Добавил:

Upload Опубликованный материал нарушает ваши авторские права? Сообщите нам.

Вуз:

Национальный авиационный университет

Предмет:

[НЕСОРТИРОВАННОЕ]

Файл:

Higher_Mathematics_Part_1

.pdf

Скачиваний:

Добавлен:

19.02.2016

Размер:

9 Mб

Скачать

☆

<<< < Предыдущая 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 2425 / 2825 26 27 28 > Следующая >>>

3) if the derivative at some point is positive (negative), then in the

neighbourhood of this point the function increases (decreases);
	4) if the functions f1 (x) and				f2 (x)	are differentiable on the interval (a; b) ,
f ′ (x) =		f ′ (x) , at		the points a	and	b functions are	continuous,	then the
1		2
difference between these functions is
				f1 (x) − f2 (x) = const.

				(Cauchy’s theorem). If the functions			f (x) and	g(x) are
	Theorem 3.14
			′	continuous on the segment [a; b] , differentiable at the
interval

		(a; b) , g (x) ≠ 0 , x (a; b) , then there’s at least one point c (a; b) ,

where the following statement takes place :

f (b) − f (a)	=	f ′(c) .
g(b) − g(a)		g ′(c)

21.2. Taylor’s and Maclaurin’s formulas

In some cases Taylor’s formula is used for function investigation (calculation the value of a function, boundaries of a function).

Theorem 3.15	Let the function f (x) at the point	x0 has the derivatives to
	the (n + 1) -th order inclusively and	x is arbitrary value from

the given neighbourhood (x ≠ x0 ) . Then, there’s point c between points x0 and x , so that Taylor’s formula is true

	f	′			f	′′(x0 )
f (x) = f (x0 ) +	f	(x0 )	(x − x0 ) +		f	′′(x0 )	(x − x0 )2	+ …
f (x) = f (x0 ) +		1!	(x − x0 ) +			2!	(x − x0 )2	+ …
		1!				2!
	f	(n) (x )		(x − x )n +
... +		0				R (x) ,
... +						R (x) ,
		n!		0		n
		n!

where R	(x) =	f (n+1) (c)	(x − x		)n+1	is remainder term in the Lagrange’s form,
where R	(x) =		(x − x		)n+1	is remainder term in the Lagrange’s form,
n		(n + 1)!		0
		(n + 1)!
c = x0 + θ(x − x0 ) , 0 < θ < 1.
The expression							f (n) (x )
				f	′(x )		f (n) (x )	(x − x0 )n
	Pn (x) = f (x0 ) +				0	(x − x0 ) + …+	0
	Pn (x) = f (x0 ) +				1!	(x − x0 ) + …+	n!
					1!		n!

is called Taylor’s polynomial.

241

http://vk.com/studentu_tk, http://studentu.tk/

If x0 = 0 then Taylor’s formula is called Maclaurin’s formula:

f (x) = f (0) +

f ′(0)

x +

f ′′(0)

+ …+

f (n) (0)

f (n+1) (c)

n+1

(n + 1)!

where c = θ x, 0 ≤ θ ≤ 1.

Let us consider some examples of expansions of elementary functions using Maclaurin’s formula:

ex = 1+

+ …+

+ Rn (x) ;

sin x = x −

− …+ (−1)

x2n+1

+ R2n+1 (x) ;

(2n + 1)!

cos x = 1−

− …+

(−1)

x2n

+ R2n (x) ;

(2n)!

ln(1+ x) = x −

− …+ (−1)

n−1

+ Rn (x) ;

(3.12)

arctg x = x −

− …+

(−1)

x2n+1

+ R2n+1 (x) ;

2n + 1

(1+ x)m = 1+

x +

m(m − 1)

x2 +

m(m − 1)(m − 2)

+ …+

(3.13)

m(m −1)(m − 2)

(m − (n −1))

xn + R

(x) ,

If m = −1 then

+ Rn (x) ,

(3.14)

= 1− x + x − x

+…+ (−1)

1+ x

= 1+ x + x2 + x3 + …+ xn + Rn (x) .

(3.15)

1− x

21.3. L’Hospital’s rule

We use L’Hospital’s rule to disclosing indeterminate form of such kind as

0		∞
	or		.

0		∞

242

http://vk.com/studentu_tk, http://studentu.tk/

Theorem 3.16 (L’Hospital’s rule). Let functions ƒ(x) and g(x) be:

1) determined and differentiable in a neighbourhood of a point x0 , except,

probably, the point x0 , and

g′(x) ≠ 0

in this neighbourhood;

2) lim

f (x) = lim

g(x) = 0

lim

f (x) = lim g(x) = ∞ ; i.e. ƒ(x), g(x)

x→ x0

are simultaneously infinitesimals or infinites as

x → x0 ;

3) there is a finite lim

f ′(x)

→ x0

g′(x)

f (x)

Then there is a limit of the quotient of functions lim

and

g(x)

x→ x0

lim

f (x)

lim

f ′(x)

x→ x0

g(x)

x→ x0

g′(x)

Let’s note, that the theorem is true and in that case, when x0 = ∞.

It should be noted, that sometimes students do a gross error, to search instead

of lim	f ′(x)	for	lim	f (x)	′ .
	g′(x)
x→ x			x→ x	g(x)
0			0

L’Hospital’s rule is named after the mathematician which for the first time has published it. But this rule was proved for the first time by I. Bernulli, therefore L’Hospital’s rule is still named Bernulli-L’Hospital’s rule.

L’Hospital’s rule directly apply to disclosing indeterminate forms of a kind

∞

which are called the basic. Other indeterminate forms [0 ∞] ,

∞

[∞ − ∞] ,

1∞

, 00

∞0

are reduced to the basic.

10. Indeterminate form [0 ∞] ( lim

f (x)g(x) , when lim f (x) = 0,

lim g(x) = ∞)

x→x0

or ∞

x→x0

is reduced to indeterminate forms 0

. Thus,

∞

f (x)g(x) =

f (x)

f (x)g(x) =

g(x) ∞

∞

g(x)

f (x)

20. Indeterminate form

[

∞ − ∞

]

( lim ( f (x) − g(x)) , when

lim

f (x) = ∞

x→ x

and lim

g(x) = ∞)

is reduced to the indeterminate form 0

so:

x→ x

243

http://vk.com/studentu_tk, http://studentu.tk/

−

f (x) − g(x) =

g(x)

f (x)

f (x)g(x)

30. Indeterminate forms

1∞ ,

∞0

are reduced to the indeterminate

form 0 ∞ with

the

help

taking

logarithm or representation of

function f (x)g(x)

as eg(x) ln f (x)

(using the basic logarithmic identity).

Micromodule 21

EXAMPLES OF PROBLEMS SOLUTION

Example 1. Expand functions using the Maclaurin’s formula:

f (x) =

;

b) f (x) = ln(2x

+ 7x + 3).

4 + x 2

Solution. а) we shall write down

f (x)

= 4

Then using the formula (3.14) we have

f (x) =

1−

− …+ (−1)

+ Rn

(x) .

b) We use transformation

ln(2x2 + 7x + 3) = ln(1+ 2x)(x + 3) = ln(1+ 2x) + ln(x + 3) =

= ln(1+ 2x) + ln(1+

) + ln 3 .

Using the formula (3.12) twice, we shall get

ln(1+ 2x) = 2x −

(2x)2

(2x)3

n−1

(2x)n

− …+

(−1)

+ Rn (x) ,

ln(1+

)

−

− …+ (−1)n−1

n + R2 (x)

3 2

3 3

(−1)k −1

ln(2x2 + 7x + 3) = ln 3 +

∑

+ R (x) .

k =1

244

http://vk.com/studentu_tk, http://studentu.tk/

Example 2. Calculate limit, using L’Hospital’s rule.

Solution. We lim e3x − ex					have indeterminate form 0	, then lim	e3x − ex	=
Solution. We lim e3x − ex					have indeterminate form 0	, then lim	x	=
	3e3x − ex	x→0	x		0	x→0	x
= lim	3e3x − ex	= 3 −1 = 2 .
= lim	1	= 3 −1 = 2 .
x→0	1
Example 3. Calculate				lim x2 ln x.
				x → +0

Solution. In this case we have indeterminate form 0 ∞ . We shall reduce to

indeterminate form ∞

and use L’Hospital’s rule:

∞

lim x2 ln x = lim

ln x

lim

(ln x)′

lim

1/ x

= −

lim x2

= 0 .

)′

−2 / x3

x→+0

x→+0 1/ x2

x→+0 (1/ x

x→+0

Example 4. Calculate lim

x + sin x

x→∞

Solution. We have indeterminate form ∞

. We shall find

∞

lim

x + sin x

= lim (1+ sin x) = 1+ 0 = 1.

x→∞

Example 5. Calculate

lim (tgx − sec x) .

x→π

Solution.

[∞ − ∞] .

Here indeterminate form

shall

reduce

it to

indeterminate form 0

then use L’Hospital’s rule:

sin x −1

lim

(tg x − sec x) =

lim (tg x −

) =

lim

cos x

x→π / 2

→π / 2

→π

/ 2

lim

(sin x − 1)′

lim

cos x

= 0 .

(cos x)′

x→π / 2

x→π / 2 − sin x

Example 6. Calculate

lim (π − 2x)ños x .

x→π

value x = π

Solution.

Substituting

expression

the

, we

shall

receive

indeterminate form 00. For

convenience purposes

execute

replacement

π − 2x = t , then t → 0 . So,

245

http://vk.com/studentu_tk, http://studentu.tk/

lim (π − 2x)cos x

= lim tsin(t / 2) = lim esin(t / 2) ln t =

lim sin(t / 2) ln t

= eA ,

x→π / 2

t→0

et→0

ln t

A = lim sin(t / 2) ln t = [0 ∞] =

lim t ln t = [0 ∞] =

lim

t→0

∞

(ln t)′

2 t→0

2 t→0 1/ t

1/ t

lim

= −

lim t = 0 .

(1/ t)′

−1/ t

∞

2 t→0

Then

lim	(	π − 2x	)	cos x = e0	= 1.
x→π / 2	(		)

Micromodule 21

CLASS AND HOME ASSIGNMENTS

1.Expand a polynomial P4(x)=x4 – 5x3 + 5x2 + x + 2 in powers of binomial

x– 2.

Expand functions using Maclaurin’s formula:

2. 1− 2x .

5.x2 + 1 .

x− 2

8. x sin x .

3.	1	.		4.	1				.
	2 + x					(x −1)(x + 2)
6.	x cos2 x .			7.	ln		1− 2x	.

							1+ x
9.	1		.	10. ln(x2 − 3x + 2) .

	1+ 4x

Calculate limits, using L’Hospital’s rule:

11.

lim

7x − 3

12. lim (x2

+ 1)e− x . 13.

lim(tg 2x)x .

x→∞ x2 + 2

x→∞

x→0

x cos 2x − sin x

16. lim x4 3

14.

. 15. lim(ex + x)

lim

x→0

→0

17.

lim ln(7 − 2x) ln(6 − 2x) .

18. lim

ln(cos 3x)

19. lim

ln2 x

x→3

x→0

ln(cos 8x)

x→∞

arctg x

lim tg x ln

20.

x .

21.

lim

−

. 22.

lim

− 1

x→0

sin x

x→0

23. lim

ex − esin x

24.

2 − (ex

+ e− x ) cos x

. 25. lim

x + x2

lim

x→0

x − sin x

x→0

x→∞

246

http://vk.com/studentu_tk, http://studentu.tk/

		Answers
	1. (x − 2)4 + 3(x − 2)3 − 7(x − 2) . 3. The instruction. Write		down expression			as
1	1	. 5. The instruction. To reduce expression of a kind	x + 2 −		5		.
2	1+ x / 2			2(1	− x / 2)

11. ∞ . 12. 0. 13. 1. 14. –11/6. 15. e2 . 16. ∞ . 17. 0. 18. 3/8. 19. 0. 20. 0. 23. 1. 24. 1/3. 25. 0.

Micromodule 21

SELF-TEST ASSIGNMENTS

21.1. а) lim x ln2 x ;
	x→0
21.2. а) lim sin x ln2 x ;
	x→0
21.3. а) lim x2e− x ;
	x→∞
21.4. а)	lim (x3 + 1)4− x ;
	x→∞
21.5. а) lim (x3 − x − 2) 3− x ;
	x→∞
21.6. а) lim		1− cos x					)	ctg x		;
	x→0	(					)
21.7. а) lim		ln x ln			(	x −1			;
	x→1				(			)
		1
21.8. а)	lim x2 e		x2	;
	x→0
		1

21.9. а) lim x4 2 x2 ;

x→0

− 1

21.10. а) lim x−2 2 x2 ;

x→0

− 1

21.11. а) lim x−4 5 x2 ;

x→0

lim

;

c) lim(sin x)x .

x→∞ x2 − 5x + 2

x→0

b) lim

;

c) lim(tg x)x .

x→∞ x2 + 3x + 1

x→0

b) lim

tg x − sin x

;

lim (tg x)2x − π .

x→0

x − sin x

x→ 2

b) lim

ln(x2 − 8)

;

lim

(

ctg x

)

sin x

x→3 2x2 − 5x − 3

x→0

ex − e− x

b) lim

;

lim

(

ctg x

)

ln x .

x→0 sin x cos x

x→0

π x

; c)

(1− x)

cos

lim

−

lim

2 .

x→1

ln x

x→1

b) lim

x cos x − sin x

; c) lim(1− cos x)x .

x→0

b) lim

x − arctg x

;

lim

(ex + x)

x→0

b) lim

ex −cosx−xcosx

; c) lim

(

sin x

)

tg x .

x→0

b) lim

ln cos x

;

c) lim

(

cos x

)

ctg2 x

x→0 ln cos 4x

x→0

lim

2 x

;

c) lim

(

cos 2x

)

x−2

x→∞

x→0

247

http://vk.com/studentu_tk, http://studentu.tk/


	1				ln2 x
21.12. а) lim x−4 6−			;	b) lim			. ;
		x4

	x→0			x→∞ 5 x4
21.13. а) lim ln(5− 2x) ln(4 − 2x) ;				b) lim	ln x	;

	x→2			x→0 ctg x
21.14. а)	lim ln(x − 3) ln(7 − 2x) ;			b) lim	ln x	;

	x→3			x→0 ctg x
21.15. а)	lim ln(2x − 1) ln(2 − 2x) ; b) lim				ln(sin 2x)
					ln(sin 5x)
	x→1			x→0

c) lim (arcsin x)x .

x→0

tg πx

c) lim(2 − x) 2 .

x→1

c) lim (tg x)tg 2x .

x→π / 4

; c) lim x1− x .

x→1

21.16. а) lim

1− cos x

)

ctg2

x ;

b) lim

ln(sin 3x)

;

lim (

arctg x)x .

x→0

(

x→0 ln(sin 7x)

x→+∞

ln(tg3

21.17. а) lim(cos2 x − 3cos x + 2) e

; b) lim

; c)

lim

xtg x .

ln(sin 4x)

x→0

x→+0

21.18. а) lim(cos2 x − 1) e x2 ;

x→0

21.19. а) lim(1− cos2 x) e x2 ;

x→0

21.20. а)	lim arcsin x ln2 (2x) ;
	x→0
21.21. а) lim x3 ln2 x ;
	x→0
21.22. а)	lim x ln2 x ;
	x→0
21.23. а)	lim 4 x ln2 x ;
	x→0
21.24. а)	lim 3 x ln2 x ;
	x→0
	1
21.25. а)	lim x8 2	x4	;
	x→0

b) lim

−

; c)

lim

(tg x)

2x−π

− 1

x→0

x→π

/ 2

ln3 x

arctg x

b) lim

;

c) lim

x→∞

x→0

sin x

b) lim

−

;c)

lim

x→1

ln x

− 1

x→0

b) lim

2x + 3x

;

lim xarcsin x .

x→∞ x3 − x + 1

x→0

lim

5x − 3x

; c)

lim (− ln x)x .

x→∞ x2 − 6x + 4

x→+0

b) lim

3x2

+ 4x − 6

; c) lim(ctg x)sin x .

x→∞

(1,1)x

x→0

6x2 − x − 2

tg x

lim

; c)

lim

x→∞

(1, 5)

x→0

2x2 +10x−20

lim

; c)

lim(x −1)2− x .

x→∞

(2,5)x

x→2

248

http://vk.com/studentu_tk, http://studentu.tk/

21.26. а)

lim (x2

+ 4x + 5) 3− x ;

b) lim

ln(x2 + 3x + 1)

; c) lim xarc tg 2x .

x→∞

ln(2x2 −1)

x→0

21.27. а) lim (5x2

+ 3x + 2) 6− x ;

b) lim

ln(x2 + ex )

;

c) lim sin xarc tg x .

x→∞

x→0

21.28. а)

lim x ln3 x ;

b) lim

ln(3x2 + 2x )

;c)

lim xx−sin x .

x→0

x→∞

x→0

21.29. а) lim(x2 − 1) ln2 (x − 1) ;

b) lim

ln(4x2 + 3x )

;c) lim ln xln x .

x→1

x→∞

x→1

−

e− x

−6

− x

21.30. а)

lim x

;

b) lim

−

; c) lim (e

)

tg x

x→0

sin x

−1

x→∞

Micromodule 22

BASIC THEORETICAL INFORMATION.

THE USAGE OF DERIVATIVE IN INVESTIGATION OF A FUNCTION

Monotony of a function. Extremum. Intervals of convexity and concavity, inflection points. Asymptotes. The greatest and the least meanings of the function. General scheme of investigation of a function and constructing of its graph.

Literature: [2, chapter 5], [4, part 5], [6, chapter 5, § 6], [7, chapter 6, §20, 21], [9], [10, chapter 4, §§ 27—31], [11, chapter 5, §1], [12, chapter 5, § 3, 4].

22.1. Increasing and decreasing functions. Relative extremum of a function

Definition 3.25.				Function f (x) is		called increasing				(decreasing) on an
interval		(a; b) , if		for any two points		x1	and x2 from			the given interval
( x1 < x2 ), the following inequality is fulfilled							f (x1 ) < f (x2 )			( f (x1 ) > f (x2 )) .
Properties of increasing and decreasing functions:
1)	if	f	′
			(x) > 0 for all x (a; b) , then function f (x) increases on (a; b);
2)	if	f	′		x (a; b) , then function			f (x)	decreases on (a; b);
			(x) < 0 for all
3)	if	f	′		x (a; b) ,then function			f (x)	is constant on (à; b) .
			(x) = 0 for all
										249

http://vk.com/studentu_tk, http://studentu.tk/

ff((xx0))

f(x) f(x0)

	О x0–δ x x0		x0+δ x		О x0–δ x x0 x0+δ x
	x0 — point of maximum				x0 — is a point of minimum
	Fig. 3.23				Fig. 3.24
	Definition 3.26. Point x0 is			called the point of relative maximum (or
minimum) of function			f (x) , if there exists such neighbourhood 0 <				x − x0		< δ

of point x0 , which belongs to the domain of the function, and for all x								from
this neighbourhood the following inequality is fulfilled
			f (x) < f (x0 ) ( f (x) > f (x0 )) .
	Geometrical interpretation of the definition is clear from the Fig. 3.23 and 3.24.
	Let’s define the conditions of relative extreme.

	Theorem 3.17	(necessary condition			of relative extreme). If	a function
		f (x)	has at point x0		relative extremum and differentiable
	at this point, then f ′(x0 ) = 0 .
	Geometrical	interpretation of		the theorem 3.17. If function		f (x) has in

point x0 relative extremum and differentiable at this point, then in this point there exists the tangent line to the graph of function y = f (x) and this point

belongs to the x-axis.

Condition f ′(x0 ) = 0 is necessary, but is not enough for the function, which is differentiable at the point x0 , to have relative extremum. For example, the derivative of the function y = x3 at point x = 0 is equal to zero, but it does not have relative extremum at this point. So, function y =| x | has at point x = 0

minimum, but does not have the derivative at this point. Function y = 3 x is not

differentiable at point x = 0 and does not have extremum at this point.

Point, in which the derivative is equal to zero, is called stationary. Point, in which the derivative is equal to zero or does not exist are called critical. Critical point is the point of possible extremum.

250

http://vk.com/studentu_tk, http://studentu.tk/

<<< < Предыдущая 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 2425 / 2825 26 27 28 > Следующая >>>

Соседние файлы в предмете [НЕСОРТИРОВАННОЕ]

#
30.04.2019424.45 Кб11Graphics software.doc
#
05.09.2019478.78 Кб6gtfnmТ.docx
#
15.08.2019637.53 Кб4Gulenko.docx
#
12.11.2018765.44 Кб6Hagakure_-_Sokrytoe_v_listve.doc
#
19.02.20166.95 Mб201Higher Mathematics. Part 3.pdf
#
19.02.20169 Mб95Higher_Mathematics_Part_1.pdf
#
19.02.20166.63 Mб1052Higher_Mathematics_Part_1.pdf
#
19.02.20167.48 Mб408Higher_Mathematics_Part_2.pdf
#
19.02.20169.73 Mб45Higher_Mathematics_Part_2.pdf
#
19.02.20167.31 Mб123Higher_Mathematics_Part_2.pdf
#
19.02.20169.11 Mб304Higher_Mathematics_Part_3.pdf