TRAPEZOIDAL METHOD: ERROR FORMULA

Theorem Assume f (x) twice continuously differentiable on theinterval [a, b]. Then

ETn (f ) :=

∫ b

af (x) dx − Tn(f ) = −h2 (b − a)

12f ′′ (cn)

for some cn in the interval [a, b].

Later we will say something about the proof of this result, as itleads to some other useful formulas for the error.

The above formula says that the error decreases in a manner thatis roughly proportional to h2. Thus doubling n (and halving h)should cause the error to decrease by a factor of approximately 4.This is what we observed with some past examples from thepreceding section.

Example

Consider evaluating

I =

∫ 2

0

dx

1 + x2

using the trapezoidal method Tn(f ). Let us bound the error

ETn (f ) = −h2 (b − a)

12f ′′ (cn)

Here, b − a = 2. We bound |f ′′ (cn)| by max0≤x≤2 |f ′′(x)|.Calculate the derivatives:

f ′(x) =−2x

(1 + x2)2, f ′′(x) =

−2 + 6x2

(1 + x2)3, f ′′′(x) =

24 x(1− x2

)(1 + x2)4

For x ∈ (0, 2), f ′′′(x) = 0 only when x = 1. So

max0≤x≤2

∣∣f ′′(x)∣∣ = max

{∣∣f ′′(0)∣∣ , ∣∣f ′′(1)

∣∣ , ∣∣f ′′(2)∣∣} = 2

Therefore, ∣∣∣ETn (f )

∣∣∣ ≤ h2 (2)

12· 2 =

h2

3

∣∣∣ETn (f )

∣∣∣ ≤ h2

3

How large should n be chosen in order to ensure that∣∣∣ETn (f )

∣∣∣ ≤ 5× 10−6 (1)

To ensure this, we choose h so small that

h2

3≤ 5× 10−6

This is equivalent to choosing h and n to satisfy

h ≤ .003873

n =2

h≥ 516.4

Thus n ≥ 517 will imply (1).

DERIVING THE ERROR FORMULA

There are two stages in deriving the error:(1) Obtain the error formula for the case of a single subinterval(n = 1);(2) Use this to obtain the general error formula given earlier.

For the trapezoidal method with only a single subinterval, we have∫ α+h

αf (x) dx − h

2[f (α) + f (α + h)] = −h3

12f ′′(c)

for some c in the interval [α, α + h].

This error formula can be derived through an application of Taylor’stheorem or the Newton form of the linear interpolation error.

The general trapezoidal rule Tn(f ) was obtained by applying thesimple trapezoidal rule to a subdivision of the original interval ofintegration. Recall the notation

h =b − a

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

I (f ) =

∫ xn

x0

f (x) dx

=

∫ x1

x0

f (x) dx +

∫ x2

x1

f (x) dx + · · ·+∫ xn

xn−1

f (x) dx

Tn(f ) =h

2[f (x0) + f (x1)] +

h

2[f (x1) + f (x2)]

+ · · ·+ h

2[f (xn−1) + f (xn)]

Then the error

ETn (f ) ≡

∫ b

af (x) dx − Tn(f )

can be analyzed by adding together the errors over the subintervals[x0, x1], [x1, x2], ..., [xn−1, xn]. Recall∫ α+h

αf (x) dx − h

2[f (α) + f (α + h)] = −h3

12f ′′(c)

Then on [xj−1, xj ],∫ xj

xj−1

f (x) dx − h

2[f (xj−1) + f (xj)] = −h3

12f ′′(γj)

with xj−1 ≤ γj ≤ xj . Combining these errors, we obtain

ETn (f ) = −h3

12f ′′(γ1)− · · · − h3

12f ′′(γn)

This formula can be further simplified, and we will do so in twoways.

Rewrite this error as

ETn (f ) = −h3n

12

[f ′′(γ1) + · · ·+ f ′′(γn)

n

]Denote the quantity inside the brackets by ζn. This numbersatisfies

mina≤x≤b

f ′′(x) ≤ ζn ≤ maxa≤x≤b

f ′′(x)

Since f ′′(x) is a continuous function (by original assumption), wehave some number cn in [a, b] for which

f ′′(cn) = ζn

Recall also that hn = b − a. Then

ETn (f ) = −h3n

12

[f ′′(γ1) + · · ·+ f ′′(γn)

n

]= −h2 (b − a)

12f ′′ (cn)

This is the error formula given on the first slide.

AN ERROR ESTIMATE

We now obtain a way to estimate the error ETn (f ). Return to the

formula

ETn (f ) = −h3

12f ′′(γ1)− · · · − h3

12f ′′(γn)

and rewrite it as

ETn (f ) = −h2

12

[f ′′(γ1)h + · · ·+ f ′′(γn)h

]The quantity

f ′′(γ1)h + · · ·+ f ′′(γn)h

is a Riemann sum for the integral∫ b

af ′′(x) dx = f ′(b)− f ′(a)

By this we mean

limn→∞

[f ′′(γ1)h + · · ·+ f ′′(γn)h

]=

∫ b

af ′′(x) dx

Thusf ′′(γ1)h + · · ·+ f ′′(γn)h ≈ f ′(b)− f ′(a)

for larger values of n. Combining this with the earlier error formula

ETn (f ) = −h2

12

[f ′′(γ1)h + · · ·+ f ′′(γn)h

]we have

ETn (f ) ≈ −h2

12

[f ′(b)− f ′(a)

]≡ ET

n (f )

This is a computable estimate of the error in the numericalintegration. It is called an asymptotic error estimate.

Example

Consider evaluating

I (f ) =

∫ π

0ex cos x dx = −eπ + 1

2.

= −12.0703463163896

In this case,

f ′(x) = ex (cos x − sin x) , f ′′(x) = −2ex sin x

max0≤x≤π

∣∣f ′′(x)∣∣ =

∣∣f ′′ (.75π)∣∣ = 14. 921

From

ETn (f ) = −h2 (b − a)

12f ′′ (cn)

we have the error bound∣∣∣ETn (f )

∣∣∣ ≤ h2π

12· 14.921 = 3.906h2 (2)

Also, we have the error estimate

ETn (f ) = −h2

12

[f ′(π)− f ′(0)

]=

h2

12(eπ + 1)

.= 2.012h2 (3)

n Tn ETn Ratio Bound (2) ET

n

2 -17.38925933 5.319 9.638 4.9644 -13.33602285 1.266 4.20 2.409 1.2418 -12.38216243 3.118E−1 4.06 6.024E−1 3.103E−1

16 -12.14800410 7.766E−2 4.02 1.506E−1 7.757E−232 -12.08974212 1.940E−2 4.00 3.765E−2 1.939E−264 -12.07519410 4.848E−3 4.00 9.412E−3 4.848E−3

128 -12.07155818 1.212E−3 4.00 2.353E−3 1.212E−3256 -12.07064928 3.030E−4 4.00 5.822E−4 3.030E−4

In looking at the table, we see that the error ETn (f ) and the error

estimate ETn (f ) are quite close. Therefore

I (f )− Tn(f ) ≈ h2

12(eπ + 1)

I (f ) ≈ Tn(f ) +h2

12(eπ + 1)

This last formula is called the corrected trapezoidal rule, CTn(f ).

The corrected trapezoidal rule is illustrated in the following table.

n I − Tn Ratio I − CTn Ratio

2 5.319 3.552E−14 1.266 4.20 2.474E−2 14.48 3.118E−1 4.06 1.583E−3 15.6

16 7.766E−2 4.02 9.949E−5 15.932 1.940E−2 4.00 6.227E−6 16.064 4.848E−3 4.00 3.893E−7 16.0

128 1.212E−3 4.00 2.433E−8 16.0256 3.030E−4 4.00 1.521E−9 16.0

The corrected trapezoidal rule

In general,

I (f )− Tn(f ) ≈ −h2

12

[f ′(b)− f ′(a)

]I (f ) ≈ CTn(f ) := Tn(f )− h2

12

[f ′(b)− f ′(a)

]This is the corrected trapezoidal rule. It is easy to obtain from thetrapezoidal rule, and in most cases, it converges more rapidly thanthe trapezoidal rule.

ERROR FORMULA FOR SIMPSON’S RULE

Recall the general Simpson’s rule∫ b

af (x) dx ≈ Sn(f ) :=

h

3[f (x0) + 4f (x1) + 2f (x2)

+ 4f (x3) + 2f (x4) + · · ·+ 2f (xn−2) + 4f (xn−1) + f (xn)]

For its error, we have

ESn (f ) ≡

∫ b

af (x) dx − Sn(f ) = −h4 (b − a)

180f (4)(cn)

where a ≤ cn ≤ b. For an asymptotic error estimate,∫ b

af (x) dx − Sn(f ) ≈ ES

n (f ) ≡ − h4

180

[f ′′′(b)− f ′′′(a)

]

DISCUSSION

For Simpson’s error formula, both formulas assume that theintegrand f (x) has four continuous derivatives on the interval[a, b]. What happens when this is not valid? We return later tothis question.

Both formulas also say the error should decrease by a factor ofaround 16 when n is doubled.

Compare these results with those for the trapezoidal rule errorformulas:

ETn (f ) ≡

∫ b

af (x) dx − Tn(f ) = −h2 (b − a)

12f ′′ (cn)

ETn (f ) ≈ −h2

12

[f ′(b)− f ′(a)

]≡ ET

n (f )

EXAMPLE

Consider evaluating

I =

∫ 2

0

dx

1 + x2

using Simpson’s rule Sn(f ). Let us bound the error.Begin by noting that

f (4)(x) = 245x4 − 10x2 + 1

(1 + x2)5

max0≤x≤1

∣∣∣f (4)(x)∣∣∣ = f (4)(0) = 24

Then

ESn (f ) = −h4 (b − a)

180f (4)(cn)∣∣∣ES

n (f )∣∣∣ ≤ h4 · 2

180· 24 =

4h4

15

How large should n be chosen in order to ensure that∣∣∣ESn (f )

∣∣∣ ≤ 5× 10−6

This is true if

4h4

15≤ 5× 10−6

h ≤ .0658

n ≥ 30.39

Therefore, choosing n ≥ 32 will give the desired error bound.Compare this with the earlier trapezoidal example in whichn ≥ 517 was needed.

For the asymptotic error estimate, we have

f ′′′(x) = −24xx2 − 1

(1 + x2)4

ESn (f ) ≡ − h4

180

[f ′′′(2)− f ′′′(0)

]=

h4

180· 144

625=

4

3125h4

INTEGRATING√x

Consider the numerical approximation of∫ 1

0

√x dx =

2

3

In the following table, we give the errors when using both thetrapezoidal and Simpson rules.

n ETn Ratio ES

n Ratio

2 6.311E − 2 2.860E − 24 2.338E − 2 2.70 1.012E − 2 2.828 8.536E − 3 2.74 3.587E − 3 2.83

16 3.085E − 3 2.77 1.268E − 3 2.8332 1.108E − 3 2.78 4.485E − 4 2.8364 3.959E − 4 2.80 1.586E − 4 2.83

128 1.410E − 4 2.81 5.606E − 5 2.83

The rate of convergence is slower because the function f (x) =√x

is not sufficiently differentiable on [0, 1]. Both methods convergewith a rate proportional to h1.5.

ASYMPTOTIC ERROR FORMULAS

For a numerical integration formula,∫ b

af (x) dx ≈

n∑j=0

wj f (xj)

let En(f ) denote its error,

En(f ) =

∫ b

af (x) dx −

n∑j=0

wj f (xj)

We say another formula En(f ) is an asymptotic error formula thisnumerical integration if it satisfies

limn→∞

En(f )

En(f )= 1 or lim

n→∞

En(f )− En(f )

En(f )= 0

These conditions say that En(f ) looks increasingly like En(f ) as nincreases, and thus En(f ) ≈ En(f ).

Examples

For the trapezoidal rule, assuming f (x) has two continuousderivatives on [a, b],

ETn (f ) ≈ ET

n (f ) ≡ −h2

12

[f ′(b)− f ′(a)

]For Simpson’s rule, assuming f (x) has four continuous derivativeson [a, b],

ESn (f ) ≈ ES

n (f ) ≡ − h4

180

[f ′′′(b)− f ′′′(a)

]

Example (cont.)

Note that both of these formulas can be written in an equivalentform as

En(f ) =c

np

for appropriate constant c and exponent p. With the trapezoidalrule, p = 2 and

c = −(b − a)2

12

[f ′(b)− f ′(a)

]and for Simpson’s rule, p = 4 and

c = −(b − a)4

180

[f ′′′(b)− f ′′′(a)

]

The formulaEn(f ) =

c

np(4)

occurs for many other numerical integration formulas that we havenot yet defined or studied. In addition, if we use the trapezoidal orSimpson rules with an integrand f (x) which is not sufficientlydifferentiable, then (4) may hold with a p less than the ideal.Example. Consider

I =

∫ 1

0xβf (x) dx

in which 0 < β < 1 and f (x) is a smooth function with f (0) 6= 0 .Then the convergence of the trapezoidal rule can be shown to havean asymptotic error formula

En ≈ En =c

nβ+1(5)

for some constant c dependent on β. A similar result holds forSimpson’s rule, with 0 < β < 3, β not an integer. We can actuallyspecify a formula for c ; but the formula is often less importantthan knowing that (4) is valid for some c.

APPLICATION OF ASYMPTOTIC ERROR FORMULAS

Assume we know that an asymptotic error formula

I − In ≈c

np

is valid for some numerical integration rule denoted by In. Initially,assume we know the exponent p. Then imagine calculating both Inand I2n. With I2n, we have

I − I2n ≈c

2pnp

This leads to

I − In ≈ 2p (I − I2n)

I ≈ 2pI2n − In2p − 1

= I2n +I2n − In2p − 1

The formula

I ≈ I2n +I2n − In2p − 1

(6)

is called Richardson’s extrapolation formula.

Examples

With the trapezoidal rule and with the integrand f (x) having twocontinuous derivatives, p = 2 and

I ≈ T2n +1

3(T2n − Tn)

With Simpson’s rule and with the integrand f (x) having fourcontinuous derivatives, p = 4 and

I ≈ S2n +1

15(S2n − Sn)

Richardson’s error estimate

We can also use the formula (4)

En(f ) =c

np

to obtain error estimation formulas:

I − I2n ≈I2n − In2p − 1

(7)

This is called Richardson’s error estimate. For example, with thetrapezoidal rule,

I − T2n ≈1

3(T2n − Tn)

and for Simpson’s rule,

I − S2n ≈1

15(S2n − Sn)

AITKEN EXTRAPOLATION

In this case, we again assume

I − In ≈c

np

But in contrast to previously, we do not know either c or p.Imagine computing In, I2n, and I4n. Then

I − In ≈c

np, I − I2n ≈

c

2pnp, I − I4n ≈

c

4pnp

We can directly try to estimate I . Dividing

I − InI − I2n

≈ 2p ≈ I − I2nI − I4n

Solving for I , we obtain

(I − I2n)2 ≈ (I − In) (I − I4n)

I (In + I4n − 2I2n) ≈ InI4n − I 22n

I ≈ InI4n − I 22nIn + I4n − 2I2n

This can be improved computationally, to avoid loss of significanceerrors.

I ≈ I4n +

(InI4n − I 22n

In + I4n − 2I2n− I4n

)= I4n −

(I4n − I2n)2

(I4n − I2n)− (I2n − In)

This is called Aitken’s extrapolation formula.

To estimate p, we use

I2n − InI4n − I2n

≈ 2p

To see this, write

I2n − InI4n − I2n

=(I − In)− (I − I2n)

(I − I2n)− (I − I4n)

Then substitute from the following and simplify:

I − In ≈c

np, I − I2n ≈

c

2pnp, I − I4n ≈

c

4pnp

Example

Consider the following table of numerical integrals.

n In In − In/2 Ratio

2 .284517796864 .28559254576 1.075E − 38 .28570248748 1.099E − 4 9.78

16 .28571317731 1.069E − 5 10.2832 .28571418363 1.006E − 6 10.6264 .28571427643 9.280E − 8 10.84

What is its order of convergence? It appears

2p.

= 10.84, p.

= log2 10.84 = 3.44

We could now combine this with Richardson’s error formula:

I − In ≈1

2p − 1

(In − In/2

)For example,

I − I64 ≈1

9.84[9.280E − 8] = 9.43E − 9

PERIODIC FUNCTIONS

A function f (x) is periodic if the following condition is satisfied.There is a smallest real number τ > 0 for which

f (x + τ) = f (x), −∞ < x <∞ (8)

The number τ is called the period of the function f (x). Theconstant function f (x) ≡ 1 is also considered periodic, but itsatisfies this condition with any τ > 0. Basically, a periodicfunction is one which repeats itself over intervals of length τ .

The condition (8) implies

f (m)(x + τ) = f (m)(x), −∞ < x <∞ (9)

for the mth-derivative of f (x), provided there is such a derivative.Thus the derivatives are also periodic.

Periodic functions occur very frequently in applications ofmathematics, reflecting the periodicity of many phenomena in thephysical world.

PERIODIC INTEGRANDS

Consider the special class of integrals

I (f ) =

∫ b

af (x) dx

in which f (x) is periodic, with b − a an integer multiple of theperiod τ for f (x). In this case, the performance of the trapezoidalrule and other numerical integration rules is much better than thatpredicted by earlier error formulas. To hint at this improvedperformance, recall∫ b

af (x) dx − Tn(f ) ≈ En(f ) ≡ −h2

12

[f ′(b)− f ′(a)

]With our assumption on the periodicity of f (x), we have

f (a) = f (b), f ′(a) = f ′(b)

Therefore, En(f ) = 0 and we should expect improved performancein the convergence behaviour of the trapezoidal sums Tn(f ).

If in addition to being periodic on [a, b], the integrand f (x) alsohas m continous derivatives, then it can be shown that

I (f )− Tn(f ) =cmnm

+ smaller terms

By “smaller terms”, we mean terms which decrease to zero morerapidly than n−m.

Thus if f (x) is periodic with b − a an integer multiple of theperiod τ for f (x), and if f (x) is infinitely differentiable, then theerror I − Tn decreases to zero more rapidly than n−m for anym > 0. For periodic integrands, the trapezoidal rule is an optimalnumerical integration method.

Example

Consider evaluating

I =

∫ 2π

0

sin x

1 + esin xdx

Using the trapezoidal rule, we have the results in the followingtable. In this case, the formulas based on Richardson extrapolationare no longer valid.

n Tn Tn − T 12n

2 0.04 −0.72589193317292 −7.259E − 18 −0.74006131211583 −1.417E − 2

16 −0.74006942337672 −8.111E − 632 −0.74006942337946 −2.746E − 1264 −0.74006942337946 0.0