南开大学2011年数学分析考研试题参考解答

x→0 x4 6
x4 3
.
−
x6 90
+ o(x6 ) − 1 − x2 + x6
x4 2
−
x6 6
+ o(x6 ) +
x4 3
1 1 + 90 6
7 . 45
(b) OŽ I =
L
−y dx + x dy , L • x2 + y 2 = 1, 4x2 + y 2 −y , + y2
_ž ••.
)‰. P P = K ∂P −4x2 + y 2 ∂Q = = , 2 2 2 ∂y (4x + y ) ∂x d Green úª• I =
é¡5
(d) ¦¼ê f (x, y ) = 2x2 − 7y 2 34«• ¯ = (x, y ); x2 + 2xy + 4y 2 ≤ 13 D •ŒŠ†• Š. )‰. d ¦f 3 ∂D = (x, y ); ϕ(x, y ) ≡ x2 + 2xy + 4y 2 − 13 = 0 þ •ŒŠ†• Š. fx = 4x fy = −14y •f 4Š:• (0, 0), … f (0, 0) = 0.
min f = −26,
¯ D
max f =
¯ D
91 . 3
1
d (1)1,2 • (4 + 2λ)x + 2λy = 0 2λx + (−14 + 8λ)y = 0.
Ùkš") (ÄK† (1)3 gñ). 4 + 2λ 2λ =k 7 λ=− , 3 òþã λ Š“\ (1)1 , 2éá (1)3 =•(Ø. λ = 2. 2λ −14 + 8λ = 0,
|f (t)| dt,
x∈
a+b ,b 2
.
x
F (x) =
a+b 2
|f (t)| dt,
K F (x) = |f (t)| , l
b
F
a+b 2
= 0,
|f (x)| ≤
b −F (x), x ∈ a, a+ , 2
F (x),
x∈
a+b ,b 2
.
|f (x)f (x)| dx ≤
a a
a+b 2
b
¼ê, … f b−a 4
b a
a+b 2
2
= 0. Áy²:
|f (x)f (x)| dx ≤
a
|f (x)| dx.
y². d f
a+b 2
=0•
x
f (x) =
a+b 2
f (t) dt,
|f (x)| ≤ P

a+b 2
x x
a+b 2
b |f (t)| dt, x ∈ a, a+ , 2
k+1 k
n−1
≥ nan
k=[ln n] n−1
{nan } 4~ dx x ln x
≥ nan
k=[ln n]
= nan (ln ln n − ln ln [ln n]) ≥ nan (ln ln n − ln ln ln n) .
n→∞
lim nan ln ln n =
n→∞
lim [nan (ln n ln n − ln ln ln n)] ·
The Cozy Hut of Dr. Zhang738
(n ≥ 3),
{ cn } ∞ n=2 üN4~…ke. l
n→∞
√ 3.
lim cn = c > 0 •3. u (2) ü>- n → ∞, c= √ c2 + 3 ⇒ c = 3. 2c
3. (15 )
f (x) 3 [a, b] þkëY
Ly = −14y + λ(2x + 8y ) = 0 L = x2 + 2xy + 4y 2 − 13 = 0 λ •1 7 √ x=∓ 3
1 y = ±√ 3 λ = −7 3
(1)
½
x = ±1
y = ∓2 . λ=2
† OŽk f 7 1 ∓√ , ±√ 3 3 = 91 , 3 f (±1, ∓2) = −26.
K p2 = 3q 2 ,
p = 3k (3k + 1, 3k + 2
²•þØU Ø 3). u´ q 2 = 3k 2 . Ó
3|p, 3|q , ù† (p, q ) = 1 gñ], ´˜‡gñ. d = b, a = c.
[pìŒÆêÆ,“ 2011 c 1 2 ò 1 60 Ï 734–742 5¿ cn − cn−1 ·‚k −c2 n−1 + 3 = ≤0 2cn−1
x2 + y 2 x2 + y 2
2
1− 1− dx dy
2
· 1+ 1 = √ 2 √ −x
x2 +y 2
+
√ −y
x2 +y 2 3
x3 + y 3 + 1 −
x2 +y 2 ≤1 3
x2 + y 2 dx dy
x2 + y 2 1− x2 + y 2 dx dy
1 = √ 2 x2 +y2 ≤1 x2 + y 2 1 √ (1 − r)3 2π · r dr = r 0 √ 2π = . 4
∞
{nan } üN4~. Áy²: lim nan ln ln n = 0.
n→∞
y². ´• an ≥ 0. d?ê
n=2
an Âñ• ln n
n−1
n→∞ k=[ln n]
lim
ak = 0. ln k
Ï
n−1
k=[ln n]
ak = ln k
n−1
k=[ln n]
kak k ln k 1 k ln k
2
2 an−1 + bn−1 n−1 2a bn−1
3 .
(n ≥ 2).
(2)
√ √ a+b 3=c+d 3 √ ⇒ a − c = (d − b) 3. e d = b, K √ •knê [^‡y{. e √ 3= p , q p, q þ• 3= a−c d−b
a, b, c, d ´
ê
ê… (p, q ) = 1. n`² q = 3m,
[pìŒÆêÆ,“ 2011 c 1 2 ò 1 60 Ï 734–742 •d, |^ Lagrange ¦ê{. P
The Cozy Hut of Dr. Zhang736
L(x, y, λ) = f (x, y ) + λϕ(x, y ) = 2x2 − 7y 2 + λ x2 + 2xy + 4y 2 − 13 . Kd Lx = 4x + λ(2x + 2y ) = 0
[pìŒÆêÆ,“ 2011 c 1 2 ò 1 60 Ï 734–742 2. {an }, {bn } þ• êê , …·Ü an +
The Cozy Hut of Dr. Zhang737
a1 = b1 = 1, y²: ê y². d an bn
√ √ 3bn = an−1 + 3bn−1
b
[−F (x)] · F (x) dx +
a+b 2
F (x)F (x) dx
[pìŒÆêÆ,“ 2011 c 1 2 ò 1 60 Ï 734–742 F 2 (x) =− 2 1 = 2 1 ≤ 2 =
a+b 2 a+b 2
The Cozy Hut of Dr. Zhang739 F 2 (x) 2
2 b
a+b 2
+
a
|f (t)| dt
2
a
a+b 2
1 + 2
a+b 2
b
2
|f (t)| dt
a+b 2
|f (t)| dt ·
a b a
12 dt +
a
1 2
b
a+b 2
|f (t)| dt ·
a+b 2
2
b
12 dt
b−a 4
|f (t)| dt.
2
∞
4. (20 )
?ê
n=2
an Âñ, ê ln n
du
x → 0+ ž, esin x |sin 2x| 1 † Ó ; xp (1 + xq ) xp−1
x → +∞ ž, e esin x |sin 2x| ≤ p+q , p q x (1 + x ) x e−1 sin2 x 1 cos 4x esin x |sin 2x| ≥ = − , p q p + q p + q x (1 + x ) 2x 4ex 4exp+q esin x |sin 2x| x−p−q ≥ xp (1 + xq ) 2e ·‚k i. ii. l p < 2 ž, I1 Âñ; {); p ≥ 2 ž, I1 uÑ; 0 < p + q ≤ 1 ž, I2 uÑ (Dirichlet O π kπ 5π kπ + ≤ + 2 12 2 12
p + q > 0 ž, J2 Âñ3 ; …=
−q < p < 2 ž, 2ÂÈ©Âñ. nþ¤ã, ·‚ (a) (b) Ñ(Ø: 2ÂÈ©ýéÂñ; 2ÂÈ©^‡Âñ;
1 − q < p < 2 ž,
p < 2 … 0 < p + q ≤ 1 ž,
(c) Ù¦œ¹ž,
2ÂÈ©ýéuÑ.
+∞
6. (20 )
,
p + q > 1 ž, I2 Âñ; …=
p + q ≤ 0 ž, I2 •uÑ.
1−q <p<2 ž, (b) •Ä
∞ 0 1 sin x esin x sin 2x e sin 2x dx = dx + p q p q x (1 + x ) 0 x (1 + x ) ≡ J1 + J2 . ∞ 1
HmŒÆ2011cêÆ©Û•ïÁKë•)‰
Üy<
Á‡
©‰Ñ HmŒÆ 2011 cêÆ©Û•ïÁK êÆ©Û •ïÁK
˜‡ë•)‰.
Ì• http://bbs.sciencenet.cn/home.php?mod=space&uid=287000 '…c HmŒÆ
1. OŽK (4 × 15 = 60 ). √ 2 cos 2x − e−x + (a) ¦4• lim x→0 x6 )‰. 4• 1 − x2 + = lim =− =
L
4x2
Байду номын сангаас
Q=
4x2
x , + y2
(x, y ) = (0, 0),
P dx + Q dy 734
[pìŒÆêÆ,“ 2011 c 1 2 ò 1 60 Ï 734–742 = = 1 ε2 = π.
4x2 +y 2 =ε2 2π 0
The Cozy Hut of Dr. Zhang735 −y dx + x dy 4x2 + y 2 0<ε 1, _ž ••
2ÂÈ©ýéÂñ.
esin x sin 2x dx xp (1 + xq )
´•
[pìŒÆêÆ,“ 2011 c 1 2 ò 1 60 Ï 734–742 i. ii. l p < 2 ž, J1 Âñ;
The Cozy Hut of Dr. Zhang741 p ≥ 2 ž, J1 uÑ; p + q ≤ 0 ž, I2 uÑ4 .
5. (20 )
p, q •¢ê, Á?Ø2ÂÈ©
0
Âñ, ÛžuÑ, ¿`²nd. )‰. (a) •ÄÈ©
∞ 0
1 sin x esin x |sin 2x| e |sin 2x| d x = dx + p q p x (1 + x ) x (1 + xq ) 0 ≡ I1 + I2 .
∞ 1
esin x |sin 2x| dx xp (1 + xq )
∞
F (y ) = −
0
√ x sin xy dx. x(1 + x)
∞
F (y ) − F (y ) = −
dx x ∞ sin t 2t dt = − · t2 /y 2 y 2 0 ∞ sin t = −2 dt t 0 = −π.
−ε sin t ·
−ε sin t ε cos t + · ε cos t dt 2 2
(c) OŽ I =
S
x3 + y 3 + z 3 dS , S • x2 + y 2 = (1 − z )2 , 0 ≤ z ≤ 1. 1−z
3
)‰. x3 + y 3 + 1 − I =
x2 +y 2 ≤1
F (y ) =
0
√ sin xy dx, y > 0. ®• x(1 + x)
+∞ 0
sin x π dx = . x 2
(a) Áy²: F (y ) − F (y ) + π = 0. (b) ¦Ñ F (y ) Ð ¼êLˆª.
y². (a) dƒA ˜—Âñ5• √ ∞ √ x cos xy dx, F (y ) = x(1 + x) 0
2
.
4••3, ¿¦T4•Š.
√ an + b n 3 =
an−1 +
√
2
3bn−1
√ 2 = a2 n−1 + 3bn−1 + 2an−1 bn−1 3 9 an , bn þ• ê• 2
2 an = a2 n−1 + 3bn−1 ,
bn = 2an−1 bn−1 .
u´ a2 + 3 b 2 an n−1 = n−1 = bn 2an−1 bn−1 - cn = an ,K bn c2 +3 √ cn = n−1 ≥ 3 2cn−1
x→∞
ln ln n ln ln n − ln ln ln n
= 0
lim
x =1 . x − ln x
[pìŒÆêÆ,“ 2011 c 1 2 ò 1 60 Ï 734–742
+∞
The Cozy Hut of Dr. Zhang740 esin x sin 2x ÛžýéÂñ, Ûž^‡ xp (1 + xq )