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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 DifferentialEquations

Euler's Method

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

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Linear Regression

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3

What is Regression?What is regression? Given ndata points ),(,...),,(),,( 2211 nnyxyxyx

best fit )(xfy to the data. The best fit is generally based onminimizing the sum of the square of the residuals, rS

Residual at a point is

)( iii xfy

n

i

iir xfyS1

2))(( ),( 11yx

),( nn yx

)(xfy

Figure. Basic model for regression

Sum of the square of the residuals

.

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4

Least Squares CriterionThe least squares criterion minimizes the sum of the square of theresiduals in the model, and also produces a unique line.

2

1

10

1

2

n

i

ii

n

i

ir xaayS

x

iiixaay10

11

, yx

22, yx

33, yx

nnyx ,

iiyx ,

iiixaay10

y

Figure. Linear regression of y vs. x data showing residuals at a typical point, xi .

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5

Finding Constants of Linear Model

2

1

10

1

2

n

i

ii

n

i

ir xaayS Minimize the sum of the square of the residuals:

To find

0121

10

0

n

i

iir xaay

aS

021

10

1

n

i

iiir xxaay

a

S

giving

0a and 1a we minimize with respect to 1a 0aandrS .

yxanayxaa 10n

1iii

n

1i1

n

1i0

xyxaxaxyxaxan

1i

210i

n

1ii

2i

n

1i1i

n

1i0

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6

Finding Constants of Linear Model0aSolving for

2

2

1

xxn

yxxyn

a

and

or

xxn

xyxyx

a2

2

2

0

1aand directly yields,

)xaya( 10

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Example 1The torque, T needed to turn the torsion spring of a mousetrap throughan angle, is given below.

Angle, Torque, T

Radians N-m

0.698132 0.188224

0.959931 0.209138

1.134464 0.230052

1.570796 0.250965

1.919862 0.313707

Table: Torque vs Angle for atorsional spring

Find the constants for the model given by

10 aaT

Figure. Data points for Angle vs. Torque data

0.1

0.2

0.3

0.4

0.5 1 1.5 2

(radians)

Torque(N-m

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Example 1 cont.

1a

The following table shows the summations needed for the calculations ofthe constants in the regression model.

2 T

Radians N-m Radians 2 N-m-Radians

0.698132 0.188224 0.487388 0.131405

0.959931 0.209138 0.921468 0.200758

1.134464 0.230052 1.2870 0.260986

1.570796 0.250965 2.4674 0.394215

1.919862 0.313707

3.6859 0.602274

6.2831 1.19218.8491 1.5896

Table. Tabulation of data for calculation of important

5

1i

5n

Using equations described for

2

2

1

5

TT5

a

228316849185

1921128316589615

..

...

2

1060919

.N-m/rad

summations

0aT

and with

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Example 1 cont.

n

T

T i

i

5

1_

Use the average torque and average angle to calculate

_

1

_

0 aTa

n

i

i

5

1_

5

1921.1

1103842.2

5

2831.6

2566.1

Using,

)2566.1)(106091.9(103842.2 21 1

101767.1 N-m

0a

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Example 1 Results

Figure. Linear regression of Torque versus Angle data

Using linear regression, a trend line is found from the data

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Nonlinear Regression

Some popular nonlinear regression models:

1. Exponential model: )( bxaey

2. Power model: )( baxy

3. Saturation growth model:

xb

axy

4. Polynomial model: )( 10m

mxa...xaay

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Exponential ModelGiven best fit

bxaey to the data.

bxaey

),(nn

yx

),( 11 yx

),(22

yx

),(ii

yx

)(ii

xfy

),(,...),,(),,( 2211 nn yxyxyx

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Finding constants of Exponential ModelConsider the equation Taking the logarithm on both sides, we get

bxaey

bxalnelnalnaelnyln bxbx

bxAY Then by using the linear module let then,alnA,Yyln

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Example 2-Exponential ModelMany patients get concerned when a test involves injection of a

radioactive material. For example for scanning a gallbladder, a few dropsof Technetium-99m isotope is used. Half of the techritium-99m would begone in about 6 hours. It, however, takes about 24 hours for theradiation levels to reach what we are exposed to in day-to-day activities.Below is given the relative intensity of radiation as a function of time.

Table. Relative intensity of radiation as a function of time.

t(hrs) 0 1 3 5 7 9

1.000 0.891 0.708 0.562 0.447 0.355

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Example 2-Exponential Model cont.The relative intensity is related to time by the equation

Find:

a) The value of the regression constants and

b) Radiation intensity after 24 hours

tAe

A

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t Y t

0 1.000 0 0 0

1 0.891 -0.115 1 -0.115

3 0.708 -0.345 9 -1.035

5 0.562 -0.576 25 -2.88

7 0.447 -0.805 49 -5.635

9 0.355 -1.035 81 -9.315

3.963

lnY 2t

876.2Y 25t 165t2

98.18Yt

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1151.0

365

98.41

625990

9.7188.113

25)165(6

)876.2(25)98.18(6

tt6

tYYt6

222

Calculating the Other Constant

The value of can now be calculated

and the value of A can now be calculated

9998.0A

00025.047958.047933.0)6

25(1151.0

6

876.2tYAln

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Plot of data

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Plot of data and regression curve

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Relative Intensity After 24 hrs

The relative intensity of radiation after 24 hours

241151.09998.0 e2103160.6

This result implies that only %317.61009998.0

10316.6 2

radioactive intensity is left after 24 hours.