Background Image
Previous Page  8 / 12 Next Page
Information
Show Menu
Previous Page 8 / 12 Next Page
Page Background

+

2

2

(+1)

|

x

x

|

(5

n

+ 8) 5

n

+

2

3

(+1)

|

x

x

|

3/2

(5

n

+ 9) (5

n

+ 1)

+

2

4

(+1)

|

x

x

|

2

(5

n

+ 10) (5

n

+ 2)

!

,

при этом

ρ

2

= min

 

ρ

0

,

1

4

5

q

(

M

+ 1)

2

 

,

M

= max

{

M

0

, α

}

.

J

По определению

Δ

y

N

(

x

) =

|

y

(

x

)

y

N

(

x

)

|

=

=

X

n

=0

C

n

(

x

x

)

(

n

1)

/

2

N

X

n

=0

C

n

(

x

x

)

(

n

1)

/

2

=

=

X

n

=

N

+1

C

n

(

x

x

)

(

n

1)

/

2

X

n

=

N

+1

|

C

n

| |

x

x

|

(

n

1)

/

2

.

Учитывая закономерность образования коэффициентов

C

n

из тео-

ремы 1, имеем

X

n

=

N

+1

|

C

n

| |

x

x

|

(

n

1)

/

2

=

=

X

k

=

N

+1

|

C

5

k

| |

x

x

|

(5

k

1)

/

2

+

X

k

=

N

+1

|

C

5

k

+1

| |

x

x

|

5

k/

2

+

+

X

k

=

N

+1

|

C

5

k

+2

| |

x

x

|

(5

k

+1)

/

2

+

X

k

=

N

+1

|

C

5k+3

| |

x

x

|

(5

k

+2)

/

2

+

+

X

k

=

N

+1

|

C

5

k

+4

| |

x

x

|

(5

k

+3)

/

2

,

при

N

+ 1 = 5

k

находим

Δ =

X

k

=

N

+1

|

C

5

k

| |

x

x

|

5

k

1

2

2

5

k

M

(

M

+ 1)

k

|

x

x

|

(5

k

1)/2

1

2

5

(

M

+ 1)

|

x

x

|

5/2

×

×

1

(5

k

+ 2) (5

k

6)

+

2

|

x

x

|

1/2

(5

k

+ 3) (5

k

5)

+

2

2

|

x

x

|

(5

k

+ 4) (5

k

4)

+

+

2

3

|

x

x

|

3/2

(5

k

+ 5) (5

k

3)

+

2

4

|

x

x

|

2

(5

k

+ 6) (5

k

2)

!

.

Аналогично получаем оценки для коэффициентов

N

+ 1 = 5

k

+ 1

;

N

+ 1 = 5

k

+ 2

;

N

+ 1 = 5

k

+ 3

;

N

+ 1 = 5

k

+ 4

.

I

ISSN 1812-3368. Вестник МГТУ им. Н.Э. Баумана. Сер. “Естественные науки”. 2015. № 2

33