Методы решения экзаменационных задач по матану

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x + 0, x − 0 : x R}

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−∞

b

x R x

+∞, −∞, ∞

lim

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b

R

∞

e

«¨¥

x

R

x

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−∞, ∞, a + 0, a − 0, £¤¥ a

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f (x) → b,

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a

f (a + 0) =

lim f (x), a

R

e

e

x→a+0

R

f (a

−

0) =

x

lim f (x), a

a

0

→ −

§- ªf (®¡®§x) |- §- ¥ ¥-¨¥-ª¨¬¨î-ª¬ ¨¨ f

¢ ® ª¥ x,

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lim f (x) = 1; a

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x → e

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a f (x) g(x) = o(g(x)) ¯ ¨ x a

xn yn ¯ ¨ n → ∞ xn

− yn = o(yn)

¯ ¨ n → ∞

e

f (x) = λ(x)g(x)

¨ ¯

x → a,

¯ ¨ ñ¬ lim λ(x) = 1;

a

R

x→a

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f (b) − f (a)

(2n)!! = 2 · 4 · 6· . . . ·2n = 2n(n!)

(2n)!

(2n)!

(2n − 1)!! = 1 · 3 · 5· . . . ·(2n − 1) =

=

(2n)!!

2n(n!)

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§ 1. ‚ ¨ -

54-Œ”’ˆ

1

íª§21–Œ”’ˆ¬¥¨- ¨®-54¢--|®©ª¢-ª¤®¨¬¥- 4¥-®«ì®¢2®§--•®©,¨¥©¢ë¥ ¨¨¬6¯¥¢«ï¥¬®¥-ª ¨¬:¢ë©4-«ì§¢-¨®¥¢¥¨¢ª--®«¨®2005®©¥èñ¥ £--¢®¤¨. î® §ª®¢¥¤ ¥--.

I.¨¥--«ì¥è¥¬•”

¯® -D®¡(1;(ý1)ï ®¤®¦¤¥¬o((«ìx-−¢®®¥1)¤¨+ (y¥ −¥-1)¨).®¢ -¨¥þ).

묨,--®ï¯®

¨¨ ª -

(1),

dx

¨ dy

«ë ¨

¢

¤¨®ª ¥D¥(1; 1)

f (x, y), ¥ «¨ f (x, y

=

2xy

x

y

.

f (x, y)

= yeª

−

−

¥©«® ’ «¥ ¬ ® ¯®

¢

¨î ª - ì¨ ¦¨ ª¨ ® §«® • ¨ ® -

2

2

2xy

x

−

y

2xy

x

y

f = ye2xy−x−y , - ®¤¨¬

(2dx

y + 2xdy

dx

dy) (1).

df•®¤= dy¢«ïïe

¢

−

+ ye

−

−

·

(1)

·

−

−

”® ¬ «ì-® ¤¨x =¥1,¥y- =¨ 1, ï- ®¦¤¥¤¨¬ df¢®(1;

= dx¨

+ï

2dy.

d(df ) = d2f , -

¤¨¬ d2f = dy·e2xy−x−y (2dx·y+

= 2dy”®(dx¬ +« dy’)¥©«®+ dx

2

2

2

.

+¨¬¥¥dy) +¢¨¤:4dxdy = dx

+ 8dxdy + 3dy

+

d2f (1,1)

+ o(ρ2). “ ¨ ë¢ ï, ®

f (x, y) − f (1; 1) = df (1; 1) +

dx = (x − 1), dy = (y − 1),

2!

ρ2

= (x −1)2 + y −1)2,

-

®¤¨¬ f (x, y) = 1 + (x −1) + 2(y −1) +

+

1

2

3

2

2

2

2 II(x −¯®1) ®¡+ 4(x − 1)(y −-1)¨¥+ 2 (y-−ë1)¯ +®¨§¢®¤o((x −-ë1) þ)+.(y − 1) ).

(ý¢ë ¨ «¥

∂f∂x = ye2xy−x−y (2y − 1);

∂f∂y = e2xy−x−y + ye2xy−x−y (2x − 1);

df =

∂f dx +

∂f dy;

∂x

∂y

df (1; 1) = dx + 2dy;

∂2f2 (1; 1) =

d

∂f

(x, 1)

x=1

=

d

ex−1 x=1 = ex−1 x=1 = 1;

∂x

dx ∂x

|

dx

|

∂2f

d ∂f

d

y 1 |

(1; 1) =

(1, y) y=1 =

ye

− (2y

1) y=1 =

∂y∂x

dy ∂x

dy

|

− |

= (4y

1)ey−1

+ (2y2

−

y)ey−1 y=1 = 4;

2 −

d

|

d

∂ f2 (1; 1) =

∂f (1, y) y=1 =

(ey−1 + yey−1) y=1 = 2ey−1 +

dy

∂y

dy ∂y

|

|

+ yey−1 y=1 = 3;

2 |

∂2f

2

∂2f

∂2f

2

d f =

∂x2 dx

+ 2

dxdy+

∂y2 dy

;

∂y∂x

2

2

2

.

d”®f (1;¬ 1)« =’dx¥©«®+ 8dxdy¨¬¥¥+ ¢¨¤:3dy

d2f (1,1)

f (x, y) − f (1; 1) = df (1; 1) +

+

+ o(ρ2). Ž î¤

f (x, y) = 1 + (x − 1) + 2(y − 1) + 21 (x −

2!

+ ve2uv+u+v = (1 + (2uv + u + v) + 21 (u2 + 2uv + v2) + o(ρ2)) +

+ (v + uv + v2 + o(ρ2)) = 1 + u + 2v + 1 u2

+ 4uv + 3 v2 + o(ρ2),

£

2

2

ρ2 = (x − 1)2 + (y − 1)2 . ’ ª¨¬ ®¡ §®¬, f (x, y) = 1 + (x − 1) +

+ 2(y − 1) +

21 (x − 1)2 + 4(x − 1)(y − 1) +

23 (y − 1)2 + o((x − 1)2 +

+¢¨¤(y-−®, 1)

2

¯®{) ¨®¡ª®¬®¥ ¥©«®¨ ¬¨®¢ ¥®¥ª §«®¤¨¦ -¥¨¥,¥ ® ª ¤ ª¦¥

2

2

‹®£IV¨ ¬¨df (1;ï 1)®(ý«®£¦¤¥= dx ¢®+ 2dy, d f (1;®¥1) = dx

+ 8-dxdy¨ ®¢+ -3dy¨¥þ). .

f = ye2xy−x−y , -

•®¤¨¬ ln f = ln y +

- ”®¤¨¬¬x

y (1). ”® ¬ «ì-® ¤¨ ¥ ¥- ¨ ï ®¦¤¥ ¢®

(1),

+ 2xy − df−

dy

¤ ¢«ïï x = 1,

f = y + 2ydx + 2xdy − dx − dy (2).

y = 1, f =«ìf (-D®)¤¨= f (1;¥ 1)¥-=¨1 -ï ®¤¨¬¦¤¥ df¢®(1;(2),1) = dx¨ + 2dy.

-묨,--®ï¯®

®¤¨¬

dx ¨ dy

f d2f −(df )2

dy2

¢«ïï•®¤

= − y2 + 4dxdy.

f 2

x = 1, y = 1, f = 1, df = dx + 2dy, -

¤¨¬ d2f (1, 1) = (dx +

2

2

2

2

.

+ 2dy”®) ¬− dy«

’+¥©«®4dxdy ¨¬¥¥= dx +¢¨¤:8dxdy + 3dy

+

d2f (1,1)

+ o(ρ2)

f (x, y) − f (1, 1) = df (1, 1) +

“ ¨ ë¢ ï, ® dx = (x − 1), dy = (y − 1),

2!

ρ2

= (x −1)2 + (y −1)2, - ¨ ®¢¤¨¬ f (x, y) = 1 + (x −1) + 2(y −1) +

1

2

3

2

2

2

+®2 Ž•¬(x −®¢¥«ì1)-®¬¯®: + ¤¨®¡4(x¢−¥1)(--¥y-®¬− 1)¯ +¨

.¥¨¨¥-¬

¨¯¨ï-«¥¨¢ë¥®é¯

2 (y − 1)

+ o((x − 1) + (y − 1) ).

df (D) = dx + 2dy; d2f (D) = dx2 + 8dxdy + 3dy2 ;

f (x, y) = 1 + (x − 1) + 2(y − 1) +

21 (x − 1)2 + 4(x − 1)(y − 1) +

3

2

2

2

®©¨-¬¥-¤®¦¨¨¨¢¨¤¥:®¬®ª¤¥-¥«¥-î饬¨¤«ïë¬--®«¥¤¥¥¢¥©«®®.-®¢’-¤¨¨,¢«¥ë«¥®®©¬--¤®®£®”®-¥¬¥¤‚¯ì¯¥.¥¢®¡®¡ë-¥-®«ìª¨¦,¥ë¬®ª¬¥-®--¥©‡-¨

+•¥¨«2 (y − 1) + o((x − 1) + (y − 1) ).

n

f =

=

P

dk f + o(ρn). ‚

¤¢¨¨ª-¤«ï8

®¢-¬¥£

k=1

k!

¢

¥ª®

M (x0, y0) :

f = f (x, y)

−

f (x0, y0), dk f =

∂

∂

k

=

∂x

dx +

∂y

dy f (x0, y0), ρ = (Δx)2 + (Δy)2,

x = x −

¥«¬®

’¥©«®= y

®y ® -.묄«ï¯¥¤«¥--®¬ª¢«¥¨¨ ¤¢® ¬¥ ‹£ ¬¥--¦®¢

− x0

= dx,

y

−

0 = dy

p

n

M (x0, y0)

ì¡ë¨¨¬®¦¬¥

-

¢¨¤¥:

f

=

dk f (x0,y0)

dn+1f (ξ,η)

=

+

.

ª’®

(ξ, η) ¤¥«¨ ® ¥§®ª ª® -¬ ¬¨

k=1

k!

(n+1)!

®¨§-,®è¥y- )«-¨¨¨,(¨x,¥-y)ª®©¢¤¨¬®£¥-®è¥-

θ®:

¤¥«¥(1 -訨.)

§ ïï¢ - ¥§ª ¨¬¥ ®•

«ë®¬ -- ¤

(x0P0

−

− θ)‡y,

¤θ (0;21). .

ξ = θx0 + (1 − θ)x, η = θy0 + (1 −

4

¨¢®©ª£¨¤-¨¢®©£¨¤«¨ª©¤•

x = cos4 t, y =

4

π