Chapter 2 Numerical integration

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Chapter 2 Numerical integration

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I. Motivation

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Example : f (x) = cos(πx) √

x

²

+ 1

Can we compute J = Z ₂

0

f(x)dx?

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

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Example : f (x) = cos(πx) √

x

²

+ 1

Can we compute J = Z ₂

0

f(x)dx?

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

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Rectangular rule

(leftpoints)

Approximation by a piecewise constant function

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

11 points

10 subintervals

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Rectangular rule (leftpoints)

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

11 points 10 subintervals

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Rectangular rule (leftpoints)

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

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Rectangular rule

(rightpoints)

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

11 points

10 subintervals

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Rectangular rule (rightpoints)

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

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Rectangular rule (rightpoints)

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

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Rectangular rule

(midpoints)

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

11 points

10 subintervals

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Rectangular rule (midpoints)

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

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Rectangular rule (midpoints)

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

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Trapezoidal rule

Approximation by a piecewise affine function

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

11 points

10 subintervals

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Trapezoidal rule

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

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Trapezoidal rule

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

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Comparison of the different rules (20 subintervals)

Leftpoint rule Rightpoint rule

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

I=0.019664765868385 I=0.143271563618364 Midpoint rule Trapezoidal rule

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

I=0.080343177070866 I=0.081468164743374

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Comparison of the different rules (100 subintervals)

Leftpoint rule Rightpoint rule

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

I=0.068388241234746 I=0.093109600784742 Midpoint rule Trapezoidal rule

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

0 0.5 1 1.5 2

−1.5

−1

−0.5 0 0.5 1 1.5 2 2.5

I=0.080704188569061 I=0.080748921009744

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II. Quadrature rules

1 Principle

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Definition of a quadrature rule

Letf : [a, b]→R(a < b) be a continuous function

a

b x f(x)

Can we compute an approximate value of J(f) =

Z b a

f(x)dx?

Aquadrature ruleis an approximation of the integral J(f) stated as a linear combination of function values at specified points in [a, b].

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Definition of a quadrature rule

a

b x f(x)

Z b a

f(x)dx?

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Definition of a quadrature rule

a

b x f(x)

Z b a

f(x)dx?

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Reduction to elementary quadrature rules

x0

x_n x f(x)

xk

xk+1

Z b a

f(x)dx?

Partition : a=x0< x1 < . . . < x_k< x_k+1< . . . < x_N =b, withh_k=x_k+1−x_k(0≤i≤N−1) andh= maxh_k.

Z _b

a

f(x)dx=

N−1

X

k=0

Z _x_k+1

xk

f(x)dx

Þwe want a quadrature rule on small elementary intervals

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Reduction to elementary quadrature rules

x0

x_n x f(x)

xk

xk+1

Z b a

f(x)dx?

Partition : a=x₀< x₁ < . . . < x_k< x_k+1< . . . < x_N =b, withh_k=x_k+1−x_k(0≤i≤N−1) andh= maxh_k.

Z _b

a

f(x)dx=

N−1

X

k=0

Z _x_k+1

xk

f(x)dx

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Reduction to elementary quadrature rules

x0

x_n x f(x)

xk

xk+1

Z b a

f(x)dx?

Z b a

f(x)dx=

N−1

X

k=0

Z xk+1

xk

f(x)dx

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Reduction to elementary quadrature rules

x0

x_n x f(x)

xk

xk+1

Z b a

f(x)dx?

Z b a

f(x)dx=

N−1

X

k=0

Z xk+1

xk

f(x)dx

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II. Quadrature rules

1 Principle

2 Examples

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Leftpoint rule

xk xk+1

Z xk+1

xk

f(x)dx≈(x_k+1−x_k)f(x_k)

Z b a

f(x)dx≈

N−1

X

k=0

(xk+1−xk)f(xk)

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Leftpoint rule

xk xk+1

Z xk+1

xk

Z b a

f(x)dx≈

N−1

X

k=0

(xk+1−xk)f(xk)

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Leftpoint rule

xk xk+1

Z xk+1

xk

Z b a

f(x)dx≈

N−1

X

k=0

(xk+1−xk)f(xk)

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Rightpoint rule

xk xk+1

Z xk+1

xk

f(x)dx≈(x_k+1−x_k)f(x_k+1)

Z b a

f(x)dx≈

N−1

X

k=0

(xk+1−xk)f(xk+1)

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Rightpoint rule

xk xk+1

Z xk+1

xk

Z b a

f(x)dx≈

N−1

X

k=0

(xk+1−xk)f(xk+1)

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Rightpoint rule

xk xk+1

Z xk+1

xk

Z b a

f(x)dx≈

N−1

X

k=0

(xk+1−xk)f(xk+1)

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Midpoint rule

xk xk+1 xk+xk+1

2

Z xk+1

xk

f(x)dx≈ (xk+1−x_k)f

xk+xk+1

2

Z _b

a

f(x)dx≈

N−1

X

k=0

(x_k+1−x_k)fx_k+x_k+1 2

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Midpoint rule

xk xk+1 xk+xk+1

2

Z xk+1

xk

xk+xk+1

2

Z _b

a

f(x)dx≈

N−1

X

k=0

(x_k+1−x_k)fx_k+x_k+1 2

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Midpoint rule

xk xk+1 xk+xk+1

2

Z xk+1

xk

xk+xk+1

2

Z b a

f(x)dx≈

N−1

X

k=0

(x_k+1−x_k)fx_k+x_k+1 2

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Trapezoidal rule

xk xk+1

Z xk+1

xk

f(x)dx≈ (x_k+1−x_k)

f(x_k) +f(x_k+1) 2

Z b a

f(x)dx≈

N−1

X

k=0

(x_k+1−x_k)

f(x_k) +f(x_k+1) 2

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Trapezoidal rule

xk xk+1

Z xk+1

xk

f(x_k) +f(x_k+1) 2

Z b a

f(x)dx≈

N−1

X

k=0

(x_k+1−x_k)

f(x_k) +f(x_k+1) 2

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Trapezoidal rule

xk xk+1

Z xk+1

xk

f(x_k) +f(x_k+1) 2

Z b a

f(x)dx≈

N−1

X

k=0

(x_k+1−x_k)

f(x_k) +f(x_k+1) 2

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II. Quadrature rules

1 Principle

2 Examples

3 Generalization

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Change of variables

From[x_k, x_k+1] to [−1,1]







x= x_k+x_k+1

2 +x_k+1−x_k

2 s

dx= x_k+1−x_k

2 ds

Z xk+1

xk

f(x)dx= xk+1−xk

2

Z 1

−1

f

xk+xk+1

2 +xk+1−xk

2 s

ds

= xk+1−xk

2

Z 1

−1

ϕ_k(s)ds.

Þquadrature rule for Z 1

−1

ϕ(s)ds?

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Change of variables

From[x_k, x_k+1] to [−1,1]







x= x_k+x_k+1

2 +x_k+1−x_k

2 s

dx= x_k+1−x_k

2 ds

Z xk+1

xk

f(x)dx= xk+1−xk

2

Z 1

−1

f

xk+xk+1

2 +xk+1−xk

2 s

ds

= x_k+1−x_k 2

Z 1

−1

ϕ_k(s)ds.

−1

ϕ(s)ds?

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Change of variables

From[x_k, x_k+1] to [−1,1]







x= x_k+x_k+1

2 +x_k+1−x_k

2 s

dx= x_k+1−x_k

2 ds

Z xk+1

xk

f(x)dx= xk+1−xk

2

Z 1

−1

f

xk+xk+1

2 +xk+1−xk

2 s

ds

= x_k+1−x_k 2

Z 1

−1

ϕ_k(s)ds.

−1

ϕ(s)ds?

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Elementary quadrature rule (on [−1, 1])

Z 1

−1

ϕ(s)ds

| {z }

≈

l

X

j=0

ωjϕ(τj)

| {z } J(ϕ) J^QR(ϕ) with

l∈N

τj ∈[−1,1]for 0≤j≤l,

l

X

j=0

ωj = 2.

Examples

Midpoint rule :l= 0,τ0= 0,ω0 = 2,

Trapezoidal rule :l= 1,τ0=−1, τ₁= 1,ω0 =ω1 = 1.

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Elementary quadrature rule (on [x

_k

, x

_k+1

])

Z xk+1

xk

f(x)dx= x_k+1−x_k 2

Z 1

−1

f

x_k+x_k+1

2 +x_k+1−x_k

2 s

ds

≈ xk+1−xk

2

l

X

j=0

ωjf(xk+xk+1

2 +xk+1−xk

2 τj) Þ let setλj = ωj

2 ,τ_kj = xk+xk+1

2 + xk+1−xk

2 τj

General form Z xk+1

xk

f(x)dx≈(x_k+1−x_k)

l

X

j=0

λ_jf(τ_kj)

with l∈N,

τ_kj ∈[x_k, x_k+1]for 0≤j≤l,

l

X

j=0

λj = 1.

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Elementary quadrature rule (on [x

_k

, x

_k+1

])

Z xk+1

xk

Z 1

−1

f

x_k+x_k+1

2 +x_k+1−x_k

2 s

ds

≈ xk+1−xk

2

l

X

j=0

ωjf(xk+xk+1

2 +xk+1−xk

2 τj)

Þ let setλj = ωj

2 ,τ_kj = xk+xk+1

2 + xk+1−xk

2 τj

General form Z xk+1

xk

l

X

j=0

λ_jf(τ_kj)

with l∈N,

l

X

j=0

λj = 1.

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Elementary quadrature rule (on [x

_k

, x

_k+1

])

Z xk+1

xk

Z 1

−1

f

x_k+x_k+1

2 +x_k+1−x_k

2 s

ds

≈ xk+1−xk

2

l

X

j=0

ωjf(xk+xk+1

2 +xk+1−xk

2 ,τ_kj = x_k+x_k+1

2 +x_k+1−x_k 2 τj

General form Z xk+1

xk

l

X

j=0

λ_jf(τ_kj)

with l∈N,

l

X

j=0

λj = 1.

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Elementary quadrature rule (on [x

_k

, x

_k+1

])

Z xk+1

xk

Z 1

−1

f

x_k+x_k+1

2 +x_k+1−x_k

2 s

ds

≈ xk+1−xk

2

l

X

j=0

ωjf(xk+xk+1

2 +xk+1−xk

2 ,τ_kj = x_k+x_k+1

2 +x_k+1−x_k 2 τj

General form Z xk+1

xk

l

X

j=0

λ_jf(τ_kj)

with l∈N,

l

X

j=0

λj = 1.

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Conclusion : general form of a quadrature rule

Z b a

f(x)dx

| {z }

≈

N−1

X

k=0

(x_k+1−x_k)

l

X

j=0

λ_jf(τ_kj)

| {z }

J(f) J_h^QR(f) with

l∈N,

τ_kj ∈[x_k, x_k+1]for 0≤j≤l,0≤i≤N−1

l

X

j=0

λ_j = 1.

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III. Order and error estimate

1 Order of a quadrature rule

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Definition of the order

Definition

A quadrature rule is of orderp if

it integrates exactly every polynomial function with a degree less than p,

it doesn’t integrate exactly at least one polynomial function with a degree p+ 1.

How to determinate the degree of a quadrature rule ? We verify that the elementary ruleJ^QR(ϕ) =

l

X

j=0

ω_jϕ(τ_j) is exact(it means that J^QR(ϕ) =J(ϕ)) for the functionsϕ: x7→1,x7→x, . . . , x7→x^p

inexact for x7→x^p+1.

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Examples

Leftpoint rule

áorder 0

J^QR(ϕ) = 2ϕ(−1)

J^QR(x7→1) = 2 =J(x7→1) J^QR(x7→x) =−2 6=J(x7→x)(= 0)

Rightpoint rule

á order 0

J^QR(ϕ) = 2ϕ(1)

J^QR(x7→1) = 2 =J(x7→1) J^QR(x7→x) = 2 6=J(x7→x)(= 0)

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Examples

Leftpoint rule áorder 0

J^QR(ϕ) = 2ϕ(−1)

J^QR(x7→1) = 2 =J(x7→1) J^QR(x7→x) =−2 6=J(x7→x)(= 0)

Rightpoint rule

á order 0

J^QR(ϕ) = 2ϕ(1)

J^QR(x7→1) = 2 =J(x7→1) J^QR(x7→x) = 2 6=J(x7→x)(= 0)

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Examples

J^QR(ϕ) = 2ϕ(−1)

J^QR(x7→1) = 2 =J(x7→1) J^QR(x7→x) =−2 6=J(x7→x)(= 0) Rightpoint rule

á order 0

J^QR(ϕ) = 2ϕ(1)

J^QR(x7→1) = 2 =J(x7→1) J^QR(x7→x) = 2 6=J(x7→x)(= 0)

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Examples

J^QR(ϕ) = 2ϕ(−1)

J^QR(x7→1) = 2 =J(x7→1) J^QR(x7→x) =−2 6=J(x7→x)(= 0) Rightpoint rule á order 0

J^QR(ϕ) = 2ϕ(1)

J^QR(x7→1) = 2 =J(x7→1) J^QR(x7→x) = 2 6=J(x7→x)(= 0)

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Examples

Midpoint rule

á order 1

J^QR(ϕ) = 2ϕ(0)

J^QR(x7→1) = 2 =J(x7→1) J^QR(x7→x) = 0 =J(x7→x)

J^QR(x7→x²) = 0 6=J(x7→x²)(= 2 3) Trapezoidal rule

á order 1

J^QR(ϕ) = ϕ(−1) +ϕ(1) J^QR(x7→1) = 2 =J(x7→1) J^QR(x7→x) = 0 =J(x7→x) J^QR(x7→x²) = 2 6=J(x7→x²)

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Examples

Midpoint rule

á order 1

J^QR(ϕ) = 2ϕ(0)

J^QR(x7→1) = 2 =J(x7→1) J^QR(x7→x) = 0 =J(x7→x) J^QR(x7→x²) = 0 6=J(x7→x²)(= 2

3)

Trapezoidal rule

á order 1

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Examples

Midpoint rule á order 1

J^QR(ϕ) = 2ϕ(0)

3)

Trapezoidal rule

á order 1

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Examples

J^QR(ϕ) = 2ϕ(0)

3) Trapezoidal rule

á order 1

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Examples

J^QR(ϕ) = 2ϕ(0)

3) Trapezoidal rule á order 1

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III. Order and error estimate

1 Order of a quadrature rule

2 Error estimate

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How to estimate the efficiency of a quadrature rule ?

Z b a

f(x)dx

| {z }

≈

N−1

X

k=0

(x_k+1−x_k)

l

X

j=0

λjf(τ_kj)

| {z }

J(f) J_h^QR(f)

We want that, for every functionf,

J_h^QR(f)→J(f)as h tends to 0 (N tends to+∞),

the error E_f(h) =|J_h^QR(f)−J(f)|tends to 0 “as rapidly as possible”.

Particularly, ifE_f(h) =C_fh^α, the quadrature rule is even more efficient thatα is bigger.

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Comparison of classical methods

Test case

f(x) =xsinx, a= 0, b= π 2, I=

Z π/2 0

f(x)dx= 1.

Convergence result

Ef(h) =Cfh^α. Leftpoint rule : α= 1, Midpoint rule :α= 2, Trapezoidal rule :α= 2.

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Comparison of classical methods

Test case

f(x) =xsinx, a= 0, b= π 2, I=

Z π/2 0

f(x)dx= 1.

Convergence result

Ef(h) =Cfh^α. Leftpoint rule : α= 1, Midpoint rule :α= 2, Trapezoidal rule :α= 2.

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General result

Theorem Let

J^QR(f) be an elementary quadrature rule of orderp J_h^QR(f) the corresponding approximation ofJ(f) =

Z b a

f(x)dx Then, iff ∈ C^p+1([a, b]), there exists a constantC_f such that

|J_h^QR(f)−J(f)|

| {z }

≤ C_fh^p+1. E_f(h)

ÞA quadrature rule of orderp converges like h^p+1.

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IV. How to obtain quadrature rules ?

1 Link between numerical integration and interpolation

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Value of the weights in an elementary quadrature rule

Problem

The points (τ_j)0≤j≤l are given.

We want to calculate the weights(ω_j)0≤j≤l such that J^QR(ϕ) =

l

X

j=0

ωjϕ(τj) has the maximal order.

Practical calculation

Writing J^QR(x7→x^k) =J(x7→x^k)for all 0≤k≤l leads to a linear system of (l+ 1)equations on (ω_j)0≤j≤l.

This system has a unique solution if τ_i 6=τ_j for all i6=j (the matrix is a Vandermonde matrix).

The obtained quadrature rule will have an order greater thanl.

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Link with the Lagrangian basis associated to the (τ

_j

)

_0≤j≤l

J^QR(ϕ) =

l

X

j=0

ω_jϕ(τ_j) Assume that the (ωj)0≤j≤l have been calculated so that the quadrature rule has an order greater thanl.

Then, for all0≤k≤l,L_k∈Rl[X]and J^QR(L_k) = J(L_k)

Reciprocally, let assume ω_k=R1

−1L_k(s)ds for all 0≤k≤l. Then, for all P ∈Rl[X],P =

l

X

j=0

P(τ_j)L_j and J(P) =

Z 1

−1

P(s)ds=

l

X

j=0

P(τ_j) Z 1

−1

L_j(s)ds

áorder≥l

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Link with the Lagrangian basis associated to the (τ

_j

)

_0≤j≤l

J^QR(ϕ) =

l

X

j=0

ω_jϕ(τ_j) Assume that the (ω_j)0≤j≤l have been calculated so that the quadrature rule has an order greater thanl.

Then, for all0≤k≤l,L_k∈Rl[X]and J^QR(L_k)

| {z }

= J(L_k)

| {z }

l

X

j=0

ω_jL_k(τ_j) = Z 1

−1

L_k(s)ds

l

X

j=0

Z 1

−1

P(s)ds=

l

X

j=0

P(τ_j) Z 1

−1

L_j(s)ds

áorder≥l

(70)

Link with the Lagrangian basis associated to the (τ

_j

)

_0≤j≤l

J^QR(ϕ) =

l

X

j=0

Then, for all0≤k≤l,L_k∈Rl[X]and J^QR(Lk)

| {z }

= J(Lk)

| {z }

ω_k =

Z ₁

−1

L_k(s)ds

l

X

j=0

Z 1

−1

P(s)ds=

l

X

j=0

P(τ_j) Z 1

−1

L_j(s)ds

áorder≥l

(71)

Link with the Lagrangian basis associated to the (τ

_j

)

_0≤j≤l

J^QR(ϕ) =

l

X

j=0

| {z }

= J(Lk)

| {z }

ω_k =

Z ₁

−1

L_k(s)ds Reciprocally, let assume ωk=R1

−1Lk(s)ds for all 0≤k≤l.

Then, for allP ∈Rl[X],P =

l

X

j=0

Z 1

−1

P(s)ds=

l

X

j=0

P(τj) Z 1

−1

Lj(s)ds

áorder≥l

(72)

Link with the Lagrangian basis associated to the (τ

_j

)

_0≤j≤l

J^QR(ϕ) =

l

X

j=0

| {z }

= J(Lk)

| {z }

ω_k =

Z ₁

−1

L_k(s)ds Reciprocally, let assume ωk=R1

−1Lk(s)ds for all 0≤k≤l.

l

X

j=0

Z 1

−1

P(s)ds=

l

X

j=0

P(τj)ωj

áorder≥l

(73)

Link with the Lagrangian basis associated to the (τ

_j

)

_0≤j≤l

J^QR(ϕ) =

l

X

j=0

| {z }

= J(L_k)

| {z }

ω_k =

Z 1

−1

L_k(s)ds Reciprocally, let assume ω_k=R1

−1L_k(s)ds for all 0≤k≤l.

l

X

j=0

P(τj)Lj and J(P) =

Z 1

−1

P(s)ds=J^QR(P)

áorder≥l

(74)

Link with the Lagrangian basis associated to the (τ

_j

)

_0≤j≤l

J^QR(ϕ) =

l

X

j=0

| {z }

= J(L_k)

| {z }

ω_k =

Z 1

−1

L_k(s)ds Reciprocally, let assume ω_k=R1

−1L_k(s)ds for all 0≤k≤l.

l

X

j=0

P(τj)Lj and J(P) =

Z 1

−1

P(s)ds=J^QR(P) áorder≥l

(75)

Theorem Let

(τj)0≤j≤l,(l+ 1)distinct points in[−1,1],

(Lk)0≤k≤l, the Lagrangian basis ofRl[X]associated to the points(τj)0≤j≤l.

Then, the quadrature rule J^QR(ϕ) =

l

X

j=0

ω_jϕ(τ_j) has an order at less equal tolif and only if

ωk= Z 1

−1

Lk(s)ds ∀0≤i≤l.

In this case, for all ϕ∈ C[−1,1], we have J^QR(ϕ) = Z 1

−1

Pϕ(s)ds, where P_ϕ is the Lagrangian interpolating polynomial of ϕ in the points(τj)0≤j≤l.

(76)

IV. How to obtain quadrature rules ?

2 Newton-Cotes quadrature

(77)

Definition

The Newton-Cotes formulas are the quadrature rules

J^QR(ϕ) =

l

X

j=0

ω_jϕ(τ_j)

of maximal order obtained for points equally spaced on[−1,1].

It means :

τj =−1 +2j

l for 0≤j≤l, ω_j =

Z 1

−1

L_j(s)ds.

(78)

Examples of Newton-Cotes formulas

l= 1 : trapezoidal rule

J^QR(ϕ) =ϕ(1) +ϕ(−1).

l= 2 : Simpson rule J^QR(ϕ) = 1

3ϕ(−1) + 4

3ϕ(0) + 1 3ϕ(1).

l= 4 : Boole-Villarceau rule J^QR(ϕ) = 7

45ϕ(−1)+32 45ϕ(−1

2)+ 4

15ϕ(0)+32 45ϕ(1

2)+ 7 45ϕ(1).

Order of these quadrature rules l, if lis odd,

l+ 1, ifl is even.

(79)

IV. How to obtain quadrature rules ?

2 Newton-Cotes quadrature

3 Gaussian quadrature

(80)

Principle

Knowing (τj)0≤j≤l, we take ωj = Z 1

−1

Lj(s)ds in order to have the maximal order associated to the given points.

The objective of Gaussian method consists now in choosing the points (τj)0≤j≤l leading to a quadrature rule of maximal order.

Example 2 points τ₀,τ₁ (l= 1), we assume τ₀ =−τ₁=−τ. J^QR(ϕ) =ω0ϕ(−τ) +ω1ϕ(τ).

J^QR(x7→1) =J(x7→1) =⇒ ω₀+ω₁ = 2 J^QR(x7→x) =J(x7→x) =⇒ −ω₀τ+ω₁τ = 0 J^QR(x7→x²) =J(x7→x²) =⇒ ω₀τ²+ω₁τ² = 2 3

=⇒ω₀ =ω₁= 1, τ =

√3 3 .

á order 3

(81)

Principle

−1

Example 2 points τ0,τ1 (l= 1), we assume τ0 =−τ₁=−τ. J^QR(ϕ) =ω0ϕ(−τ) +ω1ϕ(τ).

=⇒ω₀ =ω₁= 1, τ =

√3 3 .

á order 3

(82)

Principle

−1

=⇒ω₀ =ω₁= 1, τ =

√3 3 .

á order 3

(83)

Principle

−1

=⇒ω₀ =ω₁ = 1, τ =

√3 3 .

á order 3

(84)

Principle

−1

=⇒ω₀ =ω₁ = 1, τ =

√3 3 .

á order 3

(85)

A second example, with 3 points

We assume that the points are−τ,0 andτ.

J^QR(ϕ) =ω0ϕ(−τ) +ω1ϕ(0) +ω2ϕ(τ).

J^QR(x7→1) =J(x7→1) =⇒ ω₀+ω₁+ω₂= 2 J^QR(x7→x) =J(x7→x) =⇒ −ω₀τ +ω₂τ = 0 J^QR(x7→x²) =J(x7→x²) =⇒ ω₀τ²+ω₂τ² = 2 3 J^QR(x7→x³) =J(x7→x³) =⇒ −ω₀τ³+ω₂τ³= 0 J^QR(x7→x⁴) =J(x7→x⁴) =⇒ ω0τ⁴+ω2τ⁴ = 2 5

=⇒τ = r3

5, ω0 =ω2 = 5

9, ω1= 8 9.

á order 5

(86)

A second example, with 3 points

We assume that the points are−τ,0 andτ.

J^QR(ϕ) =ω0ϕ(−τ) +ω1ϕ(0) +ω2ϕ(τ).

J^QR(x7→1) =J(x7→1) =⇒ ω₀+ω₁+ω₂= 2 J^QR(x7→x) =J(x7→x) =⇒ −ω₀τ +ω₂τ = 0 J^QR(x7→x²) =J(x7→x²) =⇒ ω₀τ²+ω₂τ² = 2 3 J^QR(x7→x³) =J(x7→x³) =⇒ −ω₀τ³+ω₂τ³= 0 J^QR(x7→x⁴) =J(x7→x⁴) =⇒ ω0τ⁴+ω2τ⁴ = 2 5

=⇒τ = r3

5, ω0 =ω2 = 5

9, ω1= 8 9.

á order 5