010.141 NUMERICAL METHODS IN GENERAL MODULE 5 · 2018. 1. 30. · 2010 Engineering Mathematics II...

ENGINEERING MATHEMATICS II

010.141

NUMERICAL METHODSIN

GENERAL

MODULE 5

2B.D. Youn2010 Engineering Mathematics II Module 5

NUMERICAL METHODS

Ø Methods for solving problems numerically on a computer

Ø Steps: • Modeling, • Choice of a numerical method, Programming, • Doing the computation, • Interpreting the results

Ø Solution of equations by iteration

Ø Interpolation

Ø Numerical integration and differentiation


Nonlinear Equations in Engineering Fields

f(x) = wn (x) - w = 0x: bridge design variables

To design a safe bridge , a natural frequency must be higher than that of an external loading.

f(x) = T (x) - T* = 0x: battery design variables

To design a safe battery, a temperature level must be smaller than a marginal temperature.

f(x) = s (x) - S = 0x: bridge gusset plate

To design a safe bridge, a stress level at a critical bridge element must be smaller than its strength.


SOLUTION OF EQUATIONS BY ITERATION

Ø Solving equation f(x) = 0Ø Methods:

• Fixed – Point Iteration• Newton's Method• Secant Method


FIXED-POINT ITERATION FOR SOLVING f(x) = 0

Ø Idea: transform f(x) = 0 into x = g(x)

Ø Steps:1. Choose x02. Compute x1 = g(x0), x2 = g(x1), ¼, xn+1 = g(xn)

Ø A solution of x = g(x) is called a fixed point

Ø Depending on the initial value chosen (x0), the related sequences may converge or diverge



Ø Example: f(x) = x2 – 3x + 1 = 0

îíì=

381966.0618034.2

Solutions


NEWTON'S METHOD FOR SOLVING f(x) = 0

Ø f must have a continuous derivative f 'Ø The method is simple and fast

( ) ( )

( )( )0

001

10

00

x'fxfxx

xxxfx'ftan

-=

ß

-==b



( )( )

( )( )n

nn1n

1

112

x'fxfxx

x'fxfxx

-=

-=

+

M



Example: Find the positive solution of f(x) = 2 sin x – x = 0


SECANT METHOD FOR SOLVING f(x) = 0

Newton’s method is powerful but disadvantageous because it is difficult to obtain f '. The secant method approximates f '.



Ø Newton's Method

Ø Secant


진정성이 마음을 움직인다.

송나라범관, 계산행려도


INTERPOLATION

Ø Function f(x) is unknownØ Some values of f(x) are known (f1, f2, ¼, fn)Ø Idea: Find a polynomial pn(x) that is an approximation of f(x)

( ) ( ) ( ) nnn11n00n fxp ,,fxp,fxp === L

Ø Lagrange interpolation• Linear• Quadratic• General

Ø Newton's interpolation• Divided difference• Forward difference• Backward difference

Ø Splines


LINEAR LAGRANGE INTERPOLATION

Ø Use 2 known values of f(x) ® f0, f1

p1 is the linear Lagrange polynomial.



Ø p1(x) = L0(x)f0 + L1(x)f1Ø L0 and L1are linear polynomials (weight functions).

( ) ( )ïî

ïíì

=

==

ïî

ïíì

=

==

1

01

1

00

x x if 1

x x if 0xL

x x if 0

x x if 1xL

( ) ( )01

01

10

10 xx

x ,xx

x--

=--

=xxLxxL

( ) 101

00

10

11 xx

xxx

x fxfxxp--

+--

=



Example: Compute ln 9.2 from ln 9.0 = 2.1972 and ln 9.5 = 2.2513 by linear Lagrange interpolation

Solution:

( ) ( )

( )( ) ( )( ) ( )

( ) 0004.02188.22192.22.9p-2.9ln :Error2188.2

2513.24.01972.26.0

f2.9Lf2.9L

2.9p2.9ln

4.00.95.90.92.92.9L 6.0

5.90.95.92.92.9L

5.9lnf ,0.9lnf ,5.9x ,0.9x

1

1100

1

10

1010

=-=

=

+=

+=

»

=--

==--

=

====


QUADRATIC LAGRANGE INTERPOLATION

Ø Use of 3 known values of f(x) ® f0, f1, f2Ø Approximation of f(x) by a second-degree polynomial

( ) ( ) ( ) ( )( )( )( )( )( )( )( )( )( )( )( )( )1202

102

2101

201

2010

210

2211002

xxxxxxxxL

xxxxxxxxL

xxxxxxxxL

fxLfxLfxLxp

----

=

----

=

----

=

++=


QUADRATIC LAGRANGE INTERPOLATION

Example: Compute ln 9.2 for f0 (x0 = 9.0) = ln 9.0, f1 (x1 = 9.5) = ln 9.5, f2 (x2 = 11.0) = ln 11.0

Solution:

( )

( ) ( )

( ) ( )

( ) 2192.22.92.9ln

5.855.1831

992075.01

5.1045.20

2

22

21

20

=»

+-=

+--=

+-=

p

xxxL

xxxL

xxxL


GENERAL LAGRANGE INTERPOLATION

( ) ( ) ( )

( )( )

( ) ( ) ( )( ) ( )nkk

k

n

k

n

kn

xxxxxxxxxl

fxlxl

fxxpxf

----=

=

=»

+-

=

=

å

å

LL 110k

0 k k

k

0 k k

L


NEWTON'S INTERPOLATION

Ø More appropiate for computation

Ø Level of accuracy can be easily improved by adding new terms that increase the degree of the polynomial

Ø Divided difference

Ø Forward difference

Ø Backward difference


Is High-Order Polynomial a Good Idea?

f(x) = 1/(1+25x^2)


Ø Method of interpolation used to avoid numerical instability

Ø Idea: given an interval [a, b] where the high-degree polynomial can oscillate considerable, we subdivide [a, b] in several smaller intervals and use several low-degree polynomials (which cannot oscillate much)

SPLINES


SPLINES (cont)

Ø In an interval x Î [xj, xj+1], j = 0 , ¼, n – 1

where

( ) ( ) ( ) ( )3jj2jjjjjj xxaxxaxxaaxp 3210 -+-+-+=

( ) ( )

( ) ( )jjjjj

jjjjjjj

jjjj

jjjj

kkh

ffh

a

kkh

ffh

xpa

kxpa

fxpa

++-=

+--=¢¢=

=¢=

==

++

++

12133

1122

1

0

12

213)(21

)(

)(


SPLINES (cont)

Ø Let (x0,f0), (x1,f1), … , (xn,fn).Ø k0, kn are two given numbers.Ø k1, k2, ¼, kn-1 are determined by a linear system of n-1

equations:

Ø h is the distance between nodesxn = x0 + nh

( )( ) ( ) ( )1-n , 1, 0,j 1','

34

1

1111

K==+=

-=++

+

-++-

jjjjjj

jjjjj

kxpkxp

ffh

kkk


SPLINES (cont)

Ø Clamped conditionsg'(x0) = f'(x0), g'(xn) = f'(xn)

Ø Free/natural conditionsg"(x0) = 0, g"(xn) = 0

Ø Example:Interpolate f(x) = x4 on interval x Î [-1, 1] by cubic spline in partitions x0 = -1, x1 = 0, x2 = 1, satisfying clamped conditions

g'(-1) = f'(-1), g'(1) = f'(1)


SPLINES (cont)


NUMERICAL INTEGRATION

Ø Numerical evaluation of integrals whose analytical evaluation is toocomplicated or impossible, or that are given by recorded numerical values

Ø Rectangular ruleØ Trapezoidal ruleØ Simpson's ruleØ Gauss integration

( )òb

a

dxxf


RECTANGULAR RULE

Ø Approximation by n rectangular areas

where

hxx

nabh

01 +=

-=

( ) ( ) ( ) ( )[ ]nb

a

xfxfxfhdxxfJ *** 21 +++»= ò L


TRAPEZOIDAL RULE

Ø Approximation by n trapezoidal areas

( ) ( ) ( ) ( ) ( ) ( )úûù

êëé +++++»= -ò bfxfxfxfafhdxxfJ n

b

a 21

21

121 L


TRAPEZOIDAL RULE


SIMPSON'S RULE

Ø Approximation by parabolas using Lagrange polynomials p2(x)Ø Interval of integration [a, b] divided into an even number of subintervals

( )

( ) ( )

( )jj

mmm

b

a

xff

fffffffhdxxf

hxhaxfmabh

=

+++++++»

+=+==-

=

--ò 212223210

12100

424243

x xf 2

L


SIMPSON'S RULE


ADAPTIVE INTEGRATION


GAUSS INTEGRATION

where

( ) ( ) åòò¥

=-

»=1 j

jj

1

1

b

a

fAdttfdxxf

( ) ( )[ ]

( )jjn1

tfftscoefficien A , ,A

1tb1-ta21x

=Þ

++=

K

010.141 NUMERICAL METHODS IN GENERAL MODULE 5 · 2018. 1. 30. · 2010 Engineering Mathematics II...

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