Find with p x y Solution: n j=1,j6= k xj Theoremtrenchea/math1070/MATH1070_4...>...

> 5. Numerical Integration

Review of Interpolation

Find pn(x) with pn(xj) = yj , j = 0, 1, 2, . . . , n. Solution:

pn(x) = y0`0(x) + y1`1(x) + . . .+ yn`n(x), `k(x) =∏nj=1,j 6=k

x−xj

xk−xj.

Theorem

Let yj = f(xj), f(x) smooth and pn interpolates f(x) atx0 < x1 < . . . < xn. For any x ∈ (x0, xn) there is a ξ such that

f(x)− pn(x) =f (n+1)(ξ)(n+ 1)!

(x− x0)(x− x1) . . . (x− xn)︸︷︷︸ψ(x)

Example: n = 1, linear:

f(x)− p1(x) =f′′(ξ)2

(x− x0)(x− x1),

p1(x) =y1 − y0

x1 − x0(x− x0) + y0.

5. Numerical Integration Math 1070

> 5. Numerical Integration > 5.1 The Trapezoidal Rule

Mixed rule

Find area under the curve y = f(x)

∫ b

af(x)dx =

N−1∑j=0

∫ xj+1

xj

f(x)dx︸︷︷︸interpolate f(x)on (xj, xj+1) andintegrate exactly

≡N−1∑j=0

(xj+1 − xj)f(xj) + f(xj+1)

2



Trapezoidal Rule

∫ xj+1

xj

f(x) ≈∫ xj+1

xj

p1(x)dx

=∫ xj+1

xj

{f(xj+1) + f(xj)

xj+1 − xj(x− xj) + f(xj)

}dx

= (xj+1 − xj)f(xj) + f(xj+1)

2

T1(f) = (b− a)f(b) + f(a)2

(6.1)



Another derivation of Trapezoidal Rule:

Seek: ∫ xj+1

xj

f(x)dx ≈ wjf(xj) + wj+1f(xj+1)

exact on polynomials of degree 1 (1, x).

f(x) ≡ 1 :∫ xj+1

xj

f(x)dx = xj+1 − xj = wj · 1 + wj+1 · 1

f(x) = x :∫ xj+1

xj

xdx =x2j+1

2−x2j

2= wjxj + wj+1xj+1



Example

Approximate integral

I =∫ 1

0

dx

1 + x

The true value is I = ln(2) ·= 0.693147. Using (6.1), we obtain

T1 =12

(1 +

12

)=

34

= 0.75

and the error is

I − T1(f) ·= −0.0569 (6.2)



To improve on approximation (6.1) when f(x) is not a nearly linear functionon [a, b], break interval [a, b] into smaller subintervals and apply (6.1) on eachsubinterval.If the subintervals are small enough, then f(x) will be nearly linear on eachone.

Example

Evaluate the preceding example by using T1(f) on two subintervals of equallength.

For two subintervals,

I =∫ 1

2

0

dx

1 + x+∫ 1

12

dx

1 + x

·=12

1 + 23

2+

12

23 + 1

2

2

T2 =1724

·= 0.70833

and the error

I − T2·= −0.0152 (6.3)

is about 14 of that for T1 in (6.2).



We derive the general formula for calculations using n subintervals ofequal length h = b−a

n . The endpoints of each subinterval are then

xj = a+ jh, j = 0, 1, . . . , n

Breaking the integral into n subintegrals

I(f) =∫ b

af(x)dx =

∫ xn

x0

f(x)dx

=∫ x1

x0

f(x)dx+∫ x2

x1

f(x)dx+ . . .+∫ xn

xn−1

f(x)dx

≈ hf(x0) + f(x1)2

+ hf(x1) + f(x2)

2+ . . .+ h

f(xn−1) + f(xn)2

The trapezoidal numerical integration rule

Tn(f)=h(

12f(x0)+f(x1)+f(x2)+· · ·+f(xn−1)+

12f(xn)

)(6.4)



With a sequence of increasing values of n, Tn(f) will usually be anincreasingly accurate approximation of I(f).

But which sequence of n should be used?If n is doubled repeatedly, then the function values used in each T2n(f)will include all earlier function values used in the preceding Tn(f).Thus, the doubling of n will ensure that all previously computed informationis used in the new calculation, making the trapezoidal rule less expensive thanit would be otherwise.

T2(f) = h

(f(x0)

2+ f(x1) +

f(x2)2

)with

h =b− a

2, x0 = a, x1 =

a+ b

2, x2 = b.

AlsoT4(f) = h

(f(x0)

2+ f(x1) + f(x2) + f(x3) +

f(x4)2

)with

h =b− a

4, x0 = a, x1 =

3a+ b

4, x2 =

a+ b

4, x3 =

a+ 3b4

, x4 = b

Only f(x1) and f(x3) need to be evaluated.



ExampleWe give calculations of Tn(f) for three integrals

I(1) =∫ 1

0

e−x2dx

·= 0.746824132812427

I(2) =∫ 4

0

dx

1 + x2= tan−1(4) ·= 1.32581766366803

I(3) =∫ 2π

0

dx

2 + cos(x)=

2π√3

·= 3.62759872846844

n I(1) I(2) I(3)

Error Ratio Error Ratio Error Ratio

2 1.55E-2 -1.33E-1 -5.61E-14 3.84E-3 4.02 -3.59E-3 37.0 -3.76E-2 14.98 9.59E-4 4.01 5.64E-4 -6.37 -1.93E-4 195.016 2.40E-4 4.00 1.44E-4 3.92 -5.19E-9 37,600.032 5.99E-5 4.00 3.60E-5 4.00 *64 1.50E-5 4.00 9.01E-6 4.00 *128 3.74E-6 4.00 2.25E-6 4.00 *

The error for I(1), I(2) decreases by a factor of 4 when n doubles, for I(3) the answers for n = 32, 64, 128 were

correct up to the limits due to rounding error on the computer (16 decimal digits).5. Numerical Integration Math 1070

> 5. Numerical Integration > 5.1.1 Simpson’s rule

To improve on T1(f) in (6.1), use quadratic interpolation to approximatef(x) on [a, b]. Let P2(x) be the quadratic polynomial that interpolates f(x)at a, c = a+b

2 and b.

I(f) ≈∫ b

a

P2(x)dx (6.5)

=∫ b

a

((x−c)(x−b)(a−c)(a−b)

f(a) +(x−a)(x−b)(c−a)(c−b)

f(c) +(x−a)(x−c)(b−a)(b−c)

f(b))dx

This can be evaluated directly. But it is easier with a change of variables.Let h = b−a

2 and u = x− a. Then∫ b

a

(x−c)(x−b)(a−c)(a−b)

dx =1

2h2

∫ a+2h

a

(x− c)(x− b)dx

=1

2h2

∫ 2h

0

(u− h)(u− 2h)du =1

2h2

(u3

3− 3

2u2h+ 2h2u

)∣∣∣∣2h

0

=h

3

and

S2(f) =h

3

(f(a) + 4f(

a+ b

2) + f(b)

)(6.6)



Example

I =∫ 1

0

dx

1 + x

Then h = b−a2 = 1

2 and

S2(f) =12

3

(1 + 4(

23) +

12

)=

2536

·= 0.69444 (6.7)

and the error is

I − S2 = ln(2)− S2·= −0.00130

while the error for the trapezoidal rule (the number of functionevaluations is the same for both S2 and T2) was

I − T2·= −0.0152.

The error in S2 is smaller than that in (6.3) for T2 by a factor of 12,a significant increase in accuracy.



0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10.5

0.55

0.6

0.65

0.7

0.75

0.8

0.85

0.9

0.95

1

Figure: An illustration of Simpson’s rule (6.6), y = f(x), y = P2(x)5. Numerical Integration Math 1070


The rule S2(f) will be an accurate approximation to I(f) if f(x) is nearlyquadratic on [a, b]. For the other cases, proceed in the same manner as forthe trapezoidal rule.Let n be an even integer, h = b−a

n and define the evaluation points for f(x)by

xj = a+ jh, j = 0, 1, . . . , n

We follow the idea from the trapezoidal rule, but break [a, b] = [x0, xn] intolarger intervals, each containing three interpolation node points:

I(f) =∫ b

a

f(x)dx =∫ xn

x0

f(x)dx

=∫ x2

x0

f(x)dx+∫ x4

x2

f(x)dx+ . . .+∫ xn

xn−2

f(x)dx

≈ h

3[f(x0) + 4f(x1) + f(x2)] +

h

3[f(x2) + 4f(x3) + f(x4)]

+ . . .+h

3[f(xn−2) + 4f(xn−1) + f(xn)]



Simpson’s rule:

Sn(f) =h

3(f(x0)+4f(x1)+2f(x2)+4f(x3)+2f(x4) (6.8)

+· · ·+2f(xn−2)+4f(xn−1)+f(xn))

It has been among the most popular numerical integrationmethods for more than two centuries.



ExampleEvaluate the integrals

I(1) =∫ 1

0

e−x2dx

·= 0.746824132812427

I(2) =∫ 4

0

dx

1 + x2= tan−1(4) ·= 1.32581766366803

I(3) =∫ 2π

0

dx

2 + cos(x)=

2π√3

·= 3.62759872846844

n I(1) I(2) I(3)

Error Ratio Error Ratio Error Ratio

2 -3.56E-4 8.66E-2 -1.264 -3.12E-5 11.4 3.95E-2 2.2 1.37E-1 -9.28 -1.99E-6 15.7 1.95E-3 20.3 1.23E-2 11.216 -1.25E-7 15.9 4.02E-6 485.0 6.43E-5 191.032 -7.79E-9 16.0 2.33E-8 172.0 1.71E-9 37,600.064 -4.87E-10 16.0 1.46E-9 16.0 *128 -3.04E-11 16.0 9.15E-11 16.0 *

For I(1), I(2), the ratio by which the error decreases approaches 16.

For I(3), the errors converge to zero much more rapidly


> 5. Numerical Integration > 5.2 Error formulas

Theorem

Let f ∈ C2[a, b], n ∈ N. The error in integrating

I(f) =∫ b

af(x)dx

using the trapezoidal ruleTn(f) = h

[12f(x0) + f(x1) + f(x2) + · · ·+ f(xn−1) + 1

2f(xn)]

isgiven by

ETn ≡ I(f)− Tn(f) =−h2(b− a)

12f ′′(cn) (6.9)

where cn is some unknown point in a, b, and h = b−an .



Theorem

Suppose that f ∈ C2[a, b], and h = maxj(xj+1 − xj). Then∣∣∣∣∣∣∫ b

af(x)dx−

∑j

(xj+1−xj)f(xj+1)+f(xj)

2

∣∣∣∣∣∣ ≤ b− a12

h2 maxa≤x≤b

|f ′′(x)|.

Proof: Let Ij be the jth subinterval and p1 = linear interpolant on Ijat xj , xj+1.

f(x)− p1(x) =f ′′(x)

2(x− xj)(x− xj+1)︸︷︷︸

ψ(x)

.

Local error:∣∣∣∣∣∫ xj+1

xj

f(x)dx− (xj+1−xj)f(xj) + f(xj+1)

2

∣∣∣∣∣ =5. Numerical Integration Math 1070


proof

=

∣∣∣∣∣∫ xj+1

xj

f ′′(ξ)2!

(x− xj)(x− xj+1)

∣∣∣∣∣≤ 1

2

∫ xj+1

xj

|f ′′(ξ)||x− xj | |x− xj+1|dx

≤ 12

maxa≤x≤b

|f ′′(x)|∫ xj+1

xj

(x− xj)(xj+1 − x)dx

=12

maxa≤x≤b

|f ′′(x)|(xj+1 − xj)3

6.

Hence

|local error| ≤ 112

maxa≤x≤b

|f ′′(x)| · h3j , hj = xj+1 − xj .



Example

Recall the exampleI(f) =

∫ 1

0

dx

1 + x= ln 2

Here f(x) = 11+x , [a, b] = [0, 1], and f ′′(x) = 2

(1+x)3 . Then by (6.9)

ETn (f) = −h

2

12f ′′(cn), 0 ≤ cn ≤ 1, h =

1n.

This cannot be computed exactly since cn is unknown. But

max0≤x≤1

|f ′′(x)| = max0≤x≤1

2(1 + x)3

= 2

and therefore

|ETn (f)| ≤ h2

12(2) =

h2

6For n = 1 and n = 2 we have

|ET1 (f)︸︷︷︸

−0.0569

| ≤ 16

·= 0.167, |ET2 (f)︸︷︷︸

−0.0152

| ≤(

12

)26

·= 0.0417.



A possible weakness in the trapezoidal rule can be inferred from theassumption of the theorem for the error.If f(x) does not have two continuous derivatives on [a, b], thenTn(f) does converge more slowly??

YES

for some functions, especially if the first derivative is not continuous.


> 5. Numerical Integration > 5.2.1 Asymptotic estimate of Tn(f)

The error formula (6.9)

ETn (f) ≡ I(f)− Tn(f) = −h2(b−a)12 f ′′(cn)

can only be used to bound the error, because f ′′(cn) is unknown.

This can be improved by a more careful consideration of the error formula.

A central element of the proof of (6.9) lies in the local error∫ α+h

αf(x)dx− hf(α) + f(α+ h)

2= −h

3

12f ′′(c) (6.10)

for some c ∈ [α, α+ h].



Recall the derivation of the trapezoidal rule Tn(f) and use the local error(6.10):

ETn (f) =

∫ b

a

f(x)dx− Tn(f) =∫ xn

x0

f(x)dx− Tn(f)

=∫ x1

x0

f(x)dx− hf(x0) + f(x1)2

+∫ x2

x1

f(x)dx− hf(x1) + f(x2)2

+ · · ·+∫ xn

xn−1

f(x)dx− hf(xn−1) + f(xn)2

= −h3

12f ′′(γ1)−

h3

12f ′′(γ2)− · · · − −

h3

12f ′′(γn)

with γ1 ∈ [x0, x1], γ2 ∈ [x1, x2], . . . γn ∈ [xn−1, xn], and

ETn (f) = −h

2

12

(hf ′′(γ1) + · · ·+ hf ′′(γn)︸︷︷︸

=(b−a)f ′′(cn)

), cn ∈ [a, b].



To estimate the trapezoidal error, observe thathf ′′(γ1) + · · ·+ hf ′′(γn) is a Riemann sum for the integral∫ b

af ′′(x)dx = f ′(b)− f ′(a) (6.11)

The Riemann sum is based on the partition

[x0, x1], [x1, x2], . . . , [xn−1, xn] of [a, b].

As n→∞, this sum will approach the integral (6.11). With (6.11),we find an asymptotic estimate (improves as n increases)

ETn (f) ≈ −h2

12(f ′(b)− f ′(a)) =: ETn (f). (6.12)

As long as f ′(x) is computable, ETn (f) will be very easy to compute.



Example

Again consider I =∫ 1

0

dx

1 + x.

Then f ′(x) = − 1(1 + x)2

, and the asymptotic estimate (6.12) yields

the estimate

ETn =−h2

12

(−1

(1 + 1)− −1

(1 + 0)2

)=−h2

16, h =

1n

and for n = 1 and n = 2

ET1 = − 116 = −0.0625, ET2

·= −0.0156I − T1

·= −0.0569, I − T2·= −0.0152



The estimate ETn (f) = −h2

12 (f ′(b)− f ′(a)) has several practical

advantages over the earlier formula (6.9) ETn (f) = −h2(b−a)12 f ′′(cn).

1 It confirms that when n is doubled (or h is halved), the error decreasesby a factor of about 4, provided that f ′(b)− f ′(a) 6= 0. This agreeswith the results for I(1) and I(2).

2 (6.12) implies that the convergence of Tn(f) will be more rapid whenf ′(b)− f ′(a) = 0. This is a partial explanation of the very rapidconvergence observed with I(3)

3 (6.12) leads to a more accurate numerical integration formula

by taking ETn (f) into account:

I(f)− Tn(f) ≈− h2

12(f ′(b)− f ′(a))

I(f) ≈ Tn(f)− h2

12(f ′(b)− f ′(a)) := CTn(f), (6.13)

the corrected trapezoidal rule



Example

Recall the integral I(1), I =∫ 1

0e−x2

dx·= 0.74682413281243

n I − Tn(f) En(f) CTn(f) I − CTn(f) Ratio2 1.545E-4 1.533E-2 0.746698561877 1.26E-44 3.840E-3 3.832E-3 0.746816175313 7.96E-6 15.88 9.585E-4 9.580E-4 0.746823634224 4.99E-7 16.016 2.395E-4 2.395E-4 0.746824101633 3.12E-8 16.032 5.988E-5 5.988E-5 0.746824130863 1.95E-9 16.064 1.497E-5 1.497E-5 0.746824132690 2.22E-10 16.0

Table: Example of CTn(f) and En(f)Note that the estimate

ETn (f) =

h2e−1

6, h =

1n

is a very accurate estimator of the true error.Also, the error in CTn(f) converges to zero at a more rapid rate than doesthe error for Tn(f).When n is doubled, the error in CTn(f) decreases by a factor of about 16.


> 5. Numerical Integration > 5.2.2 Error formulae for Simpson’s rule

Theorem

Assume f ∈ C4[a, b], n ∈ N. The error in using Simpson’s rule is

ESn (f) = I(f)− Sn(f) = −h

4(b− a)180

f (4)(cn) (6.14)

with cn ∈ [a, b] an unknown point, and h = b−an . Moreover, this error can be

estimated with the asymptotic error formula

ESn (f) = − h4

180(f ′′′(b)− f ′′′(a)) (6.15)

Note that (6.14) says that Simpson’s rule is exact for all f(x) that arepolynomials of degree ≤ 3, whereas the quadratic interpolation on whichSimpson’s rule is based is exact only for f(x) a polynomial of degree ≤ 2.

The degree of precision being 3 leads to the power h4 in the error, rather

than the power h3, which would have been produced on the basis of the error

in quadratic interpolation.

The higher power of h4, and

the simple form of the method

that historically have caused Simpson’s rule to become the most popular numerical integration rule.



Example

Recall (6.7) where S2(f) was applied to I =∫ 1

0dx

1+x :

S2(f) =12

3

(1 + 4(

23) +

12

)=

2536

·= 0.69444

f(x) =1

1 + x, f3(x) =

−6(1 + x)4

, f (4)(x) =24

(1 + x)5

The exact error is given by

ESn (f) = − h4

180f (4)(cn), h =

1n

for some 0 ≤ cn ≤ 1. We can bound it by

|ESn (f)| ≤ h4

18024 =

2h4

15The asymptotic error is given by

ESn (f) = − h4

180

(−6

(1 + 1)4− −6

(1 + 0)4

)= −h

4

32

For n = 2, ESn

·= −0.00195; the actual error is −0.00130.



The behavior in I(f)− Sn(f) can be derived from (6.15):

ESn (f) = − h4

180(f ′′′(b)− f ′′′(a)

),

i.e.,when n is doubled, h is halved, and h4 decreases by of factor of 16.

Thus, the error ESn (f) should decrease by the same factor, providedthat f ′′′(a) 6= f ′′′(b). This is the error observed with integrals I(1) andI(2).

When f ′′′(a) = f ′′′(b), the error will decrease more rapidly, which is apartial explanation of the rapid convergence for I(3).



The theory of asymptotic error formulae

En(f) = En(f) (6.16)

such as for ETn (f) and ESn (f), says that (6.16) will vary with theintegrand f , which is illustrated with the two cases I(1) and I(2).

From (6.14) and (6.15) we infer that Simpson’s rule will not performas well if f(x) is not four times continuously differentiable, on [a, b].



Example

Use Simpson’s rule to approximate

I =∫ 1

0

√xdx =

23.

n Error Ratio

2 2.860E-2

4 1.014E-2 2.82

8 3.587E-3 2.83

16 1.268E-3 2.83

32 4.485E-4 2.83

Table: Simpson’s rule for√x

The column ”Ratio” show the convergence is much slower.



As was done for the trapezoidal rule, a corrected Simpson’s rule canbe defined:

CSn(f) = Sn(f)− h4

180(f ′′′(b)− f ′′′(a)

)(6.17)

This will usually will be more accurate approximation than Sn(f).


> 5. Numerical Integration > 5.2.3 Richardson extrapolation

The error estimates for Trapezoidal rule (6.12)

ETn (f) ≈ −h2

12(f ′(b)− f ′(a)

)and Simpson’s rule (6.15)

ESn (f) = − h4

180(f ′′′(b)− f ′′′(a)

)are both of the form

I − In ≈c

np(6.18)

where In denotes the numerical integral and h = b−an .

The constants c and p vary with the method and the function. Withmost integrands f(x), p = 2 for the trapezoidal rule and p = 4 forSimpson’s rule.There are other numerical methods that satisfy (6.18), with other value of p

and c. We use (6.18) to obtain a computable estimate of the error I − In,

without needing to know c explicitly.



Replacing n by 2n

I − I2n ≈c

2pnp(6.19)

and comparing to (6.18)

2p(I − I2n) ≈c

np≈ I − In

and solving for I gives the Richardson’s extrapolation formula

(2p − 1)I ≈ 2pI2n − In

I≈ 12p − 1

(2pI2n − In) ≡ R2n (6.20)

R2n is an improved estimate of I, based on using In, I2n, p, and theassumption (6.18). How much more accurate than I2n depends on thevalidity of (6.18), (6.19).



To estimate the error in I2n, compare it with the more accurate valueR2n

I − I2n ≈ R2n − I2n =1

2p − 1(2pI2n − In)− I2n

I − I2n ≈1

2p − 1(I2n − In) (6.21)

This is Richardson’s error estimate.



Example

Using the trapezoidal rule to approximate

I =∫ 1

0e−x

2dx

·= 0.74682413281243

we haveT2

·= 0.7313702518, T4·= 0.7429840978

Using (6.20) I ≈ 12p−1(2pI2n − In) with p = 2 and n = 2, we obtain

I ≈ R4 =13(4I4 − I2) =

13(4T4 − T2)

·= 0.7468553797

The error in R4 is −0.0000312; and from a previous Table, R4 is moreaccurate than T32. To estimate the error in T4, use (6.21) to get

I − T4 ≈13(T4 − T2)

·= 0.00387

The actual error in T4 is 0.00384; and thus (6.21) is a very accurate error

estimate.


> 5. Numerical Integration > 5.2.4 Periodic Integrands

Definition

A function f(x) is periodic with period τ if

f(x) = f(x+ τ), ∀x ∈ R (6.22)

and this relation should not be true with any smaller value of τ .

For example,f(x) = ecos(πx)

is periodic with periodic τ = 2.

If f(x) is periodic and differentiable, then itsderivatives are also periodic with period τ.



Consider integrating

I =∫ b

af(x)dx

with trapezoidal or Simpson’s rule, and assume that b− a is andinteger multiple of the period τ .Assume f(x) ∈ C∞[a, b] (has derivatives of any order).

Then for all derivatives of f(x), the periodicity of f(x) implies that

f (k)(a) = f (k)(b), k ≥ 0 (6.23)

If we now look at the asymptotic error formulae for the trapezoidaland Simpson’s rules, they become zero because of (6.23).Thus, the error formulae ETn (f) and ESn (f) should converge to zeromore rapidly when f(x) is a periodic function, provided b− a is aninteger multiple of the period of f .



The asymptotic error formulae ETn (f) and ESn (f) can be extended tohigher-order terms in h, using the Euler-MacLaurin expansion andthe higher-order terms are multiples of f (k)(b)− f (k)(a) for all oddintegers k ≥ 1. Using this, we can prove that the errors ETn (f) andESn (f) converge to zero even more rapidly than was implied by theearlier comments for f(x) periodic.

Note that the trapezoidal rule is the preferred integration rulewhen we are dealing with smooth periodic integrands.The earlier results for the integral I(3) illustrate this.



Example

The ellipse with boundary (xa

)2

+(yb

)2

= 1

has area πab. For the case in which the area is π (and thus ab = 1), we studythe variation of the perimeter of the ellipse as a and b vary.

The ellipse has the parametric representation

(x, y) = (a cos θ, b sin θ), 0 ≤ θ ≤ 2π (6.24)

By using the standard formula for the perimeter, and using the symmetry ofthe ellipse about the x-axis, the perimeter is given by

P = 2∫ π

0

√(dx

dθ

)2

+(dy

dθ

)2

dθ

= 2∫ π

0

√a2 sin2 θ + b2 cos2 θdθ



Since ab = 1, we write this as

P (b) = 2∫ π

0

√1b2

sin2 θ + b2 cos2 θdθ

=2b

∫ π

0

√(b4 − 1) cos2 θ + 1dθ (6.25)

We consider only the case with 1 ≤ b <∞. Since the perimeters forthe two ellipses(x

a

)2+(yb

)2= 1 and

(xb

)2+(ya

)2= 1

are equal, we can always consider the case in which the y-axis of theellipse is larger than or equal to its x-axis; and this also shows

P

(1b

)= P (b), b > 0 (6.26)



The integrand of P (b)

f(θ) =2b

[(b4 − 1) cos2 θ + 1

] 12

is periodic with period π. As discussed above, the trapezoidal rule is the

natural choice for numerical integration of (6.25). Nonetheless, there is a

variation in the behaviour of f(θ) as b varies, and this will affect the accuracy

of the numerical integration.

0 0.5 1 1.5 2 2.5 3 3.50

2

4

6

8

10

12

14

16

π/2

Figure:The graph of integrand f(θ) : b = 2, 5, 8



n b = 2 b = 5 b = 88 8.575517 19.918814 31.690628

16 8.578405 20.044483 31.953632

32 8.578422 20.063957 32.008934

64 8.578422 20.065672 32.018564

128 8.578422 20.065716 32.019660

256 8.578422 20.065717 32.019709

Table: Trapezoidal Rule Approximation of (6.25)Note that as b increases, the trapezoidal rule converges more slowly.This is due to the integrand f(θ) changing more rapidly as b increases.For large b, f(θ) changes very rapidly in the vicinity of θ = 1

2π; andthis causes the trapezoidal rule to be less accurate than when b issmaller, near 1. To obtain a certain accuracy in the perimeter P (b), wemust increase n as b increases.



0 1 2 3 4 5 6 7 8 9 105

10

15

20

25

30

35

40

45

z=P(b)

Figure: The graph of perimeter function P (b) for ellipseThe graph of P (b) reveals that P (b) ≈ 4b for large b. Returning to (6.25), wehave for large b

P (b) ≈ 2b

∫ π

0

(b4 cos2 θ

) 12 dθ

≈ 2bb2∫ π

0

| cos θ|dθ = 4b

We need to estimate the error in the above approximation to know when we

can use it to replace P (b); but it provides a way to avoid the integration of

(6.25) for the most badly behaved cases.


> 5. Numerical Integration > Review and more

Review

∫ xj+1

xj

f(x)dx ≈∫ xj+1

xj

pn(x)dx︸︷︷︸≡Ij

pn(x) interpolates at x(0)j , x

(1)j , . . . , x

(n)j points on [xj , xj+1]

Local error:∫ xj+1

xj

f(x)dx− Ij =∫ xj+1

xj

f (n+1)(ξ)

(n+ 1)!ψ(x)dx

(integrand is error in interpolation)

where ψ(x) = (x− x(0)j )(x− x(1)

j ) · · · (x− x(n)j ).



x_j x_{j+1}−6

−4

−2

0

2

4

6x 106

xj ≤ x ≤ x

j+1

ψ(x

)

Conclusion: exact on Pn1 |local error| ≤ Cmax |f (n+1)(x)|hn+2

2 |global error| ≤ Cmax |f (n+1)(x)|hn+1(b− a)



Observation:If ξ is a point on (xj , xj+1), then

g(ξ) = g(xj+ 12) +O(h)

(if g′ is continuous) i.e.,

g(ξ) = g(xj+ 12) + (ξ − xj+ 1

2)︸︷︷︸

≤h

g′(η)

︸︷︷︸O(h)



Local error:

1(n+ 1)!

∫ xj+1

xj

f (n+1)(ξ)︸︷︷︸=f (n+1)(x

j+12)+O(h)

ψ(x)dx

=1

(n+ 1)!f (n+1)(xj+ 1

2)∫ xj+1

xj

ψ(x)dx︸︷︷︸Dominant Term O(hn+2)

+

C(n+1)!

max |f (n+2)|hn+3︷︸︸︷1

(n+ 1)!

∫ xj+1

xj

f (n+2)(η(x))︸︷︷︸take max out

(ξ − xj+ 12)︸︷︷︸

integrate

ψ(x)dx

︸︷︷︸Higher Order Terms O(hn+3)



The dominant term: case N = 1, Trapezoidal Rule

x_j x_{j+1}−0.2

−0.15

−0.1

−0.05

0

0.05ψ(x) = (x−x

j)(x−x

j+1)



The dominant term: case N = 2, Simpson’s Rule

ψ(x) = (x− xj)(x− xj+ 12)(x− xj+1)⇒

∫ xj+1

xj

ψ(x)dx = 0

x_j x_{j+1/2} x_{j+1}−0.2

−0.15

−0.1

−0.05

0

0.05

0.1

0.15

0.2ψ(x) = (x−x

j)(x−x

j+1/2)(x−x

j+1)



The dominant term: case N = 3, Simpson’s 3/8’s Rule

Local error = O(h5)

x_j x_{j+1/3} x_{j+2/3} x_{j+1}−0.06

−0.04

−0.02

0

0.02

0.04

0.06

0.08

0.1ψ(x) = (x−x

j)(x−x

j+1/3)(x−x

j+2/3)(x−x

j+1)



The dominant term: case N = 4

∫ψ(x)dx = 0⇒ local error = O(h7)

x_j x_{j+1}−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4



Simpson’s Rule

Is exact on P2 (and P3 actually)

xj+1/2

= (xj+ x

j+1)/2

← xj+1/2

Seek:∫ xj+1

xj

f(x)dx ≈ wjf(xj) + wj+1/2f(xj+1/2) + wj+1f(xj+1)



Exact on 1, x, x2:

1 :∫ xj+1

xj

1dx = xj+1 − xj = wj · 1 + wj+1/2 · 1 + wj+1 · 1,

x :∫ xj+1

xj

xdx =x2j+1

2−x2j

2= wjxj + wj+1/2xj+1/2 + wj+1xj+1,

x2 :∫ xj+1

xj

x2dx =x3j+1

3−x3j

3= wjx

2j + wj+1/2x

2j+1/2 + wj+1x

2j+1.

3× 3 linear system:

wj =16(xj+1 − xj);

wj+1/2 =46(xj+1 − xj);

wj+1 =16(xj+1 − xj).



Theorem

h = max(xj+1 − xj), I(f) = Simpson’s rule approximation:∣∣∣∣∫ b

af(x)dx− I(f)

∣∣∣∣ ≤ b− a2880

h4 maxa≤x≤b

∣∣∣f (4)(x)∣∣∣ .

Trapezoid rule versus Simpson’s rule

Cost in TR

Cost in SR=

2 function evaluations/integral × no. intervals

3 function evaluations/integral × no. intervals=

23;

(reducible to 12 if storing the previous values)

Accuracy in TR

Accuracy in SR=h2 b−a

12

h4 b−a2880

=240h2

.

E.g. for h = 1100 ⇒ 2.4× 106, i.e. SR is more accurate than TR by factor of 2.4× 106.



What if there is round-off error?

Suppose we use the method with

f(xj)computed = f(xj)true ± εj , εj = O(machine precision︸︷︷︸=ε

)

∫ b

af(x)dx ≈

∑j

(xj+1 − xj)f(xj+1)computed + f(xj)computed

2

=∑j

(xj+1 − xj)f(xj+1)± εj+1 + f(xj)± εj

2

=∑j

(xj+1 − xj)f(xj+1) + f(xj)

2︸︷︷︸value in exact arithmetic

+∑j

(xj+1 − xj)±εj+1 ± εj

2︸︷︷︸contribution of round-off error≤ε(b−a)


> 5. Numerical Integration > 5.3 Gaussian Numerical Integration

The numerical methods studied in the first 2 sections were based onintegrating

1 linear (trapezoidal rule) and

2 quadratic (Simpson’s rule)

and the resulting formulae were applied on subdivisions of ever smallersubintervals.

We consider now a numerical method based on exact integration ofpolynomials of increasing degree; no subdivision of the integrationinterval [a, b] is used.Recall the Section 4.4 of Chapter 4 on approximation of functions.



Let f(x) ∈ C[a, b].Then ρn(f) denotes the smallest error bound that can be attained inapproximating f(x) with a polynomial p(x) of degree ≤ n on the giveninterval a ≤ x ≤ b.The polynomial mn(x) that yields this approximation is called theminimax approximation of degree n for f(x)

maxa≤x≤b

|f(x)−mn(x)| = ρn(f) (6.27)

and ρn(f) is called the minimax error.



Example

Let f(x) = e−x2

for x ∈ [0, 1]

n ρn(f) n ρn(f)1 5.30E-2 6 7.82E-6

2 1.79E-2 7 4.62E-7

3 6.63E-4 8 9.64E-8

4 4.63E-4 9 8.05E-9

5 1.62E-5 10 9.16E-10

Table: Minimax errors for e−x2, 0 ≤ x ≤ 1

The minimax errors ρn(f) converge to zero rapidly, although not at auniform rate.



If we have a

numerical integration formula to integrate low- tomoderate-degree polynomials exactly,

then the hope is that

the same formula will integrate other functions f(x) almostexactly,

if f(x) is well ”approximable” by such polynomials.



To illustrate the derivation of such integration formulae, we restrict ourselvesto the integral

I(f) =∫ 1

−1

f(x)dx.

The integration formula is to have the general form:(Gaussian numerical integration method)

In(f) =n∑j=1

wjf(xj) (6.28)

and we require that

the nodes {x1, . . . , xn} and

weights {w1, . . . , wn}be so chosen that

In(f) = I(f)for all polynomials f(x) of as large degree as possible.



Case n = 1

The integration formula has the form∫ 1

−1

f(x)dx ≈ w1f(x1) (6.29)

Using f(x) ≡ 1, and forcing (6.29) 2 = w1

Using f(x) = x 0 = w1x1

which implies x1 = 0. Hence (6.29) becomes∫ 1

−1

f(x)dx ≈ 2f(0) ≡ I1(f) (6.30)

This is the midpoint formula, and is exact for all linear polynomials.

To see that (6.30) is not exact for quadratics, let f(x) = x2. Then theerror in (6.30) is ∫ 1

−1

x2dx− 2(0)2 =236= 0,

hence (6.30) has degree of precision 1.



Case n = 2

The integration formula is∫ 1

−1f(x)dx ≈ w1f(x1) + w2f(x2) (6.31)

and it has four unspecified quantities: x1, x2, w1, w2. To determinethese, we require (6.31) to be exact for the four monomials

f(x) = 1, x, x2, x3.

obtaining 4 equations

2 = w1 + w2

0 = w1x1 + w2x223 = w1x

21 + w2x

22

0 = w1x31 + w2x

32



Case n = 2

This is a nonlinear system with a solution

w1 = w2 = 1, x1 = −√

33, x2 =

√3

3(6.32)

and another one based on reversing the signs of x1 and x2.This yields the integration formula∫ 1

−1f(x)dx ≈ f

(−√

33

)+ f

(√3

3

)≈ I2(f) (6.33)

which has degree of precision 3 (exact on all polynomials of degree 3and not exact for f(x) = x4).



Example

Approximate

I =∫ 1

−1exdx = e− e−1 ·= 2.5304024

Using ∫ 1

−1f(x)dx ≈ f

(−√

33

)+ f

(√3

3

)≈ I2(f)

we get

I2 = e−√

33 + e−

√3

3·= 2.2326961

I − I2·= 0.00771

The error is quite small, considering we are using only 2 node points.



Case n > 2

We seek the formula (6.28)

In(f) =n∑

j=1

wjf(xj)

which has 2n points unspecified parameters x1, . . . , x,w1, . . . , wn, by forcingthe integration formula to be exact for 2n monomials

f(x) = 1, x, x2, . . . , x2n−1

In turn, this forces In(f) = I(f) for all polynomials f of degree ≤ 2n− 1.This leads to the following system of 2n nonlinear equations in 2nunknowns:

2 = w1 + w2 + . . .+ wn

0 = w1x1 + w2x2 + . . .+ wnxn23 = w1x

21 + w2x

22 + . . .+ wnx

2n

...2

2n−1 = w1x2n−21 + w2x

22 + . . .+ wnx

2n−2n

0 = w1x2n−11 + w2x

22 + . . .+ wnx

2n−1n

(6.34)

The resulting formula In(f) has degree of precision 2n− 1.



Solving this system is a formidable problem. The nodes {xi} and weights{wi} have been calculated and collected in tables for most commonly usedvalues of n.

n xi wi2 ± 0.5773502692 1.03 ± 0.7745966692 0.5555555556

0.0 0.88888888894 ± 0.8611363116 0.3478548451

± 0.3399810436 0.65214515495 ± 0.9061798459 0.2369268851

± 0.5384693101 0.47862867050.0 0.5688888889

6 ± 0.9324695142 0.1713244924± 0.6612093865 0.3607651730± 0.2386191861 0.4679139346

7 ± 0.9491079123 0.1294849662± 0.7415311856 0.2797053915± 0.4058451514 0.3818300505

0.0 0.41795918378 ± 0.9602898565 0.1012285363

± 0.7966664774 0.2223810345± 0.5255324099 0.3137066459± 0.1834346425 0.3626837834

Table: Nodes and weights for Gaussian quadrature formulae



There is also another approach to the development of the numericalintegration formula (6.28), using the theory of orthogonalpolynomials.From that theory, it can be shown that the nodes {x1, . . . , xn} are thezeros of the Legendre polynomials of degree n on the interval [−1, 1].Recall that these polynomials were introduced in Section 4.7. Forexample,

P2(x) =12(3x2 − 1)

and its roots are the nodes given in (6.32)

x1 = −√

33 , x2 =

√3

3 .Since the Legendre polynomials are well known, the nodes {xj} can befound without any recourse to the nonlinear system (6.34).



The sequence of formulae (6.28) is called Gaussian numerical integrationmethod. From its definition, In(f) uses n nodes, and it is exact for all

polynomials of degree ≤ 2n− 1. In(f) is limited to∫ 1

−1f(x)dx, an integral

over [−1, 1]. But this limitation is easily removed.Given an integral

I(f) =∫ b

a

f(x)dx (6.35)

introduce the linear change of variable

x =b+ a+ t(b− a)

2, −1 ≤ t ≤ 1 (6.36)

transforming the integral to

I(f) =b− a

2

∫ 1

−1

f(t)dt (6.37)

with

f(t) = f

(b+ a+ t(b− a)

2

)Now apply In to this new integral.



Example

Apply Gaussian numerical integration to the three integralsI(1) =

∫ 10 e

−x2dx, I(2) =

∫ 40

dx1+x2 , I(3) =

∫ 2π0

dx2+cos(x) , which were

used as examples for the trapezoidal and Simpson’s rules.

All are reformulated as integrals over [−1, 1]. The error results are

n Error in I(1) Error in I(2) Error in I(3)

2 2.29E-4 -2.33E-2 8.23E-1

3 9.55E-6 -3.49E-2 -4.30E-1

4 -3.35E-7 1.90E-3 1.77E-1

5 6.05E-9 1.70E-3 -8.12E-2

6 -7.77E-11 2.74E-4 3.55E-2

7 7.89E-13 -6.45E-5 -1.58E-2

10 * 1.27E-6 1.37E-3

15 * 7.40E-10 -2.33E-5

20 * * 3.96E-7



If these results are compared to those of trapezoidal and Simpson’srule, thenGaussian integration of I(1) and I(2) is much more efficient thanthe trapezoidal rules.But then integration of the periodic integrand I(3) is not as efficientas with the trapezoidal rule.These results are also true for most other integrals.

Except for periodic integrands, Gaussian numerical integration isusually much more accurate than trapezoidal and Simpson rules.

This is even true with many integrals in which the integrand does nothave a continuous derivative.



Example

Use Gaussian integration onI =

∫ 1

0

√xdx =

23

The results are n I − In Ratio2 -7.22E-34 -1.16E-3 6.28 -1.69E-4 6.916 -2.30E-5 7.432 -3.00E-6 7.664 -3.84E-7 7.8

where n is the number of node points. The ratio column is defined asI − I 1

2 n

I − Inand it shows that the error behaves like

I − In ≈c

n3(6.38)

for some c. The error using Simpson’s rule has an empirical rate of

convergence proportional to only 1n1.5 , a much slower rate than (6.38).



A result that relates the minimax error to the Gaussian numerical integrationerror.

Theorem

Let f ∈ C[a, b], n ≥ 1. Then, if we aply Gaussian numerical integration to

I =∫ b

af(x)dx, the error In satisfies

|I(f)− In(f)| ≤ 2(b− a)ρ2n−1(f) (6.39)

where ρ2n−1(f) is the minimax error of degree 2n− 1 for f(x) on [a, b].



Example

Using the tablen ρn(f) n ρn(f)1 5.30E-2 6 7.82E-6

2 1.79E-2 7 4.62E-7

3 6.63E-4 8 9.64E-8

4 4.63E-4 9 8.05E-9

5 1.62E-5 10 9.16E-10apply (6.39) to

I =∫ 1

0e−x

2dx

For n = 3, the above bound implies

|I − I3| ≤ ρ5(e−x2) ·= 3.24× 10−5.

The actual error is 9.95E − 6.



Gaussian numerical integration is not as simple to use as are thetrapezoidal and Simpson rules, partly because the Gaussian nodes andweights do not have simple formulae and also because the error isharder to predict.Nonetheless, the increase in the speed of convergence is so rapid anddramatic in most instances that the method should always beconsidered seriously when one is doing many integrations.Estimating the error is quite difficult, and most people satisfythemselves by looking at two or more succesive values.If n is doubled, then repeatedly comparing two successive values, Inand I2n, is almost always adequate for estimating the error in In

I − In ≈ I2n − In

This is somewhat inefficient, but the speed of convergence in In is sorapid that this will still not diminish its advantage over most othermethods.


> 5. Numerical Integration > 5.3.1 Weighted Gaussian Quadrature

A common problem is the evaluation of integrals of the form

I(f) =∫ b

aw(x)f(x)dx (6.40)

with f(x) a “well-behaved” function and w(x) a possibly (and often)ill-behaved function. Gaussian quadrature has been generalized tohandle such integrals for many functions w(x). Examples include∫ 1

−1

f(x)√1− x2

,

∫ 1

0

√xf(x)dx,

∫ 1

0f(x) ln(

1x

)dx.

The function w(x) is called a weight function.



We begin by imitating the development given earlier in this section,and we do so for the special case of

I(f) =∫ 1

0

f(x)√xdx

in which w(x) = 1√x.

As before, we seek numerical integration formulae of the form

In(f) =n∑j=1

wjf(xj) (6.41)

and we require that the nodes {x1, . . . , xn} and the weights{w1, . . . , wn} be so chosen that In(f) = I(f) for polynomials f(x)of as large as possible.



Case n = 1

The integration formula has the form∫ 1

0

f(x)√xdx ≈ w1f(x1)

We force equality for f(x) = 1 and f(x) = x. This leads to equations

w1 =∫ 1

0

1√xdx = 2

w1x1 =∫ 1

0

x√xdx =

23

Solving for w1 and x1, we obtain the formula∫ 1

0

f(x)√xdx ≈ 2f(

13) (6.42)

and it has the degree of precision 1.



Case n = 2The integration formula has the form∫ 1

0

f(x)√xdx ≈ w1f(x1) + w2f(x2) (6.43)

We force equality for f(x) = 1, x, x2, x3. This leads to equations

w1 + w2 =∫ 1

0

1√xdx = 2

w1x1 + w2x2 =∫ 1

0

x√xdx =

23

w1x21 + w2x

22 =

∫ 1

0

x2

√xdx =

25

w1x31 + w2x

32 =

∫ 1

0

x3

√xdx =

27

This has the solution

x1 = 37 −

235

√30 ·= 0.11559, x2 = 3

7 + 235

√30 ·= 0.74156

w1 = 1 + 118

√30 ·= 1.30429, w2 = 1− 1

18

√30 ·= 0.69571

The resulting formula (6.43) has degree of precision 3.



Case n > 2

We seek formula (6.41), which has 2n unspecified parameters, x1, . . . , xn,w1, . . . , wn, by forcing the integration formula to be exact for the 2nmonomials

f(x) = 1, x, x2, . . . , x2n−1.

In turn, this forces In(f) = I(f) for all polynomials f of degree ≤ 2n− 1.This leads to the following system of 2n nonlinear equations in 2n unknowns:

w1 + w2 + . . .+ wn = 2w1x1 + w2x2 + . . .+ wnxn = 2

3

w1x21 + w2x

22 + . . .+ wnx

2n = 2

5...w1x

2n−11 + w2x

2n−12 + . . .+ wnx

2n−1n = 2

4n−1

(6.44)

The resulting formula In(f) has degree of precision 2n− 1.

As before, this system is very difficult to solve directly, but there are

alternative methods of deriving {xi} and {wi}. It is based on looking at the

polynomials that are orthogonal with respect to the weight function

w(x) = 1√x

on the interval [0, 1].



Example

We evaluate

I =∫ 1

0

cos(πx)√x

dx·= 0.74796566683146

using (6.42) ∫ 1

0

f(x)√xdx ≈ 2f(

12) ≡ 1.0

and (6.43) ∫ 1

0

f(x)√xdx ≈ w1f(x1) + w2f(x2)

·= 0.740519

I2 is a reasonable estimate of I, with I − I2·= 0.00745.



A general theory can be developed for the weighted Gaussian quadrature

I(f) =∫ b

a

w(x)f(x)dx ≈n∑

j=1

wjf(xj) = In(f) (6.45)

It requires the following assumptions for the weight function w(x):

1 w(x) > 0 for a < x < b;

2 For all integers n ≥ n, ∫ b

a

w(x)|x|ndx <∞

These hypotheses are the same as were assumed for the generalized leastsquares approximation theory following Section 4.7 of Chapter 4. This is notaccidental since both Gaussian quadrature and least squares approximationtheory are dependent on the subject of orthogonal polynomials. The nodepoints {xj} solving the system (6.44) are the zeros of the degree n orthogonalpolynomial on [a, b] with respect to the weight function w(x) = 1√

x.

For the generalization (6.45), the nodes {xi} are the zeros of the degree n

orthogonal polynomial on [a, b] with respect to the weight function w(x).


> 5. Numerical Integration > Supplement

Gauss’s idea:

The optimal abscissas of the κ−point Gaussian quadrature formulas are

precisely the roots of the orthogonal polynomial for the same interval and

weighting function.∫ b

af(x)dx =

∑j

∫ xj+1

xj

f(x)dx︸︷︷︸composite formula

=∑j

∫ 1

−1f

(xj+1 − xj

2t+

xj+1 + xj2

)(xj+1 − xj

2

)dt︸︷︷︸

=R 1−1 g(t)dt ≈

κ∑`=1

w`g(q`)︸︷︷︸κ point Gauss Rule for max accuracy

w1, . . . , wκ: weights, q1, . . . , qκ: quadrature points on (−1, 1).Exact on polynomials p(x) ∈ P2κ−1, i.e., 1, t, t2, . . . , t2κ−1.



Example: 3 point Gauss, exact on P5 ⇔ exact on 1, t, t2, t3, t4, t5

∫ +1

−1g(t)dt ≈ w1g(q1) + w2g(q2) + w3g(q3)∫ 1−1 1dt = 2 = w1 · 1 + w2 + w3∫ 1−1 t dt = 0 = w1q1 + w2q2 + w3q3∫ 1−1 t

2dt = 23 = w1q

21 + w2q

22 + w3q

23∫ 1

−1 t3dt = 0 = w1q

31 + w2q

32 + w3q

33∫ 1

−1 t4dt = 2

5 = w1q41 + w2q

42 + w3q

43∫ 1

−1 t5dt = 0 = w1q

51 + w2q

52 + w3q

53

Guess: q1 = −q3, q2 = 0(q1 ≤ q2 ≤ q3), w1 = w3.



Example: 3 point Gauss, exact on P5 ⇔ exact on 1, t, t2, t3, t4, t5

With this guess: 2w1 + w2 = 22w1q

21 = 2/3

2w1q41 = 2/5

,

hence

q1 = −√

35, q3 =

√35

w1 =59, w3 =

59, w2 =

89

A. H. Stroud and D. Secrest: ”Gaussian Quadrature Formulas”.Englewood Cliffs, NJ: Prentice-Hall, 1966.



1 The idea of GaussGauss-Lobatto∫ 1

−1g(t)dt = w1g(−1) + w2g(q2) + · · ·+ wk−1g(k−1) + wkg(1)

k − 2 nodes locates as k-point formula; is accurate P2k−3. (Orderdecreased by 2 beside the Gauss quadrature formula).



Adaptive Quadrature

Problem

Given∫ ba f(x)dx and ε−preassigned tolerance

compute

I(f) ≈∫ b

af(x)dx

with

(a) to assured accuracy ∣∣∣∣∫ b

af(x)dx− I(f)

∣∣∣∣ < ε

(b) at minimal / near minimal cost (no. function evaluations)

Strategy: LOCALIZE!



Localization Theorem

Let I(f) =∑

j Ij(f) where Ij(f) ≈∫ xj+1

xjf(x)dx.

If

∣∣∣∣∣∫ xj+1

xj

f(x)dx− Ij(f)

∣∣∣∣∣ < ε(xj+1 − xj)b− a

(= local tolerance),

then

∣∣∣∣∫ b

af(x)dx− I(f)

∣∣∣∣ < ε(= tolerance)

Proof:

∣∣∣∣∫ b

af(x)dx− I(f)

∣∣∣∣ = |∑j

∫ xj+1

xj

f(x)dx−∑j

I(f)|

= |∑j

(∫ xj+1

xj

f(x)dx− Ij(f)

)| ≤

∑j

|∫ xj+1

xj

f(x)dx− Ij(f)|

=∑j

ε(xj+1 − xj)b− a

=ε

b− a∑j

(xj+1 − xj) =ε

b− a(b− a) = ε.



Need:

Estimator for local error

and

Strategy

when to cut h to ensure accuracy?when to increase h to ensure minimal cost?

One approach: halving and doubling!

Recall: Trapezoidal rule

Ij ≈ (xj+1 − xj)f(xj)+f(xj+1)2 . A priori estimate:∫ xj+1

xj

f(x)dx− Ij =(xj+1 − xj)3

12f ′′(sj)

for some sj in (xj , xj+1).



Step 1: compute Ij

Ij = f(xj+1)+f(xj)2 (xj+1 − xj)

Step 2: cut interval in half + reuse trapezoidal rule

Ij =f(xj) + f(xj+ 1

2)

2(xj+ 1

2−xj)+

f(xj+1) + f(xj+ 12)

2(xj+1−xj+ 1

2)

Error estimate:∫ xj+1

xj

f(x)dx− Ij =h3j

12f ′′(ξj) = ej 1st use of trapezoid rule∫ xj+1

xj

f(x)dx− Ij =(hj/2)3

12f ′′(η1) +

(hj/2)3

12f ′′(η2) 2nd use of TR

=14h3j

12f ′′(ξj)︸︷︷︸ej

+O(h4j )



ej = 4ej + Higher Order Terms

Substracting

Ij − Ij = 3ej +O(h4) =⇒ ej =Ij − Ij

3+ Higher Order Terms︸︷︷︸

O(h4)



4 points Gauss: exact on P7

local error: O(h9)global error: O(h8)A priori estimate:∫ xj+1

xj

f(x)dx− Ij = C(xj+1 − xj)9f (8)(ξj)∫ xj+1

xj

f(x)dx− Ij = Ch9jf

(8)(ξj)∫ xj+1

xj

f(x)dx− Ij = C

(hj2

)9

f (8)(ξ′j) + C

(hj2

)9

f (8)(ξ′′j )︸︷︷︸= C

28h9

jf(8)(ξj)+O(h10)

⇒ Ij − Ij = 225ej +O(h10)

⇒ ej =Ij − Ij

225+ Higher Order Terms︸︷︷︸

O(h10)



Algorithm

Input: a, b, f(x)upper error tolerance: εmax

initial mesh width: h

Initialize: Integral = 0.0xL = aεmin = 1

2k+3 εmax

* xR = xL + h( If xR > b, xR ← b

do integral 1 more time and stop)

Compute on xL, xR:I, I and EST

EST =∣∣∣∣ I − I2k+1 − 1

∣∣∣∣ (if exact on Pk)



’error is just right’:If εmin

hb−a < EST < εmax

hb−a

Integral← Integral + IxL ← xRgo to *

’error is too small’:If EST ≤ εmin

b−a h

Integral← Integral + IxL ← xRh← 2hgo to *

’error is too big’:If EST ≥ εmax

b−a hh← h/2.0go to *

STOPEND



Trapezium rule∫ xj+1

xj

f(x)dx− (xj+1 − xj)f(xj+1) + f(xj)

2∫ xj+1

xj

f(x)− p1(x)dx =∫ xj+1

xj

f ′′(ξ)2

(x− xj)(x− xj+1)︸︷︷︸ψ(x)

dx

=f ′′(x)

2

∫ xj+1

xj

ψ(x)dx︸︷︷︸integrate exactly

+O(h)∫ xj+1

xj

ψ(x)dx



The mysteries of ψ(x)

x_j q_1 q_2 q_3 q_4 q_5 q_6 q_7−1.5

−1

−0.5

0

0.5

1

1.5x 105

xj ≤ x ≤ x

j+1

ψ(x

)

ψ(x) = (x− q1)(x− q2) · · · (x− q7)



Error in k + 1 point quadrature

pk(x) interpolates f(x) =⇒ f(x)− pk(x) = fk+1(ξ)(k+1)! ψ(x)

(xj ≤) q1 < q2 < . . . < qk+1 (≤ xj+1)

∫ xj+1

xj

f(x)dx︸︷︷︸true

−∫ xj+1

xj

pk(x)dx︸︷︷︸approx

=∫ xj+1

xj

ψ(x)(k + 1)!

f (k+1)(ξ)dx



1. A simple error bound

Ignoring oscillation of ψ(x):

|error| ≤ max |fk+1|(k + 1)!

∫ xj+1

xj

|ψ(x)|dx︸︷︷︸=

R xj+1xj

|x−q1|···|x−qk+1|≤hk+1R xj+1

xjdx

≤ max |fk+1|(k + 1)!

|xj+1−xj |k+2

x_j q_1 q_2 q_3 q_4 q_5 q_6 q_7−1.5

−1

−0.5

0

0.5

1

1.5x 105

xj ≤ x ≤ x

j+1

ψ(x

) an

d ab

s(ψ

(x))



2. Analysis without cancelation

xj < ξ < x < xj+1

Lemma

Let ξ, x ∈ (xj , xj+1). Then

f (k+1)(ξ) = f (k+1)(x) + (ξ − x)fk+2(η) MVT

for some η between ξ and x, and |ξ − x| ≤ xj+1 − xj ≤ h.



2. Analysis without cancelation

|error| ≤ |true− approx|

= | 1(k + 1)!

∫ xj+1

xj

ψ(x)[f (k+1)(x)︸︷︷︸

fixed

+(ξ − x︸︷︷︸O(h)

)f (k+2)(η)]dx|

≤ f (k+1)(x)(k + 1)!

|∫ xj+1

xj

ψ(x)dx︸︷︷︸=0 if

R xj+1xj

ψ(x)dx=0

|

+1

(k + 1)!|∫ xj+1

xj

f (k+2)(η)(ξ − x)ψ(x)|

≤ max |f (k+2)|(k + 1)!

∫ xj+1

xj

|ξ − x|︸︷︷︸≤h

|ψ(x)|︸︷︷︸≤hk+1

dx

≤ hk+3 max |f (k+2)(x)|(k + 1)!

The error for Simpson’s rule, i.e. cancelation.



ψ(x) interpolates zero at k + 1 points (degψ(x) = k + 1)

Lemma

The general result of p(q`) = 0, ` = 1, . . . , k + 1, p ∈ Pk+1 isp(x) = Constant · ψ(x).



Questions:

1) How to pick the points q1, . . . , qk+1 so that∫ 1

−1g(x)dx ≈ w1g(q1) + . . .+ wk+1g(qk+1) (6.46)

integrates Pk+m exactly?

2) What does this imply about the error?

Remark

If m < 1, pick q1, . . . , qk+1 so that∫ 1−1 ψ(x)dx = 0 and then the error

converges O(hk+3).

m=2



Step 1

Let r1 be some fixed point on [−1, 1]:−1 < q1 < q2 < . . . < r1 < . . . < qk < qk+1

pk+1(x) = pk(x) + ψ(x)g(r1)− pk(r1)

ψ(r1)(6.47)

pk interpolates g(x) at q1, . . . , qk+1.Claim: pk+1 interpolates g(x) at k + 2 points q1, . . . , qk+1, r1.Suppose now that (6.46) is exact on Pk+1, then from (6.47)∫ 1

−1g(x)dx︸︷︷︸true

−∫ 1

−1pk+1(x)︸︷︷︸

substitute (6.47)

dx error in k + 2 quadrature rule, Ek+2

=∫ 1

−1g(x)dx−

∫ 1

−1pk(x)dx︸︷︷︸

error in k + 1 quadrature rule ≡ Ek+1

−∫ 1

−1ψ(x)

f(r1)− pk(r1)ψ(r1)

dx



Step 1

So Ek+2 = Ek+1 −f(r1)− pk(r1)

ψ(r1)

∫ 1

−1ψ(x)dx

Conclusion 1

If∫ 1−1 ψ(x)dx = 0, then error in k + 1 point rule is exactly the same as

if we had used k + 2 points.



Step 2

Let r1, r2 be fixed points in [−1, 1]. So interpolate at k + 3 points:q1, . . . , qk+1, r1, r2

pk+2(x) = pk(x) + ψ(x)(x− r1)g(r2)− pk(r2)(r2 − r1)ψ(r2)

(6.48)

+ ψ(x)(x− r2)g(r1)− pk(r1)(r1 − r2)ψ(r1)

Consider error in a rule with k + 1 + 2 points:

error in k + 3 p. r. =∫ 1

−1g(x)dx−

∫ 1

−1pk+2(x)dx

=∫ 1

−1g(x)dx−

∫ 1

−1pk(x)dx−

g(r2)− pk(r2)(r2 − r1)ψ(r2)

∫ 1

−1ψ(x)(x− r1)dx

− g(r1)− pk(r1)(r1 − r2)ψ(r1)

∫ 1

−1ψ(x)(x− r2)dx.



Step 2

So Ek+3 = Ek+1 + Const∫ 1−1 ψ(x)(x− r1)dx+ Const

∫ 1−1 ψ(x)(x− r2)dx

Conclusion 2

If∫ 1−1 ψ(x)dx = 0 and

∫ 1−1 xψ(x)dx = 0, then error in k + 1 point rule

has the same error as k + 3 point rule.



. . .

Ek+1+m = Ek+1 + C0

∫ 1

−1ψ(x)dx+ C1

∫ 1

−1ψ(x)x1dx+ . . . (6.49)

+ Cm

∫ 1

−1ψ(x)xm−1dx (6.50)

So

Conclusion 3

If∫ 1−1 ψ(x)xjdx = 0, j = 0, . . . ,m− 1, then error is as good as using

m extra points.



Overview

Interpolating QuadratureInterpolate f(x) at q0, q1, q2, . . . , qk ⇒ pk(x)

f(x)− pk(x) =f (k+1)(ξ)(k + 1)!

(x− q0)(x− q1) . . . (x− qk)∫ 1

−1f(x)dx−

∫ 1

−1pk(x)dx =

1(k + 1)!

∫ 1

−1f (k+1)(ξ)ψ(x)dx

Gauss rules

pick q` to maximize exactness

what is the accuracy

what are the q`’s?



Overview

Interpolate at k + 1 +m points Interpolate at k + 1 pointsq0, . . . , qk, r1 . . . , rm q0, . . . , qk

error error⇓ ⇓

Ek+m = Ek + c0∫ 1−1 ψ(x) · 1dx

+c1∫ 1−1 ψ(x) · xdx+ . . .

+cm∫ 1−1 ψ(x) · xm−1dx

Definition

p(x) is the µ+ 1st orthogonal polynomial on [−1, 1] (weightw(x) ≡ 1) if p(x) ∈ Pµ+1 and

∫ 1−1 p(x)x

`dx = 0, ` = 0, . . . , µ, i.e.,∫ 1−1 p(x)q(x)dx = 0∀q ∈ Pµ.



Overview

Pick q0, q1, . . . , qk so∫ 1−1 ψ(x) · 1dx = 0∫ 1−1 ψ(x) · xdx = 0

...∫ 1−1 ψ(x) · xm−1dx = 0

⇔∫ 1

−1ψ(x)︸︷︷︸

deg k+1

degm−1︷︸︸︷q(x) dx = 0, ∀q ∈ Pm−1

So, maximum accuracy if ψ(x) is the orthogonal polynomial of degreek + 1:∫ 1

−1ψ(x)q(x)dx = 0 ∀q ∈ Pk ⇒ m− 1 = k,m = k + 1

So, the Gauss quadrature points are the roots of the orthogonalpolynomial.



Overview

Adaptivity

I = I1 + I2

Trapezium rule’s local error = O(h3)∫ xj+1

xj

f(x)dx− I = 4e⇐ e ≈ 8e1, e ≈ 8e2 so, e ≈ 4e∫ xj+1

xj

f(x)dx− I = e = e1 + e2

I − I = 3e (+ Higher Order Terms)⇒ e =I − I

3(+ Higher Order Terms)



Overview

Final Observation

True−Approx ≈ 4e

True−Approx ≈ e

}2 equations, 2 unknowns: e, True

So, can solve for e+ TrueSolving for True:

True ≈ I + e ≈ I +I − I

3≈ 4

3I − 1

3I (+ Higher Order Terms)


> 5. Numerical Integration > 5.4 Numerical Differentiation

To numerically calculate the derivative of f(x), begin by recalling thedefinition of derivative

f ′(x) = limx→0

f(x+ h)− f(x)h

This justifies using

f ′(x) ≈f(x + h) − f(x)

h≡ Dhf(x) (6.51)

for small values of h. Dhf(x) is called a numerical derivative of f(x)with stepsize h.



Example

Use Dhf to approximate the derivative of f(x) = cos(x) at x = π6 .

h Dn(f) Error Ratio

0.1 -0.54243 0.04243

0.05 -0.52144 0.02144 1.98

0.025 -0.51077 0.01077 1.99

0.0125 -0.50540 0.00540 1.99

0.00625 -0.50270 0.00270 2.00

0.003125 -0.50135 0.00135 2.00

Looking at the error column, we see the error is nearly proportional toh; when h is halved, the error is almost halved.



To explain the behaviour in this example, Taylor’s theorem can be usedto find an error formula. Expanding f(x+ h) about x, we get

f(x+ h) = f(x) + hf ′(x) +h2

2f ′′(c)

for some c between x and x+ h. Substituting on the right side of(6.51), we obtain

Dhf(x) =1h

{[f(x) + hf ′(x) +

h2

2f ′′(c)

]− f(x)

}= f ′(x) +

h

2f ′′(c)

f ′(x)−Dhf(x)= −h2f ′′(c) (6.52)

The error is proportional to h, agreeing with the results in the Tableabove.



For that example,f ′(π

6

)−Dhf

(π6

)=h

2cos(c) (6.53)

where c is between 16π and 1

6π + h.Let’s check that if c is replaced by 1

6π, then the RHS of (6.53) agrees withthe error column in the Table.As seen in the example, we use the formula (6.51) with a positive stepsizeh > 0. The formula (6.51) is commonly known as the forward differenceformula for the first derivative.We can formally replace h by −h in (6.51) to obtain the formula

f ′(x) ≈f(x) − f(x − h)

h, h > 0 (6.54)

This is the backward difference formula for the first derivative.A derivation similar to that leading to (6.52) shows that

f ′(x)− f(x)− f(x− h)h

=h

2f ′′(c)) (6.55)

for some c between x and x− h.Thus, we expect the accuracy of the backward difference formula to be

almost the same as that of the forward difference formula.


> 5. Numerical Integration > 5.4.1 Differentiation Using Interpolation

Let Pn(x) denote the degree n polynomial that interpolates f(x) atn+ 1 node points x0, . . . , xn.

To calculate f ′(x) at some point x = t, use

f ′(t) ≈ P ′n(t) (6.56)

Many different formulae can be obtained by

1 varying n and by

2 varying the placement of the nodes x0, . . . , xn relative to thepoint t of interest.



As an especially useful example of (6.56), take

n = 2, t = x1, x0 = x1 − h, x2 = x1 + h.

Then

P2(x) =(x− x1)(x− x2)

2h2f(x0) +

(x− x0)(x− x2)−h2

f(x1)

+(x− x0)(x− x1)

2h2f(x2)

P ′2(x) =2x− x1 − x2

2h2f(x0) +

2x− x0 − x2

−h2f(x1)

+2x− x0 − x1

2h2f(x2)

P ′2(x1) =x1 − x2

2h2f(x0) +

2x1 − x0 − x2

−h2f(x1) +

x1 − x0

2h2f(x2)

=f(x2)− f(x0)

2h(6.57)



The central difference formula

Replacing x0 and x2 by x1 − h and x1 + h, from (6.56) and (6.57) weobtain central difference formula

f ′(x1) ≈f(x1 + h) − f(x1 − h)

2h≡ Dhf(x1), (6.58)

another approximation to the derivative of f(x).

It will be shown below that this is a more accurate approximation tof ′(x) than is the the forward difference formula Dhf(x) of (6.51), i.e.,

f ′(x) =f(x+ h)− f(x)

h≡ Dhf(x).



Theorem

Assume f ∈ Cn+2[a, b].Let x0, x1, . . . , xn be n+ 1 distinct interpolation nodes in [a, b], andlet t be an arbitrary given point in [a, b].Then

f ′(t) − P ′n(t) = Ψn(t)

f (n+2)(c1)

(n + 2)!+ Ψ′

n(t)f (n+1)(c2)

(n + 1)!(6.59)

withΨn(t) = (t− x0)(t− x1) · · · (t− xn)

The numbers c1 and c2 are unknown points located between themaximum and minimum of x0, x1, . . . , xn and t.



To illustrate this result, an error formula can be derived for the centraldifference formula (6.58).Since t = x1 in deriving (6.58), we find that the first term on the RHSof (6.59) is zero. Also n = 2 and

Ψ2(x) = (x− x0)(x− x1)(x− x2)Ψ′

2(x) = (x− x0)(x− x1) + (x− x0)(x− x2) + (x− x0)(x− x1)Ψ′

2(x1) = (x1 − x0)(x1 − x2) = −h2

Using this in (6.59), we get

f ′(x1)−f(x1 + h)− f(x1 − h)

2h= −h2

6f ′′′(c2) (6.60)

with x1 − h ≤ c2 ≤ x1 + h. This says that for small values of h, thecentral difference formula (6.58) should be more accurate that theearlier approximation (6.51), the forward difference formula, becausethe error term of (6.58) decreases more rapidly with h.



Example

The earlier example f(x) = cos(x) is repeated using the centraldifference fomula (6.58) (recall x1 = π

6 ).

h Dn(f) Error Ratio

0.1 -0.49916708 -0.0008329

0.05 -0.49979169 -0.0002083 4.00

0.025 -0.49994792 -0.00005208 4.00

0.0125 -0.49998698 -0.00001302 4.00

0.00625 -0.49999674 -0.000003255 4.00

The results confirm the rate of convergence given in (6.58), and theyillustrate that the central difference formula (6.58) will usually will besuperior to the earlier approximation, the forward difference formula(6.51).


> 5. Numerical Integration > 5.4.2 The Method of Undetermined Coefficients

The method of undetermined coefficients is a procedure used inderiving formulae for numerical differentiation, interpolation andintegration. We will explain the method by using it to derive anapproximation for f ′′(x).To approximate f ′′(x) at some point x = t, write

f ′′(t) ≈ D(2)h f(t) ≡ Af(t+ h) +Bf(t) + Cf(t− h) (6.61)

with A,B and C unspecified constants. Replace f(t− h) and f(t+ h)by the Taylor polynomial approximations

f(t− h) ≈ f(t)− hf ′(t) + h2

2 f′′(t)− h3

6 f′′′(t) + h4

24f(4)(t)

f(t+ h) ≈ f(t) + hf ′(t) + h2

2 f′′(t) + h3

6 f′′′(t) + h4

24f(4)(t)

(6.62)

Including more terms would give higher powers of h; and for smallvalues of h, these additional terms should be much smaller that theterms included in (6.62).



Substituting these approximations into the formula for D(2)h f(t) and

collecting together common powers of h give us

D(2)h f(t) ≈ (A+B + C)︸︷︷︸

=0

f(t) + h(A− C)︸︷︷︸=0

f ′(t) +h2

2(A+ C)︸︷︷︸

=1

f ′′(t)

+h3

6(A− C)f ′′′(t) +

h4

24(A+ C)f (4)(t) (6.63)

To haveD

(2)h f(t) ≈ f ′′(t)

for arbitrary functions f(x), it is necessary to require

A+B + C = 0; coefficient of f(t)h(A− C) = 0; coefficient of f ′(t)h2

2 (A+ C) = 1; coefficient of f ′′(t)

This system has the solution

A = C =1h2, B = − 2

h2.



This determines

D(2)h f(t) =

f(t + h) − 2f(t) + f(t − h)

h2(6.64)

To determine an error error formula for D(2)h f(t), substitute

A = C = 1h2 , B = − 2

h2 into (6.63) to obtain

D(2)h f(t) ≈ f ′′(t) +

h2

12f (4)(t).

The approximation in this arises from not including in the Taylorpolynomials (6.62) corresponding higher powers of h. Thus,

f ′′(t)− f(t+ h)− 2f(t) + f(t− h)h2

≈ −h2

12f (4)(t) (6.65)

This is accurate estimate of the error for small values of h. Of course,in a practical situation we would not know f (4)(t). But the errorformula shows that the error decreases by a factor of about 4when h is halved. This can lead to justify Richardson’s extrapolationto obtain an even more accurate estimate of the error and of f ′′(t).



Example

Let f(x) = cos(x), t = 16π, and use (6.64) to calculate

f ′′(t) = − cos(16π).

h D2h(f) Error Ratio

0.5 -0.84813289 -1.789E-2

0.25 -0.86152424 -4.501E-3 3.97

0.125 -0.86489835 -1.127E-3 3.99

0.0625 -0.86574353 -2.819E-4 4.00

0.03125 -0.86595493 -7.048E-5 4.00

The results shown (see the ratio column) are consistent with the errorformula (6.65)

f ′′(t)− f(t+ h)− 2f(t) + f(t− h)h2

≈ −h2

12f (4)(t).



In the derivation of (6.65), the form (6.61)

f ′′(t) ≈ D(2)h f(t) ≡ Af(t+ h) +Bf(t) + Cf(t− h)

was assumed for the approximate derivative. We could equally well havechosen to evaluate f(x) at points other than those used there, for example,

f ′′(t) ≈ Af(t+ 2h) +Bf(t+ h) + Cf(t)

Or, we could have chosen more evaluation points, as in

f ′′(t) ≈ Af(t+ 3h) +Bf(t+ 2h) + Cf(t+ h) +Df(t)

The extra degree of freedom could have been used to obtain a more accurateapproximation to f ′′(t), by forcing the error term to be proportional to ahigher power of h.

Many of the formulae derived by the method of undetermined coefficients can

also be derived by differentiating and evaluating a suitably chosen

interpolation polynomial. But often, it is easier to visualize the desired

formula as a combination of certain function values and to then derive the

proper combination, as was done above for (6.64).


> 5. Numerical Integration > 5.4.3 Effects of error in function evaluation

The formulae derived above are useful for differentiating functions thatare known analytically and for setting up numerical methods for solvingdifferential equations. Nonetheless, they are very sensitive to errors inthe function values, especially if these errors are not sufficiently smallcompared with the stepsize h used in the differentiation formula.To explore this, we analyze the effect of such errors in the formula

D(2)h f(t) approximating f ′′(t).

Rewrite (6.64), D(2)h f(t) = f(t+h)−2f(t)+f(t−h)

h2 , as

D(2)h f(x1) =

f(x2)− 2f(x1) + f(x0)h2

≈ f ′′(x1)

where x2 = x1 + h, x0 = x1 − h. Let the actual function values usedin the computation be denoted by f0, f1, and f2 with

f(xi)− fi = εi, i = 0, 1, 2

the errors in the function values.



Thus, the actual quantity calculated is

D2hf(x1) =

f2 − 2f1 + f0

h2

For the error in this quantity, replace fj by f(xj)− εj , j = 0, 1, 2 toobtain

f ′′(x1)−D2hf(x1) = f ′′(x1)−

[f(x2)− ε2]− 2[f(x1)− ε1] + [f(x0)− ε0]h2

=[f ′′(x1)−

f(x2)− 2f(x1) + f(x0)h2︸︷︷︸

(6.65)= −h2

12f (4)(t)

]+ε2 − 2ε1 + ε0

h2

≈ −h2

12f (4)(x1) +

ε2 − 2ε1 + ε0

h2(6.66)



The errors ε0, ε1, ε2 are generally random in some interval [−δ, δ].If the values f0, f1, f2 are experimental data, then δ is a bound onthe experimental data.

Also, if these function values fi are obtained from computingf(x) on a computer, then the errors εj are combination ofrounding or chopping errors and δ is a bound on these errors.

In either case, (6.66)

f ′′(x1)−D2hf(x1) ≈ −

h2

12f (4)(x1) +

ε2 − 2ε1 + ε0

h2

yields the approximate inequality∣∣∣f ′′(x1)− D(2)h f(x1)

∣∣∣ ≤ h2

12

∣∣∣f (4)(x1)∣∣∣+ 4δ

h2(6.67)

This error bound suggests that as h→ 0,the error will eventually increase, because of the final term 4δ

h2 .



Example

Calculate D(2)h (x1) for f(x) = cos(x) at x1 = 1

6π. To show the effect of

rounding errors, the values fi are obtained by rounding f(xi) to sixsignificant digits; and the errors satisfy

|εi| ≤ 5.0× 10−7 = δ, i = 0, 1, 2

Other than these rounding errors, the formula D(2)h f(x1) is calculated exactly.

The results are

h D2h(f) Error

0.5 -0.848128 -0.0178970.25 -0.861504 -0.0045210.125 -0.864832 -0.0011930.0625 -0.865536 -0.0004890.03125 -0.865280 -0.0007450.015625 -0.860160 -0.0058650.0078125 -0.851968 -0.0140570.00390625 -0.786432 -0.079593



In this example, the bound (6.67), i.e.,∣∣∣f ′′(x1)− D(2)h f(x1)

∣∣∣ ≤ h2

12

∣∣f (4)(x1)∣∣+ 4δ

h2 ,

becomes∣∣∣f ′′(x1)− D(2)h f(x1)

∣∣∣ ≤ h2

12cos(

16π

)+(

4h2

)(5× 10−7)

·= 0.0722h2 +2× 10−6

h2≡ E(h)

For h = 0.125, the bound E(h) ·= 0.00126, which is not too far offfrom actual error given in the table.

The bound E(h) indicates that there is a smallest value of h, call ith∗, below which the error will begin to increase.To find it, let E′(h) = 0, with its root being h∗.

This leads to h∗·= 0.00726, which is consistent with the behaviour of

the errors in the table.



One must be very cautious in using numerical differentiation,because of the sensitivity to errors in the function values.

This is especially true if the function values are obtained empiricallywith relatively large experimental errors, as is common in practice. Inthis latter case, one should probably use a carefully prepared packageprogram for numerical differentiation. Such programs take intoaccount the error in the data, attempting to find numerical derivativesthat are as accurate as can be justified by the data.

In the absence of such a program, one should consider producing acubic spline function that approximates the data, and then use itsderivative as a numerical derivative for the data. The cubic splinefunction could be based on interpolation; or better for data with largerelatively errors, construct a cubic spline that is a least squaresapproximation to the data. The concept of least squaresapproximation is introduced in Section 7.1.


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