... B
j,j
and C
n,n
are relatively prime, we can choose integers C
j,n
and B
j,n
such that this equation is
satisfied. Doing this step by step for all j = n − 1, n − 2, . . . , 1, we finally get B and C such ... 3.
(20 points)
Solution. We note that the problem is trivial if A
j
= λI for some j, so suppose this is not the case.
Consider then first the situation where some A
j
, say A
3
, has two distinct ... minimum of f on [x, y]. Then f([x, y]) = [f(b), f (a)]; hence
y − x = f (a) − f(b) ≤ |a − b| ≤ y − x
This implies {a, b} = {x, y}, and therefore f is a monotone function. Suppose f is increasing. Then
f(x)...
... using only a translation or a rotation. Does this imply that
f(x) = ax + b for some real numbers a and b ?
Solution. No. The function f(x) = e
x
also has this property since ce
x
= e
x+log c
.
Problem ...
≥|ξ−b|
≥ m|ξ − b|
m−1
≥
d
m−1
m
m−2
we get
f(ξ) + f
′
(ξ) ≥ f(b) + ε.
Together with (2) this shows (3). This finishes the proof of Lemma 2.
b
a
ξ
f
′
f
f + f
′
4
For every element X = {x
1
, x
2
, ... 0,
since A
2
y
= 0. So, A
¯
X
annihilates the span of all the v
Y
with X Y . This implies that v
X
does not
lie in this span, because A
¯
X
v
X
= v
{1,2, ,k}
= 0. Therefore, the vectors v
X
(with...
... required properties. For an arbitrary rational q, consider the
function g
q
(x) = f(x+q)−f(x). This is a continuous function which attains only rational values, therefore
g
q
is constant.
Set ... b))
we can assume that P (X) = 0.
Now we are going to prove that P (X
k
) = 0 for all k ≥ 1. Suppose this is true for all k < n. We know
that P (X
n
+ e) = 0 for e = −P (X
n
). From the induction ... . 0 b
2k−2,2k−1
0 b
2k−1,2
0 . . . b
2k−1,2k−2
0
.
Note that every matrix of this form has determinant zero, because it has k columns spanning a vector
space of dimension at...
... VnMath.Com
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Thi lớp 10
Olympic
Giáo án
các ... nghịch và vì vậy
I = (A + B)A
−1
(I − C) = A
−1
(A + B)(I − C),
4
Dịch Vụ Toán Học
Đáp án ĐềthiOlympicToánSinh viên
năm 2010
Đại số và Giải tích
WWW.VNMATH.COM
... HỘI TOÁN HỌC VIỆT NAM BỘ GIÁO DỤC VÀ ĐÀO TẠO
ĐÁP ÁN OLYMPICTOÁNSINHVIÊN LẦN THỨ XVIII
Môn : Đại số
Câu 1. Cho A, B là các ma trận...
... 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
= ... y) such that f(c) = 0 and f (x) > 0 for x ∈ (c, y]. For x ∈ (c, y] we
have |f
(x)| ≤ λf(x). This implies that the function g(x) = ln f (x) − λx is
not increasing in (c, y] because of g
(x) ... The number of indices (i, j)
for which a
ii
= a
jj
= c
m
for some m = 1, 2, . . . , k is d
2
m
. This gives the
desired result.
Problem 5. (18 points)
Let x
1
, x
2
, . . . , x
k
be vectors of...
... −
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|
p
+ |y|
p
= 2}.
Since D
δ
is compact it is enough to show that f is continuous on D
δ
.
For this we show that the denominator of f is different from zero. Assume
the contrary. Then |x + y|...
... 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, ... and T
is its reflexion about the x-axis, then C(E) = 8 > K(E).
Remarks: All distances used in this problem are Euclidian. Diameter
of a set E is diam(E) = sup{dist(x, y) : x, y ∈ E}. Contraction ... < ε. So the sequence
cannot come into the interval (x − δ, x + δ), but also cannot jump over this
interval. Then all cluster points have to be at most x − δ (a contradiction
with L being a...
... 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
= ... number and
det ω(BA − AB) = ω
n
det(BA − AB) and det(BA − AB) = 0, then ω
n
is a
real number. This is possible only when n is divisible by 3.
2
For each k we write
θ
k
=
p
k
q
k
as a fraction ... 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
FOURTH INTERNATIONAL...
... 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
....
... 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 ... that
x is uniquely determined by the triple u, v, w; since the set of such triples is countable, this will finish
the proof.
To prove the claim, suppose, that from some x
∈ M we arrived to the ... induction that
(1) f
n
(x) =
1
2
− 2
2
n
−1
x −
1
2
2
n
holds for n = 1, 2, . . . . For n = 1 this is true, since f(x) = 2x(1 − x) =
1
2
− 2(x −
1
2
)
2
. If (1) holds for
some n = k, then we...
... 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 ... contains at most one marked point, delete it. This decreases n + k by 1 and the
number of the marked points by at most 1, so the condition remains true. Repeat this step until each row
and column contains ... was
bcdedef = xyzyxzxyzxyxzxzyxyzyzxyzxyxzxzyxyzyzyxyxzyxzxzxzxyxyzxyz,
which is of length 46. This is not the shortest way: reducing the length of word a can be done for example
by the following...
... 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 ... c
0
I + CA. It is well-known that the characteristic
polynomials of AC and CA are the same; denote this polynomial by f(x). Then the
characteristic polynomials of matrices q(e
AB
) and q(e
BA
) are ... = 0 is x
m
,
so the same holds for the matrix q(e
BA
). By the theorem of Cayley and Hamilton, this
implies that
q(e
BA
)
m
=
p(e
BA
)
km
= 0. Thus the matrix q(e
BA
) is nilpotent, too.
4
...
... 0, we have
lim
x→∞
f(x) ·
g(x)
A
g(x)
A+1
=
B
A + 1
.
By l’Hospital’s rule this implies
lim
x→∞
f(x)
g(x)
= lim
x→∞
f(x) ·
g(x)
A
g(x)
A+1
=
B
A + 1
.
4
8
th
IMC ... 4/x
2
. It is a matter of simple manipulation to prove that
2f(x) > x for all x ∈ (0, 2), this implies that the sequence (2
n
a
n
) is strictly
1
increasing. The inequality 2g(x) < x ... 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...
... it converges to zero. Therefore a
n
≤
4c
2
l
n+1
for n ≥ 2
l+1
,
meaning that c
2
l+1
≤ 4c
2
l
. This implies that a sequence ((4c
l
)
2
−l
)
l≥0
is non-
increasing and therefore bounded from above ... 1
n−1
k=0
n−k
n
+
k+1
n
n−1
k
+
2
n+1
n + 1
=
2
n
n
n−1
k=0
1
n−1
k
+
2
n+1
n + 1
= x
n
+
2
n+1
n + 1
.
This implies (2) for n + 1.
Problem 4. Let f : [a, b] → [a, b] be a continuous function and let...
... b
n
is an integer.
Solution. We prove by induction on n that a
n
/e and b
n
e are integers, we
prove this for n = 0 as well. (For n = 0, the term 0
0
in the definition of the
sequences must be replaced ... subsets of slice OBC without
common interior point, and they do not cover the whole slice OBC; this
implies the statement. In cases (b) and (c) where at least one of the signs
is negative, projections ... x
2
.
Prove that
∀x
1
, x
2
∈ R
n
∇f(x
1
) − ∇f(x
2
)
2
≤ L∇f(x
1
) − ∇f(x
2
), x
1
− x
2
. (1)
In this formula a, b denotes the scalar product of the vectors a and b.
Solution. Let g(x) = f(x)−f(x
1
)−∇f...