2018北京大学数学分析试题及其解答

0
dt
0
0
K4(20©): f (x)(0,0)NC3 ëYŒ‡§OŽ
1 lim R→0+ R4
(f (x, y) − f (0, 0))dxdy
x2 +y 2 ≤R2
y²: ŠâK¿k
21 f (x, y) = f (0, 0) +
∂∂ x +y
i
f (0, 0) + o x2 + y2 (x2 + y2 → 0)
dy
0
n
0
(2k + 1)!
k=1
5¿ d`?ê
1
1
y1 n
−1
d
y
=
1,
(−1)k
y2k−1+
1 n
≤
1 , y ∈ (0, 1)
n0
(2k + 1)!
(2k + 1)!
O{Ú˜—Âñ¼ê‘?êÅ‘È© 5Ÿ•
1 n
1 +∞
(−1)k
y2k−1+
1 n
dy
=
1
+∞
(−1)k
0
(2k + 1)!
k=1
√
5n
≥
6
lim (1 +
n→∞
1 0
sinn xn
x
d
x)n
=
+∞
(2)
1 sin xn d x −y−=−x→n 1
0 xn
n
1
y
1 n
−2
sin
y
d
y
=
1
0
n
1 =
n
1
+∞
y1 n
−2
(−1)k y2k+1 d y
0
(2k + 1)!
k=0
1
y
1 n
−1
d
y
+
1
1 +∞
(−1)k
y2k−1+
1 n
0 = fxx(0, 0)x + fxy(0, 0)ϕ (0)x + o(x) ⇒ 0 = fxx(0, 0) + fxy(0, 0)ϕ (0)
0 = fyx(0, 0)x + fyy(0, 0)ϕ (0)x + o(x) ⇒ 0 = fyx(0, 0) + fyy(0, 0)ϕ (0) df (x, y) 3(0,0) HessianÝ
|f (A) − f (B)| ≤ M L
y²: Ø” γ = γ(t), t ∈ [0, 1], γ(0) = B, γ(1) = A§K
1 df (γ(t))
|f (A)−f (B)| =
dt
0 dt
1 df (γ(t))
1
1
dt = |(gradf (γ(t)), γ (t))| dt M |γ (t)|dt = M L
Ù¥ξ3y0†ϕ(x0)ƒm, gñ. ¤±ϕ(x)ëY. e¡·‚ ½ε > 0, •Ä¼ê fε(x, y) = f (x, y) + εx2.
0 < x1 < δ1, du
∇fε(x1, ϕ(x1)) = ∇f (x1, y1) + (2εx1, 0) = (2εx1, 0),
=
fε,x(x1, ϕ(x1)) = 2εx1 > 0,
1
=0
n
n
qϕ
1 +∞ (−1)k
1
+∞ (−1)k
1
lim n ln(1
n→∞
+
n
k=1
(2k
+
1)!
2k
+
1
)
=
lim
n→∞ k=1
(2k
+
1)!
2k
+
1
n
n
(−1)k
1 ≤
1
+∞ (−1)k
, ∀x ∈ (0, 1) ⇒
1 ˜—Âñ
(2k + 1)! 2k + x (2k + 1)!
(2k + 1)! 2k + x
π
π
(f (x, y) − f (0, 0)) dxdy = 8 fxx(0, 0) + 8 fyy(0, 0)
x2+y2 R2
5: K¥w•f ∈ C3§¢Sþ•If ∈ C2"
K5(20©): y = ϕ(x)3x = 0 ?Œ §ϕ(0) = 0§f 3(0,0) NC2 f 3(0,0) HessianÝ Œ ½…š"Ý "¦yf 3(0,0)? 4 Š"
x ∈ (2nπ, 2nπ + π) ⇒ x sin(x) > 0 ⇒ f (x) > 0
√
√
π
3π
2
π2
3π
x ∈ ((2n − 1)π + , (2n − 1)π + ) ⇒ xsin(x) ∈ (− ((2n − 1)π + , − (2n − 1)π + ) ⇒ f (x) < 0
4
4
2
42
n→∞ 0 xn
2k(2k + 1)!
k=1
1 +∞ (−1)k
1
+∞ (−1)k
⇔
lim n ln(1
n→∞
+
n
k=1
(2k
+
1)!
2k
+
1
)
=
( 2k(2k
k=1
+
) 1)!
n
+∞ (−1)k
1
+∞
1
1 +∞ (−1)k
1
k=1 (2k + 1)! 2k +
1
≤ (2k + 1)!
k=1
< +∞ ⇒ lim n→∞ n k=1 (2k + 1)! 2k +
k2
1n
k2 − k
1n
k2
k2 − k
1 n k n+1
n
ln(1 + n2 ) − n
ln(1 +
n2
)≤ n
ln(1 + n2 ) − ln(1 +
n2
)≤ n
= n2 2n2
k=1
k=1
k=1
k=1
1n
Байду номын сангаас
k2 − k
1n
k2
π
lim n→∞ n
ln(1 +
n2
) = lim n→∞ n
ln(1 + n2 ) = ln 2 − 2 + 2
f (x, ϕ(x)) ≥ f (0, 0).
aq/éux < 0§•Ä¼êfε(x, y) = f (x, y) − εx2Œy þãØ ª•¤á" • Šâfy(x, ϕ(x)) = 0±9fyy(x, y) > 0 Œ
f (x, y) ≥ f (x, ϕ(x)) ≥ f (0, 0).
f 3(0, 0)? 4 Š"
4
…
π x ∈ ((2n − 1)π, (2n − 1)π + ), f (x) < 0
4
3π x ∈ ((2n − 1)π + , 2nπ), f (x) > 0
4
25¿ f ((2n−1)π) > 0, f (2nπ) > 0§Œ•f (x)3z‡((2n−1)π, (2n+1)π) STkü‡Šx2n−1 < x2n§
k=1
d˜—Âñ¼ê‘?ê4•†¦ÚŒ † 5Ÿ±9Heine½n•
+∞ (−1)k
1
+∞
(−1)k 1
+∞ (−1)k
lim n→∞ k=1 (2k + 1)! 2k +
1
= lim k=1 n→∞ (2k + 1)! 2k +
1
=(
)
2k(2k + 1)!
k=1
n
n
(3)dRiemann驥
1n
x5 = (1 − t0)x1 + t0x3, x6 = (1 − t0)x2 + t0x4
∃t0 ∈ (0, 1), s.t.λ =
=k
λ
=
G(t0)
=
f (x6) x6
− −
f (x5) x5
K3(10©): A, B´R3 ü:§γ ´±A, B•à: 1w-‚§l••L§U ´˜‡•¹γ m8§ f ´U þëYŒ‡ ¼ê§§ FÝ•þ þ.´M§¦y
i! ∂x ∂y
i=1
l
Ïd
1 (f (x, y) − f (0, 0))dxdy =
2
(fxx(0, 0)x2 + fyy(0, 0)y2)dxdy + o(R4)
x2+y2 R2
x2+y2 R2
=
πR4 8 fxx(0, 0)
+
πR4 8 fyy(0, 0)
+
o(R4)(R
→
0+)
1 lim R→0+ R4
•3 •U (x1, ϕ(x1)) = (x1 − δx1 , x1 + δx1 ) × (ϕ(x1) − δx1 , ϕ(x1) + δx1 ), ¦ ?¿(x, y) ∈ U (x1, ϕ(x1)), þk
fε,x(x, y) > 0.
δx1 > 0, ¦ |x − x1| < δx1 žk
|ϕ(x) − ϕ(x1)| < δx1 .
y²: df (x, y)3(0,0)NC ëYŒ‡•
ëYŒ‡§∇f (x, ϕ(x)) = 0§
fx(x, y) − fx(0, 0) = fxx(0, 0)x + fxy(0, 0)y + o( x2 + y2)
)‰ ngƬêÆ[
3
‡&ú¯Òµ"‰„
fy(x, y) − fy(0, 0) = fyx(0, 0) + fyy(0, 0)y + o( x2 + y2) dy = ϕ(x)3x = 0 ?Œ ±9∇f (x, ϕ(x)) = 0•
Ïd, 0 < x1 < δ1ž§•3δx1 > 0, éux1 < x2 < x1 + δx1§¤á
fε(x2, ϕ(x2)) > fε(x1, ϕ(x1)). ddØJ x > 0žfε(x, ϕ(x))´O¼ê§u´
fε(x, ϕ(x)) > fε(0, ϕ(0)) = f (0, 0).
-ε → 0+Œ
fxx(0, 0) fxy(0, 0) = fyy(0, 0)[ϕ (0)]2 −fyy(0, 0)ϕ (0)
fyx(0, 0) fyy(0, 0)
−fyy(0, 0)ϕ (0)
fyy(0, 0)
š ½…š"•
fyy(0, 0) > 0
dy = ϕ(x)3x = 0?Œ •ëY, Œ 0 < δ1 < δ, ¦ |x| < δ1žk
…
π
3π
x2n−1 ∈ ((2n − 1)π, (2n − 1)π + 4 ), x2n ∈ ((2n − 1)π +
, 2nπ) 4
?˜Ú5¿
lim
n→∞
f (2nπ
−
1 2nπ
)
=
0§…
nv
Œž
f (2nπ −
|ϕ(x)| < δ.
Äky²ϕ(x)3|x| < δ1ž´ëY . ÄK, •3ê xn → x0÷v^‡
ϕ(xn) → y0 = ϕ(x0),
u´k ,˜•¡
∇f (x0, y0) = lim ∇f (xn, ϕ(xn)) = 0. n→∞
fy(x0, y0) = fy(x0, ϕ(x0)) + fyy(x0, ξ)(y0 − ϕ(x0)) = fyy(x0, ξ)(y0 − ϕ(x0)) = 0,
n (2k + 1)! k=1
1
y2k−1+
1 n
0
dy
=
1 n
+∞ k=1
(−1)k (2k + 1)!
2k
1 +
1
n
1 sin xn
1 +∞ (−1)k
1
0 xn d x = 1 + n k=1 (2k + 1)! 2k + 1
n
)‰ ngƬêÆ[
1
‡&ú¯Òµ"‰„
d
lim ( 1 sin xn d x)n = +∞ exp( (−1)k )
‡&ú¯Òµ"‰„
2018PKUa¬ïÄ)ç)êÆ©ÛÁK9)‰
"‰„ ‰l 2018 c 8 19 F
K1(30©): y²±eS 4•:
(1) lim (1 + n→∞
1 0
sinn (x) xn
d x)n
=
+∞;
(2) lim ( n→∞
1 0
sin(xn ) x
d
x)n
=
∞
exp(
(−1)k 2k(2k+1)!
k2
lim n→∞ n
ln(1 + n2 ) =
k=1
1
ln(1 + x2) d x = x ln(1 + x2)
1 0
−
0
1 2x2
π
0 1 + x2 dx = ln 2 − 2 + 2
d
x−y
|ln(1 + x) − ln(1 + y)| =
≤ |x − y| , ∀x, y ≥ 0
1+ξ
K
1n
)
k=1
(3) lim n→∞
1 n
n k=1
ln(1
+
) k2−k n2
=
ln
2
+
2
−
π 2
.
y²:
(1)5¿
sin x
≥
x
−
x3 6
,
∀x
≥
0.Œ
(1 +
1 0
sinn xn
x
d
x)n
≥
(1
+
1
(1
−
x2
)n
d
x)n
0
6
≥ (1 +
√1 n
(1
−
x2 )n
d x)n
0
6
≥ (1 +
5 √
)n
6n
K6(20©): y ² • §e−x + cos(2x) + x sin(x) = 0 3 z ‡((2n − 1)π, (2n + 1)π) S T k ü ‡ Šx2n−1 < x2n§¿y²e 4••3¿¦ƒ
y²:
lim (−1)nn(xn − nπ)
n→∞
f (x) = e−x + cos(2x) + x sin(x)§5¿ éf (x) Ù̇Š^ ´xsin(x)ù˜‘"Kk
k=1
k=1
K2(10©):
f
∈
C(0, 1), α
=
f (x2)−f (x1) x2 −x1
<
β
=
, x f (x4)−f (x3)
x4 −x3
i
∈
(0, 1)(i
=
1, 2, 3, 4)" ¦yéz
‡λ ∈ (α, β)§•3x5, x6 ∈ (0, 1) ¦
λ
=
" f (x6)−f (x5) x6 −x5
u´éux1 < x2 < x1 + δx1 , Šâ
fε,y(x1, ϕ(x1)) = 0
±9
fε,yy(x, y) = fyy(x, y) > 0
)‰ ngƬêÆ[
4
‡&ú¯Òµ"‰„
Œ fε(x1, ϕ(x2)) ≥ fε(x1, ϕ(x1)),
2Šâ fε,x(x, y) > 0
Œ fε(x2, ϕ(x2)) > fε(x1, ϕ(x2)).
y²: Ø” x1 < x2, x3 < x4"-
G(t) = f ((1 − t)x2 + tx4) − f ((1 − t)x1 + tx3) (1 − t)(x2 − x1) + t(x4 − x3)
)‰ ngƬêÆ[
2
‡&ú¯Òµ"‰„
KG(t)3[0, 1]þëY§…α = G(0) < λ < G(1) = β,ŠâëY¼ê 0Ÿ5½nŒ G(t0)§-