Chapter 1- Modelling, Computers and Error Analysis

8/11/2019 Chapter 1- Modelling, Computers and Error Analysis

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Numerical Methodsfor Engineers

Chapter 1

Modeling, Computers and

Error Analysis


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1.1 Introduction

1.2 Mathematical Modeling andEngineering Problem Solving

1.3 Programming and Software

1.4 Errors

Chapter Outline


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Mathematical Modeling and

Engineering Problem Solving

What is the relationshipbetween mathematic

modelingand engineering

problem-solving?


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What is mathematic model?

Equation that expresses the features of system

or process in mathematical terms.

Dependent

independent forcingvariable = f variables,parameters,functions{

A mathematical model is a functional relationshipof the form:

{


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Simple example: Newtons 2ndlaw of motion

F = m a

F= net force (kgm/s2)

m

= mass of the object (kg)a= acceleration (m/s2)


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Some mathematical models more complex.

Complex example:Model for falling parachute,

vm

c

gdt

dv

t = time (s)

g= gravitational constant (9.8 m/s2)

m= mass of the object (kg)

v= terminal velocity (m/s)

c = drag coefficient (kg/s)


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ConservationLaws and Engineering

Conservation laws: fundamental laws that

are used in engineering.

Change = increasesdecreases

If the no change or steady-state, the

increases and decreases must be balance.Increases =Decreases

(Steady-sate)


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For steady-statefluid flow in pipe,

Flow in = Flow out


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What have you learned


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Familiar yourself with MATLAB!!

1) Install MATLABsoftware

in your notebook.

How to do it??

2) Explore yourself the

Appendix Bin Chapra

and Canale (2006).3) Print out your work as

Assignment 1.


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Errors

Why errors are concerned??For many engineering problems, we

cannot obtain exact solutions.

Numerical methods yield approximateresults, results that are close to the exact

solution.

The question is How much error is presentin our calculation and is it tolerable?


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Accuracy?

Precision?

Inaccuracy?

Imprecision?


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In numerical methods, we use approximation

to represent the exact mathematical operations.

Numerical errors rise

Numerical error equal to discrepancy betweenthe truth and approximation:

ionapproximatvaluetrueEt

True percent relative error,t

100%

valuetrue

errortruet


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If we cannot solved the problem

analytically to get the true value,

how to calculate its true error?

We normalized the error to approximate value.

Numerical methods use iterative approachtocompute answers. A present approximation is

made on the basis of a previous approximation.

Percent relative error,

100%approx.current

approx.previous-approx.currenta

a


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The may be in +veorvesigns. But

the most important is its absolute value.

a

The calculation should proceed until the

absolute value of lower thanpercenttolerancegiven, .a

s

sa


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Result is correct/almost exact after theiteration to at least nsignificant figures.

n= 1,2,3.

%100.5 2 ns

(See Example 3.2)


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What is significant figures?

Significant digits of a number are thosethat can be usedwith confidence, e.g.,

the number of certain digits plus one

estimated digit.

53,800 How many significant figures?

5.38 x 104 3

5.380 x 104 4

5.380 x 104 5


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Types of errors- Truncation error

Truncation errorsare those that result from

using an approximation in place of an exact

mathematical procedure.


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The Taylor series

A mathematical formulation that usedwidely in numerical methods to predict a

function value in approximate fashion.

Why it is called in series?

Its build term by term, started with zero-order

approximation. The higher the order ofapproximation applied, the lower the

truncation error.


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)()( 1 ii xfxf

Zero-order approximation:

Additional terms of the Taylor series are

required to provide a better estimate.

First-order approximation:

))((')()( 11 iiiii xxxfxfxf

The additional term consist

of a slope multiply the distance

between and .i

x1

i

x


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21111 )(

!2)(''))((')()( xxxfxxxfxfxf iiiiiii

Second-order approximation:

and so on

n

n

iii

n

iii

iii

iiiii

Rxxnxfxxxf

xxxf

xxxfxfxf

)(!

)(..........)(!3

)(

)(!2

)(''))((')()(

1

)(3

1

)3(

2

111

where is a remainder termnR


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n

ni

n

i

i

iii

Rhn

xfh

xf

hxf

hxfxfxf

!

)(..........

!3

)(

!2

)('')(')()(

)(3

)3(

2

1

1

11

1

)()!1(

)(

n

i

n

n xxn

f

R

Remainder term:

where is a value lies between and ix 1ix

)( 1 ii xxh Simplifying , hence,

(See Example 4.2)


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How to solve the derivatives of an

equation given using Taylor series?

We use an approximation using

numerical differentiation with:

a) Forward divided difference

b) Backward divided difference

c) Centered divided difference

They are developed from the Taylor series

to approximate derivatives numerically.


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a) Forward divided difference (1stderivative)

)()()(

)('1

1 hOxx

xfxfxf

ii

iii

!2

))(()( 1

)2(iii xxxfhO

where

)(hO is an truncation error.


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b) Backward divided difference (1stderivative)

)(

)()()('1

1

ii

iii

xx

xfxfxf

c) Centered divided difference (1stderivative)

)()(2

)()()(' 2

1

11 hOxx

xfxfxf

ii

iii

!3

))(()(

2

1

)3(2 iii xxxfhO

where


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What have you learned


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Tutorial 1

Chapra and Canale (2006):

Problem 4.1(a)(b)

Problem 4.2Problem 4.4

Problem 4.6

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Chapter 1- Modelling, Computers and Error Analysis