四川大学2011年数学分析考研试题解答

1
∀ ε > 0, ∃ N ∈ N , s.t. n ≥ N ⇒ n
0
xn−1 f (x)dx − f (1) < ε.
ùÒy 8.
(Ø. ëY ê, …
¼ê f (x, y ) 3«• D : x2 + y 2 ≤ 1 þk ∂ 2f ∂ 2f 2 2 + = e−(x +y ) . 2 2 ∂x ∂y y²: x
)‰. 1 4• = lim n→∞ n
2 2n
k=1
1 1 + k/n
=
1 dx 0 1+x = ln 3.
1
XLDD (c) e lim 1 1+ x
ax
2 √ x+1−1 = lim arccos , ¦ a. x→0 sin x
ax
x→∞
)‰. d
x→∞
lim
1 1+ x
= ea ,
√
x→0
lim arccos
x+1−1 x 1 = arccos lim ·√ x→0 sin x sin x x+1+1 1 π = arccos = 2 3
• ea =
x→ 0
π , 3
π a = ln . 3
1 2x
(d) lim ex + x2 + 3 sin x )‰.
.
4• = exp lim
ln (ex + x2 + 3 sin x) x→0 2x 1 1 = exp lim · x · (ex + 2x + 3 cos x) x→0 2 e + x2 + 3 sin x = e2 .
9 Green úª• I =
xy =6
f rac1 + y 2 f (xy )ydx +
(2,3) → (3,2)
x 2 y f (xy ) − 1 dy y2
XLDD
3
5 x 6 x3 6 + f (6) + xf (6) − · − 2 6 x 36 x x dx 3
=
2 3
dx
=
2
5 . = 6 xdydz + z 2 dxdy
6
4. ?ؼê x sin )‰. d
1 1 Ú sin 3 (0, ∞) þ x x
˜—ëY5, `²nd.
x→0
lim x sin
1 = 0, x
x→∞
lim x sin
sin 1 1 = lim 1 x = 1 x x→∞ x
• x sin
1 3 (0, ∞) þ˜—Âñ. qd x 1 1 − →0 1 nπ π n+ 2 (n → ∞),
D
∂f ∂f +y ∂x ∂y
dxdy =
π . 2e
y². d Green úª ∂f ds = ∂n f dx
Ω
∂Ω
• x
D
∂f ∂f +y ∂x ∂y
1
dxdy =
0
rdr
x2 +y 2 =r2
x ∂f y ∂f + r ∂x r ∂y
ds
XLDD
1
9 ∂ 2f ∂ 2f + 2 ∂x2 ∂y e−(x
2. OŽe È©. (a) ¦ )‰. cos(ln x)dx. I= cos(ln x)dx, K [− sin(ln x)] sin(ln x)dx 1 · xdx x
I = x cos(ln x) − = x cos(ln x) +
XLDD
3
= x cos(ln x) + x sin(ln x) − = x [cos(ln x) + sin(ln x)] − I.
cos(ln x)dx
I=
x [cos(ln x) + sin(ln x)] + C. 2
∞
(b)
0
1 dx. 1 + x4
∞
)‰.
I=
0
1 dx, K 1 + x4
0
I=
∞
1 1+
1 t
1 4d = t
∞ 0
t2 dt. 1 + t4
I =
1 2 1 = 2 1 = 2 = =
∞ 0 ∞ 0 ∞ 0
1 + x2 dx 1 + x4 1 +1 x2 dx 1 + x2 x2 1 x−
∞ 0 1 2 x
d x−
+2
1 x− x √ 2
1 x
1 √ 2 2 π √ . 2 2
1 1+
1 x− x √ d 2 2
(c) ¦ I = ‚.
L
|y | ds, Ù¥ L ´¥¡ x2 + y 2 + z 2 = 2 †²¡ x = y
XLDD )‰. dé¡5• I = 4 = 4
y =x≥0,z ≥0 2x2 +z 2 =2 π 2
4
yds |cos θ|2 + |cos θ|2 + √
2
cos θ 0 π √ = 4· · 2 √2 = 2 2π.
2 sin θ dθ
(d) ¦ I =
Σ
(x + y + z )2 dS , Ù¥ Σ : x2 + y 2 + z 2 = R2 .
1−δ n x∈[1−δ,1]
≤
n
0 [0,1]
x
[f (x) − f (1)] dx + n
≤ 2 max |f | · (1 − δ ) + max |f (x) − f (1)| =: I1 + I2 . u´é?¿ ε > 0, d f 3 1 ? ëY5• ε ∃ δ ∈ (0, 1), s.t. I2 < , 2 yéT δ , ε ∃ N ∈ N , s.t. n ≥ N ⇒ I1 < . 2 u´·‚k
sin
1
1 nπ
− sin
1
1 π (n+ 1 2)
=1
• sin
1 3 (0, ∞) þؘ—Âñ. x
5. ó¼ê f (x)
ê f (x) 3 x = 0 ,‡ •SëY, … f (0) = ∞ 1 1, f (0) = 2. y²?ê f − 1 ýéÂñ. n n=1 f (x) •Û¼ê, f (0) = 0, u´d T aylor f (ξ ) 2 x, 2
y². Ï f (x) ´ó¼ê, Ъ•
f (x) = f (0) + f (0)x +
x ∈ U (0).
XLDD =k f 1 n −1 =
∞
7
f (ξn ) 1 2 ≤ 2, 2 2 n n f 1 n
n¿©Œ.
d'
O{=•?ê
n=1
− 1 ýéÂñ. ,… ¼êØ 1. y ²: • §
f (x) 3 [0, 1] þŒÈ, 3 x = 1 ?ëY y²:
1来自百度文库n→∞
lim n
0
xn−1 f (x)dx = f (1).
)‰. [ÜXe
1 1
n
0
xn−1 f (x)dx − f (1) = n
0
xn−1 [f (x) − f (1)] dx
XLDD
1−δ 1 n−1
8 xn−1 [f (x) − f (1)] dx
)‰. dé¡5• I=
Σ
(x2 + y 2 + z 2 )dS = R2 · 4πR2 = 4πR4 .
(e) ®•¼ê f (x) 3 R þëYŒ , ¦ I=
L
x 1 + y 2 f (xy ) dx + 2 y 2 f (xy ) − 1 dy, y y k
Ù¥ L ´þŒ²¡ {y > 0} S± (2, 3) •å:, (3, 2) •ª: •©¡1w-‚. )‰. d ∂ 1 + y 2 f (xy ) 1 = − 2 + f (xy ) + xyf (xy ), ∂y y y ∂ ∂x x 2 y f (xy ) − 1 y2 = f (xy ) + xyf (xy ) − 1 y2
oAŒÆ 2011 cêÆ©Û•ïÁK‰Y
1. OŽ. (a) lim n+ √ √ n+2 n− n .
n→∞
)‰. 4• =
n→∞
lim
n+
√ n+2 n √ √ n+2 n+ n 1+
2 √ n 2 n3/2
=
n→∞
lim
1+ = 1 . 2
1 n
+
+1
2n
(b) lim
n→∞
k=1
1 . n+k
6. ¼ ê f : [0, 1] → (0, 1) 3 [0, 1] þ Œ f (x) = x 3 (0, 1) Sk•˜ ¢Š. )‰. F (x) = f (x) − x, K F (0) = f (0) > 0, dëY¼ê0Š½n,
F (1) = f (1) − 1 < 0,
∃ ξ ∈ (0, 1), s.t. F (ξ ) = 0, = f (ξ ) = ξ . ye„k˜ ξ = η ∈ (0, 1) ÷v F (η ) = 0, Kd Rolle ½n, ∃ ζ 3 ξ † η ƒm , s.t. F (ζ ) = 0, f (ζ ) = 1, ù´˜‡gñ. o( k f (x) = x 3 (0, 1) Sk•˜ 7. ¢Š.
x2 +y 2 <r2 r − s2
2 +y 2 )
=
0 1
rdr
x2 +y 2 <r2
dxdy
=
0 1
rdr rdr π . 2e
0 0
dxdy
= =
2πse
ds
2
= − = −
x fxx + fxy − f f
(fy fy
2
)2
fxx − 2fx fy fxy + (fx )2 fyy (fy )3
XLDD • f (x, y ) = c ´†‚ dy ´~ê ⇔ Ç dx 2 2 ⇔ fy fxx − 2fx fy fxy + (fx ) fyy = 0.
ëY
ê … fy = 0. y ²: é ˜ ƒ c ∈ R,
f (x, y ) = c ´˜^†‚ ¿‡^‡´ fy y². d d dx dy dx = d dx − fx fy
y x · fy − fx · fyx + fyy − f f y
2
fxx − 2fx fy fxy + (fx ) fyy = 0.
Σ
(f) OŽ I = þý.
x2 + y 2 + z 2
Ù¥ Σ •eŒ¥¡ z = −
1 − x2 − y 2
)‰. dé¡5• I =
Σ
z2 x2 + y 2 + z 2
x2 + y 2 ≤ 1
dxdy
= −
1 3 π = − . 6
(1 − x2 − y 2 )dxdy
3. ¼ ê z = f (x, y ) k