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ScheduleWeek Date Topic Classification of Topic

1 9 Feb. 2010 Introduction toNumerical Methodsand Type of Errors

Measuring errors, Binaryrepresentation, Propagation of errorsand Taylor series

2 14 Feb. 2010 Nonlinear Bisection Method

3 21 Feb. 2010 Newton-Raphson Method

4 28 Feb. 2010 Interpolation Lagrange Interpolation

5 7 March 2010 Newton's Divided Difference Method

6 14 March 2010 Differentiation Newton's Forward and BackwardDifference

7 21 March 2010 Regression Least squares

8 28 March 2010 Systems of Linear

Equations

Gaussian Jordan

9 11 April 2010 Gaussian Seidel

10 18 April 2010 Integration Composite Trapezoidal and SimpsonRules

11 25 April 2010 Ordinary Differential

Equations

Euler's Method

12 2 May 2010 Runge-Kutta 2nd and4th order Method

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2

Trapezoidal Rule of

Integration

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3

What is IntegrationIntegration:

b

a

dx)x(fI

The process of measuringthe area under a function

plotted on a graph.

Where:

f(x) is the integrand

a= lower limit of integration

b= upper limit of integration

f(x)

a b

b

a

dx)x(f

y

x

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4

Method Derived From GeometryThe area under thecurve is a trapezoid.

The integral

trapezoidofAreadxxf

b

a

)(

)height)(sidesparallelofSum(2

1

)ab()a(f)b(f 2

1

2

)b(f)a(f)ab(

Figure 2: Geometric Representation

f(x)

a b

b

a

dx)x(f1

y

x

f1(x)

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5

Example 1The vertical distance covered by a rocket from t=8 tot=30 seconds is given by:

30

8

892100140000

1400002000 dtt.

tlnx

a) Use single segment Trapezoidal rule to find the distance covered.

b) Find the true error, for part (a).c) Find the absolute relative true error, for part (a).

tEa

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6

Solution

2

)b(f)a(f)ab(Ia)

8

a 30bt.

tln)t(f 89

2100140000

1400002000

)(.)(ln)(f 88982100140000

140000

20008

)(.)(

ln)(f 3089302100140000

140000200030

s/m.27177

s/m.67901

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Solution (cont)

2

6790127177830

..)(I

m11868

b) The exact value of the above integral is

30

8

892100140000

1400002000 dtt.

tlnx m11061

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Solution (cont)ValueeApproximatValueTrueEt

1186811061

m807

c) The absolute relative true error, , would bet

10011061

1186811061

t %.29597

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Multiple Segment Trapezoidal Rule

f(x)

a b

y

x

4

aba

42

aba

43

aba

Figure 3: Multiple (n=4) Segment Trapezoidal Rule

Divide into equal segmentsas shown in Figure 3. Thenthe width of each segment is:

n

abh

The integral I is:

b

a

dx)x(fI

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b

h)n(a

h)n(a

h)n(a

ha

ha

ha

a dx)x(fdx)x(f...dx)x(fdx)x(f 1

1

2

2

10

Multiple Segment Trapezoidal RuleThe integral Ican be broken into hintegrals as:

b

a dx)x(f

Applying Trapezoidal rule on each segment gives:

b

a

dx)x(f

1

1

)(2

)(

2

)( n

i

ihafbfaf

h

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Example 2The vertical distance covered by a rocket from to seconds isgiven by:

30

8

892100140000

1400002000 dtt.t

lnx

a) Use two-segment Trapezoidal rule to find the distance covered.

b) Find the true error, for part (a).c) Find the absolute relative true error, for part (a).a

tE

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Solutiona) The solution using 2-segment Trapezoidal rule is

1

1)(2

)()( n

iihaf

bfaf

hI

2n 8a 30b

2

830

n

abh

11

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Solution (cont)

12

1

)(2

)30()8(11

i

ihafff

I

)19(

2

)30()8(11 f

ff

m11266

Then:

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Solution (cont)

30

8

892100140000

1400002000 dtt.t

lnx m11061

b) The exact value of the above integral is

so the true error is

ValueeApproximatValueTrueEt

1126611061

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Solution (cont)c) The absolute relative true error, , would bet

100ValueTrueErrorTrue t

10011061

1126611061

%8534.1

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Solution (cont)Table 1 gives the valuesobtained using multiple

segment Trapezoidal rulefor:

n Value 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

30

8

892100140000

1400002000 dtt.

tlnx

Table 1: Multiple Segment Trapezoidal Rule Values

%t %a

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The error Bounds in MultipleSegment Trapezoidal Rule

The error bounds are now obtained by taking the largest value for

Say, and the smallest value, , in the interval of integration by

"f

2M

*

2M

.1212

)( 22

3*

22 hab

n

abkwherekMkM

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Simpsons 1/3rd Rule of

Integration

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Basis of Simpsons 1/3rd

Rule Trapezoidal rule was based on approximating the integrand by a first

order polynomial, and then integrating the polynomial in the intervalof integration.

Simpsons 1/3rd rule is an extension of Trapezoidal rule where theintegrand is approximated by a second order polynomial.

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]24[3

)( 210 sssh

dxxfb

a

Multiple Segment Simpsons 1/3rd

Rulesuch that

2

2

2

1

1

1

00

)(

)(

),()(

n

evenii

i

n

oddii

i

n

xfs

xfs

xfxfs

,n

abh

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Example 3 Use 4-segment Simpsons 1/3rd Rule to approximate the

distance Use 4-segment Simpsons 1/3rd Rule to approximate the

distance

covered by a rocket from t= 8 to t=30 as given by

30

8

dtt8.9t2100140000

140000ln2000x

a) Use four segment Simpsons 1/3rd Rule to find the approximatevalue of x.b) Find the true error, for part (a).c) Find the absolute relative true error, for part (a).

tE

a

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Solution

4

22

4

83024

3) 210

hsss

hxa

)30(f)t(f2)t(f4)t(f4)8(f12

22231

)(

)()(

)30()8()()(

22

311

400

tfs

tftfs

fftftfs

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23

Solution (cont.)cont.

)30(f)19(f2)5.24(f4)5.13(f4)8(f611

6740.901)7455.484(2)0501.676(4)2469.320(42667.1776

11

m64.11061

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24

Solution (cont.)b) In this case, the true error is

64.1106134.11061Et

m30.0

c) The absolute relative true error

%10034.11061

64.1106134.11061t

%0027.0

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Solution (cont.)

Table 1: Values of Simpsons 1/3rd Rule for Example 3 with multiple segments

n Approximate Value Et |t |

246

810

11065.7211061.6411061.40

11061.3511061.34

4.380.300.06

0.010.00

0.0396%0.0027%0.0005%

0.0001%0.0000%