解答：浙江大学 数学分析 2007

an (Rn − xn ) +
n=1 N
an Rn
> >
M +1−
n=1
an (Rn − xn )
M
∞
=
x→R
∞
lim−
n=1
an xn = +∞ =
n=1
an Rn
yy²3),
x→1 x→1 ∞ 1 n=1 n ln(1+n)
lim− f (x) = lim−
1 xn−1 = n ln(1 + n) n=1 n ln(1 + n) n=1
f (x) =
Âñ,l
3)
∞ n=1 n
xn−1 n ln(1 + n) n=1
∞
(−1)n n ln(1 + n) n=1 (−1)n n ln(1 + n) n=1 ≥ 0,
∞ ∞
∞
f+ (−1) = lim− f (x) =
x→1
ky²Ún: ?ê a x 'Âñ»R, a
n ∞ x→R−
5:¨,\±|^Heine8(¦ny².
3
5
5.1
¼ê{f (x)}{g (x)}Q«mI þ©yÂñuf (x)g(x), b½f (x), g(x)Ñ QI þ k..Áy²: f (x) · g (x) ⇒ f (x) · g (x)(QI þ) y²: dK¿,
n n n n
1) ∃M > 0, s.t. x ∈ M ⇒ |f (x)| ≤ M, |g (x)| ≤ M 2) ∀ε > 0, ∃δ1 > 0, s.t. x , x ∈ I, |x − x | < δ1 ⇒ |f (x ) − f (x )| < ε/(2M ) ∃δ2 > 0, s.t. x , x ∈ I, |x − x | < δ2 ⇒ |g (x ) − g (x )| < ε/(2M )
n
N
∀M > 0, ∃N > 0, s.t.
an Rn > M + 1
n=1
d
x→R−
N
lim
an (Rn − xn ) = 0
n=1
7

∃δ > 0, s.t. x ∈ [R − δ, R) ⇒
N
an (Rn − xn ) < 1
n=1
l
∞ N
an xn
n=1
≥
n=1
an xn
N N
= −
n=1
qf Qx = b?ëY,
∀ε > 0, ∃δ ∈ (0, b − a), s.t. x ∈ [b − δ, b] ⇒ |f (x) − f (b)| <
5¿ ·k
n+1 (b − a)n+1
b a
ε 2
n+1 (b − a)n+1
b
(x − a)n dx = 1
a
(x − a)n f (x)dx − f (b)
úôÆ2007êÆ©Û
David Zhang Department of Mathematics Sun Yat-Sen Univsersity QQ: 745657812 E-mail: zhangzujin360732@ November 13, 2007
1
1.1
y²
y²:
1 + x + x2 /2 + o(x2 ) x /3 + o(x ) x3
|an+1 − an | = |f (an ) − f (an−1 )| <
3 |an − an−1 | < · · · < 4
3 4
n−1
|a2 − a1 |
n+p−1
|an+p − an+1 |
≤
k=n+1
|ak+1 − ak |
n+p−1
< |a2 − a1 |
k=n+1
3 4
k−1
fk{a }´Cauchy,Âñua.Qvª¥-n → ∞,ka = 2.
S´kþ.'ê8,^S v«S'²£,= S ¥a´¢ê.Áy²:
a
3.2
a
= {s + a | s ∈ S },
Ù
.:
y²:duSkþ.,d(.¦n,supS´k½Â'. ysupS + a´S 'þ(
a
supSa = supS + a
1) s = s + a ∈ Sa ⇒ s ≤ supS + a 2) ∀ε > 0, ∃sε ∈ S, s.t. sε > supS − ε,

a1 > 0, an+1 = 1 +
3an , n = 1, 2, 3, · · · 4 + an
5
y²:ê{a }k4,¿¦Ù. y²:¼ê
n
f (x) = 1 +
u
f (x) =
3x 12 =4− 4+x x+4 0, 3 4 (∀x ∈ (0, +∞))
u´ l
12 ∈ (x + 4)2
= = π
dxdy dz
f (z )(1 − z 2 )dz
−1 1
= π
−1
f (u)(1 − u2 )du
3
3.1
Qãê8'þ(.9e(.'½Â. ):S´ê8,êξ¡S'þ(e)(.,XJ
1) s ∈ S ⇒ s ≤ (≥)ξ 2) ∀ε > 0, ∃sε ∈ S, s.t. sε > (<)ξ − ε
n
Q þ, f g ØÂñuf g.
n n
f (x)Q[a, b]þÈ,¿ Qx = b?ëY.y²:
n+1 n→∞ (b − a)n+1 lim
b
(x − a)n f (x)dx = f (b)
a
4
y²:duf È,l k.,
∃M > 0, s.t. x ∈ [a, b] ⇒ |f (x)| ≤ M
lim (−1)2k
∃N > 0, s.t. n ≥ N ⇒ (−1)2k
éue(.´3/2'y²aq.
4
Dirichlet
¼ê
D(x) =
Á©y^ 1) 4½Â 2) Cauchy ÂñOu y²¨x → 1,D(z)'4ØQ. y²: 1) ^4½Ây:
2) Cauchy
1, x 0, x
n
3 < |a2 − a1 |4( )n → 0 (n → ∞) 4
8

f (x) =
y²:
1 xn 2 ln(1 + n) n n=1
∞
Q þëY 2) f (x)Qx = −1?
1) f (x) [−1, 1]
3) limx→1− f (x) = +∞
y²: 1)
4) f (x) x = 1
knê Ãnê
∃ε0 > 0, ∀δ > 0, ∃u ∈ Qc ∩ U (1, δ ), s.t. |D(u) − D(1)| = |0 − 1| = 1 ≥ ε0
ÂñOuy:
∃ε0 > 0, ∀δ > 0, ∃r ∈ Q∩U (1, δ ), u ∈ Qc ∩U (1, δ ), s.t. |D(r)−D(u)| = |1−0| = 1 ≥ ε0
y²: u l
f (x) = cos x + sin x − 1 − x + x2
f (x) = − sin x + cos x − 1 + 2x f (x) = − cos x − sin x + 2 > 0 x > 0 ⇒ f (x) > f (0) = −0 + 1 − 1 + 0 = 0 x > 0 ⇒ f (x) > f (0) = 1 + 0 − 1 − 0 + 0 = 0
= ≤
n+1 (b − a)n+1 n+1 (b − a)n+1 + n+1 (b − a)n+1
b
(x − a)n [f (x) − f (b)]dx
a b−δ
(x − a)n |f (x) − f (b)|dx
a b
(x − a)n |f (x) − f (b)|dx
b−δ b−δ
≤ 2M
n+1 (x − a)n dx (b − a)n+1 a (b − a − δ )n+1 ε 1− + 2 (b − a)n 1− δ b−a
δ = min{δ , δ } > 0,u¨x , x
1 2
∈ I, |x − x | < δ
,
|f (x )g (x ) − f (x )g (x )| = |f (x )g (x ) − f (x )g (x ) + f (x )g (x ) − f (x )g (x )| ≤ |f (x )| · |g (x ) − g (x )| + |g (x )| · |f (x ) − f (x )| ε ε +M < M 2M 2M = ε
n+1
≤ 2M
+
ε 2
duéT½'δ,k l =k
7
n+1 (b − a)n+1
b n→∞