Tài liệu Đề thi Olympic sinh viên thế giới năm 2001 ppt
Tài liệu Đề thi Olympic sinh viên thế giới năm 2001 ppt
... 8
th
IMC 2001
July 19 - July 25
Prague, Czech Republic
First day
Problem 1.
Let n be a positive integer. ... integers x and y,
b = b
xr+ys
= (b
r
)
x
(b
s
)
y
= e.
It follows similarly that a = e as well.
2. This is not true. Let a = (123) and b = (34567) be cycles of the permu-
tation group S
7
of order ... lim
t1
(1 − t)
∞
n=1
t
n
1 + t
n
, where t 1 means that t ap-
proaches 1 from...
Tài liệu Đề thi Olympic sinh viên thế giới năm 1994 pptx
... at least
one is negative. Hence we have at least two non-zero elements in every
column of A
−1
. This proves part a). For part b) all b
ij
are zero except
b
1,1
= 2, b
n,n
= (−1)
n
, b
i,i+1
= ... −
bk/n
b(k−1)/n
f(x)dx
b
0
g(x)dx + O(ω(f, b/n)g
1
)
=
1
b
b
0
f(x)dx
b
0
g(x)dx + O(ω(f, b/n)g
1
).
This proves a). For b) we set b = π, f(x) = sin x, g(x) = (1 + 3cos
2
x)
−1
.
From...
Tài liệu Đề thi Olympic sinh viên thế giới năm 1998 ppt
... contains
only powers of ¯y = yK, i.e., S
4
/K is cyclic. It is easy to see that this factor-group is not comutative
(something more this group is not isomorphic to S
3
).
3) n = 5
a) If x = (12), then for ... vector space.
Solution First choose a basis {v
1
, v
2
, v
3
} of U
1
. It is possible to extend this basis with vectors v
4
,v
5
and
v
6
to get a basis of U
2
. In the same way we can e...
Tài liệu Đề thi Olympic sinh viên thế giới năm 1995 docx
... −
1
3
.
From the hypotheses we have
1
0
1
x
f(t)dtdx ≥
1
0
1 − x
2
2
dx or
1
0
tf(t)dt ≥
1
3
. This completes the proof.
Problem 3. (15 points)
Let f be twice continuously differentiable on (0, ... x tends to 0+ we obtain
−x
0
≤ lim inf
x→0+
f(x)
f
(x)
≤ lim sup
x→0+
f(x)
f
(x)
≤ 0.
Since this happens for all x
0
∈ (0, r) we deduce that lim
x→0+
f(x)
f
(x)
exists and
lim
x→0+
f(x...
Tài liệu Đề thi Olympic sinh viên thế giới năm 1996 doc
... consider k ≥ 2.
For any m we have
(2)
cosh θ = cosh ((m + 1)θ − mθ) =
= cosh (m + 1)θ.cosh mθ − sinh (m + 1)θ .sinh mθ
= cosh (m + 1)θ.cosh mθ −
cosh
2
(m + 1)θ − 1.
√
cosh
2
mθ − 1
Set cosh kθ = a, ... the first and the third integral tend to −
1
f
(0)
as n → ∞, hence
so does the second.
Also n
1
A
(f(x))
n
dx ≤ n(f(A))
n
−→
n→∞
0 (f (A) < 1). We get L = −
1
f
(0)
in this case....
Tài liệu Đề thi Olympic sinh viên thế giới năm 1997 ngày 1 pptx
... suffices
to show that this sequence is strictly decreasing. Now,
p
k
− q
k
− (p
k−1
− q
k−1
) = n
k
p
k−1
− (n
k
+ 1)q
k−1
− p
k−1
+ q
k−1
= (n
k
− 1)p
k−1
− n
k
q
k−1
and this is negative because
p
k−1
q
k−1
= ... is the rational number
p
q
. Our aim is to show
that for some m,
θ
m−1
=
n
m
n
m
− 1
.
Suppose this is not the case, so that for every m,
(3) θ
m−1
<
n
m
n
m
− 1
.
4
For each k we...
Tài liệu Đề thi Olympic sinh viên thế giới năm 1997 ngày 2 docx
... then there
are axes making k
2π
n
angle). If A is infinite then we can think that A = Z
and f (m) = m + 1 for every m ∈ Z. In this case we define g
1
as a symmetry
relative to
1
2
, g
2
as a symmetry ... It is enough to prove the theorem for every such set. Let A = T (x).
If A is finite, then we can think that A is the set of all vertices of a regular
n polygon and that f is rotation by
2π
n
.
Tài liệu Đề thi Olympic sinh viên thế giới năm 1999 doc
... (1)
From this recursion we can compute the probabilities for small values of n and can conjecture that p
(r)
n
=
1
5
+
4
5·6
n
if n ≡ r (mod )5 and p
(r)
n
=
1
5
−
1
5·6
n
otherwise. From (1), this ... opposite inequality holds ∀m 1. This
contradiction shows that g is a constant, i.e. f(x) = Cx, C > 0.
Conversely, it is easy to check that the functions of this type verify the conditions...
Tài liệu Đề thi Olympic sinh viên thế giới năm 2000 doc
... e, this yields e + 2ef = 0, and similarly f + 2ef = 0,
so that f = −2ef = e, hence e = f = g by symmetry. Hence, finaly, 3e = e + f + g = 0,
i.e. e = f = g = 0.
For part (i) just omit some of this.
Problem ... into n smaller cubes may be refined to
give a dissection into n + (a
d
− 1) cubes, for any a ≥ 1. This refinement is achieved by
picking an arbitrary cube in the dissection, and cutting it...
Tài liệu Đề thi Olympic sinh viên thế giới năm 2002 ngày 1 doc
... are pairwise disjoint,
so there are at most countably many of the third type.
b. Let f be such a map. Then for each value y of this map there is an x
0
such
that y = f(x) and f
(x) = 0, because ... δ
n
centered at p
n
such that there are there are infinitely many k’s such
that p
k
/∈
n
j=0
I
j
, this can be done by induction. Let n
0
= 0 and n
m+1
be the
smallest integer k > n
m
such th...