ГОСЫ Методы вычислений 18 стр Январь 2012

- .

.

–

-5

2012 .

1. Решение задачи Коши для обыкновенного дифференциального уравнения первого порядка методом Эйлера, Эйлера с пересчетом, Коши. Локальные погрешности этих методов. Геометрическая интерпретация метода Эйлера.

.

1- ! " #

!$ # ! # .

y′ = f ( x, y)
	– ! % " 1- .
y(x0 ) = y0

# f ( x, y) # # #$ #$ !$ y

$ # # $ x0 , & ! $ ! [ x0 ; x0 + H ]
" ! % " # ' # # ( # ' #
# # ! % " 1- ).

: & # # " y(x) y′ = f ( x, y) ,

(' # y( x0 ) = y0 xi		[ x0 ; x0 + H ] .
) " ' #	# # !$ y1, y2 , K, yn ,	yi – &
! "	y( x) xi .
	y = y(x)
	y2
	y1
	y0

	x0	x1	x2
x1 = x0 + h, h = H/n
y1	= y(x1) – & ! "			y(x) x1
x2	= x1 + h
y2	= y(x2) – & ! "			y(x) x2
…

* # & ! ( (

x0, x1, …, xn y0, y1, …, yn

y′ = f ( x, y)

y( x0 ) = y0

dy = f ( x, y) dx

dy = f ( x, y)dx

, .

! .

2

xi+1 xi+1

∫dy = ∫ f ( x, y)dx

xi

xi

y( xi +1 ) − y( xi ) = h f ( xi , y( xi ))

1442443

yi = y( xi ) – ! ! &.

yi+1 = yi + h f ( xi , yi ) – .

" # &, : " h2 (O(h2) –

| R[f]

| ≤ const· h2) (

" #). " # – h. , !,

- # # h.

! .

y2

N

α

h·f(x1,y1)

y1

1

y=y(x) (! #)

h·f(x0,y0)

y0

α

M

y0

x0

x1

x2

h

!

x1 = x0 + h

y1 = y0 + h f ( x0 , y0 )

tgα = f ( x0 , y0 )

!

h tgα = MN

y2 = y1 + h f ( x1, y1 )

tgα1 = f ( x1, y1 )

! &

L

, .

y′

= f ( x, y)

y( x0 ) = y0

dy

= f ( x, y)

dx

dy = f ( x, y)dx

xi+1

xi+1

∫dy = ∫ f ( x, y)dx

xi

xi

y( x

) − y( x ) =

h

( f ( x , y( x )) + f ( x

, y( x

)))

i +1

i

i

i

i +1

i +1

2

14444424444443

3

.$ y( xi +1 ) $ : y( xi +1 ) = y( xi ) + h f ( xi , y( xi )) yi = y( xi ) – ! ! &.

y

= y +

h

( f ( x , y ) + f ( x + h, y + h f ( x , y )))

– .

i +1

i

2

i

i

i

i

i

i

" #

&, $: " h3 (O(h3) – | R[f] | ≤ const· h3)

(

" #). " # – h2. , !, -

# # h.

", .

y′

= f ( x, y)

y( x0 ) = y0

dy

= f ( x, y)

dx

dy = f ( x, y)dx

xi+1

xi+1

∫dy = ∫ f ( x, y)dx

xi

xi

h

h

y( x

) − y( x ) = h

f x +

, y x

+

i +1

i

i

i

2

2

14444244443

h

h

h

.$

y x

+

$ :

y x +

= y( x ) +

f ( x , y( x ))

i

2

i

2

i

2

i

i

yi = y( xi ) – ! ! &.

			h			h
y	= y	+ h f x +		, y	+		f ( x , y )	– .
i +1	i	i	2	i		2	i i
" #			&, # : " h3 (O(h3) –
| R[f] | ≤ const· h3) (							" #). " # – h2. , !,

- # # h.

4

+ A2 (b − a)2 ,

2. Формула Симпсона. Ее погрешность (без вывода). Составные формулы.

.

	b
, # # $ ∫ f (x)dx , f(x) –		[a, b] .
	a
! f(x) Ln(x) ,			! #
b	b
$: ∫ f (x)dx ≈ ∫ Ln ( x)dx .
a	a
# $%% !. & ' .
/ 0 #	( ) #	!.	!

( ! [a, b] 1 & $ # # ! x0 = a, x1 = (a + b)/2, x2 = b. / # .

b	a + b
∫ f ( x)dx = A0 f (a) + A1 f			+ A2 f (b)
a		2

%- ! , , # f(x) –

& #$ #.

1. ,, ( $ #. # f(x) ≡ 1

b

∫1dx = b − a

a

f (a) = f ((a + b) / 2) = f (b) = 1

I = A0 + A1 + A2 , b − a = A0 + A1 + A2

2.,, ( $ #. # f(x) = x – a

	b	( x − a)2			b		(b − a)2
	∫x − a dx =				=
		2					2
	a				a
							+ b
					a					a + b
	f (a) = a − a = 0,		f						=
							2			2

− a = b − a , f (b) = b − a

2

(b − a)2	= A	b − a	+ A	(b − a),	b − a	=	1	A	+ A
2	1	2	2		2		2	1	2

3.,, ( $ #.

# f(x) = (x – a)2

b

( x − a)3

b

(b − a)3

∫( x − a)2 dx =

=

3

a

a

3

2

a + b

a

+ b

f (a) = 0,

f

=

− a

2

2

(b − a)3

(b − a)2

= A

3

1

4

= (b − a)2 , f (b) = (b − a)2 4

b − a = 1 A1 + A2

3 4

5

A + A + A = b − a

0

1

2

1

+ A2

=

b − a

A1

2

2

1

A1

+ A2

=

b − a

4

3

! $ # (

1

1

1

A1

= (b − a)

−

4

3

2

A =

2

(b − a)

1

3

A =

b − a

−

b − a

=

b − a

2

2

3

6

A = b − a −

2

(b − a) −

b − a

=

b − a

0

3

6

6

:

b

h

a + b

∫

f ( x)dx =

f

(a) + 4 f

+ f (b)

,

a

6

2

f(x) ' # (

f IV (ξ)

" #: R[ f ] = −

(b − a)5 ,

R[ f ]

≤

M 4

h5

2880

2880

% ' . % '.

! # % '.

b

, #

# $ ∫ f ( x)dx , f(x) – [a, b] . .

a

! [a,

b]

! ( n #$ # "

h = (b − a) / n : a = x0 < x1 < L < xn = b, xi +1 − xi = h, i = 0, n − 1 .

y=f(x)

		h
		a=x0	x1	x2	xn–1 b=xn
b	x1	xn	n −1	xi+1
,: ∫ f ( x)dx = ∫ f ( x)dx + K+ ∫ f ( x)dx = ∑ ∫ f ( x)dx .
a	x0	xn−1	i =0	xi
#	& & [xi, xi+1] # - #

. # # (' # # .

6

1. % ' # '

b

∫ f ( x)dx = h f ( x0 ) + h f ( x1 ) + K+ h f ( xn−1 )

a

b	n−1
∫ f ( x)dx = h∑ f ( xi )
a	i=0

" #:

	y=f(x)
a	b

n −1

n −1

h

2

h

2

f ′(c)

M1

R[ f ] = ∑Ri [ f ] = ∑ f ′( xi )

= n f ′(c)

=

(b − a) h,

R[ f ]

≤

(b − a)h

i =0

i =0

2

2

2

2

14243

const

2.

% ' # '

b

y=f(x)

∫ f ( x)dx = h f ( x1 ) + h

f ( x2 ) + K+ h f ( xn )

a

b

n

R[ f ] ≤ M1 (b − a)h

∫ f ( x)dx = h∑ f ( xi ),

a

i=1

2

a

b

3.

% ' # '

b

h

h

h

y=f(x)

∫

f ( x)dx = h f x0 +

+ h

f x1

+

+ K+ h f xn−1

+

2

a

2

2

b

n−1

h

M

∫ f ( x)dx = h∑ f xi

+

,

R[ f ]

≤

2 (b

− a)h2

a

i=0

2

24

a

b

4.

% ' !

b

f ( x)dx = h ( f ( x0 ) + f ( x1 ))+ h ( f ( x1 ) + f ( x2 ))+ K+ h ( f ( xn−1 ) + f ( xn ))

∫

a

2

2

2

b

n −1

R[ f ] ≤ M 2 (b − a)h2

∫ f (x)dx = h ( f (a) + f (b))

+ h∑ f (xi ),

a

2

i =1

12

y=f(x)

5.

% '

b

h

h

∫

f ( x)dx =

f ( x ) + 4 f x +

+ f ( x )

+

0

0

1

a

b

a

6

2

h

+ f ( x )

1

6

			h				h
+ 4 f x		+		+ f (x	2	) + K+		f (x	n −1	) + 4 f
	1		2
							6

			h
x	n −1	+		+ f ( x	n	)
			2

b

h

2

n −1

h

1

n−1

M

∫ f ( x)dx =

( f (a) + f (b))+

h∑ f xi

+

+

h∑ f ( xi ),

R[ f ]

≤

4

(b − a)h 4

a

6

3

i =0

2

3

i =1

2880

7

3. Приближенное вычисление определенного интеграла. Формулы прямоугольников, трапеций. Погрешности этих формул (вывод).

.

b

, # # $ I = ∫ f ( x)dx , f(x) – [a, b] .

a

! f(x) Ln(x) , ! #

:

b	b	n
∫ f (x)dx ≈ ∫ Ln ( x)dx = ∑ Ak f ( xk ) ,
a	a	k =0

xk – ! , Ak – - $ , ! #,

	! # ' !,					f ( x) . 4 ! ! R[ f ]
								b
	" # #$			,				R[ f ] = ∫( f (x) − Ln ( x))dx . , !
								a
	b	n	b	n	x	− xi
	∫ f (x)dx ≈ ∑Ak f ( xk ) . # Ak		= ∫∏					– # # (' ,
							dx
					x	− x
	a	k =0	a i = 0		k	i

k ≠i

# ' # $ !

! xk, ! # .

' % ' ' ' ' # .

! ( ! [a, b] 1 & $ #

# ! x0 – #$ f(x0)., #$ , $ ' $ $

, & '			# #$ f(x0)			#
h = b – a. * ! # # x0 & # (' .
1. & ' # ' . x0 = a, f0 = f(a)							y=f(x)
	b		f(b)
	∫ f ( x)dx = h f (a)		f(a)
	a
		b
	" #: R[ f ] = ∫ f (x)dx − h f (a) = F (b) − F (a) − h f (a)
		a	a		h		b

. ! & ,$ # # x0:

g′′( x )( x

− x )2

g

′′′( x )( x − x )3

g ( x) = g ( x0 ) + g′( x0 )( x − x0 ) +

0

0

+

0

0

+ K

2!

3!

) ! & F(b) # # a: F (b) = F (a) + F ′(a)(b − a) +

F ′′(a)

(b

− a)2

+ K

123123

123

2

f ( a )

h

f ′( a )

h

2

h

2

0, R[ f ] = F (a) + f (a) h + f ′(a)

− F (a) − h

f (a)

= f ′(a)

2

2

5 # f(a) # &. 4: M

1

= max

f ′( x)

,

R[ f ]

≤ M

1

h2

2

[ a ,b]

, # " # h .

8

2. & ' # ' . x0 = b, f0 = f(b)						y=f(x)
	b		f(b)
	∫ f ( x)dx = h f (b)
			f(a)
	a
		b
	" #: R[ f ] = ∫ f ( x)dx − h f (b) = F (b) − F (a) − h f (b)
		a	a		h	b

) ! & F(a) # # b.

F (a) = F (b) +	f (b)(a − b) + f ′(b)	(a − b)2
		2
R[ f ] = F (b) − F (b) + h f (b) − f ′(b)			h2	− h f (b) = − f ′(b)		h2	,
		2		2

, # " # h .

h2 R[ f ] ≤ M1 2

a + b

a + b

3. & ' # ' .

x0 =

,

f0 =

f

2

2

y=f(x)

b

a + b

f(b)

f((a+b)/2)

∫ f ( x)dx = h f

f(a)

a

2

" #:

h

b

a + b

a + b

a

(a+b)/2

b

R[ f ] = ∫ f ( x)dx − h

f

=

F (b) − F (a) − h

f

a

2

2

) ! & F(b) F(a) # #

a + b

.

2

a + b 2

a

+ b 3

a + b

a + b

a + b

a

+ b

b

−

a + b

b −

2

2

F (b) = F

+ f

b

−

+

f ′

+ f ′′

2

2

2

2

2

2

6

a + b 2

a

+ b 3

a + b

a

+ b

a + b

a

+ b

a

−

a + b

a −

2

2

F (a) = F

+ f

a

−

+

f ′

+ f ′′

2

2

2

2

2

2

6

a + b

a

+ b

h3

F (b) − F (a) =

f

h +

f ′′

2

2

24

a + b

a + b

a + b

a + b

h3

h3

h3

R[ f ] = f

h + f ′′

− h

f

=

f

′′

,

R[ f ]

≤ M

2 24

2

2

24

2

2

24

, # " # h ( ').

& ' ! .

! ( ! [a, b] 1 & $ # # ! x0 = a, x1 = b. # - # # ! $ # ' (. , !, #$ , $ ' $ $ , & '

$ # #$ h = b – a # f(a) f(b). ! #

# &$ " $:

9

b

h

∫

f ( x)dx =

( f (a) + f (b))

a

2

" #:

b

h

h

R[ f ] = ∫ f ( x)dx −

( f (a) + f (b)) = F (b) − F (a) −

( f (a) + f (b))

a

2

2

) ! & F(b) # # a.

(b − a)2

(b − a)3

F (b) = F (a) + f (a) (b − a) + f ′(a)

+ f ′′(a)

2

6

F (b) − F (a) = h f (a) + f ′(a)

h2

+ f ′′(a)

h3

2

6

) ! & f(b) # # a.

(b − a)2

(b − a)3

f (b) = f (a) + f ′(a) (b − a) + f ′′(a)

+ f ′′(a)

2

6

h

( f (a) + f (b)) =

h

h

2

2 f (a) + f ′(a) h + f

′′(a)

2

2

2

h

2

h

3

h

h

2

R[ f ] = h f (a) + f ′(a)

+ f ′′(a)

−

2 f (a)

+ f ′(a) h + f ′′(a)

2

6

2

2

y=f(x)

f(b)

f(a)

a	h	b

		1								h	3
	= −		f ′′(a) h3 ,		R[ f ]		≤ M
	12							2	12

, # " # h ( & '$ ).

10