] ⊂ R and

I (f ) :=

∫ b

a

w(x) f (x) dx ,

and w(x) is a weight function. Given x0 < x1 · · · < xn, xi ∈ [a, b],consider the Hermite interpolation polynomial for f on these nodes:

p2n+1(x) =n∑

i=0

Hi (x)f (xi ) +n∑

i=0

Ki (x)f ′(xi ),

Hi (x) = (Li (x))2(1− 2L′i (xi )(x − xi )),

Ki (x) = (Li (x))2(x − xi ),

Li (x) :=n∏

j=0,j 6=i

x − xjxi − xj

L0(x) ≡ 1,

putting p2n+1 in place of f in the integral gives an approximation of theintegral:∫ b

a

w(x) f (x) dx ≈n∑

i=0

Wi f (xi )+n∑

i=0

Vi f′(xi ),

Wi :=∫ b

aw(x)Hi (x) dx ,

Vi :=∫ b

aw(x)Ki (x) dx .

Elena Celledoni NA

Page 4: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

Gauss quadrature

Consider [a, b] ⊂ R and

I (f ) :=

∫ b

a

w(x) f (x) dx ,

and w(x) is a weight function. Given x0 < x1 · · · < xn, xi ∈ [a, b],consider the Hermite interpolation polynomial for f on these nodes:

p2n+1(x) =n∑

i=0

Hi (x)f (xi ) +n∑

i=0

Ki (x)f ′(xi ),

Hi (x) = (Li (x))2(1− 2L′i (xi )(x − xi )),

Ki (x) = (Li (x))2(x − xi ),

Li (x) :=n∏

j=0,j 6=i

x − xjxi − xj

L0(x) ≡ 1,putting p2n+1 in place of f in the integral gives an approximation of theintegral:∫ b

a

w(x) f (x) dx ≈n∑

i=0

Wi f (xi )+n∑

i=0

Vi f′(xi ),

Wi :=∫ b

aw(x)Hi (x) dx ,

Vi :=∫ b

aw(x)Ki (x) dx .

Elena Celledoni NA

Page 5: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

Gauss quadrature

∫ b

a

w(x) f (x) dx ≈n∑

i=0

Wi f (xi )+n∑

i=0

Vi f′(xi ),

Wi :=∫ b

aw(x)Hi (x) dx ,

Vi :=∫ b

aw(x)Ki (x) dx .

Finally we choose x0, . . . , xn such that Vi = 0.We found that this is possible iff

0 = Vi =

∫ b

a

w(x) (Li (x))2(x − xi ) dx = cj

∫ b

a

w(x)ω(x) Li (x) dx

so0 = Vi ⇔ 〈ω, q〉w = 0, ∀q ∈ Πn

which means we should choose x0, . . . , xn to be the zeros of a polynomialω(x) ∈ Πn+1 belonging to a system of orthogonal polynomials w.r.t.〈·, ·〉w .

Elena Celledoni NA

Page 6: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

Gauss quadrature

Summarizing∫ b

aw(x) f (x) dx ≈ Gn(f ) :=

n∑i=0

Wi f (xi )

where

Wi :=

∫ b

aw(x) [Li (x)]2 dx

x0, . . . , xn to be the zeros of a polynomial of degree n + 1 belongingto a system of orthogonal polynomials w.r.t. 〈·, ·〉w

Elena Celledoni NA

Page 7: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

Ch. 10.4 Error estimation for Gauss quadrature

Theorem 10.1. Let w (weight function) be defined, integrable,continuous and positive on (a, b) and f ∈ C (2n+2)[a, b] (continuousdifferentiable with 2n + 2 continuous derivatives) and n ≥ 0. Then forthe Gauss quadrature ∃ η ∈ (a, b) s.t.∫ b

a

w(x)f (x) dx −n∑

k=0

Wk f (xk) = Kn f(2n+2)(η), (1)

and

Kn =1

(2n + 2)!

∫ b

a

w(x)[ω(x)]2 dx .

So the the Gauss quadrature is exact for polynomials of degree 2n + 1.

Rough estimate:

maxk=0,..,n

|x − xk | = b − a, |ω(x)|2 ≤((n + 1)(b − a)

)2Kn ≤

1(2n + 2)!

(n + 1)2(b − a)2∫ b

a

w(x) dx .

Elena Celledoni NA

Page 8: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

Ch. 10.4 Error estimation for Gauss quadrature

Theorem 10.1. Let w (weight function) be defined, integrable,continuous and positive on (a, b) and f ∈ C (2n+2)[a, b] (continuousdifferentiable with 2n + 2 continuous derivatives) and n ≥ 0. Then forthe Gauss quadrature ∃ η ∈ (a, b) s.t.∫ b

a

w(x)f (x) dx −n∑

k=0

Wk f (xk) = Kn f(2n+2)(η), (1)

and

Kn =1

(2n + 2)!

∫ b

a

w(x)[ω(x)]2 dx .

So the the Gauss quadrature is exact for polynomials of degree 2n + 1.Rough estimate:

maxk=0,..,n

|x − xk | = b − a, |ω(x)|2 ≤((n + 1)(b − a)

)2Kn ≤

1(2n + 2)!

(n + 1)2(b − a)2∫ b

a

w(x) dx .

Elena Celledoni NA

Page 9: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

Convergence of Gauss quadrature to the integral

Let us denote the Gauss quadrature formula with

Gn(f ) :=n∑

k=0

Wk f (xk)

Theorem 10.2. Let w (weight function) be defined, integrable,continuous and positive on (a, b) and f ∈ C 0[a, b] (continuous in theclosed interval [a, b]). Then

limn→∞

Gn(f ) =

∫ b

a

w(x)f (x) dx .

Proof. Weierstrass theorem: ∀ε0 > 0, ∃ pN (polynomial) such that

|f (x)− pN(x)| ≤ ε0, ∀x ∈ [a, b], pN ∈ ΠN

∫ b

aw(x)f (x) dx − Gn(f ) =

∫ b

aw(x)(f (x)− pN(x)) dx

+

∫ b

aw(x)pN dx − Gn(pN)

+Gn(pN)− Gn(f ).

Elena Celledoni NA

Page 10: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

Convergence of Gauss quadrature to the integral

Let us denote the Gauss quadrature formula with

Gn(f ) :=n∑

k=0

Wk f (xk)

Theorem 10.2. Let w (weight function) be defined, integrable,continuous and positive on (a, b) and f ∈ C 0[a, b] (continuous in theclosed interval [a, b]). Then

limn→∞

Gn(f ) =

∫ b

a

w(x)f (x) dx .

Proof. Weierstrass theorem: ∀ε0 > 0, ∃ pN (polynomial) such that

|f (x)− pN(x)| ≤ ε0, ∀x ∈ [a, b], pN ∈ ΠN

∫ b

aw(x)f (x) dx − Gn(f ) =

∫ b

aw(x)(f (x)− pN(x)) dx

+

∫ b

aw(x)pN dx − Gn(pN)

+Gn(pN)− Gn(f ).

Elena Celledoni NA

Page 11: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

Ch 10.5 Composite Gauss formulae

∫ b

af (x) dx =

m∑j=1

∫ xj

xj−1

f (x) dx ,

xj − xj−1 = h = b−am , xj = a + j h.

Change of variables: [xj−1, xj ]→ [−1, 1]:

x =12

(xj−1 + xj) +12ht, t ∈ [−1, 1],

∫ b

af (x) dx =

h

2

m∑j=1

∫ 1

−1f (

12

(xj−1 + xj) +12ht) dt =

h

2

m∑j=1

Ij

approximate each Ij with a Gauss quadrature formula:∫ b

af (x) dx ≈ h

2

m∑j=1

n∑k=0

Wk f (12

(xj−1 + xj) +12hξk)

Wk and ξk weights and nodes of Gauss quadrature on [−1, 1].

Elena Celledoni NA

Page 12: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

Ch 10.5 Composite Gauss formulae

∫ b

af (x) dx =

m∑j=1

∫ xj

xj−1

f (x) dx ,

xj − xj−1 = h = b−am , xj = a + j h.

Change of variables: [xj−1, xj ]→ [−1, 1]:

x =12

(xj−1 + xj) +12ht, t ∈ [−1, 1],

∫ b

af (x) dx =

h

2

m∑j=1

∫ 1

−1f (

12

(xj−1 + xj) +12ht) dt =

h

2

m∑j=1

Ij

approximate each Ij with a Gauss quadrature formula:∫ b

af (x) dx ≈ h

2

m∑j=1

n∑k=0

Wk f (12

(xj−1 + xj) +12hξk)

Wk and ξk weights and nodes of Gauss quadrature on [−1, 1].

Elena Celledoni NA

Page 13: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

Ch 10.5 Composite Gauss formulae

∫ b

af (x) dx =

m∑j=1

∫ xj

xj−1

f (x) dx ,

xj − xj−1 = h = b−am , xj = a + j h.

Change of variables: [xj−1, xj ]→ [−1, 1]:

x =12

(xj−1 + xj) +12ht, t ∈ [−1, 1],

∫ b

af (x) dx =

h

2

m∑j=1

∫ 1

−1f (

12

(xj−1 + xj) +12ht) dt =

h

2

m∑j=1

Ij

approximate each Ij with a Gauss quadrature formula:∫ b

af (x) dx ≈ h

2

m∑j=1

n∑k=0

Wk f (12

(xj−1 + xj) +12hξk)

Wk and ξk weights and nodes of Gauss quadrature on [−1, 1].

Elena Celledoni NA

Page 14: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

Ch 10.5 Composite Gauss formulae

∫ b

af (x) dx =

m∑j=1

∫ xj

xj−1

f (x) dx ,

xj − xj−1 = h = b−am , xj = a + j h.

Change of variables: [xj−1, xj ]→ [−1, 1]:

x =12

(xj−1 + xj) +12ht, t ∈ [−1, 1],

∫ b

af (x) dx =

h

2

m∑j=1

∫ 1

−1f (

12

(xj−1 + xj) +12ht) dt =

h

2

m∑j=1

Ij

approximate each Ij with a Gauss quadrature formula:∫ b

af (x) dx ≈ h

2

m∑j=1

n∑k=0

Wk f (12

(xj−1 + xj) +12hξk)

Wk and ξk weights and nodes of Gauss quadrature on [−1, 1].Elena Celledoni NA

Page 15: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

Exercise on Lobatto quadrature: exercise 10.7

a) Show that ∀p2n−1 ∈ Π2n−1 on the interval, we have

p2n−1(x) = (x − a)(b − x)q2n−3(x) + r(x − a) + s(b − x),

q2n−3 ∈ Π2n−3, a, b, r , s ∈ R, with a and b notsimultaneously zero.Note: (x − a), (x − b), (x − a)(b− x)xk and k = 0, . . . , 2n− 3with a and b not simultaneously zero is a basis for Π2n−1.

Solution: using the given basis we can write

p2n−1(x) = (x−a)(b−x)2n−3∑k=0

λkxk+λ2n−2(x−a)+λ2n−1(b−x),

and so taking q2n−3(x) :=∑2n−3

k=0 λkxk and r := λ2n−2,

s := λ2n−1, we see that there existq2n−3 ∈ Π2n−3, a, b, r , s ∈ R, such that any p2n−1 ∈ Π2n−1can be written in the given form.

Elena Celledoni NA

Page 16: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

Exercise on Lobatto quadrature: exercise 10.7

a) Show that ∀p2n−1 ∈ Π2n−1 on the interval, we have

p2n−1(x) = (x − a)(b − x)q2n−3(x) + r(x − a) + s(b − x),

q2n−3 ∈ Π2n−3, a, b, r , s ∈ R, with a and b notsimultaneously zero.Note: (x − a), (x − b), (x − a)(b− x)xk and k = 0, . . . , 2n− 3with a and b not simultaneously zero is a basis for Π2n−1.Solution: using the given basis we can write

p2n−1(x) = (x−a)(b−x)2n−3∑k=0

λkxk+λ2n−2(x−a)+λ2n−1(b−x),

and so taking q2n−3(x) :=∑2n−3

k=0 λkxk and r := λ2n−2,

s := λ2n−1, we see that there existq2n−3 ∈ Π2n−3, a, b, r , s ∈ R, such that any p2n−1 ∈ Π2n−1can be written in the given form.

Elena Celledoni NA

Page 17: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

Exercise on Lobatto quadrature: exercise 10.7

a) Show that ∀p2n−1 ∈ Π2n−1 on the interval, we have

p2n−1(x) = (x − a)(b − x)q2n−3(x) + r(x − a) + s(b − x),

q2n−3 ∈ Π2n−3, a, b, r , s ∈ R, with a and b notsimultaneously zero.Note: (x − a), (x − b), (x − a)(b− x)xk and k = 0, . . . , 2n− 3with a and b not simultaneously zero is a basis for Π2n−1.Solution: using the given basis we can write

p2n−1(x) = (x−a)(b−x)2n−3∑k=0

λkxk+λ2n−2(x−a)+λ2n−1(b−x),

and so taking q2n−3(x) :=∑2n−3

k=0 λkxk and r := λ2n−2,

s := λ2n−1, we see that there existq2n−3 ∈ Π2n−3, a, b, r , s ∈ R, such that any p2n−1 ∈ Π2n−1can be written in the given form.

Elena Celledoni NA

Page 18: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

Exercise on Lobatto quadrature: exercise 10.7

b) Construct the Lobatto quadrature formula

∫ b

aw(x)f (x) dx ≈W0f (a) +

n−1∑k=1

Wk f (xk ) + Wnf (b)

which is exact when f ∈ Π2n−1. Here w(x) is a weight function.

Solution: using the formula obtained in a), w(x) := w(x)(x − a)(b − x)∫ b

aw(x)p2n−1(x)dx =

∫ b

aw(x)q2n−3(x)dx+r

∫ b

aw(x)(x−a)dx+s

∫ b

aw(x)(b−x)dx ,

and using the Gauss quadrature with n − 1 weights and nodes (W ∗k , x

∗k ) wrt

w(x) one gets,

∫ b

aw(x)p2n−1(x)dx =

n−1∑k=1

W ∗k q2n−3(x∗k )+r

∫ b

aw(x)(x−a)dx+s

∫ b

aw(x)(b−x)dx ,

for an arbitrary polynomial p2n−1 ∈ Π2n−1. Since

q2n−3(x∗k ) =p2n−1(x∗k )− r(x∗k − a)− s(b − x∗k )

(x∗k − a)(b − x∗k )

r = p2n−1(b)/(b − a), s = p2n−1(a)/(b − a).

Elena Celledoni NA

Page 19: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

Exercise on Lobatto quadrature: exercise 10.7

b) Construct the Lobatto quadrature formula

∫ b

aw(x)f (x) dx ≈W0f (a) +

n−1∑k=1

Wk f (xk ) + Wnf (b)

which is exact when f ∈ Π2n−1. Here w(x) is a weight function.Solution: using the formula obtained in a), w(x) := w(x)(x − a)(b − x)∫ b

aw(x)p2n−1(x)dx =

∫ b

aw(x)q2n−3(x)dx+r

∫ b

aw(x)(x−a)dx+s

∫ b

aw(x)(b−x)dx ,

and using the Gauss quadrature with n − 1 weights and nodes (W ∗k , x

∗k ) wrt

w(x) one gets,

∫ b

aw(x)p2n−1(x)dx =

n−1∑k=1

W ∗k q2n−3(x∗k )+r

∫ b

aw(x)(x−a)dx+s

∫ b

aw(x)(b−x)dx ,

for an arbitrary polynomial p2n−1 ∈ Π2n−1. Since

q2n−3(x∗k ) =p2n−1(x∗k )− r(x∗k − a)− s(b − x∗k )

(x∗k − a)(b − x∗k )

r = p2n−1(b)/(b − a), s = p2n−1(a)/(b − a).

Elena Celledoni NA

Page 20: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

Exercise on Lobatto quadrature: exercise 10.7

b) Construct the Lobatto quadrature formula

∫ b

aw(x)f (x) dx ≈W0f (a) +

n−1∑k=1

Wk f (xk ) + Wnf (b)

which is exact when f ∈ Π2n−1. Here w(x) is a weight function.Solution: using the formula obtained in a) and the Gauss quadrature with n− 1weights and nodes (W ∗

k , x∗k ) wrt w(x) = w(x)(x − a)(b − x) one gets,

∫ b

aw(x)p2n−1(x)dx = W0p2n−1(a) +

n−1∑k=1

Wkp2n−1(xk ) + Wnp2n−1(b)

x0 := a, xn := b, xk := x∗k , Wk :=W ∗

k

(x∗k − a)(b − x∗k ), k = 1, . . . , n − 1,

W0 :=1

b − a

(∫ b

aw(x)(b − x)dx −

n−1∑k=1

W ∗k

x∗k − a

),

Wn :=1

b − a

(∫ b

aw(x)(x − a)dx −

n−1∑k=1

W ∗k

b − x∗k

).

Elena Celledoni NA

Page 21: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

Exercise on Lobatto quadrature: exercise 10.7

c) Prove that the weights are positive.

Solution: Wk :=W∗

k(x∗

k−a)(b−x∗

k)k = 1, . . . , n − 1 are positive because W ∗

k are

the Gaussian weights which are known to be positive (see beginning of chapter10). We now prove that W0 is positive, the proof is similar for Wn.

W0 :=1

b − a

(∫ b

aw(x)(b − x)

(x − a)

(x − a)dx −

n−1∑k=1

W ∗k

x∗k − a

),

with w(x) = w(x)(x − a)(b − x)

W0 =1

b − a

(∫ b

aw(x)

1(x − a)

dx −n−1∑k=1

W ∗k

x∗k − a

),

this is the error for the Gauss quadrature on m + 1 = n − 1 nodes for 1x−a

wrtw . Using theorem 10.1 (with n in the theorem replaced by m and m = n − 2)we get

W0 =1

b − aKm

d2m+2

dx2m+2

(1

x − a

).

Km = 1(2m+2)!

∫ ba w(x)(ω(x))2 dx (see theorem 10.1) is always positive, and the

even derivatives of 1x−a

are always positive.

Elena Celledoni NA

Page 22: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

Exercise on Lobatto quadrature: exercise 10.7

c) Prove that the weights are positive.

Solution: Wk :=W∗

k(x∗

k−a)(b−x∗

k)k = 1, . . . , n − 1 are positive because W ∗

k are

the Gaussian weights which are known to be positive (see beginning of chapter10).

We now prove that W0 is positive, the proof is similar for Wn.

W0 :=1

b − a

(∫ b

aw(x)(b − x)

(x − a)

(x − a)dx −

n−1∑k=1

W ∗k

x∗k − a

),

with w(x) = w(x)(x − a)(b − x)

W0 =1

b − a

(∫ b

aw(x)

1(x − a)

dx −n−1∑k=1

W ∗k

x∗k − a

),

this is the error for the Gauss quadrature on m + 1 = n − 1 nodes for 1x−a

wrtw . Using theorem 10.1 (with n in the theorem replaced by m and m = n − 2)we get

W0 =1

b − aKm

d2m+2

dx2m+2

(1

x − a

).

Km = 1(2m+2)!

∫ ba w(x)(ω(x))2 dx (see theorem 10.1) is always positive, and the

even derivatives of 1x−a

are always positive.

Elena Celledoni NA

Page 23: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

Exercise on Lobatto quadrature: exercise 10.7

c) Prove that the weights are positive.

Solution: Wk :=W∗

k(x∗

k−a)(b−x∗

k)k = 1, . . . , n − 1 are positive because W ∗

k are

the Gaussian weights which are known to be positive (see beginning of chapter10). We now prove that W0 is positive, the proof is similar for Wn.

W0 :=1

b − a

(∫ b

aw(x)(b − x)

(x − a)

(x − a)dx −

n−1∑k=1

W ∗k

x∗k − a

),

with w(x) = w(x)(x − a)(b − x)

W0 =1

b − a

(∫ b

aw(x)

1(x − a)

dx −n−1∑k=1

W ∗k

x∗k − a

),

this is the error for the Gauss quadrature on m + 1 = n − 1 nodes for 1x−a

wrtw .

Using theorem 10.1 (with n in the theorem replaced by m and m = n − 2)we get

W0 =1

b − aKm

d2m+2

dx2m+2

(1

x − a

).

Km = 1(2m+2)!

∫ ba w(x)(ω(x))2 dx (see theorem 10.1) is always positive, and the

even derivatives of 1x−a

are always positive.

Elena Celledoni NA

Page 24: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

Exercise on Lobatto quadrature: exercise 10.7

c) Prove that the weights are positive.

Solution: Wk :=W∗

k(x∗

k−a)(b−x∗

k)k = 1, . . . , n − 1 are positive because W ∗

k are

the Gaussian weights which are known to be positive (see beginning of chapter10). We now prove that W0 is positive, the proof is similar for Wn.

W0 :=1

b − a

(∫ b

aw(x)(b − x)

(x − a)

(x − a)dx −

n−1∑k=1

W ∗k

x∗k − a

),

with w(x) = w(x)(x − a)(b − x)

W0 =1

b − a

(∫ b

aw(x)

1(x − a)

dx −n−1∑k=1

W ∗k

x∗k − a

),

this is the error for the Gauss quadrature on m + 1 = n − 1 nodes for 1x−a

wrtw . Using theorem 10.1 (with n in the theorem replaced by m and m = n − 2)we get

W0 =1

b − aKm

d2m+2

dx2m+2

(1

x − a

).

Km = 1(2m+2)!

∫ ba w(x)(ω(x))2 dx (see theorem 10.1) is always positive, and the

even derivatives of 1x−a

are always positive.

Elena Celledoni NA

Page 25: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

Exercise relevant for the project

Show that

|f (x)− s(x)| ≤ 78h2 ‖f ′′‖∞, ∀x ∈ [a, b],

where f ∈ C 2[a, b] and s is the natural cubic spline on theequidistant knots a = x0 < x1 < · · · < xn = b, xi − xi−1 = h,i = 1, . . . , n.Plan:a) Show first: |f (x)− s(x)| ≤ h2

8 (‖f ′′‖∞ + maxi |s ′′i |);b) show then that |s ′′i | ≤ 6 ‖f ′′‖∞ to obtain the result. (For the

solution see problem set 5 exercise 3.)

Elena Celledoni NA

Page 26: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

Solution of point a)

Recall f (xj) = s(xj) and f (xj−1) = s(xj−1). Fix a x ∈ [xj−1, xj ] andconsider

g(x) := f (x)− s(x)− (x − xj−1)(x − xj)

(x − xj−1)(x − xj)(f (x)− s(x)), ∀x ∈ [xj−1, xj ],

and 0 = g(x) = g(xj−1) = g(xj). So there exists ξj(x) ∈ (xj−1, xj) suchthat

0 = g ′′(ξj) = f ′′(ξj)− s ′′(ξj)−2

(x − xj−1)(x − xj)(f (x)− s(x)),

this imples

f (x)− s(x) = (x − xj−1)(x − xj)f ′′(ξj)− s ′′(ξj)

2,

and |(x − xj−1)(x − xj)| ≤ h2/4,

|f (x)− s(x)| ≤ h2

8(|f ′′(ξj)|+ |s ′′(ξj)|),

leading to the proof of point a).Elena Celledoni NA

Page 27: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

Adaptive quadrature: Adaptive Simpson

Adaptive quadrature: given a tolerance TOL find I ≈ I s.t.

|I − I | ≤ TOL.

Consider

I =

∫ b

a

f (x) dx ,

Simpson:

S(a, b) :=b − a

6

[f (a) + f

(a + b

2)

+ f (b)

]error

E (a, b) = − 190

(b − a

2

)5

f (4)(ξ), ξ ∈ (a, b),

I = S + E

Plan:

1 Subdivide [a, b] recursively in disjoint subintervals;

2 apply S on each subinterval;

3 stop when |I − I | ≤ TOL is satisfied.

Elena Celledoni NA

Page 28: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

Error estimate

Case of two subintervals [a, b] = [a, c] ∪ [c, b] c := a+b2 , h = b − a.

LetI0 = S(a, b), E0 = E (a, b), I = I0 + E0

I = S(a, c) + S(c , b), E = E (a, c) + E (c , b), I = I + E

E = − 190

(12h

2

)5 (f (4)(ξ1) + f (4)(ξ2)

), ξ1 ∈ (a, c), ξ2 ∈ (c , b)

by the intermediate value theorem, since f 4 is assumed continuous,we have

E = − 190

116

(h

2

)5 (f (4)(ξ)

), ξ ∈ (a, b),

we will assume that for h small f (4)(ξ) ≈ f (4)(ξ) this gives

E0 ≈ 16 E , |I − I0| = |E0 − E | ≈ 15|E | ⇒ |E | ≈ |I − I0|15

Elena Celledoni NA

Page 29: Elena Celledoni - NTNU...Exercise on Lobatto quadrature: exercise 10.7 a)Showthat8p 2n 1 2 2n 1 ontheinterval,wehave p 2n 1(x) = (x a)(b x)q 2n 3(x) + r(x a) + s(b x); q 2n 3 2 2n

adaptiveS(f , a, b,TOL)

I0 = S(a, b)

c := b+a2

I = S(a, c) + S(c , b)

IF 115 |I − I0| ≤ TOL then I = I + 1

15(I − I0)

ELSE

I =adaptiveS(f , a, c , TOL2 )+adaptiveS(f , c , b, TOL

2 )

END

RETURN IElena Celledoni NA


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