ee 3561_unit_1(c)al-dhaifallah 14351 number representation normalized floating point...

EE 3561_Unit_1 (c)Al-Dhaifallah 1435 1

Number Representation Normalized Floating Point Representation Significant Digits Accuracy and Precision Rounding and Chopping

Reading assignment: Chapter 2

Lecture 2 Number Representation and

accuracy


Representing Real Numbers You are familiar with the decimal system

Decimal System Base =10 , Digits(0,1,…9) Standard Representations

21012 10510410210110345.312

part part

fraction integralsign

54.213


Normalized Floating Point Representation Normalized Floating Point Representation

No integral part, Advantage Efficient in representing very small or very large numbers

integer:,0

exponent mantissa sign

10.0

1

4321

nd

dddd n


Binary System

Binary System Base=2, Digits{0,1}

exponent mantissasign

21.0 432nbbb

10 11 bb

1010321

2 )625.0()212021()101.0(


7-Bit Representation(sign: 1 bit, Mantissa 3bits,exponent 3 bits)


Fact Number that have finite expansion in one numbering

system may have an infinite expansion in another numbering system

You can never represent 0.1 exactly in any computer

210 ...)011000001100110.0()1.0(


Representation Hypothetical Machine (real computers use ≥ 23 bit

mantissa)

Example:

If a machine has 5 bits representation distributed as follows

Mantissa 2 bits exponent 2 bit sign 1 bit

Possible machine numbers

(0.25=00001) (0.375= 01111) (1.5=00111)


Representation

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Gap near zero


Remarks

Numbers that can be exactly represented are called machine numbers

Difference between machine numbers is not uniform. So, sum of machine numbers is not necessarily a machine number

0.25 + .375 =0.625 (not a machine number)


Significant Digits Significant digits are those digits that can be

used with confidence.

0 1 2 3 4

Length of green rectangle = 3.45significant


Loss of Significance Mathematical operations may lead to

reducing the number of significant digits

0.123466 E+02 6 significant digits

─ 0.123445 E+02 6 significant digits

──────────────

0.000021E+02 2 significant digits

0. 210000E-02Subtracting nearly equal numbers causes loss of significance


Accuracy and Precision Accuracy is related to closeness to the true value

Precision is related to the closeness to other estimated

values


Better Precision

Better accuracy

Accuracy is related to closeness to the true value Precision is related to the closeness to other estimated values

Accuracy and Precision


Rounding and Chopping Rounding: Replace the number by the nearest machine number Chopping: Throw all extra digits

0 1 2

True 1.1681

Rounding (1.2) Chopping (1.1)


Error DefinitionsTrue Error

can be computed if the true value is known

100* valuetrue

ionapproximat valuetrue

Errorelative Percent RAbsolute

ionapproximat valuetrue

ErrorTrue Absolute

t

tE


Error DefinitionsEstimated error

Used when the true value is not known

100*estimatecurrent

estimate prevoius estimatecurrent

Errorelative Percent RAbsolute Estimated

estimate prevoius estimatecurrent

ErrorAbsolute Estimated

a

aE


Notation

We say the estimate is correct to n decimal digits if

We say the estimate is correct to n decimal digits rounded if

n10Error

n 102

1Error


Summary Number Representation

Number that have finite expansion in one numbering system may

have an infinite expansion in another numbering system.

Normalized Floating Point Representation Efficient in representing very small or very large numbers Difference between machine numbers is not uniform Representation error depends on the number of bits used in the

mantissa.


Summary Rounding Chopping Error Definitions:

Absolute true error True Percent relative error Estimated absolute error Estimated percent relative error


Lecture 3Taylor Theorem

Motivation Taylor Theorem Examples

Reading assignment: Chapter 4


Motivation

We can easily compute expressions like

?)6.0sin(,4.1 computeyou do HowBut,

)4(2

103 2

x

way?practical a thisis

?)6.0sin(

compute todefinition theusecan We

0.6

ab


Taylor Series

∑∞

000

)(

000

)(

30

0)3(

20

0)2(

00'

0

0

)()(!

1)(

can write weconverge series theif

)()(!

1

...)(!3

)()(

!2

)()()()(

about )( ofexpansion seriesTaylor The

k

kk

k

kk

xxxfk

xf

xxxfk

SeriesTaylor

or

xxxf

xxxf

xxxfxf

xxf


Taylor SeriesExample 1

∞xfor converges series The

!)()(

!

1

11)0()(

1)0()(

1)0(')('

1)0()(

∑∑∞

0

∞

000

)(

)()(

)2()2(

k

k

k

kkx

kxk

x

x

x

k

xxxxf

ke

kforfexf

fexf

fexf

fexf

0about )( ofexpansion seriesTaylor Obtain 0 xexf x



-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 10

0.5

1

1.5

2

2.5

3

1

1+x

1+x+0.5x2

exp(x)




....!7!5!3

)(!

)()sin(

1)0()cos()(

0)0()sin()(

1)0(')cos()('

0)0()sin()(

753∞

00

0)(

)3()3(

)2()2(

∑

xxxxxx

k

xfx

fxxf

fxxf

fxxf

fxxf

k

kk

0about )sin()( ofexpansion seriesTaylor Obtain 0 xxxf


-4 -3 -2 -1 0 1 2 3 4-4

-3

-2

-1

0

1

2

3

4

x

x-x3/3!

x-x3/3!+x5/5!

sin(x)


Convergence of Taylor Series(Observations, Example 1)

The Taylor series converges fast (few terms are needed) when x is near the point of expansion. If |x-c| is large then more terms are needed to get good approximation.



....11

1 ofExpansionSeriesTaylor

6)0(1

6)(

2)0(1

2)(

1)0(1

1)('

1)0(1

1)(

01

1f(x) ofexpansion seriesTaylor Obtain

32

4)3(

3)2(

2

0

xxxx

fx

xf

fx

xf

fx

xf

fx

xf

xaboutx


Example 3remarks

Can we apply Taylor series for x>1??

How many terms are needed to get good approximation???

These questions will be answered using Taylor Theorem


Taylor Theorem

. andbetween x is)()!1(

)(

where

)(!

)()(

bygiven is at value then the, and

containing intervalan on continuous

are sderavative 1first its andfunction a If

1)1(

1

1

n

0

)(

∑

candcxn

fE

Ecxk

cfxf

xxc

nf(x)

nn

n

nk

kk

(n+1) terms Truncated Taylor Series

Reminder


Taylor Theorem

.applicablenot is TheoremTaylor

defined.not are sderivative

its andfunction then the1],0[

1||if0expansion ofpoint with 1

1

for oremTaylor theapply can We

includesxif

xcx

f(x)


Error Term

. and between allfor

)()!1(

)(

on boundupper an derivecan We

errorion approximat about the ideaan get To

1)1(

1

xcofvalues

cxn

fE n

n

n


Example 4 (The Approximation Error)

0514268.82.0)!1(

)()!1(

)(

1≥≤)()(

41

2.0

1

1)1(

1

2.0)()(

EEn

eE

cxn

fE

kforefexf

nn

nn

n

kxk

?2.00about

expansion seriesTaylor its of3)(n terms4first the

by )( replaced weiferror theis large How

xwhenc

exf x


Example 5

? 1with xe eapproximat to

used are terms1)(n when beerror can the largeHow

expansion) ofcenter (the5.0ef(x) of

expansion seriesTaylor theObtain

12x

12x

cabout


Example 5

...!

)5.0(2...

!2

)5.0(4)5.0(2

)5.0(!

)5.0(

2)5.0(2)(

4)5.0(4)(

2)5.0('2)('

)5.0()(

22

222

∞

0

)(12

2)(12)(

2)2(12)2(

212

212

∑

k

xe

xexee

xk

fe

efexf

efexf

efexf

efexf

kk

k

kk

x

kkxkk

x

x

x

5.0,)( ofexpansion seriesTaylor Obtain 12 cexf x


Example 5Error term

)!1(

max)!1(

)5.01(2

1when

)!1(

)5.0(2

)5.0()!1(

)(

2)(

3

12

]1,5.0[

11

1121

1)1(

12)(

n

eError

en

Error

x

n

xeError

xn

fError

exf

nn

nn

nn

xkk


Alternative form of Taylor Theorem

hxxwherehn

fE

Ehk

xfhxf

[a,b]hx[a,b]x

[a,b]

xfLet

nn

n

n

n

k

kk

and between is)!1(

)(

!

)()(

then and and

, intervalan on 1)1,2,...(n orders

of sderivative continuous have)(

1)1(

1

10

)(


Taylor TheoremAlternative forms

hxxwhere

hn

fh

k

xfhxf

xchxx

cxwhere

cxn

fcx

k

cfxf

nnn

k

kk

nnn

k

kk

and between is

)!1(

)(

!

)()(

,

and between is

)()!1(

)()(

!

)()(

1)1(

0

)(

1)1(

0

)(


Derivative Mean-Value Theorem

(b-a)dx

df(ξf(a)f(b)

bhxax

(b-a)

f(a)f(b)

dx

df(ξ

baξ

)

,0,n TheoremTaylor Use:Proof

)

such that ],[point oneleast at exist then there

b)(a, intervalopen on the defined is derivativefirst its and

b][a, interval closed aon function continuous a isf(x) If


Alternating Series Theorem

termomittedFirst :

n terms)first theof (sum sum partial:

converges series The

0lim

S

series galternatin heConsider t

1

1

4321

4321

n

n

nnnn

a

S

aSS

andthen

a

and

aaaa

If

aaaa

Alternating Series is special case of Taylor Series.


Alternating SeriesExample 6

!7

1

!5

1

!3

11)1(s

!5

1

!3

11)1(s

0lim

since series galternatin convergent a is This!7

1

!5

1

!3

11)1(susingcomputed becansin(1)

4321

in

in

Then

aandaaaa

in

nn


Remark In this course all angles are assumed to

be in radian unless you are told otherwise


Maclurine series Find Maclurine Maclurine series expansion

of cos (x)

Maclurine series is a special case of Taylor series with the center of expansion c = 0




....!6!4!2

1)(!

)0()cos(

0)0()sin()(

1)0()cos()(

0)0(')sin()('

1)0()cos()(

642∞

0

)(

)3()3(

)2()2(

∑

xxxx

k

fx

fxxf

fxxf

fxxf

fxxf

k

kk

)cos()( ofexpansion series MaclurineObtain xxf



)!1(

12

)!1(

)5.0(

01.0)!1(

)5.0(

!

)5.0(...

!2

)5.0(5.01)(

:

2?x when0.01error truncationthatso)5.0exp()(

eapproximat toneeded is series Maclurine theof many terms How

11

15.01

22

nnError

xn

eErrorTruncation

xn

xxxf

Solution

xxf

nn

nn

nn



432

384

1

48

1

8

1

2

11)(

0083.01667.0

43

)!1(

1

......

xxxxxf

boundError

n

trialBy

nError

Solution


Summary Taylor series expansion is very important in

most numerical methods applications

approximation remainder

Remainder can be used to estimate approximation error or estimate the number of terms to achieve desirable

accuracy

1)1(n

0

)(

)()!1(

)()(

!

)()( ∑

nn

k

kk

cxn

fcx

k

cfxf

ee 3561_unit_1(c)al-dhaifallah 14351 number representation normalized floating point...

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