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Chung minh cac cong thuc the tich


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#1 cunbin

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Đã gửi 25-10-2010 - 00:24

Cac ban oi, giup minh chung minh cac cong thuc tinh the tich cua hinh lang tru,hinh chop,hinh tru,hinh non voi.dung phuong phap nao cug duoc.

#2 tran thai son

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Đã gửi 05-05-2011 - 19:04

thay cho em hoi may van de ve dinh thuc jacobi mot tj nha thay
son

#3 tran thai son

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Đã gửi 05-05-2011 - 19:06

ai co tai lieu toan cua nhom chua cho minh xin voi
son

#4 tran thai son

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Đã gửi 05-05-2011 - 19:19

0=∫_0^∞▒∫_(-∞)^∞▒[uv+F(u)vx]dxdt


=∫_0^∞▒∫_(-∞)^∞▒[ut+F(u)x]vdxdt
son

#5 tran thai son

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Đã gửi 05-05-2011 - 19:20

Cần chứng minh

*n=1 thì (1) đúng.
*giả sử (1) đúng đến n=k, ta chứng minh (1) đúng với n=k+1(k là số nguyên dương)
Cần chứng minh


Cần chứng minh

Suy ra, (2) đúng. Vậy (1) đã được chứng minh.
son

#6 tran thai son

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Đã gửi 05-05-2011 - 19:22

1415926535 8979323846 2643383279 5028841971 6939937510 : 50
5820974944 5923078164 0628620899 8628034825 3421170679 : 100
2962457053 9070959679 6673211870 6342459769 2128529850 : 999,999,999,950
2976735807 0882130902 2460461146 5810642210 6680122702 : 1,000,000,000,000
9354516713 6069123212 1286195062 3408400370 1793492657 : 1,999,999,999,950
8386341797 9368318191 5708299469 1313121384 3887908330 : 2,000,000,000,000
3840840269 5893047555 2627475826 8598006396 3215856883 : 2,699,999,989,950
9256371619 3901058063 3448436720 6294374587 7597230153 : 2,699,999,990,000
8012497961 5892988915 6174704230 3863302264 3931687863 : 2,699,999,990,050
3126006397 8582637253 6739664083 9716870851 0983536511 : 2,699,999,990,100
5628334110 5221005309 8638608325 4364661745 5833914321 : 2,999,999,999,950
9150024270 6285788691 0228572752 8179710957 7137931530 : 3,000,000,000,000
5209957313 0955102183 1080456596 1489168093 0578494464 : 3,999,999,999,950
3638467628 3610607856 5071920145 5255995193 8577295739 : 4,000,000,000,000
2597691971 6538537682 7963082950 0909387733 3987211875 : 4,999,999,999,950
6399906735 0873400641 7497120374 4023826421 9484283852 : 5,000,000,000,000
son

#7 tran thai son

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Đã gửi 05-05-2011 - 19:34

Có một ngày...
Sẽ đến lúc bạn chợt nhận ra sự khác biệt tinh tế giữa việc giữ một bàn tay và sự rang buộc một tâm hồn.
Sẽ đến lúc bạn chợt nhận ra tình yêu không còn là điểm tựa và bên nhau không có nghiã là bình yên.
Sẽ đến lúc bạn chợt nhận ra nụ hôn không phải là lời cam kết và quà tặng khác với lời hứa thật lòng.
Sẽ đến lúc bạn chợt nhận ra không phải mùa nắng nào cũng đẹp.
Và bạn biết chấp nhận thất bại với tư thế ngẩng cao đầu và đôi mắt sang, với sự cao thượng của tuổi trưởng thành chứ không bi lụy, cố chấp của trẻ thơ.
Có ai đi không vấp ngã một đôi lần.
Hãy góp nhặt những mảnh vỡ của mình và bước tiếp từ đây – trên con đường đã chọn của những ngày hôm nay và không trông chờ vào những gì chưa chắc chắn của ngày mai.
Bạn hãy cho đi đừng tiếc nuối, níu kéo. Có ai cho đi mà cảm thấy mất bao giờ?
Và hãy giữ lại những điều tốt đẹp nhất, gieo hạt trồng hoa trên mảnh đất tâm hồn, hơn mòn mỏi đợi chờ ai mang đến.
Và bạn nhận ra rằng mình đã vượt qua.
Cuộc sống sẽ them phần ý nghĩa
Tự do mơ về những điều sẽ đến
Ngước mắt vượt qua khung cửa sổ - ngắm nhìn các vì sao
Cảm nhận thật rằng bạn đang sống
Bản lĩnh, mạnh mẽ và xứng đáng
Dù bất kỳ điều gì xảy ra
Tất cả là bắt đầu , với tất cả những gì vốn có
Chờ đón bạn phía truớc.
Trong ánh mắt lấp lánh niềm tin
Của ngày mới mang đến.
son

#8 tran thai son

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Đã gửi 05-05-2011 - 19:35

Let f be a C3® non-negative function, f(0)=f 0(0)=0, 0 < f00(0).
Let
g(x) =
p
f(x)
f0(x)
!0
for x 6= 0 and g(0) = 0. Show that g is bounded in some neighbourhood of 0.
Does the theorem hold for f 2 C2®?
Solution.
Let c =
1
2
f00(0). We have
g =
(f0)2 􀀀 2ff00
2(f0)2pf
;
where
f(x) = cx2 + O(x3); f0(x) = 2cx + O(x2); f00(x) = 2c + O(x):
Therefore (f0(x))2 = 4c2x2 + O(x3),
2f(x)f00(x) = 4c2x2 + O(x3)
and
2(f0(x))2
q
f(x) = 2(4c2x2 + O(x3))jxj
q
c + O(x):
g is bounded because
2(f0(x))2
p
f(x)
jxj3 􀀀x!!0
8c5=2 6= 0
and f0(x)2 􀀀 2f(x)f00(x) = O(x3).
The theorem does not hold for some C2-functions.
1
Let f(x) = (x + jxj3=2)2 = x2 + 2x2
p
jxj + jxj3, so f is C2. For x > 0,
g(x) =
1
2

1
1 + 3
2px
!0
= 􀀀
1
2 
1
(1 + 3
2px)2 
3
4 
1
px x􀀀!!0􀀀1:
Problem 2.
Let M be an invertible matrix of dimension 2n  2n, represented in
block form as
M =
"
A B
C D
#
and M􀀀1 =
"
E F
G H
#
:
Show that detM: detH = detA.
Solution.
Let I denote the identity n  n matrix. Then
detM: detH = det
"
A B
C D
#
 det
"
I F
0 H
#
= det
"
A 0
C I
#
= det A:
Problem 3.
Show that
1P
n=1
(􀀀1)n􀀀1sin (log n)
n converges if and only if > 0.
Solution.
Set f(t) =
sin (log t)
t . We have
f0(t) = 􀀀
t +1 sin (log t) +
cos (log t)
t +1 :
So jf0(t)j 
1 +
t +1 for > 0. Then from Mean value theorem for some
 2 (0; 1) we get jf(n+1)􀀀f(n)j = jf 0(n+)j 
1 +
n +1 . Since
P 1 +
n +1 < +1
for > 0 and f(n) 􀀀! n!1
0 we get that
1P
n=1
(􀀀1)n􀀀1f(n) =
1P
n=1
(f(2n􀀀1)􀀀f(2n))
converges.
Now we have to prove that
sin (log n)
n does not converge to 0 for  0.
It suces to consider = 0.
We show that an = sin (log n) does not tend to zero. Assume the
contrary. There exist kn 2 N and n 2

􀀀
1
2
;
1
2

for n > e2 such that
log n

=
kn + n. Then janj = sin jnj. Since an ! 0 we get n ! 0.
2
We have kn+1 􀀀 kn =
=
log(n + 1) 􀀀 log n
 􀀀 (n+1 􀀀 n) =
1

log

1 +
1
n

􀀀 (n+1 􀀀 n):
Then jkn+1 􀀀 knj < 1 for all n big enough. Hence there exists n0 so that
kn = kn0 for n > n0. So
log n

= kn0 + n for n > n0. Since n ! 0 we get
contradiction with log n ! 1.
Problem 4.
a) Let the mapping f : Mn ! R from the space
Mn = Rn2
of n  n matrices with real entries to reals be linear, i.e.:
(1) f(A + B) = f(A) + f(B); f(cA) = cf(A)
for any A;B 2 Mn, c 2 R. Prove that there exists a unique matrix C 2 Mn
such that f(A) = tr(AC) for any A 2 Mn. (If A = faijgn
i;j=1 then
tr(A) =
Pn
i=1
aii).
b) Suppose in addition to (1) that
(2) f(A:B) = f(B:A)
for any A;B 2 Mn. Prove that there exists  2 R such that f(A) = :tr(A).
Solution.
a) If we denote by Eij the standard basis ofMn consisting of elementary
matrix (with entry 1 at the place (i; j) and zero elsewhere), then the entries
cij of C can be de ned by cij = f(Eji). b) Denote by L the n2􀀀1-dimensional
linear subspace of Mn consisting of all matrices with zero trace. The elements
Eij with i 6= j and the elements Eii 􀀀Enn, i = 1; : : : ; n􀀀1 form a linear basis
for L. Since
Eij = Eij:Ejj 􀀀 Ejj:Eij ; i 6= j
Eii 􀀀 Enn = Ein:Eni 􀀀 Eni:Ein; i = 1; : : : ; n 􀀀 1;
then the property (2) shows that f is vanishing identically on L. Now, for
any A 2 Mn we have A 􀀀
1
n
tr(A):E 2 L, where E is the identity matrix, and
therefore f(A) =
1
n
f(E):tr(A).
3
Problem 5.
Let X be an arbitrary set, let f be an one-to-one function mapping
X onto itself. Prove that there exist mappings g1; g2 : X ! X such that
f = g1  g2 and g1  g1 = id = g2  g2, where id denotes the identity mapping
on X.
Solution.
Let fn = f  f      f | {z }
n times
, f0 = id, f􀀀n = (f􀀀1)n for every natural
number n. Let T(x) = ff n(x) : n 2 Zg for every x 2 X. The sets T(x) for
di erent x's either coinside or do not intersect. Each of them is mapped by f
onto itself. It is enough to prove the theorem for every such set. Let A = T(x).
If A is nite, then we can think that A is the set of all vertices of a regular
n polygon and that f is rotation by
2
n
. Such rotation can be obtained as a
composition of 2 symmetries mapping the n polygon onto itself (if n is even
then there are axes of symmetry making

n
angle; if n = 2k + 1 then there
are axes making k
2
n
angle). If A is in nite then we can think that A = Z
and f(m) = m + 1 for every m 2 Z. In this case we de ne g1 as a symmetry
relative to
1
2
, g2 as a symmetry relative to 0.
Problem 6.
Let f : [0; 1] ! R be a continuous function. Say that f \crosses the
axis" at x if f(x) = 0 but in any neighbourhood of x there are y; z with
f(y) < 0 and f(z) > 0.
a) Give an example of a continuous function that \crosses the axis"
in niteley often.
b) Can a continuous function \cross the axis" uncountably often?
Justify your answer.
Solution.
a) f(x) = x sin
1
x
.
b) Yes. The Cantor set is given by
C = fx 2 [0; 1) : x =
1X
j=1
bj3􀀀j; bj 2 f0; 2gg:
There is an one-to-one mapping f : [0; 1) ! C. Indeed, for x =
1P
j=1
aj2􀀀j ,
aj 2 f0; 1g we set f(x) =
1P
j=1
(2aj)3􀀀j . Hence C is uncountable.
4
For k = 1; 2; : : : and i = 0; 1; 2; : : : ; 2k􀀀1 􀀀 1 we set
ak;i = 3􀀀k
0
@6
kX􀀀2
j=0
aj3j + 1
1
A; bk;i = 3􀀀k
0
@6
kX􀀀2
j=0
aj3j + 2
1
A;
where i =
kP􀀀2
j=0
aj2j , aj 2 f0; 1g. Then
[0; 1) n C =
1[
k=1
2k􀀀1
[􀀀1
i=0
(ak;i; bk;i);
i.e. the Cantor set consists of all points which have a trinary representation
with 0 and 2 as digits and the points of its compliment have some 1's in their
trinary representation. Thus,
2k􀀀1
􀀀1
i=[0
(ak;i; bk;i) are all points (exept ak;i) which
have 1 on k-th place and 0 or 2 on the j-th (j < k) places.
Noticing that the points with at least one digit equals to 1 are every-
where dence in [0,1] we set
f(x) =
1X
k=1
(􀀀1)kgk(x):
where gk is a piece-wise linear continuous functions with values at the knots
gk

ak;i + bk;i
2

= 2􀀀k, gk(0) = gk(1) = gk(ak;i) = gk(bk;i) = 0,
i = 0; 1; : : : ; 2k􀀀1 􀀀 1.
Then f is continuous and f \crosses the axis" at every point of the
Cantor set.
5
son

#9 tran thai son

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Đã gửi 05-05-2011 - 19:38

Let f be a C3® non-negative function, f(0)=f 0(0)=0, 0 < f00(0).
Let
g(x) =
p
f(x)
f0(x)
!0
for x 6= 0 and g(0) = 0. Show that g is bounded in some neighbourhood of 0.
Does the theorem hold for f 2 C2®?
Solution.
Let c =
1
2
f00(0). We have
g =
(f0)2 􀀀 2ff00
2(f0)2pf
;
where
f(x) = cx2 + O(x3); f0(x) = 2cx + O(x2); f00(x) = 2c + O(x):
Therefore (f0(x))2 = 4c2x2 + O(x3),
2f(x)f00(x) = 4c2x2 + O(x3)
and
2(f0(x))2
q
f(x) = 2(4c2x2 + O(x3))jxj
q
c + O(x):
g is bounded because
2(f0(x))2
p
f(x)
jxj3 􀀀x!!0
8c5=2 6= 0
and f0(x)2 􀀀 2f(x)f00(x) = O(x3).
The theorem does not hold for some C2-functions.
1
Let f(x) = (x + jxj3=2)2 = x2 + 2x2
p
jxj + jxj3, so f is C2. For x > 0,
g(x) =
1
2

1
1 + 3
2px
!0
= 􀀀
1
2 
1
(1 + 3
2px)2 
3
4 
1
px x􀀀!!0􀀀1:
Problem 2.
Let M be an invertible matrix of dimension 2n  2n, represented in
block form as
M =
"
A B
C D
#
and M􀀀1 =
"
E F
G H
#
:
Show that detM: detH = detA.
Solution.
Let I denote the identity n  n matrix. Then
detM: detH = det
"
A B
C D
#
 det
"
I F
0 H
#
= det
"
A 0
C I
#
= det A:
Problem 3.
Show that
1P
n=1
(􀀀1)n􀀀1sin (log n)
n converges if and only if > 0.
Solution.
Set f(t) =
sin (log t)
t . We have
f0(t) = 􀀀
t +1 sin (log t) +
cos (log t)
t +1 :
So jf0(t)j 
1 +
t +1 for > 0. Then from Mean value theorem for some
 2 (0; 1) we get jf(n+1)􀀀f(n)j = jf 0(n+)j 
1 +
n +1 . Since
P 1 +
n +1 < +1
for > 0 and f(n) 􀀀! n!1
0 we get that
1P
n=1
(􀀀1)n􀀀1f(n) =
1P
n=1
(f(2n􀀀1)􀀀f(2n))
converges.
Now we have to prove that
sin (log n)
n does not converge to 0 for  0.
It suces to consider = 0.
We show that an = sin (log n) does not tend to zero. Assume the
contrary. There exist kn 2 N and n 2

􀀀
1
2
;
1
2

for n > e2 such that
log n

=
kn + n. Then janj = sin jnj. Since an ! 0 we get n ! 0.
2
We have kn+1 􀀀 kn =
=
log(n + 1) 􀀀 log n
 􀀀 (n+1 􀀀 n) =
1

log

1 +
1
n

􀀀 (n+1 􀀀 n):
Then jkn+1 􀀀 knj < 1 for all n big enough. Hence there exists n0 so that
kn = kn0 for n > n0. So
log n

= kn0 + n for n > n0. Since n ! 0 we get
contradiction with log n ! 1.
Problem 4.
a) Let the mapping f : Mn ! R from the space
Mn = Rn2
of n  n matrices with real entries to reals be linear, i.e.:
(1) f(A + B) = f(A) + f(B); f(cA) = cf(A)
for any A;B 2 Mn, c 2 R. Prove that there exists a unique matrix C 2 Mn
such that f(A) = tr(AC) for any A 2 Mn. (If A = faijgn
i;j=1 then
tr(A) =
Pn
i=1
aii).
b) Suppose in addition to (1) that
(2) f(A:B) = f(B:A)
for any A;B 2 Mn. Prove that there exists  2 R such that f(A) = :tr(A).
Solution.
a) If we denote by Eij the standard basis ofMn consisting of elementary
matrix (with entry 1 at the place (i; j) and zero elsewhere), then the entries
cij of C can be de ned by cij = f(Eji). b) Denote by L the n2􀀀1-dimensional
linear subspace of Mn consisting of all matrices with zero trace. The elements
Eij with i 6= j and the elements Eii 􀀀Enn, i = 1; : : : ; n􀀀1 form a linear basis
for L. Since
Eij = Eij:Ejj 􀀀 Ejj:Eij ; i 6= j
Eii 􀀀 Enn = Ein:Eni 􀀀 Eni:Ein; i = 1; : : : ; n 􀀀 1;
then the property (2) shows that f is vanishing identically on L. Now, for
any A 2 Mn we have A 􀀀
1
n
tr(A):E 2 L, where E is the identity matrix, and
therefore f(A) =
1
n
f(E):tr(A).
3
Problem 5.
Let X be an arbitrary set, let f be an one-to-one function mapping
X onto itself. Prove that there exist mappings g1; g2 : X ! X such that
f = g1  g2 and g1  g1 = id = g2  g2, where id denotes the identity mapping
on X.
Solution.
Let fn = f  f      f | {z }
n times
, f0 = id, f􀀀n = (f􀀀1)n for every natural
number n. Let T(x) = ff n(x) : n 2 Zg for every x 2 X. The sets T(x) for
di erent x's either coinside or do not intersect. Each of them is mapped by f
onto itself. It is enough to prove the theorem for every such set. Let A = T(x).
If A is nite, then we can think that A is the set of all vertices of a regular
n polygon and that f is rotation by
2
n
. Such rotation can be obtained as a
composition of 2 symmetries mapping the n polygon onto itself (if n is even
then there are axes of symmetry making

n
angle; if n = 2k + 1 then there
are axes making k
2
n
angle). If A is in nite then we can think that A = Z
and f(m) = m + 1 for every m 2 Z. In this case we de ne g1 as a symmetry
relative to
1
2
, g2 as a symmetry relative to 0.
Problem 6.
Let f : [0; 1] ! R be a continuous function. Say that f \crosses the
axis" at x if f(x) = 0 but in any neighbourhood of x there are y; z with
f(y) < 0 and f(z) > 0.
a) Give an example of a continuous function that \crosses the axis"
in niteley often.
b) Can a continuous function \cross the axis" uncountably often?
Justify your answer.
Solution.
a) f(x) = x sin
1
x
.
b) Yes. The Cantor set is given by
C = fx 2 [0; 1) : x =
1X
j=1
bj3􀀀j; bj 2 f0; 2gg:
There is an one-to-one mapping f : [0; 1) ! C. Indeed, for x =
1P
j=1
aj2􀀀j ,
aj 2 f0; 1g we set f(x) =
1P
j=1
(2aj)3􀀀j . Hence C is uncountable.
4
For k = 1; 2; : : : and i = 0; 1; 2; : : : ; 2k􀀀1 􀀀 1 we set
ak;i = 3􀀀k
0
@6
kX􀀀2
j=0
aj3j + 1
1
A; bk;i = 3􀀀k
0
@6
kX􀀀2
j=0
aj3j + 2
1
A;
where i =
kP􀀀2
j=0
aj2j , aj 2 f0; 1g. Then
[0; 1) n C =
1[
k=1
2k􀀀1
[􀀀1
i=0
(ak;i; bk;i);
i.e. the Cantor set consists of all points which have a trinary representation
with 0 and 2 as digits and the points of its compliment have some 1's in their
trinary representation. Thus,
2k􀀀1
􀀀1
i=[0
(ak;i; bk;i) are all points (exept ak;i) which
have 1 on k-th place and 0 or 2 on the j-th (j < k) places.
Noticing that the points with at least one digit equals to 1 are every-
where dence in [0,1] we set
f(x) =
1X
k=1
(􀀀1)kgk(x):
where gk is a piece-wise linear continuous functions with values at the knots
gk

ak;i + bk;i
2

= 2􀀀k, gk(0) = gk(1) = gk(ak;i) = gk(bk;i) = 0,
i = 0; 1; : : : ; 2k􀀀1 􀀀 1.
Then f is continuous and f \crosses the axis" at every point of the
Cantor set.
5
son




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