analisis numerik.pdf
DESCRIPTION
matematika teknikTRANSCRIPT
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NUMERICAL NUMERICAL
ANALYSISANALYSIS
Djoko W KarmiadjiEmail: [email protected]
“Numerical Methods in Engineering Practice” by Amir Wadi Al-Khafaji &
John R. Tooley
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COURSE GOALSCOURSE GOALS
This course has two specific goals:
i. To introduce students to basic
concepts of algebra and mathematics
using numerical analysis method.
ii. To develop analytical skills relevant to
the areas mentioned in (i) above in the
field of mechanical engineering.
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COURSE OBJECTIVESCOURSE OBJECTIVES
Upon successful completion of this course, students should be able to:
1. Determine the solution of algebra and mathematical problems.
2. Distinguish between mathematics and mechanical engineering systems.
3. Familiarize numerical solution methods.
4. Analyze the problem solution induces in mechanical engineering systems using numerical methods.
5. Apply sound analytical techniques and logical procedures in the solution of engineering problems.
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Teaching StrategiesTeaching Strategies
� The course will be taught via Lectures
and Tutorial Sessions, the tutorial
being designed to complement and
enhance both the lectures and the
students appreciation of the subject.
� Course work assignments will be
reviewed with the students.
Attendance at Lectures and Tutorials is Compulsory
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MATRICES & MATRICES &
DETERMINANTSDETERMINANTS
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LINEAR SYSTEM OF EQUATIONSLINEAR SYSTEM OF EQUATIONS
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mnmnmm
nn
nn
bxaxaxa
.
.
. ...
...
.
.
.
bxaxaxa
bxaxaxa
=+++
=+++
=+++
..........
..........
..........
2211
22222121
11212111
=
mnmnmm
n
n
b
b
b
x
x
x
aaa
aaa
aaa
.
...
..
...
...
..
..
2
1
2
1
21
22221
11211 [ ]{ } { }
[ ] { } { } 11
,.....1
,.....1
××× =
=
=
=
mnnm
ijij
bxA
n ,j
m ,i
bxa
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SPECIAL MATRICES SPECIAL MATRICES
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{ } { }
{ }
[ ] [ ]1111
1
21
11
1
112111
.
.
....
aA
a
a
a
A
aaaA
m
m
nn
=⇒
=⇒
=⇒
×
×
×
Scalar
column Vector
row Vector
Scalars&Vectors
>
>
>
[ ]
=⇒ ×
000
000000
..
...
.......
A nm Matrix Null
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SPECIAL MATRICES SPECIAL MATRICES ……continued……continued
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[ ]
=⇒
nnnn
n
n
n
aaa
aaa
aaa
A
..
...
...
..
..
21
22221
11211
Matrix Square
[ ]
=⇒
100
010001
..
...
.......
I n Matrix Identity
[ ]
=⇒
nn
n
a..
...
...
..a
..a
A
00
00
00
22
11
Matrix Diagonal
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SPECIAL MATRICES SPECIAL MATRICES ……continued……continued
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[ ]
=⇒
nnnn
n
n
n
aaa
aaa
aaa
A
..
...
...
..
..
21
22212
11211
Matrix lSymmetrica
[ ]
=⇒ ×
mnnn
m
m
t
nm
aaa
aaa
aaa
A
..
...
...
..
..
21
22212
12111
Matrix Transpose
[ ] [ ][ ][ ]
[ ] [ ]tAA
A
AAA
=⇒
⇒
=⇔⇒
−
−
−−
1
1
111
Matrix Orthogonal
no has Matrix Singular
Matrix Inverse
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SPECIAL MATRICES SPECIAL MATRICES ……continued……continued
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[ ] [ ] [ ][ ] [ ]
=⇒ ×2221
1211
AA
AAA nm Submatrix
[ ]
[ ]
[ ]
=⇒
=⇒
=⇒
1..00
...
...
..10
..1
..
...
...
0..
0..0
..00
...
...
..0
..
2
112
21
2212
11
222
11211
n
n
n
nnnn
n
nn
n
n
n
a
aa
A
aaa
aa
a
A
a
aa
aaa
A
Matrix Triangular-Unit-Upper
Matrix Triangular-Lower
Matrix Triangular-Upper
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OPERASIONAL MATRICESOPERASIONAL MATRICES
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[ ] [ ]
=
= ××
mnmm
n
n
nm
mnmm
n
n
nm
bbb
bbb
bbb
B
aaa
aaa
aaa
A
..
...
...
..
..
&
..
...
...
..
..
21
22221
11211
21
22221
11211
[ ] [ ]
[ ] [ ] [ ]
[ ] [ ] [ ]
[ ] [ ][ ]
lkmi
bac
bac
CBA
baBA
njmi
baBA
jk
n
j
ijik
jkijik
lmlnnm
ijijnmnm
ijijnmnm
....2,1;....2,1
....2,1;....2,1
1
==
=
=
=⇒
±=±⇒
===
=⇒
∑=
×××
××
××
tionmultiplica Matrix
onsubstracti& addition Matrix
equality Matrix
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ROTATION OF A COORDINATE SYSTEMROTATION OF A COORDINATE SYSTEM
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y1
x1
P
1x
1y
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ROTATION OF A COORDINATE SYSTEMROTATION OF A COORDINATE SYSTEM
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y1
x1
y2
P
x2
d1
d2
d4
d3
1x
2x
1y
2y
θ
θ
θ
θ
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ROTATION OF A COORDINATE SYSTEMROTATION OF A COORDINATE SYSTEM
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y1
x1
y2
P
x2
d1
d2
d4
d3
1x
2x
1y
2y
θ
θ
θ
θ
3x
x3
3y
y3 ββ
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DETERMINANTDETERMINANT
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[ ]
nnnn
n
n
ijn
aaa
aaa
aaa
aA
..
...
...
..
..
det
21
22221
11211
==
Properties:
1. det [A] = 0, 2 rows or 2 columns are identical
2. det [A] = - det [A], 2 rows or 2 columns are interchanged
3. det [A] = det [A]t
4. Determinant may be factored from all elements in a row or column
5. Determinant is not changed if a multiple of a row or column is added
to or subtracted from any parallel row or column
6. Determinant of a scalar matrix equals the element itself.
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DETERMINANT DETERMINANT …………continuedcontinued
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[ ]
[ ]
( ) ij
ji
ij
n
i
ijijn
n
j
ijijn
MC
CaA
CaA
+
=
=
−=
=
=
∑
∑
1
det
det
1
1
cofactor, where
or
cofactors of method Laplace
[ ] ∏=
===n
i
ii
nn
n
n
ijn uK
u
uu
uuu
KaA1
222
11211
..00
...
...
..0
..
det
Method Triangle-Upper
nnjnjnn
nijijii
nijijii
njj
ij
aaaa
aaaa
aaaa
aaaa
M
........
..
........
..
)1()1(1
)1()1)(1()1)(1(1)1(
)1()1)(1()1)(1(1)1(
1)1(1)1(111
+−
+++−++
−+−−−−
+−
=
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DETERMINANT DETERMINANT …………continuedcontinued
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[ ] ( )
nnn
n
nnnn
n
n
n
n
n
n
aa
aa
aa
aa
aa
aa
aa
aa
aa
aa
aa
aa
aa
aa
aa
aa
aa
aa
aA
1
111
31
1311
21
1211
331
111
3331
1311
3231
1211
221
111
2321
1311
2221
1211
2
11
..
...
...
..
..
det−=
onCondensati Pivotal of Method
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AREA USING DETERMINANTSAREA USING DETERMINANTS
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[ ] 1221
22
11det yxyx
yx
yxA −==
∑=
=n
i
itotal AA1
+++=1133
22
22
11
21 ......
yx
yx
yx
yx
yx
yxA
nn
+++= −−
1
1
1
......
1
1
1
1
1
1
11
11
44
33
11
33
22
11
21
nn
nn
yx
yx
yx
yx
yx
yx
yx
yx
yx
A
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VOLUME USING DETERMINANTSVOLUME USING DETERMINANTS
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333
222
111
61
zyx
zyx
zyx
V =
1
1
1
1
444
333
222
111
61
zyx
zyx
zyx
zyx
V = or
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ContohContoh SoalSoal 1:1:
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Node 1 2 3 4 5 6 7 8 9x -3 -5 -3 -6 0 4 2 5 2y 5 3 0 -3 -1 -2 3 5 7
5,55
53
72
72
55
55
32
32
24
24
10
10
36
36
03
03
35
35
53
21
=
−+++
−+
−
−+
−
−−+
−−
−+
−
−+
−
−
=A
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ContohContoh SoalSoal 2:2:
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[ ] 333,23
72
1
3
31
2
1
2 ==== ∫ xdxxI
Node 1 2 3 4 5x 2 2 1,5 1 1y 0 4 2,25 1 0
145134123
12345
2
1
2
AAA
AdxxI
++≈
≈= ∫
373,2
101
111
102
111
125,25,1
102
125,25,1
142
102
21
≈
++≈I
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ProblemsProblems
2.2 (a); (c); (e); (g)
2.6 (a); (b); (c); (d)
2.12
2,13
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MATHEMATICAL MATHEMATICAL
MODELINGMODELING
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MECHANICAL ENGINEERING SYSTEMS MECHANICAL ENGINEERING SYSTEMS
Spring Dashpot
xkFS .=
Fs = force in a spring
k = stiffness (property)
x = displacement
dt
dxcFd =
Fd = force in a dashpot
c = damping (property)
dx/dt = velocity
Mass
2
2
dt
xdmFI = FI = force in a mass (inertia)
m = mass (property)
d2x/dt2 = acceleration
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BoundaryBoundary--Value Problem Value Problem
k1 k1
k2 k2
k3 k3
P3
P2
P1
m3
m2
m1
x3
x2
x1
F1F1
F2 F2
F3 F3P3
P2
P1
m3
m2
m1
x3
x2
x1W1
W2
W3FBD
An example of a mechanical system
MECHANICAL ENGINEERING MECHANICAL ENGINEERING …………continuedcontinued
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InitialInitial--Value ProblemValue Problem
P3m3
k3
c3
P2m2
k2
c2
P1m1
k1
c1
x1 x2 x3
P1FI1
Fs1
Fd1
x1
P2FI2
Fs2
Fd2
x2
P3FI3
Fs3
Fd3
x3
MECHANICAL ENGINEERING MECHANICAL ENGINEERING …………continuedcontinued
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Eigenproblems Eigenproblems
σy
σz
σx
x
y
z
Rn
RnxRny
Rnzσy
σz
σx
x
y
z
τxz
τxy
τzy
τzx
τyz
τyx
dx
dydz
MECHANICAL ENGINEERING MECHANICAL ENGINEERING …………continuedcontinued
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BoundaryBoundary--Value Problem Value Problem …..…..lanjutanlanjutan
Newton’s Law
Mass 1: +↑∑Fx=0=2F1-P1-2F2-W1
Mass 2: +↑∑Fx=0=2F2-P2-2F3-W2
Mass 3: +↑∑Fx=0=2F3-P3-W3
F1=k1x1; W1=m1g
F2=k2(x2-x1); W2=m2g
F3=k3(x3-x2); W3=m3g
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ProblemsProblems
3.4
3.5
3.7
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LINEAR ALGEBRA LINEAR ALGEBRA
EQUATIONSEQUATIONS
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CRAMER’S RULECRAMER’S RULE
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nnnnnn
nn
nn
bxaxaxa
bxaxaxa
bxaxaxa
=+++
=+++
=+++
..........
..........
..........
2211
22222121
11212111
.
.
. ...
...
.
.
.
=
nnnnnn
n
n
b
b
b
x
x
x
aaa
aaa
aaa
.
...
..
...
...
..
..
2
1
2
1
21
22221
11211
[ ]
nnnn
n
n
aaa
aaa
aaa
A
..
...
...
..
..
det
21
22221
11211
=
[ ]
mnjnnjnnn
njj
njj
j
aabaaa
aabaaa
aabaaa
A
....
...
...
....
....
det
)1()1(21
2)1(22)1(22221
1)1(11)1(11211
+−
+−
+−
=[ ][ ]A
Ax
j
jdet
det=⇒
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GAUSS’S ELIMINATION METHODGAUSS’S ELIMINATION METHOD
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nnnnn
n
n
b
b
b
aaa
aaa
aaa
.
.
..
...
...
..
..
2
1
21
22221
11211
:matrix Augmented
n
n
n
c
c
cu
uu
.
.
1..00
...
...
..10
..1
2
1
2
112
nn
nn
nn
cx
cxux
cxuxux
=
=++
=+++
.
.
222
112121
..........
..........
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GAUSSGAUSS--JORDAN ELIMINATION METHODJORDAN ELIMINATION METHOD
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nnnnn
n
n
b
b
b
aaa
aaa
aaa
.
.
..
...
...
..
..
2
1
21
22221
11211
:matrix Augmented
nz
z
z
.
.
1..00
...
...0..100..01
2
1
nn zx
zx
zx
=
=
=
.
.
22
11
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ProblemsProblems
4.3
4.4
4.7
4.17
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MATRIX INVERSIONMATRIX INVERSION
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CRAMER’S RULECRAMER’S RULE
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[ ] [ ][ ] [ ] [ ] [ ]
[ ] matrix Inverse
→
==⇒
=
−
−−
1
11
21
22221
11211
..
...
...
..
..
A
IAAAA
aaa
aaa
aaa
A
nnnn
n
n
[ ] [ ]
( )
nnjnjnn
nijijii
nijijii
njj
ij
ij
ji
ij
t
ij
aaaa
aaaa
aaaa
aaaa
M
MC
njniCA
A
........
..
........
..
1
,.......,1,......,11
)1()1(1
)1()1)(1()1)(1(1)1(
)1()1)(1()1)(1(1)1(
1)1(1)1(111
1
+−
+++−++
−+−−−−
+−
+
−
=
−=
===
cofactor, where
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ELIMINATION METHODELIMINATION METHOD
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1..00
...
...0..100..01
..
...
...
..
..
21
22221
11211
:matrix Augmented
nnnn
n
n
aaa
aaa
aaa
[ ] [ ]
==
−
nnnn
n
n
nnnn
n
n
bbb
bbb
bbb
BA
bbb
bbb
bbb
..
...
...
..
..
..
...
...
..
..
1..00
...
...0..100..01
21
22221
11211
1
21
22221
11211
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ProblemsProblems
5.1
5.2
5.3
5.12
dwkdwkdwkdwk 38
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dwkdwkdwkdwk 39
NON LINEAR ALGEBRA NON LINEAR ALGEBRA
EQUATIONSEQUATIONS
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INTERVALINTERVAL--HALVING METHODHALVING METHOD
dwkdwkdwkdwk 40
0)(2
0)()()(
)(
0
)(
213
2122
11
≈→⇒+
=
<→→→
=
=
nn xfxxx
x
xfxfxfx
xfx
yx
xfy
............
whichfor value find to
![Page 41: Analisis Numerik.pdf](https://reader033.vdocuments.us/reader033/viewer/2022042513/5466e144af795988338b5210/html5/thumbnails/41.jpg)
FALSEFALSE--POSITION METHODPOSITION METHOD
dwkdwkdwkdwk 41
0)()()(
)()(
0)()()(
)(
0
)(
21
21123
2122
11
≈→⇒−−
=
<→→→
=
=
nn xfxxfxf
xfxxfxx
xfxfxfx
xfx
yx
xfy
............
whichfor value find to
![Page 42: Analisis Numerik.pdf](https://reader033.vdocuments.us/reader033/viewer/2022042513/5466e144af795988338b5210/html5/thumbnails/42.jpg)
NEWTONNEWTON--RAPHSON FIRST METHODRAPHSON FIRST METHOD
dwkdwkdwkdwk 42
0)()(
)(
)(
)(
)(
0
)(
1
1
112
11
≈→⇒′
−=
′−=
→
=
=
+ nn
k
kkk xfx
xf
xfxx
xf
xfxx
xfx
yx
xfy
............
whichfor value find to
![Page 43: Analisis Numerik.pdf](https://reader033.vdocuments.us/reader033/viewer/2022042513/5466e144af795988338b5210/html5/thumbnails/43.jpg)
NEWTONNEWTON--RAPHSON SECOND METHODRAPHSON SECOND METHOD
dwkdwkdwkdwk 43
0)()(
)(
)(2
)(
)(
)(
)(2
)(
)(
0
)(
1
1
1
1
1
1
112
11
≈→⇒
′−
′′′
+=
′−
′′′
+=
→
=
=
−
+
−
nn
k
k
k
kkk xfx
xf
xf
xf
xfxx
xf
xf
xf
xfxx
xfx
yx
xfy
whichfor value find to
![Page 44: Analisis Numerik.pdf](https://reader033.vdocuments.us/reader033/viewer/2022042513/5466e144af795988338b5210/html5/thumbnails/44.jpg)
ProblemsProblems
6.2
6.3
6.5
6.6
6.7
dwkdwkdwkdwk 44
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dwkdwkdwkdwk 45
EIGENPROBLEMSEIGENPROBLEMS
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INTRODUCTIONINTRODUCTION
dwkdwkdwkdwk 46
0..........
0..........
0..........
2211
2222121
1212111
=+++
=+++
=+++
nnnnn
nn
nn
xqxqxq
xqxqxq
xqxqxq
.
.
. ...
...
.
.
.
=
0..00
.
.
..
...
...
..
..
2
1
21
22221
11211
nnnnn
n
n
x
x
x
qqq
qqq
qqq
[ ] 0
..
...
...
..
..
det
21
22221
11211
==
nnnn
n
n
qqq
qqq
qqq
Q
![Page 47: Analisis Numerik.pdf](https://reader033.vdocuments.us/reader033/viewer/2022042513/5466e144af795988338b5210/html5/thumbnails/47.jpg)
CHARACTERISTIC EQUATIONSCHARACTERISTIC EQUATIONS
dwkdwkdwkdwk 47
[ ] [ ] [ ]BAQ −=
[ ]{ } [ ] [ ][ ]{ } { }
[ ]
=
=−=
)(...)(
...
...
.....
)(...)(
0
1
111
1
λλ
λλ
nnn
n
ff
ff
B
xBAxQ
where
I Tipe
[ ]{ } [ ] [ ][ ]{ } { }[ ] constants. all are of elements where,
II Tipe
C
xCAxQ 02 =−= λ
[ ]{ } [ ] [ ][ ]{ } { }03 =−= xIAxQ λ
III Tipe
![Page 48: Analisis Numerik.pdf](https://reader033.vdocuments.us/reader033/viewer/2022042513/5466e144af795988338b5210/html5/thumbnails/48.jpg)
EIGENVALUES & EIGENVECTORSEIGENVALUES & EIGENVECTORS
dwkdwkdwkdwk 48
08147)(
000
210131
012
23
3
2
1
=−+−=
=
−−−−−
−−
λλλλ
λλ
λ
f
x
x
x
{ } tor)(first vec
)eigenvalue(first
=
=
=⇒
=
111
1
3
3
3
3
3
2
1
1
1
x
x
x
x
x
x
x
x
λ
{ } vector)(second
)eigenvalue (second
−=
−=
=⇒
=
101
0
2
3
3
3
3
2
1
2
2
xx
x
x
x
x
x
λ
{ } vector)(third
)eigenvalue (third
−=
−=
=⇒
=
12
12
4
3
3
3
3
3
2
1
3
3
x
x
x
x
x
x
x
x
λ
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MECHANICAL VIBRATIONMECHANICAL VIBRATION
dwkdwkdwkdwk 49
k2=2kk1=kP(t)=0m2=2m
x2
m1=m
x1
=
+
−
−00
200
2223
2
1
2
1
x
xm
mx
xkkkk
&&
&&
( ) ( )( ) ( )
=
−
−
−→
+−=⇒+=
+−=⇒+=
00
2001
2223
sinsin
sinsin
2
12
2
1
2
2222
2
1111
A
Am
A
Ak
tAxtAx
tAxtAx
ω
θωωθω
θωωθω
&&
&&
![Page 50: Analisis Numerik.pdf](https://reader033.vdocuments.us/reader033/viewer/2022042513/5466e144af795988338b5210/html5/thumbnails/50.jpg)
dwkdwkdwkdwk 50
[ ]
eigenvalue second
eigenvaluefirst
⇒=→=+=→
⇒=→=−=→
=⇒+−=
=
−−−−
mk
mk
,
,
kmQ
A
A
93217320.332
51802679.032
14det
00
22223
22
11
22
2
1
ωλ
ωλ
ωλλλ
λλ
MECHANICAL VIBRATION MECHANICAL VIBRATION …………………..continued…………………..continued
{ }
=
=⇒
=⇒
1732,0
732,0
2679,0
1
21
1
V
AA
reigenvectoFirst λ
{ }
−=
−=⇒
=⇒
1732.2
732.2
732.3
2
21
1
V
AA
reigenvecto Second λ
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MECHANICAL VIBRATION MECHANICAL VIBRATION …………………..continued…………………..continued
dwkdwkdwkdwk 51
k2=2kk1=k
P(t)=0m2
x2
m1
x1
0.73 1.0
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MECHANICAL VIBRATION MECHANICAL VIBRATION …………………..continued…………………..continued
dwkdwkdwkdwk 52
-2.73 1.0
k2=2kk1=k
P(t)=0m2
x2
m1
x1
![Page 53: Analisis Numerik.pdf](https://reader033.vdocuments.us/reader033/viewer/2022042513/5466e144af795988338b5210/html5/thumbnails/53.jpg)
ProblemsProblems
7.1
7.3
7.5
dwkdwkdwkdwk 53
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dwkdwkdwkdwk 54
INTERPOLATIONINTERPOLATION
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dwkdwkdwkdwk 55
Interpolating Polynomials For Even IntervalsInterpolating Polynomials For Even Intervals
xi x0 x1 ... xn
f(xi) f(x0) f(xi) ... f(xn)
n data points
n
n
ii
xaxaaxf
nixxh
+++≈
=−= +
..........)(
,.......,0,
10
1
n
nnnn
n
n
n
n
xaxaaxf
xaxaaxf
xaxaaxf
+++≈
+++≈
+++≈
..........)(
.
.
..........)(
..........)(
10
11101
00100
![Page 56: Analisis Numerik.pdf](https://reader033.vdocuments.us/reader033/viewer/2022042513/5466e144af795988338b5210/html5/thumbnails/56.jpg)
dwkdwkdwkdwk 56
Forward InterpolationForward Interpolation
2
210)(
2
xaxaaxf
n
++≈
=
2
2
102
2
2
101
00
42 ahhaaf
ahhaaf
af
++=
++=
=
)(
)(
)(
22
11
00
xff
xff
xff
=
=
=
x
y
0 h 2h
f0
f1 f2
=
2
2
1
0
2
1
0
421
111
001
ha
ha
a
f
f
f
−
−−=
2
1
0
2
2
1
0
121
143
002
2
1
f
f
f
ha
ha
a( )
( )21022
2101
00
22
1
432
1
fffh
a
fffh
a
fa
+−=
−+−=
=
( ) ( ) 2
21022100 22
143
2
1)( xfff
hxfff
hfxf +−+−+−+=
2
1
0
)2(2
)(
)0(0
fhfhx
fhfhx
ffx
=⇒=
=⇒=
=⇒=
![Page 57: Analisis Numerik.pdf](https://reader033.vdocuments.us/reader033/viewer/2022042513/5466e144af795988338b5210/html5/thumbnails/57.jpg)
dwkdwkdwkdwk 57
Backward InterpolationBackward Interpolation
2
210)(
2
xaxaaxf
n
++≈
=
2
2
102
2
2
101
00
42 ahhaaf
ahhaaf
af
+−=
+−=
=
−
−x
y
-2h -h 0
f-2
f-1
f0
−
−=
−
−
2
1
0
2
2
1
0
121
143
002
2
1
f
f
f
ha
ha
a
( ) ( ) 2
01220120 22
134
2
1)( xfff
hxfff
hfxf +−++−+= −−−−
2
1
0
)2(2
)(
)0(0
−
−
=−⇒−=
=−⇒−=
=⇒=
fhfhx
fhfhx
ffx
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dwkdwkdwkdwk 58
Central InterpolationCentral Interpolation
2
210)(
2
xaxaaxf
n
++≈
=
2
2
101
00
2
2
101
ahhaaf
af
ahhaaf
++=
=
+−=−
x
y
-h 0 h
f-1
f0
f1
−
−=
−
1
0
1
2
2
1
0
121
101
020
2
1
f
f
f
ha
ha
a
( ) ( ) 2
1012110 22
1
2
1)( xfff
hxff
hfxf +−++−+= −−
![Page 59: Analisis Numerik.pdf](https://reader033.vdocuments.us/reader033/viewer/2022042513/5466e144af795988338b5210/html5/thumbnails/59.jpg)
dwkdwkdwkdwk 59
Different Form InterpolationDifferent Form Interpolation
{ }
=
2
1
0
21)(
a
a
a
xxxf
−
−−=
2
1
0
222
2
1
0
/4/2/1
/1/4/3
002
2
1
f
f
f
hhh
hhh
a
a
a
Forward interpolation
{ }
22
2
12
2
02
2
2
1
0
222
2
2
12
32
2
1
/4/2/1/1/4/3
0021
2
1)(
fh
x
h
xf
h
x
h
xf
h
x
h
x
f
f
f
hhhhhhxxxf
+−+
−+
+−=
−−−=
xh 2h
f(x) Interpolating function
![Page 60: Analisis Numerik.pdf](https://reader033.vdocuments.us/reader033/viewer/2022042513/5466e144af795988338b5210/html5/thumbnails/60.jpg)
dwkdwkdwkdwk 60
Different Form Interpolation Different Form Interpolation .......continue.......continue
xh 2h
N0
Plot of N0
1.0
h 2hx
N1
Plot of N1
1.0
h 2hx
N2
Plot of N2
1.0
)(2
1
)(2
)(3
22
1
)(
22
2
2
12
2
1
02
2
0
2
0
221100
xNh
x
h
xN
xNh
x
h
xN
xNh
x
h
xN
fNfNfNfNxfi
ii
=
+=
=−=
=
+−=
=++= ∑=
where
N0, N1 & N2 = shape function
at x=0; N0=1, N1=0, N2=0
at x=h; N0=0, N1=1, N2=0
at x=2h; N0=0, N1=0, N2=1
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dwkdwkdwkdwk 61
Difference OperatorsDifference Operators
xi x0 x1 ... xn
f(xi) f(x0) f(x1) ... f(xn) n
ii
xxx
nixxh
<<<
=−= +
.........
,.......,0,
10
1
nifff
fff
fff
fff
iii
nnn
,.......,0,
.
.
1
1
121
010
=−=∆
−=∆
−=∆
−=∆
+
+
or
Forward Differences
( )( )
( )
( ) nifff
fff
fff
fff
iii
nnn
,.......,0,
.
.
1
2
1
2
121
2
010
2
=−∆=∆
−∆=∆
−∆=∆
−∆=∆
+
+
or
( ) nifff ii
n
i
n ,.......,0,1
1 =−∆=∆ +−
nki
n
k
k
i
nf
knk
nf +−
=∑ −
−=∆0 )!(!
!)1(
![Page 62: Analisis Numerik.pdf](https://reader033.vdocuments.us/reader033/viewer/2022042513/5466e144af795988338b5210/html5/thumbnails/62.jpg)
dwkdwkdwkdwk 62
Difference Operators Difference Operators ……….continue……….continue
nifff iii ,.......,0,1 =−=∇ −
nifff iii ,.......,0,12/1 =−= ++ δ
( )ii
n
i
nfff −= +
−+ 1
1
2/1 δδ
Backward Differences
( ) nifff ii
n
i
n ,.......,0,1
1 =−∇=∇ −−
ki
n
k
k
i
nf
knk
nf −
=∑ −
−=∇0 )!(!
!)1(
Central Differences
nki
n
k
k
i
nf
knk
nf +−
=+ ∑ −
−=0
2/1)!(!
!)1(δ
n
nn
n
n
h
h
h
xfxf
dx
xfd
h
xf
dx
xdf
h
xf
dx
xdf
h
xfhxf
dx
xdf
)()(
)(
)()(
)(lim
)(
)()(lim
)(
0
0
∆==
∆≈
∆=
−+=
→
→
![Page 63: Analisis Numerik.pdf](https://reader033.vdocuments.us/reader033/viewer/2022042513/5466e144af795988338b5210/html5/thumbnails/63.jpg)
dwkdwkdwkdwk 63
Relationship between Difference OperatorsRelationship between Difference Operators
nn
i
n
i
n
ii
iiii
iii
E
fEf
fEf
E
fEfEff
fff
)1(
)1(
)1(
1
)1(
22
1
−=∆
−=∆
−=∆
−=∆
−=−=∆
−=∆ +
n
n
n
nn
iii
iiii
iii
E
E
E
E
fE
ff
fE
fEff
fff
)1(
)1(
1
1
1
1, 11
1
−∆
∆=
−=∇
−∆∆
=−
=∇
−=∇
==
−=∇
−−
−
f(x+h) = Ef(x)
fi+1=Efi, i= 0,…..,n
Forward Operator Backward Operator Central Operator
2/2/
2/1
2/12/1
2/12/1
2/1
12/1
)1()1(
)1(
)1()1(
n
n
n
nn
iii
iii
EE
fEffE
fff
∇−∇
=∆+∆
=
∆+
∆=
∆+−∆+=
−=
−=
−=
−
−
++
δ
δ
δ
δ
δ
![Page 64: Analisis Numerik.pdf](https://reader033.vdocuments.us/reader033/viewer/2022042513/5466e144af795988338b5210/html5/thumbnails/64.jpg)
dwkdwkdwkdwk 64
Differences and Interpolating PolynomialsDifferences and Interpolating Polynomials
)()(
)()()2(
)()(
2
xfEhxf
xfEhxEfhxf
xEfhxf
αα =+
=+=+
=+
)(!
)1)......(1(....
!3
)2)(1(
!2
)1(1)(
)()1()()(
32
i
n
i
iii
xfn
n
hxf
xfxfEhxf
∆+−−
++
∆−−
+∆−
+∆+=+
∆+==+
ααα
ααααααα
α αα
Newton’s Forward Interpolating Formula
Newton’s Backward Interpolating Formula
)(!
)1)......(1(....
!3
)2)(1(
!2
)1(1)(
)()1()()(
32
i
n
i
iii
xfn
n
hxf
xfxfEhxf
∇−++
++
∇
+++∇
++∇+=+
∇−==+ −
ααα
ααααααα
α αα
f(x)
f(x)f(xi)f(xi-h)
f(xi+h)
xxxixi-h xi+h
αh
f(x)
f(x)f(xi)f(xi-h) f(xi+h)
xx xixi-h xi+h
αh
![Page 65: Analisis Numerik.pdf](https://reader033.vdocuments.us/reader033/viewer/2022042513/5466e144af795988338b5210/html5/thumbnails/65.jpg)
ProblemsProblems
8.1
8.3
8.5
dwkdwkdwkdwk 65
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dwkdwkdwkdwk 66
CURVE FITTINGCURVE FITTING
![Page 67: Analisis Numerik.pdf](https://reader033.vdocuments.us/reader033/viewer/2022042513/5466e144af795988338b5210/html5/thumbnails/67.jpg)
dwkdwkdwkdwk 67
LINEAR REGRESSIONLINEAR REGRESSION
xaaxG
Nifx ii
21)(
,.......,1),(
+=
=⇒
∑∑∑
∑∑
===
==
+=
+=
N
i
i
N
i
i
N
i
ii
N
i
i
N
i
i
xaxaxf
xaNaf
1
2
2
1
1
1
1
21
1
d1= deviation
G(x)→Assumed
linear equation
(xi,fi)
f(x)
fi
Gi
xxi
( )
( )222
22
2
1
∑∑∑∑∑∑∑∑∑∑∑
−
−=
−
−=
ii
iiii
ii
iiiii
xxN
fxxfNa
xxN
xfxxfa
![Page 68: Analisis Numerik.pdf](https://reader033.vdocuments.us/reader033/viewer/2022042513/5466e144af795988338b5210/html5/thumbnails/68.jpg)
dwkdwkdwkdwk 68
LINEARIZATIONLINEARIZATION
xaaG
xaxG
Nifx
a
ii
lnlnln
)(
,.......,1),(
21
12
+=
=
=⇒
xaaG
ff
xx
aa
GG
21
11
ln
ln
ln
ln
+=
=
=
=
=
( )
( )222
22
2
1
∑∑∑∑∑∑∑∑∑∑∑
−
−=
−
−=
ii
iiii
ii
iiiii
xxN
fxxfNa
xxN
xfxxfa
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CorrelationCorrelation
−
−
−
=
∑∑∑∑
∑∑
====
==
2
11
2
2
11
2
22
112
N
i
i
N
i
i
N
i
i
N
i
i
N
i
i
N
i
ii
ffNxxN
fxfN
R
R2 = 1 Perfect correlation
R2 = 0 No correlation
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Exponential EquationExponential Equation
y= C x p
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Contoh Soal 1.Contoh Soal 1.Satu contoh numerik, misalnya sebuah grafik yang menunjukkan
konsentrasi tegangan (Kts) sebagai fungsi a/d, dimana pendekatan
dilakukan terhadap fungsi eksponensial Kts = f(a/d) dengan parameter
0.04 ≤ (a/d) ≤ 0.3 dan data-data point ditunjukkan pada grafik berikut:
i (a/d)i (Kts)i
1 0.05 1.78
2 0.10 1.65
3 0.15 1.57
4 0.20 1.50
5 0.25 1.45
6 0.30 1.41
Dengan menggunakan kalkulator atau
komputer didasarkan pada persamaan
didepan, maka dapat ditentukan
p = -0.1305; C = 1.2138; r = 0.9929Dari parameter ini, solusi persamaan
adalah sebagai berikut:
( ) 1305.0
2138.1
da
K ts =
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a/d = x (kts) = y ln x ln y (ln x)(ln y) (ln y)/n (ln x)^2 (ln y)^2
0.05 1.78
0.1 1.65
0.15 1.57
0.2 1.5
0.25 1.45
0.3 1.41
Σ
p =
C =
r =
ContohContoh SoalSoal 1.1. ……..……..lanjutanlanjutan
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a/d = x (kts) = y ln x ln y (ln x)(ln y) (ln y)/n (ln x)^2 (ln y)^2
0.05 1.78 -2.99573 0.576613 -1.72738 0.096102 8.974412 0.332483
0.1 1.65 -2.30259 0.500775 -1.15308 0.083463 5.301898 0.250776
0.15 1.57 -1.89712 0.451076 -0.85574 0.075179 3.599064 0.203469
0.2 1.5 -1.60944 0.405465 -0.65257 0.067578 2.59029 0.164402
0.25 1.45 -1.38629 0.371564 -0.5151 0.061927 1.921812 0.138059
0.3 1.41 -1.20397 0.34359 -0.41367 0.057265 1.449551 0.118054
Σ -11.3951 2.649083 -5.31754 0.441514 23.83703 1.207243
p = -0.13
C = 1.2138
r = -0.996
ContohContoh SoalSoal 1.1. ……..……..lanjutanlanjutan
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NONLINEAR REGRESSIONNONLINEAR REGRESSION
2
321)(
,.......,1),(
xaxaaxG
Nifx ii
++=
=⇒ ( )
( )
( )
( ) )(20
)(20
)1(20
2
1
2
321
3
1
2
321
2
1
2
321
1
1
22
321
i
N
i
iii
i
N
i
iii
N
i
iii
N
i
iii
xxaxaafa
D
xxaxaafa
D
xaxaafa
D
xaxaafD
−−−−==∂∂
−−−−==∂∂
−−−−==∂∂
−−−=
∑
∑
∑
∑
=
=
=
=
∑∑∑∑
∑∑∑∑
∑∑∑
====
====
===
++=
++=
++=
N
i
i
N
i
i
N
i
i
N
i
ii
N
i
i
N
i
i
N
i
i
N
i
ii
N
i
i
N
i
i
N
i
i
xaxaxaxf
xaxaxaxf
xaxaNaf
1
4
3
1
3
2
1
2
1
1
2
1
3
3
1
2
2
1
1
1
1
2
3
1
21
1
?&, 321 aaa
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PolynomialPolynomial FunctionFunctiony=ao+a1x+a2x
2+a3x3+…+anx
n
A number of (n+1) point data match the equation can be expresed as:
(x1,y1),(x2,y2),(x3,y3),…,(xn,yn),(xn+1,yn+1)
y1 =ao + a1x1 + a2x12 + a3x1
3 + … + anx1n
y2 =ao + a1x2 + a2x22 + a3x2
3 + … + anx2n
y3 =ao + a1x3 + a2x32 + a3x3
3 + … + anx3n
.
.
.
.
yn+1 =ao + a1xn+1 + a2x2
n+1 + a3 x3n+1 + … + anx
nn+1
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Solusi simultan dengan metode determinan:
n
nnn
n
n
n
nnnn
n
n
xxx
xxx
xxx
xxxy
xxxy
xxxy
a
1
2
11
2
2
22
1
2
11
1
2
111
2
2
222
1
2
111
0
.1
.....
.1
.1
.
.....
.
.
+++
++++=
Untuk koefisien umum ai dimana i =
0. 1, 2, ..., n maka
n
n
i
nnn
ni
ni
n
nnnn
n
n
i
xxxx
xxxx
xxxx
xyxx
xyxx
xyxx
a
11
2
11
22
2
22
11
2
11
11
2
11
22
2
22
11
2
11
....1
.......
....1
....1
....1
.....
....1
....1
++++
++++=
PolynomialPolynomial Function Function ……….continue
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Contoh Soal 2.Contoh Soal 2.Satu contoh numerik, misalnya sebuah grafik yang menunjukkan
konsentrasi tegangan (Kts) sebagai fungsi a/d, dimana pendekatan
dilakukan terhadap fungsi polinomial dengan 4 data sebagai berikut:
(a/d) Kts
0.0 2.0
0.1 1.65
0.2 1.50
0.3 1.41
Bentuk umum persamaan
polinomial:
32 )/()/()/( dadadaK ts δγβα +++=
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Contoh Soal 1. Contoh Soal 1. ……..lanjutan……..lanjutan
Jika a/d = 0 maka Kts = 2.0 = α, persamaan menjadi
32 )/()/()/(0.2 dadadaK ts δγβ +++=
Dengan memasukkan data kedalam persamaan
32
32
32
)3.0()3.0()3.0(0.241.1
)2.0()2.0()2.0(0.250.1
)1.0()1.0()1.0(0.265.1
δγβ
δγβ
δγβ
+++=
+++=
+++=
Menggunakan metode solusi aljabar linear
3.23
0.17
97.4
−=
+=
−=
δ
γβ
Persamaan polynomial menjadi:
32 )/(3.23)/(0.17)/(97.40.2 dadadaK ts −+−=
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MULTIPLE REGRESSIONMULTIPLE REGRESSION
yaxaaxG
yx
f
ii
i
321)(
),(
++=
⇒
⇒
variable tindependen
variable dependent ( )
( )
( )
( ) )(20
)(20
)1(20
1
321
3
1
321
2
1
321
1
1
2
321
i
N
i
iii
i
N
i
iii
N
i
iii
N
i
iii
yyaxaafa
D
xyaxaafa
D
yaxaafa
D
yaxaafD
−−−−==∂∂
−−−−==∂∂
−−−−==∂∂
−−−=
∑
∑
∑
∑
=
=
=
=
∑∑∑∑
∑∑∑∑
∑∑∑
====
====
===
++=
++=
++=
N
i
i
N
i
ii
N
i
i
N
i
ii
N
i
ii
N
i
i
N
i
i
N
i
ii
N
i
i
N
i
i
N
i
i
yayxayayf
xyaxaxaxf
yaxaNaf
1
2
3
1
2
1
1
1
1
3
1
2
2
1
1
1
1
3
1
21
1
?&, 321 aaa
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ProblemsProblems
9.11
9.12
dwkdwkdwkdwk 80
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NUMERICAL DIFFERENTIATIONNUMERICAL DIFFERENTIATION
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Review of Taylor SeriesReview of Taylor Series
∑
∑∞
=
∞
=
++
+
+
±=±
±=±
+≤≤+
=
+++++=+
0
)(
0
0
)(
0
00
)1(1
1
10
)(
0
)2(2
0
)1(
00
)0(!
)()(
)(!
)()(
),()!1(
)(!
.....)(!2
)()()(
n
nn
n
nn
nn
n
n
nn
fn
hhf
xfn
hhxf
hxsxsfn
hR
Rxfn
hxf
hxhfxfhxf
Error
Slope at x0f(x)
G(x0+h)
f(x0+h)
f(x0) True function
h
x0+hx0
x ERROR)()()(
ERROR)()(
)()()(
)()()(
00
)1(
0
00
00
)1(
0
000
)1(
±+=+
±+=+
+=+
−+=
xfxhfhxf
hxGhxf
xfxhfhxG
h
xfhxGxf
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Review of Taylor Series Review of Taylor Series ……..continue……..continue
3
)3(
32
)2(
2
321
)1(
3
3
2
210
6)(
62)(
32)(
)(
axf
xaaxf
xaxaaxf
xaxaxaaxf
=
+=
++=
+++=
=
=
+=
++=
+++=
3
2
1
0
0
2
00
3
0
2
00
0
)3(
0
)2(
0
)1(
0
30
)3(
0320
)2(
2
030210
)1(
3
03
2
020100
6000
6200
3210
1
)(
)(
)(
)(
6)(
62)(
32)(
)(
a
a
a
a
x
xx
xxx
xf
xf
xf
xf
axf
xaaxf
xaxaaxf
xaxaxaaxf
Assumed:
hxsxsfn
hR
fh
fh
hffhf
xfh
xfh
xhfxfhxf
nn
n +≤≤+
=
+++=
+++=+
++
+ 00
)1(1
1
)3(3
)2(2
)1(
0
)3(3
0
)2(2
0
)1(
00
),()!1(
)0(6
)0(2
)0()0()(
)(6
)(2
)()()(
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Taylor Series MethodTaylor Series Method
h h h h
xxi xi+1 xi+2xi-1xi-2
fi+2
fi+1fi
fi-1
fi-2
f
First forward derivative
( )21
)1(
2
)2(2
)1(
2
1
)2(2
)1(
1
432
1
ERROR,!2
)2()2(
ERROR,!2
++
++
++
−+−≅
=±++=
=±++=
iiii
iiiii
iiiii
fffh
f
xxfh
fhff
xxfh
hfff
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Taylor Series Method Taylor Series Method ……continue……continue
( )21
)1(
1
)2(2
)1(
1
2
)2(2
)1(
2
432
1
ERROR,!2
)(
ERROR,!2
)2(2
−−
−−
−−
+−≅
=±−
+−=
=±−
+−=
iiii
iiiii
iiiii
fffh
f
xxfh
hfff
xxfh
hfff
First backward derivative
Central first derivative
( )11
)1(
2
1+− +−≅ iii ff
hf
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Forward derivative for a third-order polynomial
3
)3(3
)2(2
)1(
3
2
)3(3
)2(2
)1(
2
1
)3(3
)2(2
)1(
1
,6
)3(
2
)3()3(
,6
)2(
2
)2()2(
,62
++
++
++
=+++=
=+++=
=+++=
iiiiii
iiiiii
iiiiii
xxfh
fh
fhff
xxfh
fh
fhff
xxfh
fh
hfff
Taylor Series Method Taylor Series Method ……continue……continue
xi xi+2 xi+3xi+1
h h h
x
fi+3
fi+2fi
fi+1
f
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Forward derivative for a third-order polynomial
Taylor Series Method Taylor Series Method ……continue……continue
−
−
−
≅
+
+
+
3
2
1
)3(3
)2(2
)1(
29
29
34
61
21
1001
0101
0011
3
22
1
i
i
i
i
i
i
i
f
f
f
f
fh
fh
hf
( )
( )
( )3213
)3(
3212
)2(
321
)1(
331
4521
2918116
1
+++
+++
+++
+−+−≅
−+−≅
+−+−≅
iiiii
iiiii
iiiii
ffffh
f
ffffh
f
ffffh
f
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Undetermined Coefficients MethodUndetermined Coefficients Method
2
2
1022
2
2
1011
0
42,2
,
,0
ahhaafhx
ahhaafhx
afx
ii
ii
ii
++=⇒=
++=⇒=
=⇒=
++
++
2
)2(
21
)1(
2
210
2)(
2)(
)(
axf
xaaxf
xaxaaxf
=
+=
++=
2
)2(
1
)1(
2
0
af
af
xi
=
=
=
First & second forward derivative
h h
xxi xi+1 xi+2
fi+2
fi+1
fi
f
−
−−=
+
+
2
1
2
2
1
0
121
143
002
2
1
i
i
i
f
f
f
ha
ha
a
( )
( )212
)2(
21
)1(
21
432
1
++
++
+−=
−+−=
iiii
iiii
fffh
f
fffh
f
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Undetermined Coefficients Undetermined Coefficients ……..continue……..continue
0
2
2
1011
2
2
1022
,0
,
42,2
afx
ahhaafhx
ahhaafhx
ii
ii
ii
=⇒=
+−=⇒−=
+−=⇒−=
−−
−−
First & second backward derivative
h h
xxi-2 xi-1 xi
fi-2
fi-1
fi
f
−
=
−
−
i
i
i
f
f
f
ha
ha
a
1
2
2
2
1
0
121
341
100
2
1
( )
( )iiii
iiii
fffh
f
fffh
f
+−=
+−=
−−
−−
122
)2(
12
)1(
21
342
1
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Undetermined Coefficients Undetermined Coefficients ……..continue……..continue
2
2
1011
0
2
2
1011
,
,0
,
ahhaafhx
afx
ahhaafhx
ii
ii
ii
++=⇒=
=⇒=
+−=⇒−=
++
−−
First & second central derivative
h h
xxi-1 xi xi+1
fi-1
fi
fi+1
f
−
−=
+
−
1
1
2
2
1
0
121
101
020
2
1
i
i
i
f
f
f
ha
ha
a
( )
( )112
)2(
11
)1(
21
2
1
+−
+−
+−≅
+−≅
iiii
iii
fffh
f
ffh
f
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ErrorsErrors
)4(4
)3(3
)2(2)1(
2
)4(4
)3(3
)2(2
)1(
1
3
2
3
42)2(
2462
iiiiii
iiiiii
fh
fh
fhfhff
fh
fh
fh
hfff
++++=
++++=
+
+
−
−
−
−
−
−≅
+
+
)4(4
)3(3
2
1
32
241
34
61
)2(2
)1(
21
1
0
01
11
22
1
i
i
i
i
i
i
i
fh
fh
f
f
f
fh
hf
−−
−
−
−≅
+
+
)4(4
)3(3
2
1
127
41
31
21
23
)2(2
)1(
1121
2
i
i
i
i
i
i
i
fh
fh
f
f
f
fh
hf
First & second forward derivative
)4(2
)3(
212
)2(
)4(3
)3(2
21
)1(
12
7)2(
1
43)43(
2
1
iiiiii
iiiiii
fh
hffffh
f
fh
fh
fffh
f
−−+−=
++−+−=
++
++
hOfffh
f
hOfffh
f
iiii
iiii
)()2(1
)()43(2
1
212
)2(
2
21
)1(
++−=
+−+−=
++
++
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Mixed DerivativesMixed Derivativesy
yj
xxi
i, j
i, j-1
i, j+1
i+1, ji-1, j
i-1, j-1
i-1, j+1 i+1, j+1
i+1, j-1
∆x∆x
∆y
∆y
( ) ( )
( )1,11,11,11,1
1,11,11,11,1
,
2
4
1
2
1
2
1
2
1
+++−−+−−
+++−−+−−
+−−∆∆
≅
+−∆
++−∆
−∆
≅
∂∂
∂∂
=∂∂
∂
jijijiji
jijijiji
yx
ffffyx
ffx
ffxy
y
f
xyx
f
ji
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ProblemsProblems
1. (10.1)Approximate the square root of 19
using Taylor series and retaining the first five
terms of series. Expand about x0 = 16.
2. (10.5)Evaluate the first, second, third, and
fourth derivatives of the following
polynomial and h = 0,1: f(x) = x4-3x3+4x2+10
Evaluate at x = 1, then compare your results
with the exact values.
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NUMERICAL INTEGRATIONNUMERICAL INTEGRATION
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Trapezoidal RuleTrapezoidal Rule
∫
∫∫
+=
+=
≅≅
2
1
2
1
2
1
)(
)(
)()(
10
10
x
x
x
x
x
x
dxxaaI
xaaxF
dxxfdxxFI
xx1 x2
f(x)
Actual function
Linear function
f(x1)f(x2)
h
[ ]
[ ]points. base adjacent two between increment the is where
or
h
xfhxfhxfxfhI
xfxfhI
)(....)2(2)(2)(
)()(
211121
2121
++++++=
+=
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Simpson’s Simpson’s 11//33 RuleRule
∫
∫∫
−
−−
++=
++=
≅≅
h
h
h
h
h
h
dxxaxaaI
xaxaaxF
dxxfdxxFI
)(
)(
)()(
2
210
2
210
[ ])()(4)(3
321 xfxfxfh
I ++=
x
f(x)
x1=-h x2= 0 x3= h
f(x1)f(x2)
f(x3)
[ ])(....)3(4)2(2)(4)(3
1111 nxfhxfhxfhxfxfh
I ++++++++=
or
where:
x1 = lower limit of integration; h = increment
xn = upper limit of integration; n = number of base points
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Special Integration FormulaSpecial Integration Formula
4
5
52
2
3
32
0
4
4
3
3
2
210
4
4
3
3
2
210
2
)(
)(
)()(
ahahhaI
dxxaxaxaxaaI
dxxFI
xaxaxaxaaxfxF
h
h
h
h
++≅
++++≅
≅
++++=≅
∫
∫
−
−
x
f(x)
-h 0 h
f(x1)f(x2)
f(x3)f(x4)
f(x5)
2h-2h
[ ])()(34)(114)(34)(90
54321 xfxfxfxfxfh
I −+++−=
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Unevenly Space Base PointsUnevenly Space Base Points
∫−
−
++=
++=
−≠≠−≠−
kh
h
nn
dxxaxaaI
xaxaaxF
xxxxxx
)(
)(
.....
2
210
2
210
13221
x
f(x)
-h 0 kh
f(-h)
f(0)
f(kh)
[ ])()132()()1()()23()(6
3
23
2
4
1
24
2xfkkxfkxfkkk
kk
hI −++++++−
+=
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ProblemsProblems1. (11.1)Use the trapezoidal rule of integration to determine the
value of the following integrals. Assume h = 0,1.
∫∫
∫∫
==
==
2
1
1
0
1
5,0
1
5,0
log (d) (c)
(b) (a)2
xdxxIdxxeI
dxeIdxeI
x
xx
2. (11.3)Use Simpson’s 1/3 rule to evaluate the following
integrals. Assume h = 1. (Check the accuracy of (a) & (b)
using direct integration)
∫∫
∫∫
==
−+==
5
1
2
0
5
1
233
1
2
sin
(d) sin (c)
)5( (b) (a)
dxx
xIxdxxI
dxxxIdxxI
3. (11.6) 4. (11.11)
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NUMERICAL SOLUTION OF NUMERICAL SOLUTION OF
ORDINARY DIFFERENTIAL ORDINARY DIFFERENTIAL
EQUATIONSEQUATIONS
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Taylor Series MethodTaylor Series Method
1)0( ,021 ==− yy
dx
dy
2 ln21
21
x
eyxy
dxy
dy
=⇒=
= ∫∫
161
81
21)4()3(
21)4(
81
41
21)3()2(
21)3(
41
21
21)2()1(
21)2(
21
21)1(
21)1(
)4(4
)3(3
)2(2
)1(
)()0( )(
)()0( )(
)()0( )(
)1()0( )(
......)0(24
)0(6
)0(2
)0()0()(
0 ;1
==⇒=
==⇒=
==⇒=
==⇒=
+++++=
==
yyxy
yyxy
yyxy
yyxy
yh
yh
yh
hyyxy
xy
Suppose given:
Direct integration:
......1)(
;1)0(
4
38413
4812
81
21 +++++=
==
xxxxxy
hxy
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Euler and Modified Euler MethodEuler and Modified Euler Method
),( yxfdx
dy=
2
1
2)1(
1
0
)1(
00
)(),(
)(
)()()(
hOyxhfyy
hOhyyy
xhyxyhxy
iiii
iii
++=
++=
+=+
+
+
),( );,( 11
)1(
1
)1(
+++ == iiiiii yxfyyxfy
Suppose given:
Basic Euler:
Modified Euler:
( ) 3)1(
1
)1(
1
3
)1()1(
1
2)1(
1
)1()1(
1)2(
)2(2
)1(
1
)(2
)(2
2
hOyyh
yy
hOh
yyhhyyy
h
yyy
yh
hyyy
iiii
iiiii
iii
iiii
+++=
+−
++=
−=
++=
++
++
+
+
),(
)(
1
2
11
ii
ii
yxhfk
hOkyy
=
++=+
),(
),(
)()(
112
1
3
2121
1
kyxhfk
yxhfk
hOkkyy
ii
ii
ii
+=
=
+++=
+
+
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RungeRunge--Kutta MethodsKutta Methods
),( yxfdx
dy=
),(
),(
),(
),(
)22(
34
223
222
1
432161
1
2
1
kyhxhfk
yxhfk
yxhfk
yxhfk
kkkkyy
ii
k
ih
i
k
ih
i
ii
ii
++=
++=
++=
=
++++=+
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ProblemsProblems
1. (12.1) Given the following ordinary differential equation.
1)0( ,02 ==− yyxdx
dy
Evaluate the solution at x = 0,1 using the method of
(a) Taylor Series – include the sixth derivative term
(b) Euler – use h = 0,05
(c) Modified Euler – use h = 0,05
(d) Runge-Kutta (fourth order) – use h = 0,05