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Numerical Integration 1.0.1 Error Estimations for Trapezoidal Rule

In the preceding section, numerical results for all integrands but one showed a

regular behavior in the error for both trapezoidal and Simpson rules. To explain

this regular behavior we consider error formulas for these integration methods.

These formulas will lead to a better understanding of the methods, showing

both their weakness and strengths, and they will allow improvements of the

methods. We begin by examining the error of the trapezoidal rule.

Theorem 0.0. Trapezoidal Error Estimation

Let f+x/ have two continuous derivatives on #a, b', and let n be a positiveinteger. Then for the error in integrating

I+f/ a

bf+x/ x

using the trapezoidal rule Tn+f/ of (0.XXX), we have

En

T

+f

/ I

+f

/ T

n+f

/

h2+b a/12

f''

+cn/

The numbercn is some unknown point in #a, b', and h +b a/ sn.Formula (0.0) can be used to bound the error in Tn+f/, generally by boundingthe term f'' +cn/ by its largest possible value on the interval #a, b'. This will beillustrated in the following example. Also note that the formula forEn

T+f/ isconsistent with the behavior of the error observed in the calculations (examine

this).

Example 0.0. Error Estimation

Recall the examples we discussed in the previous section, with

I 0

1 1

1 xx ln+2/.

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Solution 0.7. Here f+x/ 1 s +1 x/, #a, b' #0, 1', and f'' +x/ 2 t +1 x/3.Substituting into the error formula (0.0), we obtain

En

T

+f/ h2

12 f'' +cn/ 0 cn 1, h 1 s n.The formula cannot be computed exactly because cn is not known. But we can

bound the error by looking at the largest possible value for f'' +cn/ . Boundf'' +x/ on #a, b' #0, 1'

max0x12

+1 x/3 2.

Then

EnT+f/ h

2

12+2/ h

2

6.

Forn 1 and n 2, we have

E1T+f/ 1

6and E2

T+f/ +1 s 2/2

6.

Comparing these results with the true errors, we see that these bounds are two

to three times the actual errors.

A possible weakness in the trapezoidal rule can be inferred from the

assumptions of Theorem 0.0. Iff+x/ does not have two continuous derivativeson #a, b', then does Tn+f/ converge more slowly? The answer is yes for somefunctions, especially if the first derivative is not continuous.

The error formula (0.0) can only be used to bound the error, because f'' +cn/ isunknown. This will be improved on by a more careful consideration of the error

formula.

A central element of our proof of (0.0) leis in being able to demonstrate the

n 1 case for an interval #, h':

hf+x/ xh% f+/ f+ h/

2) h

3

12f'' +c/

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for some c in #, h'. We will use this formula to obtain the general formula(0.0) in Theorem 0.0.

Recall the derivation of the trapezoidal rule Tn+f/ as given in (0.XXX). ThenEnT+f/

a

bf+x/ x Tn+f/

x0

xnf+x/ xTn+f/

i0

n1

xi

xi1f+x/ xh f+xi/ f+xi1/.

Apply (0.0) to each of the terms on the right side of (0.0), to obtain

EnT+f/ h

3

12i1

n

f'' +i/.

The unknown constants 1, 2, , n are located in the respective subintervals

#x0, x1', #x1, x2', #xn1, xn'. By factoring (0.0), we obtain

EnT+f/ h

2

12i1

n

h f'' +i/.

The sum in (0.0) is equal to +b a/ f'' +cn/ for some cn #a, b', thus obtainingthe general case of (0.0).

To estimate the trapezoidal error, observe that the term of the sum in (0.0) is a

Riemann sum for the integral

a

bf'' +x/ x f' +b/ f' +a/.

The Riemann sum is based on the partition of#a, b' as n , this sum willapproach the integral (0.0). Using (0.0) to estimate the right side of (0.0), we

find that

EnT+f/ h2

12#f' +b/ f' +a/'.

This error estimate will be denoted EnT+f/. It is called the asymptotic estimate of

the error because it improves as n increases. As long as f' +x/ is computableEnT+f/ will be very easy to be derived.

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Example 0.0. Error Estimation for Trapezoidal Rule

Again consider the case

I 01 1

1 xx.

Solution 0.8. Then f' +x/ 1 t +1 x/2 and (0.0) yields the estimate

EnT+f/ h

2

12% 1+1 1/2

1

+1 0/2 ) h2

16with h

1

n.

Forn 1 and n 2

E

1T+f/ 1

16 0.0625

E

2T+f/ 1

64 0.0156

These compare quite closely to the true errors calculated above.

The estimate EnT+f/ has several practical advantages over the earlier error

formula (0.0). First, it confines that when n is doubled the error decreases by a

factor of about 4, provided that f' +b/ f' +a/ 0. This agrees with the results forour calculations in Example 0.XXX. Second (0.0) implies that the convergenceofTn+f/ will be more rapid when f' +b/ f' +a/ 0. This is a partial explanation ofthe very rapid convergence observed in the third integral in Example 0.XXX.

Finally, (0.0) leads to a more accurate numerical integration formula by taking

EnT+f/ into account:

I+f/ Tn+f/ h2

12+f' +b/ f' +a//

which can be write as

I+f/ Tn+f/ h2

12+f' +b/ f' +a//.

This formula is called the corrected trapezoidal rule, and it will be denoted by

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CTn+f/. This correction needs an extension of our formula defined above. Thefollowing lines define the extension of the formula.

correctedTrapezoidalMethod#f_, x_, a_, b_, n_' :Block%h,

h +b a/sn;fp xf;

trapezoidalMethod#f, x, a, b, n' h2

12++fp s. x! b/ +fp s. x ! a//

)

Example 0.0. Application of CTn+f/Let us evaluate the integral

I 0

1

x2x 0.746824

and compare it with different orders of approximations by Tn+f/ and CTn+f/.Solution 0.9. The exact value of this integral is

I0 0

1

x2

x

1

2 erf+1/

where Erf+1/ is the so called error function at 1. Using the numerical proceduresforTn+f/ and CTn+f/ for orders of

n 2, 4, 8, 16, 32, 64, 128, 256, 512;then the trapezoidal method delivers the results

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Tn -trapezoidalMethod-x2 , x, 0, 1, 11 &1 s n

0.73137, 0.742984, 0.745866, 0.746585,0.746764, 0.746809, 0.74682, 0.746823, 0.746824

For the corrected trapezoidal method the results are

CTn -correctedTrapezoidalMethod-x2, x, 0, 1, 11 &1 s n

0.746699, 0.746816, 0.746824, 0.746824,0.746824, 0.746824, 0.746824, 0.746824, 0.746824

The errors of the two methods are

1 I0 Tn

0.0154539, 0.00384004, 0.000958518, 0.000239536, 0.0000598782,0.0000149692, 3.74227 106, 9.35566 107, 2.33891 107

2 I0 CTn

0.000125571, 7.9575 106, 4.98589 107, 3.11797 108, 1.949 109,1.21817 1010, 7.61358 1012, 4.75842 1013, 2.9643 1014

The following table collects these information

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EnT+f/ I+f/ Tn+f/ h

2+b a/12

f'' +cn/

The numbercn is some unknown point in #a, b', and h +b a/ sn.Formula (0.0) can be used to bound the error in Tn+f/, generally by boundingthe term f'' +cn/ by its largest possible value on the interval #a, b'. This will beillustrated in the following example. Also note that the formula forEn

T+f/ isconsistent with the behavior of the error observed in the calculations (examine

this).

Example 0.0. Error Estimation

Recall the examples we discussed in the previous section, with

I 0

1 1

1 xx ln+2/.

Solution 0.7. Here f+x/ 1 s +1 x/, #a, b' #0, 1', and f'' +x/ 2 t +1 x/3.Substituting into the error formula (0.0), we obtain

EnT+f/ h

2

12f'' +cn/ 0 cn 1, h 1 s n.

The formula cannot be computed exactly because cn is not known. But we canbound the error by looking at the largest possible value for f'' +cn/ . Boundf'' +x/ on #a, b' #0, 1'

max0x12

+1 x/3 2.

Then

EnT+f/ h2

12 +2/ h2

6.

Forn 1 and n 2, we have

E1T+f/ 1

6and E2

T+f/ +1 s 2/2

6.

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Comparing these results with the true errors, we see that these bounds are two

to three times the actual errors.

A possible weakness in the trapezoidal rule can be inferred from the

assumptions of Theorem 0.0. Iff

+x

/does not have two continuous derivatives

on #a, b', then does Tn+f/ converge more slowly? The answer is yes for somefunctions, especially if the first derivative is not continuous.

The error formula (0.0) can only be used to bound the error, because f'' +cn/ isunknown. This will be improved on by a more careful consideration of the error

formula.

A central element of our proof of (0.0) leis in being able to demonstrate the

n 1 case for an interval #, h':

hf+x/ xh% f+/ f+ h/

2) h

3

12f'' +c/

for some c in #, h'. We will use this formula to obtain the general formula(0.0) in Theorem 0.0.

Recall the derivation of the trapezoidal rule Tn+f/ as given in (0.XXX). ThenEnT+f/

a

bf+x/ x Tn+f/

x0

xnf+x/ xTn+f/

i0

n1

xi

xi1f+x/ xh f+xi/ f+xi1/.

Apply (0.0) to each of the terms on the right side of (0.0), to obtain

EnT+f/ h

3

12i1

n

f'' +i/.

The unknown constants 1, 2, , n are located in the respective subintervals

#x0, x1', #x1, x2', #xn1, xn'. By factoring (0.0), we obtainEnT+f/ h

2

12i1

n

h f'' +i/.

The sum in (0.0) is equal to +b a/ f'' +cn/ for some cn #a, b', thus obtainingthe general case of (0.0).

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To estimate the trapezoidal error, observe that the term of the sum in (0.0) is a

Riemann sum for the integral

abf''

+x

/x f'

+b

/ f'

+a

/.

The Riemann sum is based on the partition of#a, b' as n , this sum willapproach the integral (0.0). Using (0.0) to estimate the right side of (0.0), we

find that

EnT+f/ h

2

12#f' +b/ f' +a/'.

This error estimate will be denoted EnT

+f

/. It is called the asymptotic estimate of

the error because it improves as n increases. As long as f' +x/ is computableEnT+f/ will be very easy to be derived.

Example 0.0. Error Estimation for Trapezoidal Rule

Again consider the case

I 0

1 1

1 xx.

Solution 0.8. Then f' +x/ 1 t +1 x/2 and (0.0) yields the estimate

EnT+f/ h

2

12% 1+1 1/2

1

+1 0/2 ) h2

16with h

1

n.

Forn 1 and n 2

E

1T+f/ 1

16 0.0625

E

2T+f/ 1

64 0.0156

These compare quite closely to the true errors calculated above.

The estimate EnT+f/ has several practical advantages over the earlier error

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formula (0.0). First, it confines that when n is doubled the error decreases by a

factor of about 4, provided that f' +b/ f' +a/ 0. This agrees with the results forour calculations in Example 0.XXX. Second (0.0) implies that the convergence

ofTn+f/ will be more rapid when f' +b/ f' +a/ 0. This is a partial explanation ofthe very rapid convergence observed in the third integral in Example 0.XXX.Finally, (0.0) leads to a more accurate numerical integration formula by taking

EnT+f/ into account:

I+f/ Tn+f/ h2

12+f' +b/ f' +a//

which can be write as

I+f/ Tn+f/ h2

12 +f' +b/ f' +a//.This formula is called the corrected trapezoidal rule, and it will be denoted by

CTn+f/. This correction needs an extension of our formula defined above. Thefollowing lines define the extension of the formula.

correctedTrapezoidalMethod#f_, x_, a_, b_, n_' :Block%h,

h

+ba/sn

;

fp xf;

trapezoidalMethod#f, x, a, b, n' h2

12++fp s. x! b/ +fp s. x ! a//

)

Example 0.0. Application of CTn+f/

Let us evaluate the integral

I 0

1

x2x 0.746824

and compare it with different orders of approximations by Tn+f/ and CTn+f/.Solution 0.9. The exact value of this integral is

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I0 0

1

x2

x

1

2 erf+1/where Erf+1/ is the so called error function at 1. Using the numerical proceduresforTn+f/ and CTn+f/ for orders of

n 2, 4, 8, 16, 32, 64, 128, 256, 512;then the trapezoidal method delivers the results

Tn -trapezoidalMethod-x2 , x, 0, 1, 11 &1 s n

0.73137, 0.742984, 0.745866, 0.746585,0.746764, 0.746809, 0.74682, 0.746823, 0.746824

For the corrected trapezoidal method the results are

CTn

-correctedTrapezoidalMethod-x2

, x, 0, 1,

11 &1 s

n

0.746699, 0.746816, 0.746824, 0.746824,0.746824, 0.746824, 0.746824, 0.746824, 0.746824

The errors of the two methods are

1 I0 Tn

0.0154539, 0.00384004, 0.000958518, 0.000239536, 0.0000598782,0.0000149692, 3.74227 106, 9.35566 107, 2.33891 107

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2 I0 CTn

0.000125571, 7.9575 106, 4.98589 107, 3.11797 108, 1.949 109,1.21817 1010, 7.61358 1012, 4.75842 1013, 2.9643 1014

The following table collects these information

Prepend#Transpose#n, H1, H2, H2 s RotateLeft#H2'',"n", "H1", "H2", "Ratio"' ss TableForm

n 1 2 Ratio

2 0.0154539 0.000125571 15.7802

4 0.00384004 7.9575 106 15.968 0.000958518 4.98589 107 15.9908

16 0.000239536 3.11797 108 15.9978

32 0.0000598782 1.949 109 15.9994

64 0.0000149692 1.21817 1010 16.

128 3.74227 106 7.61358 1012 16.0002

256 9.35566 107 4.75842 1013 16.0524

512 2.33891 107 2.9643 1014 2.36065 1010

The table shows that the corrected trapezoidal converges quite rapidly

compared with the conventional method Tn+f/. When n is doubled the error inCTn+f/ decreases by a factor of about 16.1.0.2.0 Error Formulas for Simpson's Rule

The type of analysis used in the preceding discussion can also be used to

derive corresponding error formulas for Simpson's rule. These are stated in the

following theorem, with the proof omitted.

Theorem 0.0. Error for Simpson's Rule

Assume f+x/ has four continuous derivatives on #a, b', and let n be an evenpositive integer. Then the error in using Simpson's rule is given by

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EnS+f/ I+f/ Sn+f/ h

4+b a/180

f+4/+cn/

with cn an unknown point in #a, b' and h +b a/ s n. Moreover, this error can beestimated with the asymptotic error formula

EnS+f/ h

4

180+f''' +b/ f''' +a//.

Note that (0.0) say that Simpson's rule is exact for all f+x/ that are polynomialsof degree 3, whereas the quadratic interpolation on which Simpson's rule is

based is exact only forf+x/ a polynomial of degree 2. The degree of precisionbeing 3 leads to the powerh4 in the error, rather than the powerh3, which

would have been produced on the basis of the error in quadratic interpolation. It

is this higher powerh4 in the error and the simple form of the method that

historically have caused Simpson's rule to be the most popular numerical

integration rule.

Example 0.0. Error in Simpson's Rule

Estimate the error for the integral

I 0

1 1

1 xx

by applying Simpson's error formula.

Solution 0.10. First let us define the function and the derivatives of third and

fourth order

f+x_/ : 1x 1

f3

3 f+x/xxx

6

+x 1/4

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f4 4 f+x/

xxxx

24

+x 1/5The exact error is given by

EnS+f/ h

4

180f+4/+cn/, h 1

n

for some 0 cn 1. We can bound it by

1

180,h40 f4 s.x 0

2 h415

The asymptotic error is given by

a 1

180,h40 ++f3 s.x 1/ +f3 s.x 0//

h4

32

Now if we set n 2, E

2S+f/ 0.00195; for comparison the actual error resulting

from the calculations by applying Simpson's rule is 0.001308.

The behavior in I+f/ Sn+f/ can be derived from (0.0). When n is doubled, h ishalved, and h4 decreases by a factor of 16. Thus, the errorEn

S+f/ shoulddecrease by the same factor, provided that f''' +b/ f''' +a/. This is the errorbehavior observed in Example 0.0.The theory of asymptotic error formulas

En+f/ En+f/

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such as forEnT+f/ and EnT+f/, says that (0.0) is valid provided that

limnE

n+f/En

+f

/ 1.

The needed size ofn in (0.0) will vary with the integrand f, which is illustrated

by Example 0.0. From (0.0) and (0.0), we also are lead to infer that Simpson's

rule will not perform as well iff+x/ is not four times continuously differentiableon #a, b'. This is correct for most such functions, and other numerical methodsare often necessary for integrating them.

Example 0.0. Simpson's Rule with Slow Convergence

Use Simpson's Rule to approximate

I 0

1x x

2

3.

Solution 0.11. The integration for different orders of iteration is done by

S2 .simpsonMethod. x , x, 0 , 1, 12 &2 s2, 4, 8, 16, 32, 64, 128, 256, 512, 1024

{0.638071, 0.656526, 0.663079, 0.665398, 0.666218, 0.666508, 0.666611,0.666647, 0.66666, 0.666664}

The error of the numerical integration with respect to the exact value is

2

3 S2

0.0285955, 0.0101404, 0.00358739, 0.00126848, 0.000448484, 0.000158564,0.0000560607, 0.0000198205, 7.00759 106, 2.47756 106

The ratio of two consequent values is determined by

ratio

RotateLeft#'{2.81996, 2.82668, 2.8281, 2.82837, 2.82842, 2.82843, 2.82843, 2.82843,

2.82843, 0.0000866416}

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The error and the ratio of the errors is collected in the following table

Prepend[Transpose[{{2, 4, 8, 16, 32, 64, 128, 256, 512, 1024}, , ratio}], {"n",

"", "Ratio"}] // TableForm

"n" "" "Ratio"

2 0.0285955 2.81996

4 0.0101404 2.82668

8 0.00358739 2.8281

16 0.00126848 2.82837

32 0.000448484 2.82842

64 0.000158564 2.82843

128 0.0000560607 2.82843

256 0.0000198205 2.82843

512 7.00759 106 2.82843

1024 2.47756 106 0.0000866416

The column Ratio shows that the convergence is much slower.

As was done by the trapezoidal rule, a corrected Simpson's rule can be defined:

CSn+f/ Sn+f/ h4

180+f''' +b/ f''' +a//.

This will be usually a more accurate approximation than Sn+f/. The correctedSimpson's rule can be implemented by the following lines

correctedSimpsonsMethod#f_, x_, a_, b_, n_' :Block%h,

h +b a/sn;fp x,x,xf;

simpsonMethod

#f,

x, a, b

, n

'

h4

180++fp s. x! b/ +fp s. x ! a//

)

Example 0.0. Application of CSn+f/

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Let us evaluate the integral

I 0

1

x2x 0.746824

and compare it with different orders of approximations by Sn+f/ and CSn+f/.Solution 0.12. The exact value of this integral formula

I0 0

1

x2x

1

2 erf+1/

where Erf+1/ is the so called error function at 1. Using the numerical proceduresforSn+f/ and CSn+f/ for orders ofn = {2, 4, 8, 16, 32, 64, 128, 256, 512};

Then the trapezoidal method delivers the results

Sn .simpsonMethod.x2 , x, 0 , 1, 12 &2 s n{0.74718, 0.746855, 0.746826, 0.746824, 0.746824, 0.746824, 0.746824,

0.746824, 0.746824}

For the corrected trapezoidal method the results are

CSn .correctedSimpsonsMethod.x2 , x, 0 , 1, 12 &2 s n{0.746669, 0.746823, 0.746824, 0.746824, 0.746824, 0.746824, 0.746824,

0.746824, 0.746824}

The errors of the two methods are

1 = I0 - Sn

0.000356296, 0.000031247, 1.98772 106,1.24623 107, 7.79456 109, 4.87245 1010,

3.04541 1011, 1.90337 10

12, 1.19016 1013

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2 = I0 - CSn

0.000154648, 6.87001 107, 8.15866 109, 1.18803 1010,1.82354 1012, 2.84217 1014, 5.55112 1016, 0., 0.

The following table collects these information

Prepend[Transpose[{n, 1, 2, 2/RotateLeft[2]}], {"n", "1", "2", "Ratio"}] //

TableForm

"n" "1" "2" "Ratio"

2 0.000356296 0.000154648 225.105

4 0.000031247 6.87001 107 84.205

8 1.98772 106 8.15866 109 68.6738

16 1.24623 107 1.18803 1010 65.1497

32 7.79456 109 1.82354 1012 64.1602

64 4.87245 1010 2.84217 1014 51.2

128 3.04541 1011 5.55112 1016 ComplexInfinity

256 1.90337 1012 0. Indeterminate

512 1.19016 1013 0. 0.

The table shows that the corrected Simpson's converges quite rapidly

compared with the conventional method Sn+f/. When n is doubled the error inCSn+f/ decreases by a factor of about 68.

1.1 Gaussian Numerical Integration

The numerical methods studied in the last section were based on integrating

linear and quadratic interpolating polynomials, and the resulting formulas were

applied on subdivisions of ever smaller subintervals. In this section, weconsider a numerical method that is based on the exact integration of

polynomials of increasing degree; no subdivision of the integration interval is

used. To motivate this approach, recall from Section 2.4 of Chapter 2 the

material on approximation of functions.

Let f+x/ be continuous on #a, b'. Then n+f/ denotes the smallest error boundthat can be attained in approximating f+x/ with a polynomial pn+x/ of degree n

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on the given interval a x b. The polynomial pn+x/ that yields thisapproximation is called the minimax approximation of degree n forf+x/,maxaxb f+x/ pn+x/ n+f/and n+f/ is called the minimax error. From Theorem 3.1 of Chapter 3, it can beseen that n+f/ will often converge to zero quite rapidly.If we have a numerical integration formula to integrate low- to moderate-degree

polynomials exactly, then the hope is that the same formula will integrate other

functions f+x/ almost exactly, iff+x/ is well approximated by such polynomials.To illustrate the derivation of such integration formulas, we restrict our attention

to the integral

I+f/ 11

f+x/ x.Its relation to integrals over other intervals #a, b' will be discussed later.The integration formula is to have the general form

In+f/ j1

n

wj f,xj0

and we require that the nodes

x

1, , x

nand weights

w

1, , w

nbe so

chosen that In+f/ I+f/ for all polynomials f+x/ of as large a degree as possible.Case n 1 The integration formula has the form

1

1f+x/ x w1 f+x1/.

It is to be exact for polynomials of as large a degree as possible.

Using f+x/ 1 and forcing equality in (0.0) gives us2 w1.

Now use f+x/ xand again force equality in (0.0). Then0 w1 x1

which implies x1 0. Thus (0.0) becomes

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1

1f+x/ x 2 f+0/ I1+f/.

This is the midpoint formula from the trapezoidal approximation. The formula

(0.0) is exact for all linear polynomials.

To see that (0.0) is not exact for quadratics, let f+x/ x2. Then the error in (0.0)is given by

1

1x

2x 2 +0/2 2

3 0.

Case n 2 The integration formula is

11

f+x/ x w1 f+x1/ w2 f+x2/.and it has four unspecified quantities: x1, x2, w1, and w2. To determine these,

we require it to be exact for the four monomials

f+x/ 1, x, x2, x3.This leads to the four equations

2 w1 w2

0 w1 x1 w2 x22

3 w1 x1

2 w2 x22

0 w1 x13 w2 x2

3

This is a nonlinear system in four unknowns;

Lecture_009.nb 21

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calculation to not be exact for the degree 4 polynomial f+x/ x4. Thus I2+f/ hasdegree of precision 3.

For cases n 3 there occurs a problem in the solution for the weights and the

interpolation points because the determining system of equations becomes

nonlinear. The following function generates the determining equations for a

Gau integration.

gaussIntegration#f_, x_, a_, b_, n_' : Block%,varsX

Table#ToExpression#StringJoin#"x", ToString#i''',i, 1, n';

varsW

Table#ToExpression#StringJoin#"w", ToString#i''',i, 1, n';

vec1 Table$varsXi, i, 0, 2 n 1(;

vecB Table%If%EvenQ#i', Abs% 22 i 1

), 0),

i, 0, 2 n 1);equations Thread#Map#varsW. &, vec1' vecB'

)

soli gaussIntegration#f, x, a, b, 3' ss TableForm

w1 w2 w3 m 2

w1 x1 w2 x2 w3 x3 m 0

w1 x12 w2 x2

2 w3 x3

2m

2

3

w1 x13 w2 x2

3 w3 x3

3m 0

w1 x14 w2 x2

4 w3 x3

4m

2

7

w1 x15 w2 x25 w3 x35 m 0

We clearly observe that the equations are nonlinear due to the fact that the xiare not specified yet. The problem here is that there exist no reliable procedure

to find the solutions for nonlinear algebraic equations.

Lecture_009.nb 23