• 검색 결과가 없습니다.

ABSTRACT OF CHAP. 19

N/A
N/A
Protected

Academic year: 2022

Share "ABSTRACT OF CHAP. 19"

Copied!
53
0
0

로드 중.... (전체 텍스트 보기)

전체 글

(1)

ENGINEERING MATHEMATICS II

010.141

Byeng Dong Youn (윤병동윤병동윤병동)윤병동

CHAP. 19

Numerical Methods in General

(2)

ABSTRACT OF CHAP. 19

 Part E: Numerical methods provide the transition from the mathematical model to an algorithm, which is a detailed stepwise recipe for solving a problem of the indicated kind to be programmed on your computer.

 Chapter 19 on numerics begins with an explanation of some general concepts, interpolations, numerical

integration and differentiation.

 Methods for solving equations (19.2), interpolation methods

including splines (19.3 and 19.4), and numerical integration and

differentiation (19.5)

(3)

CHAP. 19.1

INTRODUCTION

Steps and important issues of numerical methods.

(4)

STEPS IN NUMERICAL METHODS

 Methods for solving problems numerically on a computer

 Steps :

• Modeling

• Choice of a numerical method, Programming

• Doing the computation

• Interpreting the results

(5)

STABILITY

(6)

ACCURACY

(7)

ACCURACY

(8)

ERROR PROPAGATION

(9)

HOMEWORK IN 19.1

 HW1. Problems 14

(10)

CHAP. 19.2

SOLUTION OF EQUATIONS BY ITERATION

Finding solution of an equation using iterative steps.

(11)

Nonlinear Equations in Engineering Fields

f(x) = ω

n

(x) − ω ≠ 0

x: bridge design variables To design a safe bridge , an excitation frequency

f(x) = T (x) − T* < 0

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

f(x) = σ (x) − S < 0

x: bridge gusset plate

To design a safe bridge,

a stress level at a critical

(12)

SOLUTION OF EQUATIONS BY ITERATION

 Solving equation f(x) = 0

 Methods :

• Fixed – Point Iteration

• Newton's Method

• Secant Method

(13)

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

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

 Steps :

1. Choose x 0

2. Compute x 1 = g(x 0 ), x 2 = g(x 1 ), …, x n+1 = g(x n )

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

 Depending on the initial value chosen (x 0 ), the related

sequences may converge or diverge

(14)

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

 Example : f(x) = x 2 – 3x + 1 = 0

 

= 

381966 .

0

618034 .

Solutions 2

(15)

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

(16)

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

 f must have a continuous derivative f '

 The method is simple and fast

( ) ( )

( ) ( ) 0 0

0 1

1 0

0 0

x ' f

x x f

x

x x

x x f

' f tan

=

= −

=

β

(17)

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

( ) ( )

( ) ( ) n

n n 1

n

1 1 1

2

x ' f

x x f

x

x ' f

x x f

x

=

=

+

M

(18)

ALGORITHM

(19)

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

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

(20)

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 '.

(21)

SECANT METHOD FOR SOLVING f(x) = 0

 Newton's Method

 Secant

(22)

SECANT METHOD FOR SOLVING f(x) = 0

(23)

HOMEWORK IN 19.2

 HW1. Problems 2

 HW2. Problems 10

 HW3. Problems 11

 HW4. Problems 21

(24)

CHAP. 19.3

INTERPOLATION

Finding (approximate) values of a function f(x) for an x

between different x-values x 0 , x 1 , ! , x n at which the

values of f(x) are given.

(25)

INTERPOLATION

 Function f(x) is unknown

 Some values of f(x) are known (f

1

, f

2

, …, f

n

)

 Idea : Find a polynomial p

n

(x) that is an approximation of f(x)

( ) 0 0 n ( ) 1 1 n ( ) n n

n x f , p x f , , p x f

p = = L =

 Lagrange interpolation

• Linear

• Quadratic

• General

 Newton's interpolation

• Divided difference

(26)

LINEAR LAGRANGE INTERPOLATION

 Use 2 known values of f(x) → f 0 , f 1

p 1 is the linear Lagrange polynomial.

(27)

LINEAR LAGRANGE INTERPOLATION

 p 1 (x) = L 0 (x)f 0 + L 1 (x)f 1

 L 0 and L 1 are linear polynomials (weight functions).

( ) ( )



 

=

= =



 

=

= =

1 0 1

1 0

0 1 if x x

x x if 0 x

L x

x if 0

x x if 1 x

L

( ) ( )

0 1

0 1

1 0

1

0 x x

x x ,

x

x

= −

= − x

x x L

x

L

(28)

LINEAR LAGRANGE INTERPOLATION

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

Solution:

( ) ( )

( )

( ) ( )

( ) ( )

2188 .

2

2513 .

2 4 . 0 1972

. 2 6 . 0

f 2 . 9 L f

2 . 9 L

2 . 9 p 2

. 9 ln

4 . 0 0

. 9 5 . 9

0 . 9 2 . 2 9

. 9 L 6

. 5 0

. 9 0 . 9

5 . 9 2 . 2 9

. 9 L

5 . 9 ln f

, 0 . 9 ln f

, 5 . 9 x

, 0 . 9 x

1 1

0 0

1

1 0

1 0

1 0

=

+

=

+

=

− =

= −

− =

= −

=

=

=

=

(29)

QUADRATIC LAGRANGE INTERPOLATION

 Use of 3 known values of f(x) → f 0 , f 1 , f 2

 Approximation of f(x) by a second-degree polynomial

( ) ( ) ( ) ( )

( )( )

( )( )

( )( )

( 1 0 )( 1 2 )

2 1 0

2 0

1 0

2 0 1

2 2

1 1

0 0

2

x x

x x

x x

x L x

x x

x x

x x

x L x

f x L f

x L f

x L x

p

= −

= −

+ +

=

(30)

QUADRATIC LAGRANGE INTERPOLATION

Example: Compute ln 9.2 for f

0

(x

0

= 9.0) = ln 9.0, f

1

(x

1

= 9.5) = ln 9.5, f

2

(x

2

= 11.0) = ln 11.0

Solution:

( )

( ) ( )

( ) ( )

( ) 9 . 2 2 . 2192

2 . 9 ln

5 . 85 5

. 3 18

1

99 75 20

. 0

1

5 . 104 5

. 20

2 2 2

2 1

2 0

=

+

=

+

=

+

=

p

x x

x L

x x

x L

x x

x

L

(31)

GENERAL LAGRANGE INTERPOLATION

( ) ( ) ( )

( ) ( )

( ) ( ) ( k )( k ) ( n )

k n

k n

k n

x x x

x x

x x

x x

l

x f l

x l

f x x

p x

f

=

=

=

+

=

=

L

L 1 1

0 k

0 k k

k 0 k

k

L

(32)

NEWTON'S INTERPOLATION

 More appropiate for computation

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

 Divided difference

 Forward difference

 Backward difference

(33)

Is High-Order Polynomial a Good Idea?

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

(34)

HOMEWORK IN 19.3

 HW1. Problems 3

 HW2. Problems 8

(35)

CHAP. 19.4

SPLINE INTERPOLATION

The undesirable oscillations are avoided by the method of splines initiated by I.J. Schoenberg in 1946 (Applied Mathematics 4, pp45-99, 112-141). This method is

widely used in practice. It also laid the foundation for

much of modern CAD (Computer Aided Design).

(36)

 Method of interpolation used to avoid numerical instability

 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

SPLINES

(37)

SPLINES (cont)

 In an interval x ∈ [x

j

, x

j+1

], j = 0 , …, n – 1

where

( ) j j ( j ) j ( j ) 2 j ( j ) 3

j x a a x x a x x a x x

p =

0

+

1

− +

2

− +

3

(

j j

) (

j j

)

j j j

j j

j j

j j

j j

k h k

f h f

x p a

k x

p a

f x

p a

+

′′ =

=

′ =

=

=

=

+

+1 1

2 2 1 0

1 2 ) 3

2 ( 1

) (

)

(

(38)

SPLINES (cont)

 Let (x 0 ,f 0 ), (x 1 ,f 1 ), … , (x n ,f n ).

 k 0 , k n are two given numbers.

 k 1 , k 2 , …, k n-1 are determined by a linear system of n-1 equations:

 h is the distance between nodes x = x + nh

( )

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

'

4 3

1 1 1

1 1

= K

= +

=

= +

+

+

− +

+

j j

j j

j j

j j

j j

j

k x

p k

x p

f h f

k k

k

(39)

SPLINES (cont)

 Clamped conditions

g'(x 0 ) = f'(x 0 ), g'(x n ) = f'(x n )

 Free/natural conditions

g"(x 0 ) = 0, g"(x n ) = 0

 Example :

Interpolate f(x) = x 4 on interval x ∈ [-1, 1] by cubic

spline in partitions x 0 = -1, x 1 = 0, x 2 = 1, satisfying

clamped conditions

(40)

SPLINES (cont)

(41)

SPLINES (cont)

(42)

HOMEWORK IN 19.4

 HW1. Problems 11

 HW2. Problems 12

(43)

CHAP. 19.5

NUMERICAL INTEGRATION AND DIFFERENTIATION

Evaluating the numerical integration.

(44)

NUMERICAL INTEGRATION

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

 Rectangular rule

 Trapezoidal rule

 Simpson's rule

b ( )

a

dx

x

f

(45)

RECTANGULAR RULE

 Approximation by n rectangular areas

where

h x

x

n a h b

0

1 = +

= −

( ) [ ( ) ( ) ( n ) ]

b

a

x f x

f x

f h dx

x f

J = ∫ ≈ * 1 + * 2 + L + *

(46)

TRAPEZOIDAL RULE

 Approximation by n trapezoidal areas

( ) ( ) ( ) ( ) ( ) ( )

 

 + + + + +

= ∫ f x dx h f a f x f x f x f b

J n

b

a 2

1 2

1

1 2

1 L

(47)

TRAPEZOIDAL RULE

(48)

SIMPSON'S RULE

 Approximation by parabolas using Lagrange polynomials p

2

(x)

 Interval of integration [a, b] divided into an even number of subintervals

( )

( ) ( )

( )

j j

m m

m b

a

x f f

f f

f f

f f

h f dx x f

h x

h a x

m f a h b

=

+ +

+ +

+ +

+

+

= +

=

− =

=

0 1 2 3 22 2 1 2

1 2

1 0

0

4 2

4 2

3 4

x x

f 2

L

(49)

SIMPSON'S RULE

(50)

SIMPSON'S RULE

(51)

ADAPTIVE INTEGRATION

(52)

GAUSS INTEGRATION

where

( ) ∫ ( ) ∑

− =

=

1 j

j j 1

1 b

a

f A dt

t f dx

x f

( ) ( )

[ ]

( ) j

j

n 1

t f f

ts coefficien

A , , A

1 t b 1 - t 2 a x 1

=

+ +

=

K

(53)

HOMEWORK IN 19.5

 HW1. Problem 5

 HW2. Problem 6

 HW3. Problem 21

참조

관련 문서

produces a magnetic field which can be guided by a magnetic yoke to a linear Hall sensor; the output of the sensor is proportional to the electric current..

The “Asset Allocation” portfolio assumes the following weights: 25% in the S&amp;P 500, 10% in the Russell 2000, 15% in the MSCI EAFE, 5% in the MSCI EME, 25% in the

1 John Owen, Justification by Faith Alone, in The Works of John Owen, ed. John Bolt, trans. Scott Clark, &#34;Do This and Live: Christ's Active Obedience as the

The eigenvalue problem can be used to identify the limit state of the process, in which the state vector x is reproduced under the multiplication by the.. stochastic matrix

In this study, it expect that the student's school achievement skills and the mathematical problem solving skills increase according to textbook of an

The purpose of this study is to investigate the effect of the “problem solving and programming” area of ​​information subject on perception and

The purpose of this study is to analyze the relationship between dance performances, which is a part of social contribution activities of local governments,

Step 1 Setting up a mathematical model (a differential equation) of the physical process.. By the physical law : The initial condition : Step