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Gauss quadrature
Elena Celledoni
Department of Mathematical Sciences, NTNU
October 23rd 2012
Elena Celledoni NA
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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