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Miklós Schweitzer
1961 Miklós Schweitzer
1961 Miklós Schweitzer
Part of
Miklós Schweitzer
Subcontests
(10)
10
1
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Miklós Schweitzer 1961- Problem 10
10. Given a straight line
g
g
g
in the plane and a point
O
O
O
on
g
g
g
. Construct, without making use of the Parallel Axiom, the half-line perpendicular to
g
g
g
at the point
O
O
O
and lying in one of the half-planes defined by
g
g
g
, under the following restrictions: The construction must be effected by use of a ruler and of a length standard (i.e. an etalon-segment) only; moreover, all lines and points of the construction must lie in the chosen half-plane. (G. 20)
9
1
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Miklós Schweitzer 1961- Problem 9
9. Spin a regular coin repeatedly until the heads and tails turned up both reach the number
k
k
k
(
k
=
1
,
2
,
…
k= 1, 2, \dots
k
=
1
,
2
,
…
); denote by
v
k
v_k
v
k
the number of the necessary throws. Find the distribution of the random variable
v
k
v_k
v
k
and the limit-distribution of the random variable
v
k
−
2
k
2
k
\frac {v_k -2k}{\sqrt {2k}}
2
k
v
k
−
2
k
as
k
→
∞
k \to \infty
k
→
∞
. (P. 10)
8
1
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Miklós Schweitzer 1961- Problem 8
8. Let
f
(
x
)
f(x)
f
(
x
)
be a convex function defined on the interval
[
0
,
1
2
]
[0, \frac {1}{2}]
[
0
,
2
1
]
with
f
(
0
)
=
0
f(0)=0
f
(
0
)
=
0
and
f
(
1
2
)
=
1
f(\frac{1}{2})=1
f
(
2
1
)
=
1
; Let further
f
(
x
)
f(x)
f
(
x
)
be differentiable in
(
0
,
1
2
)
(0, \frac {1}{2})
(
0
,
2
1
)
, and differentiable at
0
0
0
and
1
2
\frac{1}{2}
2
1
from the right and from the left, respectively. Finally, let
f
′
(
0
)
>
1
f'(0)>1
f
′
(
0
)
>
1
. Extend
f
(
x
)
f(x)
f
(
x
)
to
[
0.1
]
[0.1]
[
0.1
]
in the following manner: let
f
(
x
)
=
f
(
1
−
x
)
f(x)= f(1-x)
f
(
x
)
=
f
(
1
−
x
)
if
x
∈
(
1
2
,
1
]
x \in (\frac {1} {2}, 1]
x
∈
(
2
1
,
1
]
. Show that the set of the points
x
x
x
for shich the terms of the sequence
x
n
+
1
=
f
(
x
n
)
x_{n+1}=f(x_n)
x
n
+
1
=
f
(
x
n
)
(
x
0
=
x
;
n
=
0
,
1
,
2
,
…
x_0=x; n = 0, 1, 2, \dots
x
0
=
x
;
n
=
0
,
1
,
2
,
…
) are not all different is everywhere dense in
[
0
,
1
]
[0,1]
[
0
,
1
]
; (R. 10)
7
1
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Miklós Schweitzer 1961- Problem 7
7. For the differential equation
∂
2
u
∂
x
2
+
∂
2
u
∂
y
2
=
2
∂
2
u
∂
x
∂
y
\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}= 2\frac{\partial^2 u}{\partial x \partial y}
∂
x
2
∂
2
u
+
∂
y
2
∂
2
u
=
2
∂
x
∂
y
∂
2
u
find all solutions of the form
u
(
x
,
y
)
=
f
(
x
)
g
(
y
)
u(x,y)=f(x)g(y)
u
(
x
,
y
)
=
f
(
x
)
g
(
y
)
. (R. 14)
6
1
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Miklós Schweitzer 1961- Problem 6
6. Consider a sequence
{
a
n
}
n
=
1
∞
\{ a_n \}_{n=1}^{\infty}
{
a
n
}
n
=
1
∞
such that, for any convergent subsequence
{
a
n
k
}
\{ a_{n_k} \}
{
a
n
k
}
of
{
a
n
}
\{a_n\}
{
a
n
}
, the sequence
{
a
n
k
+
1
}
\{ a_{n_k +1} \}
{
a
n
k
+
1
}
also is convergent and has the same limit as
{
a
n
k
}
\{ a_{n_k}\}
{
a
n
k
}
. Prove that the sequence
{
a
n
}
\{ a_n \}
{
a
n
}
is either convergent of has infinitely many accumulation points the set of which is dense in itself. Give an example for the second case. (A sequence
x
n
→
∞
x_n \to \infty
x
n
→
∞
or
−
∞
-\infty
−
∞
is considered to be convergente, too) (S. 13)
4
1
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Miklós Schweitzer 1961- Problem 4
4. Let
f
(
x
)
f(x)
f
(
x
)
be a real- or complex-value integrable function on
(
0
,
1
)
(0,1)
(
0
,
1
)
with
∣
f
(
x
)
∣
≤
1
\mid f(x) \mid \leq 1
∣
f
(
x
)
∣≤
1
. Set
c
k
=
∫
0
1
f
(
x
)
e
−
2
π
i
k
x
d
x
c_k = \int_0^1 f(x) e^{-2 \pi i k x} dx
c
k
=
∫
0
1
f
(
x
)
e
−
2
πik
x
d
x
and construct the following matrices of order
n
n
n
:
T
=
(
t
p
q
)
p
,
q
=
0
n
−
1
,
T
∗
=
(
t
p
q
∗
)
p
,
q
=
0
n
−
1
T= (t_{pq})_{p,q=0}^{n-1}, T^{*}= (t_{pq}^{*})_{p,q =0}^{n-1}
T
=
(
t
pq
)
p
,
q
=
0
n
−
1
,
T
∗
=
(
t
pq
∗
)
p
,
q
=
0
n
−
1
where
t
p
q
=
c
q
−
p
,
t
∗
=
c
p
−
q
‾
t_{pq}= c_{q-p}, t^{*}= \overline {c_{p-q}}
t
pq
=
c
q
−
p
,
t
∗
=
c
p
−
q
. Further, consider the following hyper-matrix of order
m
m
m
:
S
=
[
E
T
T
2
…
T
m
−
2
T
m
−
1
T
∗
E
T
…
T
m
−
3
T
m
−
2
T
∗
2
T
∗
E
…
T
m
−
3
T
m
−
2
…
…
…
…
…
…
T
∗
m
−
1
T
∗
m
−
2
T
∗
m
−
3
…
T
∗
E
]
S= \begin{bmatrix} E & T & T^2 & \dots & T^{m-2} & T^{m-1} \\ T^{*} & E & T & \dots & T^{m-3} & T^{m-2} \\ T^{*2} & T^{*} & E & \dots & T^{m-3} & T^{m-2} \\ \dots & \dots & \dots & \dots & \dots & \dots \\ T^{*m-1} & T^{*m-2} & T^{*m-3} & \dots & T^{*} & E \end{bmatrix}
S
=
E
T
∗
T
∗
2
…
T
∗
m
−
1
T
E
T
∗
…
T
∗
m
−
2
T
2
T
E
…
T
∗
m
−
3
…
…
…
…
…
T
m
−
2
T
m
−
3
T
m
−
3
…
T
∗
T
m
−
1
T
m
−
2
T
m
−
2
…
E
(
S
S
S
is a matrix of order
m
n
mn
mn
in the ordinary sense; E denotes the unit matrix of order
n
n
n
). Show that for any pair
(
m
,
n
)
(m , n)
(
m
,
n
)
of positive integers,
S
S
S
has only non-negative real eigenvalues. (R. 19)
3
1
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Miklós Schweitzer 1961- Problem 3
3. Let
f
(
x
)
=
x
n
+
a
1
x
(
n
−
1
)
+
⋯
+
a
n
f(x)= x^n +a_1 x^(n-1)+ \dots + a_n
f
(
x
)
=
x
n
+
a
1
x
(
n
−
1
)
+
⋯
+
a
n
(
n
≥
1
n\geq 1
n
≥
1
) be an irreducible polynomial over the field
K
K
K
. Show that every non-zero matrix commuting with the matrix
[
0
1
0
…
0
0
0
0
1
…
0
0
…
…
…
…
…
…
0
0
0
…
0
1
−
a
n
−
a
n
−
1
−
a
n
−
2
…
−
a
2
−
a
1
<
/
b
r
>
]
\begin{bmatrix} 0 & 1 & 0 & \dots & 0 & 0 \\ 0 & 0 & 1 & \dots & 0 & 0 \\ \dots & \dots & \dots & \dots & \dots & \dots \\ 0 & 0 & 0 & \dots & 0 & 1 \\ -a_n & -a_{n-1} & -a_{n-2} & \dots & -a_2 & -a_1 </br>\end{bmatrix}
0
0
…
0
−
a
n
1
0
…
0
−
a
n
−
1
0
1
…
0
−
a
n
−
2
…
…
…
…
…
0
0
…
0
−
a
2
0
0
…
1
−
a
1
<
/
b
r
>
is invertible. (A. 4)
5
1
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Miklós Schweitzer 1961- Problem 5
5. Determine the functions
G
G
G
defined on the set of all non-zero real numbers the values of which are regular matrices of order
2
2
2
, and the functions
f
f
f
mapping the two-dimensional real vector space
E
2
E_2
E
2
into itself, such that for any vector
y
∈
E
2
y \in E_2
y
∈
E
2
and for any regular matrix
X
X
X
of order
2
2
2
,
f
(
X
y
)
=
G
(
d
e
t
X
)
X
f
(
y
)
f(X_y)= G(det X)Xf(y)
f
(
X
y
)
=
G
(
d
e
tX
)
X
f
(
y
)
(
d
e
t
X
det X
d
e
tX
denotes the determinant of
X
X
X
).(A. 5)
2
1
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Miklós Schweitzer 1961- Problem 2
2. Show that a ring
R
R
R
has a unit element if and only if any
R
R
R
-module
G
G
G
can be written as a direct sum of
R
G
RG
RG
and of the trivial submodule of
G
G
G
. (An
R
R
R
-module is a linear space with
R
R
R
as its scalar domain.
R
G
RG
RG
denotes the submodule generated by the elements of the form
r
g
rg
r
g
(
r
∈
R
,
g
∈
G
r \in R, g \in G
r
∈
R
,
g
∈
G
). The trivial submodule of
G
G
G
consists of the elements
g
g
g
of
G
G
G
for which
r
g
=
0
rg=0
r
g
=
0
holds for every
r
∈
R
r \in R
r
∈
R
.) (A. 20)
1
1
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Miklós Schweitzer 1961- Problem 1
1. Let
a
a
a
(
≠
e
\neq e
=
e
, the unit element) be an element of finite order of a group
G
G
G
and let
t
t
t
(
≥
2
\geq 2
≥
2
) be a positive integer. Show: if the complex
A
=
{
e
,
a
,
a
2
,
…
,
a
t
−
1
}
A= \{ e,a,a^2, \dots , a^{t-1} \}
A
=
{
e
,
a
,
a
2
,
…
,
a
t
−
1
}
is not a group, then for every positive integer
k
k
k
(
2
≤
k
≤
t
2 \leq k \leq t
2
≤
k
≤
t
) the complex
B
=
{
e
.
a
k
,
a
2
k
,
…
,
a
(
t
−
1
)
k
}
B= \{ e. a^k, a^{2k}, \dots , a^{(t-1)k} \}
B
=
{
e
.
a
k
,
a
2
k
,
…
,
a
(
t
−
1
)
k
}
differs from
A
A
A
. (A. 16)