标签:des style class c ext color
1 (1)Let k
}
n
k=1
?(0,π)
n
∑
k=1
n
x
i
.
k=1
n
sinx
k
x
k
≤(sinx
x
)
n
.
Proof. Direct computations show
x
)
′′
=(lnsinx?lnx)
′′
=?1
sin
2
x
+1
x
2
<0,
x
n
∑
k=1
n
lnsinx
k
x
k
≤lnsinx
x
.
(2)From
∞
0
e
?x
2
dx=π
√
2
,
∞
0
sin(x
2
)dx
Proof. Consider the sector in 2
?
?
?
?
?
?
?
?
I:
II:
III:
0≤z≤R,
Re
iθ
, 0≤θ≤π
4
,
re
iπ
4
, 0≤r≤R.
I
+∫
II
+∫
III
]e
iz
2
dz.
(1)
(a)I
e
iz
2
dz=∫
R
0
e
ix
2
dx
(b)
∣
∣
∫
II
e
iz
2
dz∣
∣
∣
=
≤
≤
=
→
∣
∣
∣
∫
π
4
0
e
iR
2
e
2iθ
?iRe
iθ
dθ∣
∣
∣
R∫
π
4
0
e
?R
2
sin2θ
dθ
R∫
π
4
0
e
?R
2
?2
π
?2θ
dθ
π
4R
(1?e
?R
2
)
0, as R→∞,
(c)III
e
iz
2
dz=?∫
R
0
e
ir
2
e
iπ
2
?e
iπ
4
dr=e
iπ
4
∫
R
0
e
?r
2
dr
we have, by sending ∞
0
e
ix
2
dx=e
iπ
4
∫
∞
0
e
?r
2
dr.
2 Let 0
∈R; f(x
0
)=lim
x→x
0
f(x)}
δ
Proof. By definition,
∞
k=1
C
k
,
k
={x
0
∈R; ? δ
x
0
>0, s.t. |x?x
0
|<δ
x
0
?|f(x)?f(x
0
)|<1
k
}
0
∈C
k
?U(x
0
,δ
x
0
)?C
k
.
3 Consider the ODE
˙
=?x+f(t,x),
?
?
?
?
|f(t,x)|≤φ(t)|x|, (t,x)∈R×R,
∫
∞
φ(t)dt<∞.
Proof. For all >
≥
∫
t
0
φ(s)ds≥∫
t
0
∣
∣
∣
x
˙
(s)+x(s)
x(s)
∣
∣
∣
ds=∫
t
0
∣
∣
∣
(e
s
x(s))
′
e
s
x(s)
∣
∣
∣
ds
∣
∣
∣
∫
t
0
d(e
s
x(s))∣
∣
∣
=∣
∣
e
t
x(t)?x(0)∣
∣
.
t→∞
x(t)=lim
t→∞
e
?t
?[e
t
x(t)]=0.
4 Solve the PDE
u=g,
in R
+
×R,
on {x
1
=0}×R,
2
)={1,
?1,
if x
2
>0,
if x
2
<0.
Proof. It is standard (easy to verfiy) that
{y
1
=0}×R
u(y)?G
?n
(x,y)dS(y),
2π
[ln|y?x|?ln|y?x
~
|]
1
>0}
~
1
=0}
?n
(x,y)
=
=
=
??G
?y
1
(x,y)=?1
2π
[y
1
?x
1
|y?x|
2
?y
1
+x
1
|y?x
~
|
]
?1
2π
?2x
1
|y?x
1
|
2
(|y?x|=|y?x
~
|)
x
1
π|y?x|
2
.
=
=
=
=
=
∫
{y
1
=0}×R
u(y)x
1
π|y?x|
2
dS(y)
?x
1
π
∫
∞
?∞
g(y
2
)
x
2
1
+(y
2
?x
2
)
2
dy
2
?x
1
π
?
?
?
?
1
x
1
∫
0
?∞
?1
1+∣
∣
y
2
?x
2
x
1
∣
∣
2
dy
2
?x
2
x
1
+1
x
1
∫
∞
0
1
1+(y
2
?x
2
x
1
)
2
dy
2
?x
2
x
1
?
?
?
?
?1
π
[?arctany
2
?x
2
x
1
∣
∣
∣
y
2
=0
y
2
=?∞
+arctany
2
?x
2
x
1
∣
∣
∣
y
2
=∞
y
2
=0
]
2
π
arctanx
2
x
1
, x=(x
1
,x
2
)∈R
2
.
5 Let 1
0
K(x,y)f(y)dy.
sup
≤1}
Proof.
(1)1
)?Tf(x
2
)|
≤∫
1
0
|K(x
1
,y)?K(x
2
,y)||f(y)|dy
→0, as |x
1
?x
2
|→0,
(2)
(2)
(a)the unform boundedness of sup
≤1,
(b)the equicontinuity of
(c)and the Ascoli-Azer\‘a theorem.
6 Prove the Poisson
summation formula
n=?∞
∞
f(x+2nπ)=1
2π
∑
k=?∞
∞
f
^
(k)e
ikx
,
1
loc
(R); (1+|x|
m
)∣
∣
f
(n)
(x)∣
∣
≤C
m,n
, ? m,n≥0}.
^
(ξ)=∫
R
f(x)e
?ixξ
dx.
Proof. Define
n=?∞
∞
f(x+2nπ).
k
=
=
=
1
2π
∫
2π
0
h(x)e
?ikx
dx=1
2π
∑
n=?∞
∞
∫
2π
0
f(x+2nπ)e
?ikx
dx
1
2π
∑
n=?∞
∞
∫
2(n+1)π
2nπ
f(y)e
?ik(y?2nπ)
dy
∫
∞
?∞
f(x)e
?ikx
dx=f
^
(k).
n=?∞
∞
f(x+2nπ)=h(x)=∑
k=?∞
∞
a
k
e
ikx
=∑
k=?∞
∞
f
^
(k)e
ikx
.
丘成桐大学生数学竞赛2010年分析与方程个人赛试题参考解答,布布扣,bubuko.com
标签:des style class c ext color
原文地址:http://www.cnblogs.com/zhangzujin/p/3735236.html