ECIV 301 Programming & Graphics Numerical Methods for Engineers REVIEW III.

ECIV 301

Programming & Graphics

Numerical Methods for Engineers

REVIEW III

Topics• Regression Analysis

– Linear Regression– Linearized Regression– Polynomial Regression

• Numerical Integration– Newton Cotes– Trapezoidal Rule– Simpson Rules– Gaussian Quadrature

Topics• Numerical Differentiation

– Finite Difference Forms

• ODE – Initial Value Problems– Runge Kutta Methods

• ODE – Boundary Value Problems– Finite Difference Method

Regression

Often we are faced with the problem…

x y0.924 -0.003880.928 -0.00743

0.93283 0.005690.93875 0.00188

0.94 0.01278

-0.01

-0.005

0

0.005

0.01

0.015

0.92 0.925 0.93 0.935 0.94 0.945

what value of y corresponds to x=0.935?

Curve FittingQuestion 2: Is it possible to find a simple and convenient formula that represents data approximately ?

-0.01

-0.005

0

0.005

0.01

0.015

0.92 0.925 0.93 0.935 0.94 0.945

e.g. Best Fit ?

Approximation

Experimental Measurements

Strain

Str

ess

BEST FIT CRITERIA

Strain

y S

tres

s

xaaxl 10)(

ii

iii

xaay

xlye

10

)(

Error at each Point

Best Fit => Minimize Error

n

iii

n

ielimeasuredi

n

ii

xaay

yye

1

210

1

2mod,,

1

2

Best Strategy

Best Fit => Minimize Error

n

iii

n

ii xaaye

1

210

1

2

Objective:

What are the values of ao and a1

that minimize ?

n

iie

1

2

Least Square Approximation

101

210

1

2 ,aaSxaaye r

n

iii

n

ii

In our case

Since xi and yi are known from given data

02,

110

0

10

n

iii

r xaaya

aaS

02,

110

1

10

n

iiii

r xxaaya

aaS


n

ii

n

i

n

ii

r xaaya

aaS

11

10

10

10 ,

n

ii

n

ii

n

iii

r xaxaxya

aaS

1

21

10

11

10 ,


n

iii

n

ii

n

ii xyxaxa

11

21

10

n

ii

n

ii yxana

1110

2 Eqtns 2 Unknowns


xaya 10

2

11

2

1111

n

ii

n

ii

n

ii

n

ii

n

iii

xxn

yxyxna

n

xx

n

ii

1

n

yy

n

ii

1

Example

y = 0.8393x + 0.0714

0

1

2

3

4

5

6

7

0 2 4 6 8

Series1

Linear (Series1)

Quantification of Error

0

1

2

3

4

5

6

7

0 1 2 3 4 5 6 7 8

Exper 1

Average

42.37

241

n

yy

n

ii


0

1

2

3

4

5

6

7

0 1 2 3 4 5 6 7 8

Exper 1

Average

n

iit yyS

1

2


0

1

2

3

4

5

6

7

0 1 2 3 4 5 6 7 8

Exper 1

Average

n

iit yyS

1

2

1

n

Ss t

y


n

iit yyS

1

2

1

n

Ss t

y

0

1

2

3

4

5

6

7

0 1 2 3 4 5 6 7 8

Exper 1

Average

Standard Deviation Shows Spread Around mean Value

y = 0.8393x + 0.0714

0

1

2

3

4

5

6

7

0 2 4 6 8


n

iii

n

iir xaayeS

1

210

1

2


n

iii

n

iir xaayeS

1

210

1

2

2/

n

Ss r

xy

“Standard Deviation” for Linear Regressiony = 0.8393x + 0.0714

0

1

2

3

4

5

6

7

0 2 4 6 8


0

1

2

3

4

5

6

7

0 1 2 3 4 5 6 7 8

Exper 1

Average

n

iit yyS

1

2

y = 0.8393x + 0.0714

0

1

2

3

4

5

6

7

0 2 4 6 8

n

iiir xaayS

1

210Better Representation

Less Spread


0

1

2

3

4

5

6

7

0 1 2 3 4 5 6 7 8

Exper 1

Average

n

iit yyS

1

2

y = 0.8393x + 0.0714

0

1

2

3

4

5

6

7

0 2 4 6 8

n

iiir xaayS

1

210

t

rt

S

SSr

2

Coefficient of Determination

t

rt

S

SSrr

2

Correlation Coefficient

Linearized Regressionxbeay 1

1

xba

eay xb

11

1

ln

lnln 1

BxA

1

1ln

bB

aA

The Exponential Equation

Linearized Regression2

2bxay

xba

xay b

2210

21010

log

loglog 2

BxA

2

210log

bB

aA

The Power Equation

Linearized Regression

xb

xay

33

33

3 111

axa

b

y

BxA

3

3

3

1

a

bB

aA

The Saturation-Growth-Rate Equation

Polynomial Regression

exaxaay 2210

A Parabola is Preferable


210r

n

1i

222i10i

n

1i

2i

a,a,aS

xaxaay

e

Minimize


0xaxaay2a

a,a,aS n

1i

2i2i10i

0

210r

0xxaxaay2a

a,a,aS n

1ii

2i2i10i

1

210r

0xxaxaay2a

a,a,aS n

1i

2i

2i2i10i

2

210r


i22i1i0 yaxaxa)n(

ii23i1

2i0i yxaxaxax

i2i2

4i1

3i0

2i yxaxaxax

3 Eqtns 3 Unknowns


i2i

ii

i

2

1

0

4i

3i

2i

3i

2ii

2ii

yx

yx

y

a

a

a

xxx

xxx

xxn

Use any of the Methods we Learned


n

1i

222i10i210r xaxaaya,a,aS

With a0, a1, a2 known the Total Error

3n

Ss r

xy Standard Error

t

rt2

S

SSr

Coefficient of

Determination


n

1i

2mmi10im10r xaxaaya,a,aS

For Polynomial of Order m

1mn

Ss r

xy Standard Error

t

rt2

S

SSr

Coefficient of

Determination

Numerical Integration & Differentiation

Motivation

x

xfxxf

x

y ii

Motivation

x

xfxxf

dx

dy ii

x

0lim

Motivation

b

a

dxxfI

AREA BETWEEN a AND b

Motivation

)()( tydt

dtv

Motivation

b

a

dxtvty

Motivation

Motivation

Calculate Derivative

Given

MotivationGiven

Calculate

Think as Engineers!

In Summary

INTERPOLATE

In SummaryNewton-Cotes Formulas

Replace a complicated function or tabulated data with an approximating

function that is easy to integrate

b

a

n

b

a

dxxfdxxfI

nn

nnon xaxaxaaxf

111

In Summary

Also by piecewise approximation

b

ax

x

x

n

b

a

i

i

i

dxxf

dxxfI

1

Closed/Open Forms

CLOSED OPEN

Trapezoidal RuleLinear Interpolation

12

3hOError

Trapezoidal Rule Multiple Application

Trapezoidal Rule Multiple Application

n

xfxfxfabI

n

n

ii

2

22

10

x a=xo x1 x2 … xn-1 b=xn

f(x) f(x0) f(x1) f(x2) f(xn-1) f(xn)

Simpson’s 1/3 Rule

Quadratic Interpolation

22102 )( xaxaaxf

90

5hOError

Simpson’s 3/8 Rule

Cubic Interpolation

33

22102 xaxaxaa)x(f

80

3 5hOError

Gauss Quadrature

x1 x2

2211 xfwxfwI

General Case

2211

1

1

xfwxfwdx)x(fI

Gauss Method calculates pairs of wi, xi for the Integration limits

-1,1

For Other Integration LimitsUse Transformation

Gauss Quadrature

b

a

dx)x(fIGxaax 10

10 aaa

10 aab

For xg=-1, x=a

For xg=1, x=b

20

aba

21

aba

Gauss Quadrature

b

a

dx)x(fI

2

Gxababx

Gdx

abdx

2

1

12dx)x(f

abdx)x(fI

b

a

Gauss Quadrature

1

12dx)x(f

abdx)x(fI

b

a

n

ii xfwab

I12

Gauss Quadrature

Points

Weighting Factors wi

Function Arguments

Error

2 W0=1.0 X0=-0.577350269 F(4)()

W1=1.0 X1= 0.577350269

3 W0=0.5555556 X0=-0.77459669 F(6)()

W1=0.8888888 X1=0.0

W2=0.5555556 X2=0.77459669

Gaussian Points

Points

Weighting Factors wi

Function Arguments

Error

4 W0=0.3478548 X0=-0.861136312 F(8)()

W1=0.6521452 X1=-339981044

W2=0.6521452 X2=- 339981044

W3=0.3478548 X3=0.861136312

Gaussian Quadrature

Not a good method if function is not available

Fig 23.1FORWARD FINITE DIFFERENCE

Fig 23.2BACKWARD FINITE DIFFERENCE

Fig 23.3CENTERED FINITE DIFFERENCE

Data with Errors

ODE IVP, BVP

Pendulum

W=mg

02

2

l

sinmg

dt

dm

02

2

l

sing

dt

d

OrdinaryDifferentialEquation

ODEs

02

2

l

sing

dt

dNon Linear

Linearization

Assume is small

sin 02

2

l

g

dt

d

ODEs

02

2

l

g

dt

dSecond Order

ydt

d

Systems of ODEs

0

l

g

dt

dy

ODE

15810450 234 x.xxx.y

5820122 23 .xxxdx

dy

ODE - OBJECTIVES

Cx.xxx.y 5810450 234

5820122 23 .xxxdx

dy

dx.xxxy 5820122 23

15810450 234 x.xxx.y

Undetermined

ODE- Objectives

15810450 234 x.xxx.y

Initial Conditions

10 y

ODE-Objectives

y,xfdx

dy

Given

.C.Iknowny,f 0

Calculate

xy

Runge-Kutta MethodsNew Value = Old Value + Slope X Step Size

hyy ii 1

Runge Kutta Methods

hyy ii 1

Definition of yields different Runge-Kutta Methods

Euler’s Method

hyy ii 1

y,xfdx

dy

ii y,xfLet

Sources of Error

Truncation: Caused by discretization

• Local Truncation• Propagated Truncation

Roundoff: Limited number of significant digits

Sources of Error

Propagated

Local

Euler’s Method

Heun’s Method

Predictor Corrector

2-Steps

Heun’s Method

Predict

Predictor-CorrectorSolution in 2 steps

hyy ii 10

ii y,xf

Let

Heun’s Method

Correct

Corrector

hyy ii 1

01ii y,xf

Estimate

2

01

iiii y,xfy,xfLet

Error in Heun’s Method

The Mid-Point Method

hyy ii 1

Remember:Definition of yields different Runge-Kutta Methods

Mid-Point Method

Predictor Corrector

2-Steps

Mid-Point Method

Predictor

Predict

22

1

hyy i

i

ii y,xf

Let

Mid-Point Method

Corrector

Correct

hyy ii 1

2

1

2

1 ,ii

yxf

Estimate

2

1

2

1 ,ii

yxfLet

Runge Kutta – 2nd Order

hyy ii 1

21 3

2

3

1kk

y,xfdx

dy .C.Iknowny,f 0

ii y,xfk 1

hky,hxfk ii 12 4

3

4

3

Runge Kutta – 3rd Order

hyy ii 1 321 46

1kkk

y,xfdx

dy .C.Iknowny,f 0

ii y,xfk 1

hky,hxfk ii 12 2

1

2

1

hkhky,hxfk ii 213 2

Runge Kutta – 4th Order

hyy ii 1 4321 226

1kkkk

y,xfdx

dy .C.Iknowny,f 0

ii y,xfk 1

hky,hxfk ii 12 2

1

2

1

hky,hxfk ii 34

hky,hxfk ii 23 2

1

2

1

Boundary Value Problems

Fig 23.3CENTERED FINITE DIFFERENCE

xo


x1 x2 x3 xn-1 xn...


xo x1 x2 x3 xn-1 xn...

),(2 112

012 yxfhyyy



),(2 222

123 yxfhyyy



),(2 332

234 yxfhyyy



),(2 112

21 nnnnn yxfhyyy

Boundary Value ProblemsCollect Equations:

),(2 112

012 yxfhyyy

),(2 222

123 yxfhyyy

),(2 112

21 nnnnn yxfhyyy

BOUNDARY CONDITIONS

T0 T5T0 T5

Example

02

2

TTcdx

Tda

x1 x2 x3 x4

Example

02

12012

TTc

h

TTTa

aTchTchTT 20

212 2

T0 T5T0 T5

x1 x2 x3 x4x1 x2 x3 x4

Example

02

22123

TTc

h

TTTa

aTchTchTT 21

223 2

T0 T5T0 T5

x1 x2 x3 x4x1 x2 x3 x4

Example

02

32234

TTc

h

TTTa

aTchTchTT 22

234 2

T0 T5T0 T5

x1 x2 x3 x4x1 x2 x3 x4

Example

02

42345

TTc

h

TTTa

aTchTchTT 22

234 2

T0 T5T0 T5

x1 x2 x3 x4x1 x2 x3 x4

Example

52

2

20

2

4

3

2

1

2

2

2

2

2100

1210

0121

0012

TTch

Tch

Tch

TTch

T

T

T

T

ch

ch

ch

ch

a

a

a

a

T0 T5T0 T5

x1 x2 x3 x4x1 x2 x3 x4

ECIV 301 Programming & Graphics Numerical Methods for Engineers REVIEW III.

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Transcript of ECIV 301 Programming & Graphics Numerical Methods for Engineers REVIEW III.