0761214: numerical analysisjamhuri.lecturer.uin-malang.ac.id/.../sites/15/2013/09/pertemuan2.pdf ·...
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0761214_Topic1 1
0761214: Numerical AnalysisTopic 1:
Introduction to Numerical Methods and Taylor Series
Lectures 1-4:
0761214_Topic1 2
Lecture 1Introduction to Numerical Methods
What are NUMERICAL METHODS?
Why do we need them?
Topics covered in 0761214.
Reading Assignment: Pages 3-10 of textbook
0761214_Topic1 3
Numerical Methods
Numerical Methods:
Algorithms that are used to obtain numerical
solutions of a mathematical problem.
Why do we need them?
1. No analytical solution exists,
2. An analytical solution is difficult to obtain
or not practical.
0761214_Topic1 4
What do we need?
Basic Needs in the Numerical Methods:
Practical:
Can be computed in a reasonable amount of time.
Accurate:
Good approximate to the true value,
Information about the approximation error (Bounds, error order,… ).
0761214_Topic1 5
Outlines of the Course
Taylor Theorem
Number Representation
Solution of nonlinear Equations
Interpolation
Numerical Differentiation
Numerical Integration
Solution of linear Equations
Least Squares curve fitting
Solution of ordinary differential equations
Solution of Partial differential equations
0761214_Topic1 6
Solution of Nonlinear Equations
Some simple equations can be solved analytically:
Many other equations have no analytical solution:
31
)1(2
)3)(1(444solution Analytic
034
2
2
xandx
roots
xx
solution analytic No052 29
xex
xx
0761214_Topic1 7
Methods for Solving Nonlinear Equations
o Bisection Method
o Newton-Raphson Method
o Secant Method
0761214_Topic1 8
Solution of Systems of Linear Equations
unknowns. 1000in equations 1000
have weif do What to
123,2
523,3
:asit solvecan We
52
3
12
2221
21
21
xx
xxxx
xx
xx
0761214_Topic1 9
Cramer’s Rule is Not Practical
this.compute toyears 10 than more needscomputer super A
needed. are tionsmultiplica102.3 system, 30by 30 a solve To
tions.multiplica
1)N!1)(N(N need weunknowns, N with equations N solve To
problems. largefor practicalnot is Rule sCramer'But
2
21
11
51
31
,1
21
11
25
13
:system thesolve toused becan Rule sCramer'
20
35
21
xx
0761214_Topic1 10
Methods for Solving Systems of Linear
Equations
o Naive Gaussian Elimination
o Gaussian Elimination with Scaled Partial Pivoting
o Algorithm for Tri-diagonal Equations
0761214_Topic1 11
Curve Fitting
Given a set of data:
Select a curve that best fits the data. One choice is to find the curve so that the sum of the square of the error is minimized.
x 0 1 2
y 0.5 10.3 21.3
0761214_Topic1 12
Interpolation
Given a set of data:
Find a polynomial P(x) whose graph passes through all tabulated points.
xi 0 1 2
yi 0.5 10.3 15.3
tablein the is)( iii xifxPy
0761214_Topic1 13
Methods for Curve Fitting
o Least Squares
o Linear Regression
o Nonlinear Least Squares Problems
o Interpolation
o Newton Polynomial Interpolation
o Lagrange Interpolation
0761214_Topic1 14
Integration
Some functions can be integrated analytically:
?
:solutions analytical no have functionsmany But
42
1
2
9
2
1
0
3
1
2
3
1
2
dxe
xxdx
a
x
0761214_Topic1 15
Methods for Numerical Integration
o Upper and Lower Sums
o Trapezoid Method
o Romberg Method
o Gauss Quadrature
0761214_Topic1 16
Solution of Ordinary Differential Equations
only. cases special
for available are solutions Analytical *
equations. thesatisfies that function a is
0)0(;1)0('
0)(3)('3)(''
:equation aldifferenti theosolution tA
x(t)
xx
txtxtx
0761214_Topic1 17
Solution of Partial Differential Equations
Partial Differential Equations are more difficult to solve than ordinary differential equations:
)sin()0,(,0),1(),0(
022
2
2
2
xxututu
t
u
x
u
0761214_Topic1 18
Summary
Numerical Methods:
Algorithms that are used to obtain numerical solution of a mathematical problem.
We need them when
No analytical solution exists or it is difficult to obtain it.
Solution of Nonlinear Equations
Solution of Linear Equations
Curve Fitting
Least Squares
Interpolation
Numerical Integration
Numerical Differentiation
Solution of Ordinary Differential Equations
Solution of Partial Differential Equations
Topics Covered in the Course
0761214_Topic1 19
Number Representation Normalized Floating Point Representation Significant Digits Accuracy and Precision Rounding and Chopping
Reading Assignment: Chapter 3
Lecture 2
Number Representation and Accuracy
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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
0761214_Topic1 21
Normalized Floating Point Representation
Normalized Floating Point Representation:
Scientific Notation: Exactly one non-zero digit appears before decimal point.
Advantage: Efficient in representing very small or very
large numbers.
exponent signed:,0
exponent mantissa sign
104321.
nd
nffffd
0761214_Topic1 22
Binary System
Binary System: Base = 2, Digits {0,1}
exponent signed mantissa sign
2.1 4321nffff
10)625.1(10)3212201211(2)101.1(
0761214_Topic1 23
Fact
Numbers that have a finite expansion in one numbering system may have an infinite expansion in another numbering system:
You can never represent 1.1 exactly in binary system.
210 ...)011000001100110.1()1.1(
IEEE 754 Floating-Point Standard
Single Precision (32-bit representation)
1-bit Sign + 8-bit Exponent + 23-bit Fraction
Double Precision (64-bit representation)
1-bit Sign + 11-bit Exponent + 52-bit Fraction
0761214_Topic1 24
S Exponent8 Fraction23
S Exponent11 Fraction52
(continued)
0761214_Topic1 25
Significant Digits
Significant digits are those digits that can be
used with confidence.
Single-Precision: 7 Significant Digits
1.175494… × 10-38 to 3.402823… × 1038
Double-Precision: 15 Significant Digits
2.2250738… × 10-308 to 1.7976931… × 10308
0761214_Topic1 26
Remarks
Numbers that can be exactly represented are called machine numbers.
Difference between machine numbers is not uniform
Sum of machine numbers is not necessarily a machine number
0761214_Topic1 27
Calculator Example
Suppose you want to compute:
3.578 * 2.139
using a calculator with two-digit fractions
3.57 * 2.13 7.60=
7.653342True answer:
0761214_Topic1 28
48.9
Significant Digits - Example
0761214_Topic1 29
Accuracy and Precision
Accuracy is related to the closeness to the true value.
Precision is related to the closeness to other estimated values.
0761214_Topic1 30
0761214_Topic1 31
Rounding and Chopping
Rounding: Replace the number by the nearest
machine number.
Chopping: Throw all extra digits.
0761214_Topic1 32
Rounding and Chopping
0761214_Topic1 33
Can be computed if the true value is known:
100* valuetrue
ionapproximat valuetrue
Error RelativePercent Absolute
ionapproximat valuetrue
Error True Absolute
t
tE
Error Definitions – True Error
0761214_Topic1 34
When the true value is not known:
100*estimatecurrent
estimate previous estimatecurrent
Error RelativePercent Absolute Estimated
estimate previous estimatecurrent
Error Absolute Estimated
a
aE
Error Definitions – Estimated Error
0761214_Topic1 35
We say that the estimate is correct to ndecimal digits if:
We say that the estimate is correct to ndecimal digits rounded if:
n10Error
n 102
1Error
Notation
0761214_Topic1 36
Summary
Number RepresentationNumbers that have a 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.
0761214_Topic1 37
Lectures 3-4
Taylor Theorem
Motivation
Taylor Theorem
Examples
Reading assignment: Chapter 4
0761214_Topic1 38
Motivation
We can easily compute expressions like:
?)6.0sin(,4.1 computeyou do HowBut,
)4(2
103 2
x
way?practicala thisIs
sin(0.6)? compute to
definition theuse weCan
0.6
ab
0761214_Topic1 39
Remark
In this course, all angles are assumed to be in radian unless you are told otherwise.
0761214_Topic1 40
Taylor Series
∑∞
0
)(
0
)(
3)3(
2)2(
'
)()( !
1 )(
: writecan weconverge, series theIf
)()( !
1
...)(!3
)()(
!2
)()()()(
:about )( of expansion seriesTaylor The
k
kk
k
kk
axafk
xf
axafk
SeriesTaylor
or
axaf
axaf
axafaf
axf
0761214_Topic1 41
Maclaurin Series
Maclaurin series is a special case of Taylor series with the center of expansion a = 0.
∑∞
0
)(
3)3(
2)2(
'
)0( !
1 )(
: writecan weconverge, series theIf
...!3
)0(
!2
)0()0()0(
:)( of expansion series n MaclauriThe
k
kk xfk
xf
xf
xf
xff
xf
0761214_Topic1 42
Maclaurin Series – Example 1
∞.xfor converges series The
...!3!2
1!
)0(!
1
11)0()(
1)0()(
1)0(')('
1)0()(
32∞
0
∞
0
)(
)()(
)2()2(
∑∑
xxx
k
xxf
ke
kforfexf
fexf
fexf
fexf
k
k
k
kkx
kxk
x
x
x
xexf )( of expansion series n MaclauriObtain
0761214_Topic1 43
Taylor SeriesExample 1
-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)
0761214_Topic1 44
Maclaurin Series – Example 2
∞.xfor converges series The
....!7!5!3!
)0()sin(
1)0()cos()(
0)0()sin()(
1)0(')cos()('
0)0()sin()(
753∞
0
)(
)3()3(
)2()2(
∑
xxxxx
k
fx
fxxf
fxxf
fxxf
fxxf
k
kk
: )sin()( of expansion series n MaclauriObtain xxf
0761214_Topic1 45
-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)
0761214_Topic1 46
Maclaurin Series – Example 3
∞.for converges series The
....!6!4!2
1)(!
)0()cos(
0)0()sin()(
1)0()cos()(
0)0(')sin()('
1)0()cos()(
642∞
0
)(
)3()3(
)2()2(
∑
x
xxxx
k
fx
fxxf
fxxf
fxxf
fxxf
k
kk
)cos()( :of expansion series Maclaurin Obtain xxf
0761214_Topic1 47
Maclaurin Series – Example 4
1 ||for converges Series
...xxx1x1
1 :ofExpansion Series Maclaurin
6)0(1
6)(
2)0(1
2)(
1)0('1
1)('
1)0(1
1)(
of expansion series n MaclauriObtain
32
)3(
4
)3(
)2(
3
)2(
2
x
fx
xf
fx
xf
fx
xf
fx
xf
x1
1f(x)
0761214_Topic1 48
Example 4 - Remarks
Can we apply the series for x≥1??
How many terms are needed to get a good approximation???
These questions will be answered using Taylor’s Theorem.
0761214_Topic1 49
Taylor Series – Example 5
...)1()1()1(1 :)1(Expansion SeriesTaylor
6)1(6
)(
2)1(2
)(
1)1('1
)('
1)1(1
)(
1at of expansion seriesTaylor Obtain
32
)3(
4
)3(
)2(
3
)2(
2
xxxa
fx
xf
fx
xf
fx
xf
fx
xf
ax
1f(x)
0761214_Topic1 50
Taylor Series – Example 6
...)1(3
1)1(
2
1)1( :Expansion SeriesTaylor
2)1(1)1(,1)1(',0)1(
2)(,
1)(,
1)(',)ln()(
)1(at )ln( of expansion seriesTaylor Obtain
32
)3()2(
3
)3(
2
)2(
xxx
ffff
xxf
xxf
xxfxxf
axf(x)
0761214_Topic1 51
Convergence of Taylor Series
The Taylor series converges fast (few terms are needed) when x is near the point of expansion. If |x-a| is large then more terms are needed to get a good approximation.
0761214_Topic1 52
Taylor’s Theorem
. and between is)()!1(
)(
:where
)(!
)( )(
:by given is )( of value the then and containing interval an on
1)( ..., 2, 1, orders of sderivative possesses )( functiona If
1)1(
0
)(
∑
xaandaxn
fR
Raxk
afxf
xfxa
nxf
nn
n
n
n
k
kk
(n+1) terms Truncated
Taylor Series
Remainder
0761214_Topic1 53
Taylor’s Theorem
.applicablenot is TheoremTaylor
defined.not are sderivative
its and function thethen,1If
.1||if0 expansion ofpoint thewith1
1
:for theoremsTaylor'apply can We
x
xax
f(x)
0761214_Topic1 54
Error Term
. and between allfor
)()!1(
)(
:on boundupper an derive can we
error, ionapproximat about theidea anget To
1)1(
xaofvalues
axn
fR n
n
n
0761214_Topic1 55
Error Term - Example
0514268.82.0)!1(
)()!1(
)(
1≥≤)()(
31
2.0
1)1(
2.0)()(
ERn
eR
axn
fR
nforefexf
nn
nn
n
nxn
?2.00at
expansion seriesTaylor its of3)( terms4first the
by )( replaced weiferror theis largeHow
xwhena
n
exf x
0761214_Topic1 56
Alternative form of Taylor’s Theorem
hxxwherehn
fR
hRhk
xfhxf
hxx
nxfLet
nn
n
n
n
k
kk
and between is)!1(
)(
size) step ( !
)( )(
: then and containing interval an on
1)( ..., 2, 1, orders of sderivative have)(
1)1(
0
)(
0761214_Topic1 57
Taylor’s Theorem – Alternative forms
. and between is
)!1(
)(
!
)()(
,
. and between is
)()!1(
)()(
!
)()(
1)1(
0
)(
1)1(
0
)(
hxxwhere
hn
fh
k
xfhxf
hxxxa
xawhere
axn
fax
k
afxf
nnn
k
kk
nnn
k
kk
0761214_Topic1 58
Mean Value Theorem
)( )('
, ,0for Theorem sTaylor' Use:Proof
)('
),( exists therethen
),( interval open theon defined is derivative its and
],[ interval closeda on function continuousa is)( If
abξff(a)f(b)
bhxaxn
ab
f(a)f(b)ξf
baξ
ba
baxf
0761214_Topic1 59
Alternating Series Theorem
termomittedFirst :
n terms)first theof (sum sum Partial:
converges series The
then
0lim
If
S
:series galternatin heConsider t
1
1
4321
4321
n
n
nnnn
a
S
aSS
and
a
and
aaaa
aaaa
0761214_Topic1 60
Alternating Series – Example
!7
1
!5
1
!3
11)1(s
!5
1
!3
11)1(s
:Then
0lim
:since series galternatin convergent a is This
!7
1
!5
1
!3
11)1(s:usingcomputed becansin(1)
4321
in
in
aandaaaa
in
nn
0761214_Topic1 61
Example 7
? 1 with eapproximat to
used are terms1)( whenbeerror thecan largeHow
expansion) ofcenter (the5.0at)( of
expansion seriesTaylor theObtain
12
12
xe
n
aexf
x
x
0761214_Topic1 62
Example 7 – Taylor Series
...!
)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,)( of expansion seriesTaylor Obtain 12 aexf x
0761214_Topic1 63
Example 7 – Error Term
)!1(
max)!1(
)5.0(2
)!1(
)5.01(2
)5.0()!1(
)(
2)(
3
12
]1,5.0[
11
1121
1)1(
12)(
n
eError
en
Error
neError
xn
fError
exf
nn
nn
nn
xkk