23 Hand-out on Romberg Integration and Improper Integrals

8/10/2019 23 Hand-out on Romberg Integration and Improper Integrals

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MATH 174

Numerical Analysis I

Section SSecond Semester A.Y. 2013-2014

Mathematics DivisionInstitute of Mathematical Sciences and Physics

University of the Philippines Los Banos

March 18, 2014

NJA Egarguin (IMSP, UPLB) MATH 174 March 18, 2014 1 / 26

So far.... the Quadrature Rules

Newton Cotes QuadraturesThe abscissas are fixed and the weights are from the auxiliary

polynomials used in interpolation of the integrandGaussian QuadraturesA fixed number of abscissas and weights are chosen to ensure theachievement of the maximum degree of precision possible


Problems with the Quadrature Rules:

Finding the number of sub-intervals to achieve better accuracy maybe cumbersome.

The error terms are quite pessimistic, thus they tend to require morework than needed.


Romberg Integration

Romberg Integration is simply repetitive extrapolation of the TrapezoidalRule.

We start with applying the Trapezoidal Rule using halved step sizes.Denote byRk,j the Romberg approximations where k controls thestep size whilej indicates the level of extrapolation.

Usually, the approximations are organized in a tableau whereRk,j isin the kth row andj th column.



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Thus, Rk,1= Traph(f), whereh = b a2k 1 .

For the succeeding column entries, we use the extrapolation formula:

Rk,j =4j1Rk,j1 Rk1,j1

4j1 1 .


Try to recreate this!!

Lets approximate ln 2 by using a5 row Romberg Integration tableau.

Solution: Since ln2 =

2

1

1

xdx then we apply Rombergs method on this

integral. Doing so will yield the tableau:

0.7500000000

0.7083333333 0.6944444444

0.6970238095 0.6932539683 0.6931746032

0.6941218504 0.6931545307 0.6931479015 0.6931474776

0.6933912022 0.6931476528 0.6931471943 0.6931471831

The last column gives the approximate value 0.6931471819ofln 2.


Shortcomings of the Quadrature rules Addressed

Estimation of the errors will result for the following termination rule.

Romberg Integration Algorithm Termination Rule

Let be the specified convergence tolerance. IfRn,n Rn1,n12n1 <

then halt the algorithm. Else, proceed with iteration.


Row by Row Computation

Suppose we set the convergence tolerance to = 2 108

R2,2 R1,121 0.0277777778.



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R3,3 R2,222 0.0003174603.




R4,4 R3,323 0.0000033907.




R5,5 R4,424 0.00000001848125.


Another Integration Anomaly!!!!!

Even in practical situations we encounter integrals like

11

ex 1x

dx,

that can make the formulas we derived fail.

In general, improper integrals can still be successfully approximated by our

developed techniques, but more caution and care is required.

Recall from elementary calculus, a definite integral isimproperif any oneof the following occurs:

the integrand is discontinuous at a number in the integration interval;or

the integration interval is unbounded.



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Consider the (NOT-so improper) integral

10

x2/3dx. If some basic

quadrature rules, we can get the following tally of errors per varyingnumber of nodes.

n Trapezoidal Simpsons Midpoint 2-pt Gaussian

2 3.502 102 1.336 102 1.117 102 1.338 1024 1.193 102 4.229 103 3.963 103 4.225 104

8 3.982 103

1.334 103

1.361 103

1.331 104

16 1.311 103 4.202 104 4.568 104 4.194 10532 4.269 104 1.324 104 1.509 104 1.321 10564 1.380 104 4.169 105 4.930 105 4.161 106EOC 1.5986 1.6651 1.5667 1.6659

TOC 2 4 2 4

Cause of under performance: discontinuity of the derivatives of theintegrand at x = 0.


Solution: Remove the Fractional Exponent

ConsiderA =

1

0

x2/3d. Notice that the integrand is not sufficiently

differentiable, that is, it lacks sufficient smoothness to guarantee

achievement of convergence.

To remedy this, let x = u3 = A= 1

0

3u4du. Re-performing the

quadrature rules earlier with this new form for A yields the following errordata:


n Trapezoidal Simpsons Midpoint 2-pt Gaussian

2 2.438 101 2.500 102 1.195 101 1.042 1034 6.211 102 1.563 103 3.091 102 6.510 1058 1.560 102 9.766 105 7.791 103 4.069 10616 3.905 103 6.104 106 1.952 103 2.543 107

32 9.765 104

3.815 107

4.882 104

1.589 108

64 2.441 104 2.384 108 1.221 104 9.934 1010EOC 1.9939 4.0000 1.9894 4.000

TOC 2 4 2 4

Lesson: Make your integrand as smooth as possible.


Improper Integrals: Removable Discontinuities

Removable DiscontinuityRecall that f(x) has aremovable discontinuityat x = a if lim

xaf(x) exists

but is not equal to f(a).

1. 11

ex

1

x dx. Note that the integrand f(x) =

ex

1

x is

discontinuous atx = 0 sincef(0) is undefined. Further, limx0

f(x) = 1, so

the discontinuity is removable. Also, note that

f(x) =1 + 1 + x + x

2

2 +

x3

6 + ...

x = 1 +

x

2+

x2

6 + ....



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Thus, f is infinitely differentiable, which will make any quadrature rulework optimally in the integrands current form, except when 0 is used asan abscissa. Thus, we may replace the integrand with

f(x) =

ex

1

x , x = 01, x= 0

.


2. B =

1

0

sin xx

dx . Notice that even iff(x) =sin x

x is undefined at

x= 0, limx0

f(x) = 0. So f has a removable discontinuity at x = 0.

Checking the McLaurin expansion off yields:

f(x) =xx

3

3! +x

5

5! + ...

x1/22 =x1/2 x

5/2

3! +

x9/2

5! + ....

This shows that quadrature rules may not perform optimally with thecurrent integrand. Hence, a change in variable is in place.

Letx = u2 = B= 1

0

2 sin(u2)du.

Using the (adaptive) three-point Gaussian quadrature rule will require 87function evaluations to get an approximate using tolerance 5 1011. Hadthe change in variable not been done, 677 function evaluations arerequired instead.


Improper Integrals: Essential Discontinuities

Case 1: Algebraic Discontinuity

f(x) tends to behave like (x a) for some >0, as x a

3. C= 1

0

(cos( 5x 1))ex3

x2(1 + 3

x)dx.

Letfbe the integrand. Then

limx0+

f(x) = .


Further, notice that

f(x) = x2/3+k=1

(1)k x2k/5

(2k)!

+k=1

xk/2

k!

+k=1

(1)kxk/3

= x4/15

2 +

x1/15

2 +

x2/15

24 x

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2 + ....

Thus, to assure convergence of quadrature rules, we letx = u30 and usethe quadrature rules on

C=

1

0

30u9(cos u6 1)eu15

1 + u10 du.



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Case 2: Logarithmic Discontinuity

f(x) tends to behave like ln(x a) as x a

4. D=

1

0

ln x

1 + x2dx. Let fbe the integrand.

Notice thatlimx0

f(x) = thusf has an essential discontinuity. Further,

f(x) = (1 x2 + x4 x6 + ...) ln x= ln xx2 ln x + x4 ln xx6 ln x + ...


However, for >0,

limx0

x ln x= 0.

Hence, it would be more stable to write the integral as

D= 1

0 ln x

1 + x2

lnx dx +

1

0

ln x

= 1

0

x2 ln x

1 + x2dx +

1

0

ln x

The first summand has a removable discontinuity, while the second caneasily be computed manually.


Improper Integrals: Unbounded Integration Intervals

Here, we consider two substitution rules that will project an infiniteintegration interval to a finite one.Case 1: The integration interval does not contain zero.

Substitution Rule

x= 1

u


5. E=

+1

ex

x dx.

If we let x = 1

u = E=

0

1

e1/u

1/u du

u2Hence,

E=

1

0

e1/u

u du.



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6. F=

+

0

t2et2

dt.

First split the integration interval to separate 0 and as:

F =

1

0

t2et2

dt +

1

t2et2

dt.

In the second integral, let x = 1

u. Then

F =

1

0

t2et2

dt +

1

0

e1/u2

u4 du.


7. G=

+

ex2

1 + x + x2dx.

To apply the previous substitution, we need to separate the infinities andzero and come up with three integrals with bounds such as,1, 1,.Another way to solve this improper integral is via the substitution rule

x= tan .

Doing so will give

G=

2

2

e tan2

1 + sin cos d.


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23 Hand-out on Romberg Integration and Improper Integrals