Analisis Numerik.pdf

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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

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

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

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

SPECIAL MATRICES SPECIAL MATRICES ……continued……continued

dwkdwkdwkdwk 8

[ ]

=⇒

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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[ ]

=⇒

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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[ ] [ ] [ ][ ] [ ]

=⇒ ×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

OPERASIONAL MATRICESOPERASIONAL MATRICES

dwkdwkdwkdwk 11

[ ] [ ]

=

= ××

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

ROTATION OF A COORDINATE SYSTEMROTATION OF A COORDINATE SYSTEM

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y1

x1

P

1x

1y


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y1

x1

y2

P

x2

d1

d2

d4

d3

1x

2x

1y

2y

θ

θ

θ

θ


dwkdwkdwkdwk 14

y1

x1

y2

P

x2

d1

d2

d4

d3

1x

2x

1y

2y

θ

θ

θ

θ

3x

x3

3y

y3 ββ

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.

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

+−

+++−++

−+−−−−

+−

=

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

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

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

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

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

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


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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


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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

ProblemsProblems

3.4

3.5

3.7

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LINEAR ALGEBRA LINEAR ALGEBRA

EQUATIONSEQUATIONS

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=⇒

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

..........

..........

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

ProblemsProblems

4.3

4.4

4.7

4.17

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MATRIX INVERSIONMATRIX INVERSION

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

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

ProblemsProblems

5.1

5.2

5.3

5.12

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NON LINEAR ALGEBRA NON LINEAR ALGEBRA

EQUATIONSEQUATIONS

INTERVALINTERVAL--HALVING METHODHALVING METHOD

dwkdwkdwkdwk 40

0)(2

0)()()(

)(

0

)(

213

2122

11

≈→⇒+

=

<→→→

=

=

nn xfxxx

x

xfxfxfx

xfx

yx

xfy

............

whichfor value find to

FALSEFALSE--POSITION METHODPOSITION METHOD

dwkdwkdwkdwk 41

0)()()(

)()(

0)()()(

)(

0

)(

21

21123

2122

11

≈→⇒−−

=

<→→→

=

=

nn xfxxfxf

xfxxfxx

xfxfxfx

xfx

yx

xfy

............


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

............


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


ProblemsProblems

6.2

6.3

6.5

6.6

6.7

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EIGENPROBLEMSEIGENPROBLEMS

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

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

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

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

&&

&&

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


dwkdwkdwkdwk 51

k2=2kk1=k

P(t)=0m2

x2

m1

x1

0.73 1.0


dwkdwkdwkdwk 52

-2.73 1.0

k2=2kk1=k

P(t)=0m2

x2

m1

x1

ProblemsProblems

7.1

7.3

7.5

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INTERPOLATIONINTERPOLATION

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

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

=⇒=

=⇒=

=⇒=

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

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 +−++−+= −−

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

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

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(

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

∆==

∆≈

∆=

−+=

→

→

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

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

ProblemsProblems

8.1

8.3

8.5

dwkdwkdwkdwk 65

dwkdwkdwkdwk 66

CURVE FITTINGCURVE FITTING

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

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

dwkdwkdwkdwk 69

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

dwkdwkdwkdwk 70

Exponential EquationExponential Equation

y= C x p

dwkdwkdwkdwk 71

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 =

dwkdwkdwkdwk 72

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

dwkdwkdwkdwk 73

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

dwkdwkdwkdwk 74

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

dwkdwkdwkdwk 75

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

dwkdwkdwkdwk 76

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

dwkdwkdwkdwk 77

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 +++=

dwkdwkdwkdwk 78

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 −+−=

dwkdwkdwkdwk 79

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

ProblemsProblems

9.11

9.12

dwkdwkdwkdwk 80

dwkdwkdwkdwk 81

NUMERICAL DIFFERENTIATIONNUMERICAL DIFFERENTIATION

dwkdwkdwkdwk 82

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

dwkdwkdwkdwk 83

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

)()()(

dwkdwkdwkdwk 84

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

dwkdwkdwkdwk 85

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

dwkdwkdwkdwk 86

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


xi xi+2 xi+3xi+1

h h h

x

fi+3

fi+2fi

fi+1

f

dwkdwkdwkdwk 87

Forward derivative for a third-order polynomial


−

−

−

≅

+

+

+

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

dwkdwkdwkdwk 88

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

dwkdwkdwkdwk 89

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

dwkdwkdwkdwk 90

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

dwkdwkdwkdwk 91

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(

++−=

+−+−=

++

++

dwkdwkdwkdwk 92

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

dwkdwkdwkdwk 93

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.

dwkdwkdwkdwk 94

NUMERICAL INTEGRATIONNUMERICAL INTEGRATION

dwkdwkdwkdwk 95

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

++++++=

+=

dwkdwkdwkdwk 96

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

dwkdwkdwkdwk 97

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 −+++−=

dwkdwkdwkdwk 98

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 −++++++−

+=

dwkdwkdwkdwk 99

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)

dwkdwkdwkdwk 100

NUMERICAL SOLUTION OF NUMERICAL SOLUTION OF

ORDINARY DIFFERENTIAL ORDINARY DIFFERENTIAL

EQUATIONSEQUATIONS

dwkdwkdwkdwk 101

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

dwkdwkdwkdwk 102

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

+=

=

+++=

+

+

dwkdwkdwkdwk 103

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

++=

++=

++=

=

++++=+

dwkdwkdwkdwk 104

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

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