Mathematics (Математика). Учебное пособие

161

Thus

¶

¶xx

¶

z

¶

×+×+-+×-=

x

¶

z

¶

×-×+×

=

3

2045

ö ø

11

10

2

2045

4

ö

æ

-+×+÷÷

÷

ç

5

ø

è

æ

¶

U

¶

l

M

3

ö

æ ç

è

ç

55

-×

÷

ç

ø

è

-=

Example 3. Find the gradient of a scalar field

a)

b)

),,( ),,( zyxzyxU -+=

22

xzyyxzyxU -×+-=

222

. Plot the level surfaces for given val-

44

2045

ö

2045

204555

ø

),,( zyxU

2045

==÷÷

55

.

æ ç

-×

ç è

ues

1,0),,( ±=zyxU

.

Solution.

a) According to the definition of a gradient of a scalar field

gradU ×

U

=

x

¶

U

i

+×

y

¶

Find particular derivatives of functions

U

¶

12 -=

,

Thus

U

j

U

y

¶

k

+×

z

¶

.

),,( zyxU

:

zy

+-=

2

,

)2()12(

U

y

=

¶

kyjzyixgradU

.

b) Similarly to a), we get:

U

x

2=

¶

,

U

2=

y

¶

Thus

Let’s plot the level surfaces:

222

0

Then

=-+ zyx

,

y

,

222

kzjyixgradU

.

zyxU

222

zyx =+

is a cone with vertex at the

U

2-=

¶

0),,(

origin of coordinates.

z

162

Fig.15

±

=

-

=

x

y z

x

Fig.15 – graphic representation of the surface

222

1

If

1),,(

zyxU

1),,(

222

=-+ zyx

, то

1

±=+ zyx

:

zyxU

222

z

222

zyx =+

1),,(

1

-=-+ zyx

y

Fig.16

Fig.16 – graphic representation of the hyperboloid rotation monopolistic around the axis

163

()()(

)

rrr

r

rrr

()()(

)

ayzixzjxyk

r

rr

r

z

-1

x

y

Fig.17

Fig.17 – graphic representation of the duopolistic hyperboloid rotation around the axis.

Example 4. Prove, that vector field

+++++=

is conservative.

Solution.

The sufficient condition of conservativeness of the field is the curl’s equality to zero. In our case

kji

r

¶

= kji

arot

¶

( ) ( ) ( )

¶

zyx

+++

yxzxzy

rr

r

0111111

=-+---=

.

Thus, the field is conservative.

Try to do these yourself:

1. Prove, that vector field

=-+-+-

is con-

servative.

kyxjzxizya

164

Section 8. Series

Topic 22. Number series

Introduction. Infinite series can be a pleasure (sometimes). They throw a beautiful light on sinx and cosx. They give famous numbers like π and e. Usually they produce total unknown functions which might be good. But on the painful side is the fact than infinite series has infinitely many terms. It is not easy to know the sum of those terms. More than that, it is not certain that there is a sum. We need tests, to decide if the series converges.

Example 1. Examine the series for convergence:

2

¥

æ ç

a) .

å

ç

n

1

=

n

è

2

b)

c)

d)

e)

f)

g)

h)

å

n

å

n

å

n

å

n

¥

=

¥

=+×1

¥

1

=

¥

=×2

¥

å

1

=n

¥

å

=1

¥

å

=1

1

æ ç ç è

2

n

sin

n

1

ln

3

n

nn

( )

n

!

n

( )

n

2

nn

n

7

æ ç

ç è

2

n

+

2

ö

34

+-

nn

÷

2

+-

13

+×

++

.

+

1ln

÷

1100

+

ø

54

.

325

n

ö

4615

++

nn

÷

.

2

÷

1223

nn

ø

.

nn

ö

1

+×

.

÷

.

÷

3

4

n

ø

5

165

Solution.

()(

)

()(

)

(

)

+

(

)

(

)

2

ö

34

+-

nn

÷

.

2

¥®¥®

÷

1100

+

ø

2

æ ç ç

n

è

2

ö

34

+-

nn

÷

2

=

÷

1100

+

ø

æ ç

è

1

100

2

ö

.0

¹

÷ ø

a) In this case

Let’s calculate

2

æ ç

=

a

n

ç

n

è

=

a

limlim

n

Therefore, the series diverges.

b) Because there is an exponential function 3

n

in the general series,

we need to use D'Alembert’s ratio test.

For this series

2

=

a

n

+-

nn

( )

13

+×

n

54

=

a

;

1

+

n

2

1

+

n

nn

( )( )

113

++×

n

5141

++-+

.

Let’s calculate

a

1n

+

=

limlim

a

n

¥®

=

lim

n

( ) ( ) ( )

¥®

2

1n

+

( )( ) ( )

51n41n

++-+

5n4n

+-

11n3

++×

×

1n

+

51n41n

++-+

:

n

+×

( )( )

2

n

1n3

5n4n

+-

=

1n3

+×

11n3

++×

1

.1

<=

3

Therefore, according to D'Alembert’s ratio test, this series converges.

c) Because there is a factorial (n!) in the general series, we need to

use D'Alembert’s ratio test.

For this series

!

d)

=

a

n

( )

;

325

+×

n

=

a

1

+

n

1

+

n

Let’s calculate

a

+

1

n

=

limlim

a

n

( ) ( )

+

1!

nn

=

lim

=

×

5!

n

+

!1

n

( )( ) ( )

n

+

1

lim

nn

n

¥®¥®

¥=

5

!

n

:

nn

+×++×

3125

.

325

n

We got to infinity in the limit, therefore, this series diverges.

e) Let’s use Cauchy’s root test. Here

a

n

!1

n

( )( )

n

=

lim

¥®¥®¥®

nnn

2

æ ç

=

ç è

.

3125

++×

n

!

+

51

××

++

11

n

ö

4615

÷

2

÷ ø

( )( )

n

.

n

!

n

5

nn

++

1223

nn

++

+

32

n

=

++

312

166

Let’s calculate

n

2

n

=

a

n

æ ç

n

limlim

ç

n

è

++

n

ö

4615

++

nn

÷

=

2

1223

nn

lim

÷

n

ø

2

æ ç ç

¥®¥®¥®

è

++

ö

4615

++

nn

÷

2

÷

1223

nn

ø

15

12

5

4

The value we got is more than 1, therefore, this series diverges.

f) Let’s explore this series by Cauchy’s root test. Let’s compose the

respective integral and calculate it

ln

=

xt

1

=

dx

¥

1

lim

b

dx

2

ln

×

xx

1

æ ç

ln

b

¥®

è

ò

2

b

=

lim

b

+-=

1

ò

2

ln

×

2

1

ö

=

÷

2ln

ø

dt

x

=

dx

xx

.

2ln2

=Þ=

tx

ln

=Þ=

btbx

ln

b

=

lim

b

1

dt

ò

2

t

2ln

lim

b

æ

1

ç

-=

ç

t

¥®¥®¥®

è

The integral converges, therefore, this series converges.

g) Let’s compose a series, equivalent to the original, leaving just the

upper power n in both numerator and denominator :

1

2

¥

1

= nn nnn

2

3

+

n

3

7

+

nn

¥

~

n

åå

3

7

1

=

n

¥

n

å å

7

=

3

¥

==

=

The series we got is equivalent to the original, because

+

3

lim

n

3

7

¥®

+

2

nn

n

Therefore, the original series and series

verge at the same time. Because the series

11

2

6

¥¹¹=

,:

0

¥

1

å

11

=1

n

¥

1

å

11

=1

n

the original series also converges.

ö

1

3

4

1

÷

~

÷ ø

,

3

4

nn

h) Because

æ ç

1ln

+ ç è

7

1

-

2

3

nn

converge and di-

6

11

æ

>1

ç

6

è

¥

=

å

=

ö

converges,

÷ ø

.1

>==

ln

ö

b

÷

=

÷

2ln

ø

11

.

11

6

11 1

n

167

¥

n

(

)

>

-

-<-

>

-

<

);();(+¥È--¥Î

=

-

=

å

= 1

æ ç

+

1ln

n

ç

1

è

The series

ö ÷

~

÷

4

ø

1

converges

65

¥

n

å

=

1

=

×

3

4

n

5

æ

<1

ç

6

è

1

3

n

¥

å

=1

n

21

¥

n

å

=

1

ö ÷

, therefore, the original series

ø

¥

=

34

11

å

=

nnnn

.

65

nn

also diverges.

i) Let’s estimate the common term of a series:

2

sin

n

5

+

1

.

3

2

11

n

3

æ

>1

ç

2

è

¥

å

=1

n

Series

1

3

n

2

n

¥

= nn

n

¥

1

å

=1

n

converges as well. Because convergence of a bigger series

+

5

01sin0

1

~

3

5

+

converges

3

2

<Þ<<

¥

åå

=

leads to the convergence of a smaller one, the original series is diverges.

Example2. Find the convergence domain of series

Solution. Let’s use the D'Alembert’s ratio test:

1

+

( )

n

¥®

n

32

×+

1

+

n

( )

12

+×-

nx

The row converges, if

or

x

32

The row converges, if

Indeterminate case:

( )

3

n

31

×+

n

2

=

n

3

×-

nx

3

2

-xx

;

32

or

5

x

3

2

-xx

3

,1

=

т.е.

2

-x

1

<

33

5

+

nn

.

ö ÷

, therefore, the equivalent series

ø

¥

n

å

( )

-

=

1

n

+

1

,

( )

or

n

321

>-Þ<

51 1

;-ÎÞ>

.

1

x

( )

¥®

n

( ) ( )

x

5

x

+

n

,

31

232

×-××+

nxn

1

3

1231

+×-×+

nxn

51x

n

×+

31

.

n

3

2

nx

3

=

2

-

x

.

.lim:lim

168

=

x

()(

)

(

)

-

=

(

)

]

[

(

)

+¥È-¥-

2

43

.

æö

ç÷

èø

(

)

-

Let

Series

x

Let

å

n

¥

+

5

¥

=+1

n

:

1

n

3

n

=

1

n

converges as an equivalent to the original.

3

n

¥

+

n

:

-

=

1

n

¥

31

3

31

n

)3(

n

=

1

n

¥

=

åå

3

=

1

n

¥

+

1

3

1

~

n

– the series converges.

ååå

2

nn

=

1

nnn

+-

)1()1(

n

.

3

This series alternating. By exploring this series for convergence (we

explore series containing absolute values), we get a series as if x = 5, which is convergent. Because a series ряд containing absolute values is convergent, this series is absolutely convergent. We get, that

;; 51

is the convergence domain.

Try to do these yourself:

3

¥

nn

Examine the series for convergence:

å

1001

1

n

=

-+

3

n

+

Topic 23. Exponential and functional series

Introduction. This workshop is devoted to the discussion of the following concepts: the convergence of functional series partial sums of functional series, the convergence of functional series convergence of functional series, the power series, the idea of matching derivatives by powers.

Example 1. Find the first three (not equal to 0) members of factorization in the power series, solving the Cauchy’s problem

-

x

¢

Solution.

To represent the solution in the form of a McLaren series, we need to find the first three non-zero values y(0), y'(0), y''(0), … Under the conditions of the problem y(0) = 1. Let’s show y'from the equation:

x

¢ ¢

xyey

0

-

ey

Let’s find y'', by differentiating both parts of the equation (*) by x:

1)0(,2 ==-

yxyey

.

;2

*+=

.1021)0(

=×+×=

169

---

(

)

54012115227720

54012115

(

)

3780

+

-

xxx

¢¢ ¢¢

¢

+=

00

--

eey

¢

;2)2(

+-=

+

yeyexyey

.2211)0(

=+×+×-=

Finally, we get:

1 xxxxy ++=+×+=

Example 2. Calculate to the

1

!1

=

e

1

!2

310-

the integral

Solution. Let’s show the decomposition of the function

22

.

1

×

sin dxxx

ò

0

.

xxy sin×=

2

into the McLaren series:

-

12753

nn

-

)1(

sin

xtx

53

!5!3!1

7

5

3 2

xx

2

!5!3!1

...

( )

!7!5!3!1

( )

1

-

...

+++-=

( )

n

( )

1

-

...

+++-=

( )

-

n

12

-

n

2

xxxx

+...

!12

-

12

-

n

1

+

n

2

xxxx

!12

-

n

ttttt

+--+-===

=+

...

!12

...

+

Let’s calculate the integral

+

1

sin dx

0

5

æ

2

ç

-=

ç

5

ç

2

è

-= ...

5

6

2

1

2

ç

dxxx

òò

ç

0

è

9

7

2

+

7

×

2

11

+

7

×

2

...

9

×

!5!3

2

1

2

++

( )

9

×××××

54321

2

...

( )

-

n

-

2

!5!3!1

n

1

( )

-

+++-=×

...

( )

+

12

n

+

1

2

xxxx

+

12

n

( )

+-

2

1

7

5

3

æ

12

n

1

xxxx

-

n

+

1!12

10

ö

2

÷

=

+

...

÷

!12

÷ ø

ö

1

÷

=

...

÷

0

÷ ø

=-

11

×××××××

7654321

2

-3

=±+-»--+-=

1339

=

71801512

3

-

10354,0

±==

.

Notice, that by calculating the integral we get an alternating series. We exclude all the summands when calculating, starting from the summand, that is less (in absolute values) than the given exactitude

æ ç

27720

è

1

<-310

ö

.

÷ ø

170

Section 9. Elements of the complex

+-=

1

2

1

(

)

1

2

(

)

2

variable function theory

Topic 24. Complex variable function theory

Introduction. In this workshop we will present some elements of the theory of functions of a complex variable. It should be noted that this theory of course transcends the limits of this lecture and the student can deepen their knowledge on the topic using additional literature.

Exercise 1.

iz

1) Find the modulus and the argument of numbers

iz 31

+=

. Represent the numbers on the complex plane represent

numbers in trigonometric and exponential form.

2) Find: а).

2

zz ×

2

1

; б).

2

z

1

; в).

3

z

1

z

Solution.

1) Let’s show numbers on the complex plane. This way the number

1;1

iz +-= 1

will have a corresponding point

iz 31

+=

– point

M

3;1

.

-M

, and the number

1

и

Fig.18

Fig.18 – graphic representation of a numbers on the complex plane.

To find the modulus and argument of the given numbers we will use the formula: