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ECEG-2112 : Computational Methods
Introduction to Computational Methods and Taylor Series
Lecture 1:
1
Lecture 1Introduction to Computational
Methods
What are COMPUTATIONAL METHODS? Why do we need them? Topics covered in ECEG-2112.
Reading Assignment: Pages 3-10 of textbook
What are COMPUTATIONAL METHODS? Why do we need them? Topics covered in ECEG-2112.
Reading Assignment: Pages 3-10 of textbook
2
Numerical MethodsNumerical Methods:
Algorithms that are used to obtain numericalsolutions 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.
Numerical Methods:Algorithms that are used to obtain numericalsolutions 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.
3
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,… ).
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,… ).
4
Outlines of the Course Taylor Theorem Number
Representation Solution of nonlinear
Equations Interpolation Numerical
Differentiation Numerical Integration
Solution of linearEquations
Least Squares curvefitting
Solution of ordinarydifferential equations
Solution of Partialdifferential equations
Taylor Theorem Number
Representation Solution of nonlinear
Equations Interpolation Numerical
Differentiation Numerical Integration
Solution of linearEquations
Least Squares curvefitting
Solution of ordinarydifferential equations
Solution of Partialdifferential equations
5
Solution of Nonlinear Equations Some simple equations can be solved analytically:
Many other equations have no analytical solution:
31
)1(2
)3)(1(444solutionAnalytic
034
2
2
xandx
roots
xx
Some simple equations can be solved analytically:
Many other equations have no analytical solution:
31
)1(2
)3)(1(444solutionAnalytic
034
2
2
xandx
roots
xx
solutionanalyticNo052 29
xex
xx
6
Methods for Solving Nonlinear Equations
o Bisection Method
o Newton-Raphson Method
o Secant Method
o Bisection Method
o Newton-Raphson Method
o Secant Method
7
Solution of Systems of Linear Equations
unknowns.1000inequations1000
have weifdoWhat to
123,2
523,3
:asitsolvecanWe
52
3
12
2221
21
21
xx
xxxx
xx
xx
unknowns.1000inequations1000
have weifdoWhat to
123,2
523,3
:asitsolvecanWe
52
3
12
2221
21
21
xx
xxxx
xx
xx
8
Cramer’s Rule is Not Practical
this.compute toyears10 thanmoreneedscomputersuperA
needed.aretionsmultiplica102.3system,30by30asolveTo
tions.multiplica
1)N!1)(N(Nneed weunknowns,N withequationsNsolveTo
problems.largeforpracticalnotisRulesCramer'But
2
21
11
51
31
,1
21
11
25
13
:system thesolve tousedbecanRulesCramer'
20
35
21
xx
this.compute toyears10 thanmoreneedscomputersuperA
needed.aretionsmultiplica102.3system,30by30asolveTo
tions.multiplica
1)N!1)(N(Nneed weunknowns,N withequationsNsolveTo
problems.largeforpracticalnotisRulesCramer'But
2
21
11
51
31
,1
21
11
25
13
:system thesolve tousedbecanRulesCramer'
20
35
21
xx
9
Methods for Solving Systems of LinearEquations
o Naive Gaussian Elimination
o Gaussian Elimination with ScaledPartial Pivoting
o Algorithm for Tri-diagonalEquations
o Naive Gaussian Elimination
o Gaussian Elimination with ScaledPartial Pivoting
o Algorithm for Tri-diagonalEquations
10
Curve Fitting Given a set of data:
Select a curve that best fits the data. Onechoice is to find the curve so that the sumof the square of the error is minimized.
x 0 1 2
y 0.5 10.3 21.3
Given a set of data:
Select a curve that best fits the data. Onechoice is to find the curve so that the sumof the square of the error is minimized.
x 0 1 2
y 0.5 10.3 21.3
11
Interpolation Given a set of data:
Find a polynomial P(x) whose graphpasses through all tabulated points.
xi 0 1 2
yi 0.5 10.3 15.3
Given a set of data:
Find a polynomial P(x) whose graphpasses through all tabulated points.
xi 0 1 2
yi 0.5 10.3 15.3
tablein theis)( iii xifxPy 12
Methods for Curve Fittingo Least Squares
o Linear Regressiono Nonlinear Least Squares Problems
o Interpolationo Newton Polynomial Interpolationo Lagrange Interpolation
o Least Squareso Linear Regressiono Nonlinear Least Squares Problems
o Interpolationo Newton Polynomial Interpolationo Lagrange Interpolation
13
Integration Some functions can be integrated
analytically:
?
:solutionsanalyticalnohavefunctionsmanyBut
42
1
2
9
2
1
0
3
1
23
1
2
dxe
xxdx
ax ?
:solutionsanalyticalnohavefunctionsmanyBut
42
1
2
9
2
1
0
3
1
23
1
2
dxe
xxdx
ax
14
Methods for Numerical Integration
o Upper and Lower Sums
o Trapezoid Method
o Romberg Method
o Gauss Quadrature
o Upper and Lower Sums
o Trapezoid Method
o Romberg Method
o Gauss Quadrature
15
Solution of Ordinary Differential Equations
only.casesspecial
foravailablearesolutionsAnalytical*
equations. thesatisfies thatfunctionais
0)0(;1)0(
0)(3)(3)(
:equationaldifferenti theosolution tA
x(t)
xx
txtxtx
only.casesspecial
foravailablearesolutionsAnalytical*
equations. thesatisfies thatfunctionais
0)0(;1)0(
0)(3)(3)(
:equationaldifferenti theosolution tA
x(t)
xx
txtxtx
16
Solution of Partial Differential Equations
Partial Differential Equations are moredifficult to solve than ordinary differentialequations:
)sin()0,(,0),1(),0(
022
2
2
2
xxututut
u
x
u
)sin()0,(,0),1(),0(
022
2
2
2
xxututut
u
x
u
17
Summary Numerical Methods:
Algorithms that areused to obtainnumerical solution of amathematical problem.
We need them whenNo analytical solutionexists or it is difficultto 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 Numerical Methods:
Algorithms that areused to obtainnumerical solution of amathematical problem.
We need them whenNo analytical solutionexists or it is difficultto 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
18
Number Representation Normalized Floating Point Representation Significant Digits Accuracy and Precision Rounding and Chopping
Reading Assignment: Chapter 3
Number Representation and Accuracy
Number Representation Normalized Floating Point Representation Significant Digits Accuracy and Precision Rounding and Chopping
Reading Assignment: Chapter 3
19
Representing Real Numbers You are familiar with the decimal system:
Decimal System: Base = 10 , Digits (0,1,…,9)
Standard Representations:
21012 10510410210110345.312
You are familiar with the decimal system:
Decimal System: Base = 10 , Digits (0,1,…,9)
Standard Representations:
partpart
fractionintegralsign
54.213
20
Normalized Floating Point Representation Normalized Floating Point Representation:
Scientific Notation: Exactly one non-zero digit appearsbefore decimal point.
Advantage: Efficient in representing very small or verylarge numbers.
exponentsigned:,0
exponentmantissasign
104321.
nd
nffffd
Normalized Floating Point Representation:
Scientific Notation: Exactly one non-zero digit appearsbefore decimal point.
Advantage: Efficient in representing very small or verylarge numbers.
exponentsigned:,0
exponentmantissasign
104321.
nd
nffffd
21
Binary System
Binary System: Base = 2, Digits {0,1}
exponentsignedmantissasign
2.1 4321nffff
Binary System: Base = 2, Digits {0,1}
exponentsignedmantissasign
2.1 4321nffff
10)625.1(10)3212201211(2)101.1(
22
Fact Numbers that have a finite expansion in one numbering
system may have an infinite expansion in anothernumbering system:
You can never represent 1.1 exactly in binary system.
210 ...)011000001100110.1()1.1(
Numbers that have a finite expansion in one numberingsystem may have an infinite expansion in anothernumbering system:
You can never represent 1.1 exactly in binary system.
210 ...)011000001100110.1()1.1(
23
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
S Exponent8 Fraction23
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
S Exponent11 Fraction52
(continued)
24
IEEE 754 Floating-Point Standard
25
Example
26
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
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
27
Remarks
Numbers that can be exactly represented are calledmachine numbers.
Difference between machine numbers is not uniform
Sum of machine numbers is not necessarily a machinenumber
Numbers that can be exactly represented are calledmachine numbers.
Difference between machine numbers is not uniform
Sum of machine numbers is not necessarily a machinenumber
28
Calculator Example Suppose you want to compute:
3.578 * 2.139using a calculator with two-digit fractions
3.57 * 2.13 7.60=
7.653342True answer:
29
48.9
Significant Digits - Example
30
Accuracy and Precision
Accuracy is related to the closeness to the truevalue.
Precision is related to the closeness to otherestimated values.
Accuracy is related to the closeness to the truevalue.
Precision is related to the closeness to otherestimated values.
31
32
Rounding and Chopping
Rounding: Replace the number by the nearestmachine number.
Chopping: Throw all extra digits.
Rounding: Replace the number by the nearestmachine number.
Chopping: Throw all extra digits.
33
Rounding and Chopping
34
Can be computed if the true value is known:
100* valuetrue
ionapproximat valuetrue
ErrorRelativePercentAbsolute
ionapproximat valuetrue
ErrorTrueAbsolute
t
tE
Error Definitions – True Error
100* valuetrue
ionapproximat valuetrue
ErrorRelativePercentAbsolute
ionapproximat valuetrue
ErrorTrueAbsolute
t
tE
35
When the true value is not known:
100*estimatecurrent
estimatepreviousestimatecurrent
ErrorRelativePercentAbsoluteEstimated
estimatepreviousestimatecurrent
ErrorAbsoluteEstimated
a
aE
Error Definitions – Estimated Error
100*estimatecurrent
estimatepreviousestimatecurrent
ErrorRelativePercentAbsoluteEstimated
estimatepreviousestimatecurrent
ErrorAbsoluteEstimated
a
aE
36
We say that the estimate is correct to ndecimal digits if:
We say that the estimate is correct to ndecimal digits rounded if:
n10Error
NotationWe say that the estimate is correct to n
decimal digits if:
We say that the estimate is correct to ndecimal digits rounded if:
n 102
1Error
37
Summary Number Representation
Numbers that have a finite expansion in one numbering systemmay 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.
Number RepresentationNumbers that have a finite expansion in one numbering systemmay 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.
38
Taylor Theorem
Motivation Taylor Theorem Examples
Reading assignment: Chapter 4
Motivation Taylor Theorem Examples
Reading assignment: Chapter 4
39
Motivation
We can easily compute expressions like:
?)6.0sin(,4.1computeyoudoHowBut,
)4(2
103 2
x
?)6.0sin(,4.1computeyoudoHowBut,
)4(2
103 2
x
way?practicala thisIs
sin(0.6)?computeto
definition theuse weCan
0.6
ab
40
Remark
In this course, all angles are assumed tobe in radian unless you are told otherwise.
41
Taylor Series
∑∞
0
)(
0
)(
3)3(
2)2(
'
)()(!
1)(
: writecan weconverge,series theIf
)()(!
1
...)(!3
)()(
!2)(
)()()(
:about)(ofexpansionseriesTaylorThe
k
kk
k
kk
axafk
xf
axafk
SeriesTaylor
or
axaf
axaf
axafaf
axf
∑∞
0
)(
0
)(
3)3(
2)2(
'
)()(!
1)(
: writecan weconverge,series theIf
)()(!
1
...)(!3
)()(
!2)(
)()()(
:about)(ofexpansionseriesTaylorThe
k
kk
k
kk
axafk
xf
axafk
SeriesTaylor
or
axaf
axaf
axafaf
axf
42
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(
:)(ofexpansionseriesn MaclauriThe
k
kk xfk
xf
xf
xf
xff
xf
∑∞
0
)(
3)3(
2)2(
'
)0(!
1)(
: writecan weconverge,series theIf
...!3
)0(!2
)0()0()0(
:)(ofexpansionseriesn MaclauriThe
k
kk xfk
xf
xf
xf
xff
xf
43
Maclaurin Series – Example 1
∞.xforconvergesseriesThe
...!3!2
1!
)0(!
1
11)0()(
1)0()(
1)0(')('
1)0()(
32∞
0
∞
0
)(
)()(
)2()2(
∑∑
xxx
kx
xfk
e
kforfexf
fexf
fexf
fexf
k
k
k
kkx
kxk
x
x
x
xexf )(ofexpansionseriesn MaclauriObtain
∞.xforconvergesseriesThe
...!3!2
1!
)0(!
1
11)0()(
1)0()(
1)0(')('
1)0()(
32∞
0
∞
0
)(
)()(
)2()2(
∑∑
xxx
kx
xfk
e
kforfexf
fexf
fexf
fexf
k
k
k
kkx
kxk
x
x
x
44
The exponential function ex (in blue), and the sum of thefirst n+1 terms of its Taylor series at 0 (in red).
45
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+0.5x2exp(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+0.5x2exp(x)
46
Maclaurin Series – Example 2
∞.xforconvergesseriesThe
....!7!5!3!
)0()sin(
1)0()cos()(
0)0()sin()(
1)0(')cos()('
0)0()sin()(
753∞
0
)(
)3()3(
)2()2(
∑
xxxxx
kf
x
fxxf
fxxf
fxxf
fxxf
k
kk
:)sin()(ofexpansionseriesn MaclauriObtain xxf
∞.xforconvergesseriesThe
....!7!5!3!
)0()sin(
1)0()cos()(
0)0()sin()(
1)0(')cos()('
0)0()sin()(
753∞
0
)(
)3()3(
)2()2(
∑
xxxxx
kf
x
fxxf
fxxf
fxxf
fxxf
k
kk
47
-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)
-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)
48
Maclaurin Series – Example 3
∞.forconvergesseriesThe
....!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
kf
x
fxxf
fxxf
fxxf
fxxf
k
kk
)cos()(:ofexpansionseriesMaclaurinObtain xxf
∞.forconvergesseriesThe
....!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
kf
x
fxxf
fxxf
fxxf
fxxf
k
kk
49
Maclaurin Series – Example 4
1||forconvergesSeries
...xxx1x1
1:ofExpansionSeriesMaclaurin
6)0(1
6)(
2)0(1
2)(
1)0('1
1)('
1)0(1
1)(
ofexpansionseriesn MaclauriObtain
32
)3(4
)3(
)2(3
)2(
2
x
fx
xf
fx
xf
fx
xf
fx
xf
x11
f(x)
1||forconvergesSeries
...xxx1x1
1:ofExpansionSeriesMaclaurin
6)0(1
6)(
2)0(1
2)(
1)0('1
1)('
1)0(1
1)(
ofexpansionseriesn MaclauriObtain
32
)3(4
)3(
)2(3
)2(
2
x
fx
xf
fx
xf
fx
xf
fx
xf
x11
f(x)
50
Example 4 - Remarks
Can we apply the series for x≥1??
How many terms are needed to get a goodapproximation???
How many terms are needed to get a goodapproximation???
These questions will be answered usingTaylor’s Theorem.
51
Taylor Series – Example 5
...)1()1()1(1:)1(ExpansionSeriesTaylor
6)1(6
)(
2)1(2
)(
1)1('1
)('
1)1(1
)(
1atofexpansionseriesTaylorObtain
32
)3(4
)3(
)2(3
)2(
2
xxxa
fx
xf
fx
xf
fx
xf
fx
xf
ax
1f(x)
...)1()1()1(1:)1(ExpansionSeriesTaylor
6)1(6
)(
2)1(2
)(
1)1('1
)('
1)1(1
)(
1atofexpansionseriesTaylorObtain
32
)3(4
)3(
)2(3
)2(
2
xxxa
fx
xf
fx
xf
fx
xf
fx
xf
ax
1f(x)
52
Taylor Series – Example 6
...)1(31
)1(21
)1(:ExpansionSeriesTaylor
2)1(1)1(,1)1(',0)1(
2)(,
1)(,
1)(',)ln()(
)1(at)ln(ofexpansionseriesTaylorObtain
32
)3()2(
3)3(
2)2(
xxx
ffff
xxf
xxf
xxfxxf
axf(x)
...)1(31
)1(21
)1(:ExpansionSeriesTaylor
2)1(1)1(,1)1(',0)1(
2)(,
1)(,
1)(',)ln()(
)1(at)ln(ofexpansionseriesTaylorObtain
32
)3()2(
3)3(
2)2(
xxx
ffff
xxf
xxf
xxfxxf
axf(x)
53
Convergence of Taylor Series
The Taylor series converges fast (few termsare needed) when x is near the point ofexpansion. If |x-a| is large then more termsare needed to get a good approximation.
The Taylor series converges fast (few termsare needed) when x is near the point ofexpansion. If |x-a| is large then more termsare needed to get a good approximation.
54
Taylor’s Theorem
.andbetweenis)()!1(
)(
:where
)(!
)()(
:bygivenis)(of value the thenandcontainingintervalanon
1)(...,2,1,ordersofsderivativepossesses)(functionaIf
1)1(
0
)(
∑
xaandaxn
fR
Raxk
afxf
xfxa
nxf
nn
n
n
n
k
kk
(n+1) terms TruncatedTaylor Series
.andbetweenis)()!1(
)(
:where
)(!
)()(
:bygivenis)(of value the thenandcontainingintervalanon
1)(...,2,1,ordersofsderivativepossesses)(functionaIf
1)1(
0
)(
∑
xaandaxn
fR
Raxk
afxf
xfxa
nxf
nn
n
n
n
k
kk
Remainder
55
Taylor’s Theorem
.applicablenotisTheoremTaylor
defined.notaresderivative
itsandfunction thethen,1If
.1||if0expansionofpoint thewith1
1
:for theoremsTaylor'applycanWe
x
xax
f(x)
.applicablenotisTheoremTaylor
defined.notaresderivative
itsandfunction thethen,1If
.1||if0expansionofpoint thewith1
1
:for theoremsTaylor'applycanWe
x
xax
f(x)
56
Error Term
.andbetweenallfor
)()!1(
)(
:onboundupperanderivecanwe
error,ionapproximatabout theideaangetTo
1)1(
xaofvalues
axn
fR n
n
n
.andbetweenallfor
)()!1(
)(
:onboundupperanderivecanwe
error,ionapproximatabout theideaangetTo
1)1(
xaofvalues
axn
fR n
n
n
57
Error Term - Example
?2.00at
expansionseriesTayloritsof3)( terms4firstthe
by)(replaced weiferror theislargeHow
xwhena
n
exf x
0514268.82.0)!1(
)()!1(
)(
1≥≤)()(
31
2.0
1)1(
2.0)()(
ERn
eR
axn
fR
nforefexf
nn
nn
n
nxn
58
Alternative form of Taylor’s Theorem
hxxwherehn
fR
hRhk
xfhxf
hxx
nxfLet
nn
n
n
n
k
kk
andbetweenis)!1(
)(
size)step(!
)()(
: thenandcontainingintervalanon
1)(...,2,1,ordersofsderivativehave)(
1)1(
0
)(
hxxwhereh
n
fR
hRhk
xfhxf
hxx
nxfLet
nn
n
n
n
k
kk
andbetweenis)!1(
)(
size)step(!
)()(
: thenandcontainingintervalanon
1)(...,2,1,ordersofsderivativehave)(
1)1(
0
)(
59
Taylor’s Theorem – Alternative forms
.andbetweenis
)!1()(
!)(
)(
,
.andbetweenis
)()!1(
)()(
!)(
)(
1)1(
0
)(
1)1(
0
)(
hxxwhere
hn
fh
kxf
hxf
hxxxa
xawhere
axn
fax
kaf
xf
nnn
k
kk
nnn
k
kk
.andbetweenis
)!1()(
!)(
)(
,
.andbetweenis
)()!1(
)()(
!)(
)(
1)1(
0
)(
1)1(
0
)(
hxxwhere
hn
fh
kxf
hxf
hxxxa
xawhere
axn
fax
kaf
xf
nnn
k
kk
nnn
k
kk
60
Mean Value Theorem
)()('
,,0forTheoremsTaylor'Use:Proof
)('
),(exists therethen
),(intervalopen theondefinedisderivativeitsand
],[intervalclosedaonfunctioncontinuousais)(If
abξff(a)f(b)
bhxaxnabf(a)f(b)ξf
baξba
baxf
)()('
,,0forTheoremsTaylor'Use:Proof
)('
),(exists therethen
),(intervalopen theondefinedisderivativeitsand
],[intervalclosedaonfunctioncontinuousais)(If
abξff(a)f(b)
bhxaxnabf(a)f(b)ξf
baξba
baxf
61
Alternating Series Theorem
termomittedFirst:
n terms)first theof(sumsumPartial:
convergesseriesThe
then
0lim
If
S
:seriesgalternatinheConsider t
1
1
4321
4321
n
n
nnnn
a
S
aSS
and
a
and
aaaa
aaaa
termomittedFirst:
n terms)first theof(sumsumPartial:
convergesseriesThe
then
0lim
If
S
:seriesgalternatinheConsider t
1
1
4321
4321
n
n
nnnn
a
S
aSS
and
a
and
aaaa
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62
Alternating Series – Example
!7
1
!5
1
!3
11)1(s
!5
1
!3
11)1(s
:Then
0lim
:sinceseriesgalternatinconvergentaisThis!7
1
!5
1
!3
11)1(s:usingcomputedbecansin(1)
4321
in
in
aandaaaa
in
nn
!7
1
!5
1
!3
11)1(s
!5
1
!3
11)1(s
:Then
0lim
:sinceseriesgalternatinconvergentaisThis!7
1
!5
1
!3
11)1(s:usingcomputedbecansin(1)
4321
in
in
aandaaaa
in
nn
63
Example 7
?1witheapproximatto
usedare terms1)( whenbeerror thecanlargeHow
expansion)ofcenter(the5.0at)(of
expansionseriesTaylortheObtain
12
12
xe
n
aexf
x
x
?1witheapproximatto
usedare terms1)( whenbeerror thecanlargeHow
expansion)ofcenter(the5.0at)(of
expansionseriesTaylortheObtain
12
12
xe
n
aexf
x
x
64
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,)(ofexpansionseriesTaylorObtain 12 aexf x
...!
)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
65
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
)!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
66
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