mn5-2013 [tryb zgodności]home.agh.edu.pl/~zak/downloads/nm5.pdf · 2013-05-07 · romberg’s...

Numerical Methods lecture 5 1

Numerical Methods

dr hab. inż. Katarzyna Zakrzewska, prof. AGH

Lecture 5.

Numerical integration

Outline

• Trapezoidal rule• Multi-segment trapezoidal rule• Richardson extrapolation• Romberg's method• Simpson's rule• Gaussian method

Numerical Methods lecture 5

Numerical integration - idea

adx)x(fI

dxxf )(

The integral can be approximated by:

∑ Δ==

iii xcfS

1+≤≤ iii xcx

Newton–Cotes methods

Newton – Cotes integration belongs to a class of methods with fixed nodes: function f(x) is interpolatedby a polynomial (e.g. Lagrange polynomial)

)x(f)x(f n≈where: n

nn xaxa...xaa)x(f ++++= −−

Then, the integral of f(x) can be approximated as an integral of the interpolated function fn(x)

dxxfdxxfIb

a∫∫ ≈= )()(

Trapezoidal rule

The trapezoidal rule assumes: n = 1, thus:

xaaxf 101 )( +=

dxxaadxxfdxxfIb

)()()( 101 +=≈= ∫∫∫

But what is a0 and a1 ?2

10abaaba −

Now if one chooses, (a,f(a)) i (b,f(b)) as the two points to approximate f(x) by a straight line from a to b. It follows that:

aaaafaf 101 )()( +==

baabfbf 101 )()( +==ab

abfbafa−−

abafbfa

−−

Trapezoidal rule

trapezoidofareadxxfb

≈∫ )(

⎥⎦⎤

⎢⎣⎡ +

−=∫ 2)()()()( bfafabdxxf

Trapezoidal rule

tv 8.92100140000

140000ln2000)( −⎥⎦⎤

⎢⎣⎡

Example 1:Velocity v(t) of a rocket from t1=8 s to t2=30 s is given by:

a) Use the single segment trapezoidal rule to find the distance covered by the rocket from t1=8 s to t2=30 s

b) Find the true relative error.

Trapezoidal rule

⎥⎦⎤

⎢⎣⎡ +

−≈2

)()()( bfafabI sa 8= sb 30=

tf 8.92100140000

140000ln2000)( −⎥⎦⎤

⎢⎣⎡

)8(8.9)8(2100140000

140000ln2000)8( −⎥⎦

⎤⎢⎣

⎡−

)30(8.9)30(2100140000

140000ln2000)30( −⎥⎦

⎤⎢⎣

⎡−

sm /27.177=

sm /67.901=

⎥⎦⎤

⎢⎣⎡ +

67.90127.177)830(I m11868=

Trapezoidal rule

∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛−⎥⎦

⎤⎢⎣⎡

−=Δ30

2100140000140000ln2000 dtt

tx m11061=

b) The true value

10011061

1186811061×

−=∈t %.29597=

The relative error:

Multi-segment trapezoidal rule

nabh −

The true error using a single segment trapezoidal rule was large. We can divide the interval from a to b into smaller n segments of equal length – h and apply the trapezoidal rule over each segment:

∫ ∫ ∫++∫ +=∫=+

adxxfdxxfdxxfdxxfdxxfI

3)()()()()(

for n=4X

aba −+ 4

2 aba −+

43 aba −

adx)x(f ⎥⎦

⎤⎢⎣⎡ +

⎭⎬⎫

⎩⎨⎧ ++

−= ∑

=)b(f)iha(f)a(f

∫∫∫∫−+

++++=b

adx)x(fdx)x(f...dx)x(fdx)x(f

adx)x(f

)2()(2

)()(+⎥⎦

⎤⎢⎣⎡ +++

+⎥⎦⎤

⎢⎣⎡ ++

=hafhafhhafafh

⎥⎦⎤

⎢⎣⎡ +−+

−+−+2

)())1((])1(([... bfhnafhnab

tv 8.92100140000

140000ln2000)( −⎥⎦⎤

⎢⎣⎡

Example 2:Velocity v(t) of a rocket from t1=8 s to t2=30 s is given by:

a) Use the complex segment trapezoidal rule to find the distance covered from t1=8 s to t2=30 s for n = 2

b) Find the true relative error.

a) ⎥⎦⎤

⎢⎣⎡ +

⎭⎬⎫

⎩⎨⎧∑ ++

=)()(2)(

1bfihafaf

n = 2 a = 8 s b = 30 s sn

abh 112830=

⎥⎦⎤

⎢⎣⎡ +

⎭⎬⎫

⎩⎨⎧∑ ++

=)30()(2)8(

)2(2830 12

1fihaffI

[ ])30()19(2)8(422 fff ++= [ ]6790175484227177

422 .).(. ++=

m11266=

∫ ⎟⎠⎞

⎜⎝⎛ −⎥⎦

⎤⎢⎣⎡

−=30

892100140000

1400002000 dtt.t

lnx m11061=

true value:

The relative error:

10011061

1126611061×

−=∈t %8534.1=

n ∆x Et

1 11868 -807 7.296 ---

2 11266 -205 1.853 5.343

3 11153 -91.4 0.8265 1.019

4 11113 -51.5 0.4655 0.3594

5 11094 -33.0 0.2981 0.1669

6 11084 -22.9 0.2070 0.09082

7 11078 -16.8 0.1521 0.05482

8 11074 -12.9 0.1165 0.03560

%t∈ %a∈

The relative error for a simple trapezoidal rule is

ba),("f)ab(Et <ζ<ζ−

The relative error in the multi-segment trapezoidal rule is a sum of errors for each segment. The relative error within the first segment is given by:

[ ] haa),("fa)ha(E +<ζ<ζ−+

)("fh1

Estimation of error

By analogy:

[ ] ihahiafhiaihaE iii +<ζ<−+ζ−+−+

= )1(),("12

))1(()( 3

)("fhiζ=

for n-th segment :

{ }[ ] bh)n(a),("fh)n(abE nnn <ζ<−+ζ−+−

)("fhnζ=

Estimation of error

The total error in the complex trapezoidal rule is a sum of the errors for a single segment:

iit EE

ii )("fh

−= 1

Formula:

yields an approximate average value of the second derivative in the range of

bxa <<

Et α∝

Estimation of error

Table below presents the results for the integral

∫ ⎟⎠⎞

⎜⎝⎛ −⎥⎦

⎤⎢⎣⎡

21001400001400002000 dtt.

as a function of the number of segments n. When n twice increases, the absolute error Et decreases four times!

tE %t∈ %a∈n Value

2 11266 -205 1.854 5.343

4 11113 -51.5 0.4655 0.3594

8 11074 -12.9 0.1165 0.03560

16 11065 -3.22 0.02913 0.00401

Estimation of error

Richardson’s extrapolation and Romberg’s methodof integration

Richardson’s extrapolation and Romberg’s method of integration constitute an extension of the trapezoidal method and give better approximation of the integral by reducing the true error.

2nCEt ≅

nt ITVE −=

Richardson extrapolation

The true error obtained when using the multi-segmenttrapezoidal rule with n segments to approximate an integral isgiven by:

where: C is an approximate constant of proportionality

Since:

true value approximate value usingn-segments

( ) nITVnC

222−≅

If the number of segments is doubled from n to 2n:

ITV−

It can be shown that:

( ) nITVnC

−≅2

( ) nITVnC

222−≅

We get:

tv 8.92100140000

140000ln2000)( −⎥⎦⎤

⎢⎣⎡

Example 3:

The velocity v(t) of a rocket from t1=8 s to t2=30 is given by:

a) Use Richardson extrapolation rule to find the distance coveredfor n = 2

b) Find the relative true error

%t∈ %a∈n ∆x Et

1 11868 -807 7.296 ---

2 11266 -205 1.853 5.343

3 11153 -91.4 0.8265 1.019

4 11113 -51.5 0.4655 0.3594

5 11094 -33.0 0.2981 0.1669

6 11084 -22.9 0.2070 0.09082

7 11078 -16.8 0.1521 0.05482

8 11074 -12.9 0.1165 0.03560

Table of results for n = 8 segments (trapezoidal rule)

a) mI 112662 = mI 111134 =

ITV−

+≅ for n=2

4IIITV −

+≅3112661111311113 −

m11062=

∫ ⎟⎠⎞

⎜⎝⎛ −⎥⎦

⎤⎢⎣⎡

−=30

21001400001400002000 dtt.

tlnx m11061=

The absolute true error:

1106211061−=tE m1−=

The true value:

c) The relative error:

10011061

1106211061×

−=∈t %.009040=

n ∆x (m)Trapezoidal

rule Trapezoidal rule

∆x (m)Richardson

extrapolation Richardson extrapolation

11868112661111311074

7.2961.8540.46550.1165

--110651106211061

--0.036160.0090410.0000

Comparison of different methods:

%t∈ %t∈

Romberg’s method uses the same pattern as Richardson extrapolation. However, Romberg used a recursive algorithm for the extrapolation as follows:

ITV−

14 122

2 −−

+= −nn

Romberg's method

The true value TV is replaced by the result of the Richardson extrapolation

Note also that the sign is replaced by the sign =

( )RnI 2≅

Esimated true value is given by: ( ) 42 ChITV Rn +≅

where: Ch4 is the value of the error of approximation

Romberg's method

Another value of integral obtained while doubling the number ofsegments from 2n to 4n:

Estimated true value is given by:

( ) ( ) ( )15

RnRnRn

−+≅

( ) ( )14

)( 1324

−+= −

RnRnRn

214 1111

11 ≥−

−−+−

+− k,II

II kj,kj,k

j,kj,k

The index k represents the order of extrapolation

k=1 represents the values obtained from the regular trapezoidal rule

A general expression for Romberg integration can be written as:

Romberg's method

k=2 represents the values obtained using the true error estimate as O(h2)

The value of an integral with for j+ 1 is more accurate than the value of the integral for j index

tv 8.92100140000

140000ln2000)( −⎥⎦⎤

⎢⎣⎡

Example 4:

Velocity v(t) of a rocket from t1=8 s to t2=30 s is given by:

a) Use Romberg’s method to find the distance covered. Usen = 1, 2, 4, and 8

b) Find the absolute true error and the relative approximate error

Romberg's method

%t∈ %a∈

Table of results for n = 8 segments (trapezoidal rule)

n ∆x Et

1 11868 -807 7.296 ---

2 11266 -205 1.853 5.343

3 11153 -91.4 0.8265 1.019

4 11113 -51.5 0.4655 0.3594

5 11094 -33.0 0.2981 0.1669

6 11084 -22.9 0.2070 0.09082

7 11078 -16.8 0.1521 0.05482

8 11074 -12.9 0.1165 0.03560

From this table, the initial values from the trapezoidal rule are:

1186811 =,I 1126621 =,I

1111331 =,I 1107441 =,I

Romberg's method

To get the first order extrapolation values:

2112,,

3118681126611266 −

Similarly,

32,13,1

3,12,2

3112661111311113 −

11062=

33,14,1

4,13,2

3111131107411074 −

11061=

Romberg's method

For the second order extrapolation:

151222

2213,,

15110651106211062 −

11062=

Similarly,15

2,23,23,22,3

15110621106111061 −

11061=

Romberg's method

For the third order extrapolation

631323

2314,,

63110621106111061 −

m11061=

Romberg's method

1-segment

2-segment

4-segment

8-segment

Appr. 1 Appr. 2 Appr. 3

Improved estimates of the value of an integral using Romberg integration

Romberg's method

Simpson's rule

The trapezoidal rule was based on approximating the integrand by a first order polynomial, and then integrating the polynomial over an interval from a to b. Simpson’s rule assumes that the integrand can be approximated by a second order polynomial.

∫∫ ≈=b

adx)x(fdx)x(fI 2

where:

22102 xaxaa)x(f ++=

Simpson's rule

A parabola passing throughthree points :

)),a(f,a(

,baf,ba⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛ ++22

))b(f,b(2

2102 aaaaa)a(f)a(f ++==2

2102 2222⎟⎠⎞

⎜⎝⎛ +

+⎟⎠⎞

⎜⎝⎛ +

+=⎟⎠⎞

⎜⎝⎛ +

=⎟⎠⎞

⎜⎝⎛ + baabaaabafbaf

22102 babaa)b(f)b(f ++==

Simpson's rule

)a(fb)a(abfbaabf)b(abf)b(faa

++⎟⎠⎞

⎜⎝⎛ +

221 22

)b(bfbabf)a(bf)b(afbaaf)a(afa

+⎟⎠⎞

⎜⎝⎛ +

−++⎟⎠⎞

⎜⎝⎛ +

−−=

222 22

)b(fbaf)a(fa

⎟⎠⎞

⎜⎝⎛ +⎟

⎠⎞

⎜⎝⎛ +

Coefficients a0,a1,a2 are:

Simpson's rule

∫≈b

adx)x(fI 2

( )∫ ++=b

adxxaxaa 2

xaxaxa ⎥⎦

⎤⎢⎣

⎡++=

10abaaba)ab(a −

Since:

Simpson's rule

⎥⎦⎤

⎢⎣⎡ +⎟

⎠⎞

⎜⎝⎛ +

=∫ )b(fbaf)a(fabdx)x(fb

2abh −

It follows that:

⎥⎦

⎤⎢⎣

⎡+⎟⎠⎞

⎜⎝⎛ +

+=∫ )(2

)(2 bfbafafhdxxfb

it is called Simpson’s 1/3 rule

Multi-segment Simpson's rule

...)x(f)x(f)x(f)xx(dx)x(fb

a+⎥⎦

⎤⎢⎣⎡ ++

−=∫ 64 210

...)x(f)x(f)x(f)xx( +⎥⎦⎤

⎢⎣⎡ ++

4 43224

x0 x2 xn-2 xn

.....dx)x(fdx)x(fdx)x(fx

a++= ∫∫∫

∫∫−

xdx)x(fdx)x(f....

...)x(f)x(f)x(f)xx(... nnnnn +⎥⎦

⎤⎢⎣⎡ ++

−+ −−−−− 6

4 23442

⎥⎦⎤

⎢⎣⎡ ++

−+ −−− 6

)x(f)x(f)x(f)xx( nnnnn

hxx ii 22 =− −

n...,,,i 42=

Simpson's rule

...)x(f)x(f)x(fhdx)x(fb

a+⎥⎦

⎤⎢⎣⎡ ++

=∫ 642 210

...)x(f)x(f)x(fh +⎥⎦⎤

⎢⎣⎡ ++

42 432

...)x(f)x(f)x(fh nnn +⎥⎦⎤

⎢⎣⎡ ++

+ −−−

642 234

⎥⎦⎤

⎢⎣⎡ ++

+ −−

642 12 )x(f)x(f)x(fh nnn

Simpson's rule

adx)x(f { }[ ]...)x(f...)x(f)x(f)x(fh

n +++++= −1310 43

{ } )}]()(...)()(2... 242 nn xfxfxfxf +++++ −

⎥⎥

⎢⎢

⎡+++= ∑∑

)()(2)(4)(3

evenii

i xfxfxfxfh

⎥⎥

⎢⎢

⎡+++

−= ∑∑

)()(2)(4)(3

evenii

i xfxfxfxfnab

Approximate values of the integral, using Simpson's rule with multiple segments

n Approximate values Et |Єt |

246810

11065.7211061.6411061.4011061.3511061.34

4.380.300.060.010.00

0.0396%0.0027%0.0005%0.0001%0.0000%

Estimation of errors in Simpson's rule

Error for one segment bafabEt <ζ<ζ−

−= ),(2880

)( )4(5

)(f)xx(E )(1

021 2880

−= ),(fh )(1

90ζ−=

)(f)xx(E )(2

242 2880

−= ),(fh )(2

90ζ−=

210 xx <ζ<

422 xx <ζ<

Simpson's rule errors

Error for the multi-segment

)(f)xx(

E i)()i(i

i ζ−

−= − 45

2880),(fh

i)( ζ−= 4

90ii)i( xx 212 <ζ<−

True error

iit EE ∑

=ζ−=

=ζ−

Simpson's rule errors

90)( f

−−=

the average value of the derivativen

Gauss-Quadrature Method

adx)x(fI )x(fc)x(fc 2211 +≈

Gaussian integral is given by:

Points x1 and x2, which define the value of the integrand are not fixed (as before), but there are a priori distributed randomly within <a,b>.

constant coefficients

.xaxaxaa)x(f 33

2210 +++=

( )∫∫ +++=b

adxxaxaxaadx)x(f 3

xaxaxaxa ⎥⎦

⎤⎢⎣

⎡+++=

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛ −+⎟⎟

⎞⎜⎜⎝

⎛ −+⎟⎟

⎞⎜⎜⎝

⎛ −+−=

10abaabaabaaba

There are four unknowns x1, x2, c1, c2. These are found by assuming that the formula gives exact results for integrating a general third order polynomial:

( ) ( )3232222102

2121101 xaxaxaacxaxaxaacdx)x(f

a+++++++=∫

( ) ( ) ( ) ( )3223113

211222111210 xcxcaxcxcaxcxcacca +++++++=

adx)x(fI )x(fc)x(fc 2211 +≈

( )∫∫ +++=b

adxxaxaxaadx)x(f 3

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛ −+⎟⎟

⎞⎜⎜⎝

⎛ −+⎟⎟

⎞⎜⎜⎝

⎛ −+−=

10abaabaabaaba

Hence:

The formula would then give:

21 ccab +=−2211

2xcxcab

3xcxcab

+=− 3

4xcxcab

( ) ( ) ( ) ( )3223113

211222111210 xcxcaxcxcaxcxcacca +++++++

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛ −+⎟⎟

⎞⎜⎜⎝

⎛ −+⎟⎟

⎞⎜⎜⎝

⎛ −+−=

10abaabaabaaba

21abc −

= 22abc −

ababx ++⎟

⎠⎞

⎜⎝⎛−⎟⎠⎞

⎜⎝⎛ −

22ababx +

+⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛ −

This gives us four equations as follows:

21 ccab +=−

2xcxcab

3xcxcab

4xcxcab

we can find that the above four simultaneous nonlinear equations have only one acceptable solution

( ) ( )

⎟⎟⎠

⎞⎜⎜⎝

⎛ ++⎟

⎞⎜⎝

⎛−−+⎟⎟

⎞⎜⎜⎝

⎛ ++⎟

⎞⎜⎝

⎛−−−

+≈∫

)( 2211

ababfabababfab

xfcxfcdxxfb

Hence:

)x(fc.......)x(fc)x(fcdx)x(f nn

a+++≈∫ 2211

General n-point rules would approximate the integral:

∫ ∑− =

iii )x(gcdx)x(g

n Coefficients Function arguments

2 c1 = 1.000000000c2 = 1.000000000

x1 = -0.577350269x2 = 0.577350269

3 c1 = 0.555555556c2 = 0.888888889c3 = 0.555555556

x1 = -0.774596669x2 = 0.000000000x3 = 0.774596669

4 c1 = 0.347854845c2 = 0.652145155c3 = 0.652145155c4 = 0.347854845

x1 = -0.861136312x2 = -0.339981044x3 = 0.339981044x4 = 0.861136312

The coefficients and arguments for n-point Gauss method are given within the range of <-1,1> :

n Coefficients Function arguments

5 c1 = 0.236926885c2 = 0.478628670c3 = 0.568888889c4 = 0.478628670c5 = 0.236926885

x1 = -0.906179846x2 = -0.538469310x3 = 0.000000000x4 = 0.538469310x5 = 0.906179846

6 c1 = 0.171324492c2 = 0.360761573c3 = 0.467913935c4 = 0.467913935c5 = 0.360761573c6 = 0.171324492

x1 = -0.932469514x2 = -0.661209386x3 = -0.2386191860x4 = 0.2386191860x5 = 0.661209386x6 = 0.932469514

So if the table is given for: ∫−

1dx)x(g integrals, how does one solve

adx)x(f ?

The answer lies in that any integral with limits of [ ]b,a

can be converted into an integral with limits: [ ]11,−

Let, cmtx +=

if ,ax = 1−=t

,bx =ir 1=t

It follows that:

2abm −

2abc +

Hence:22

abtabx ++

dtabdx2−

Substituting our values of x and dx into the integral gives us:

∫∫−

−⎟⎠⎞

⎜⎝⎛ +

1 222)( dtababtabfdxxf

tv 8.92100140000

140000ln2000)( −⎥⎦⎤

⎢⎣⎡

Example 5:Use two-point Gauss quadrature rule to approximate the distance covered by a rocket from t = 8 s to t = 30 s if the velocity is given by:

a) Use the Gauss-Quadrature Mmethod to find the distance covered from t1=8 s to t2=30 s

b) Find the true error.

First, change the limits of integration from from [8,30] to [-1,1]

∫∫−

⎟⎠⎞

⎜⎝⎛ +

+−−

8 2830

2830 dxxfdt)t(f

( )∫−

1191111 dxxf

The weighting factors and function argument values are (n=2):

00000000011 .c =

57735026901 .x −=

00000000012 .c =

57735026902 .x =

The formula is:

( ) ( ) ( )191111191111191111 2211

1+++≈+∫

−xfcxfcdxxf

( ) ( )1957735030111119577350301111 +++−= ).(f).(f

).(f).(f 350852511649151211 +=

).().( 481170811831729611 +=

m.4411058=

Since:

).(.).(

ln).(f 64915128964915122100140000

14000020006491512 −⎥⎦

⎤⎢⎣

⎡−

8317296.=

).(.).(

ln).(f 35085258935085252100140000

14000020003508525 −⎥⎦

⎤⎢⎣

⎡−

4811708.=

The relative error: t∈

..t 100

341106144110583411061

%.02620=

The absolute true error:b)

44.1105834.11061 −=tE

m9000.2=

mn5-2013 [tryb zgodności]home.agh.edu.pl/~zak/downloads/nm5.pdf · 2013-05-07 · romberg’s...

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