International Competition in Mathematics for
Universtiy Students
in
Plovdiv, Bulgaria
1995
1
PROBLEMS AND SOLUTIONS
First day
Problem 1. (10 points)
Let X be a nonsingular matrix with columns X
1
, X
2
, . . . , X
n
. Let Y be a
matrix with columns X
2
, X
3
, . . . , X
n
, 0. Show that the matrices A = Y X
−1
and B = X
−1
Y have rank n −1 and have only 0’s for eigenvalues.
Solution. Let J = (a
ij
) be the n × n matrix where a
ij

f
2
(t)dt ≥
1
3
.
Solution. From the inequality
0 ≤

1
0
(f(x) − x)
2
dx =

1
0
f
2
(x)dx − 2

1
0
xf(x)dx +

1
0
x
2
dx

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, +∞) such that
lim
x→0+
f

(x) = −∞ and lim
x→0+
f

(x) = +∞. Show that
lim
x→0+
f(x)

). Taking into account that f

is increasing, f

(x) <
f

(ξ) < 0, we get
x − x
0
<
f

(ξ)
f

(x)
(x − x
0
) =
f(x) − f(x
0
)
f

(x)
< 0.
Taking limits as x tends to 0+ we obtain
−x
0

Let F : (1, ∞) → R be the function defined by
F (x) :=

x
2
x
dt
ln t
.
Show that F is one-to-one (i.e. injective) and find the range (i.e. set of
values) of F .
Solution. From the definition we have
F

(x) =
x − 1
ln x
, x > 1.
Therefore F

(x) > 0 for x ∈ (1, ∞). Thus F is strictly increasing and hence
one-to-one. Since
F (x) ≥ (x
2
− x) min

1
ln t
: x ≤ t ≤ x
2

2
ln 2
and similarly F (x) > x ln 2. Thus F (1+) = ln 2.
Problem 5. (20 points)
Let A and B be real n × n matrices. Assume that there exist n + 1
different real numbers t
1
, t
2
, . . . , t
n+1
such that the matrices
C
i
= A + t
i
B, i = 1, 2, . . . , n + 1,
are nilpotent (i.e. C
n
i
= 0).
Show that both A and B are nilpotent.
Solution. We have that
(A + tB)
n
= A
n
+ tP
1
+ t

2
, . . . , P
n−1
, B
n
. Then the polynomial
bt
n
+ p
n−1
t
n−1
+ ··· + p
2
t
2
+ p
1
t + a
has at least n + 1 roots t
1
, t
2
, . . . , t
n+1
. Hence all its coefficients vanish.
Therefore A
n
= 0, B
n

2
≤ K
p,δ
for every (x, y) ∈ D
δ
= {(x, y) : |x −y| ≥ δ, |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| = 2, and




x + y
2




p
= 1. Since p > 1, the function
g(t) = |t|




p
<
|x|
p
+ |y|
p
2
= 1 =




x + y
2




p
. We get a contradiction.
If x and y have different signs then (x, y) ∈ D
δ
for all 0 < δ < 1 because
then |x −y| ≥ max{|x|, |y|} ≥ 1 > δ. So we may further assume without loss
of generality that x > 0, y > 0 and x
p
+ y

2
)

1/p
= 1 +
1
p

−pt −
p(p − 1)
2
t
2
+ o(t
2
)

+
1
2p

1
p
− 1

(−pt + o(t))
2
+ o(t
2
)

and
4−(x+y)
2
=4−(2−(p−1)t
2
+o(t
2
))
2
=4−4+4(p−1)t
2
+o(t
2
)=4(p−1)t
2
+o(t
2
).
So there exists δ
p
> 0 such that if |t| < δ
p
we have (x−y)
2
< 5t
2
, 4−(x+y)
2
>
3(p − 1)t

p
and x
p
+ y
p
= 2. Indeed, since x
p
+ y
p
= 2 we have
that max{x, y} ≥ 1. So let x − 1 ≥ δ
p
. Since

x + y
2

p
≤
x
p
+ y
p
2
= 1 we
get x + y ≤ 2. Then x −y ≥ 2(x − 1) ≥ 2δ
p
.
Second day
Problem 1. (10 points)

with u
i
= δ
ik
we get a
kk
= 0. If we use (1) with u
i
= δ
ik
+ δ
im
we get
a
kk
+ a
km
+ a
mk
+ a
mm
= 0
and hence a
km
= −a
mk
.
b) Set v
1
= −a

− v
2
u
1
)

= v ×u.
Problem 2. (15 points)
Let {b
n
}
∞
n=0
be a sequence of positive real numbers such that b
0
= 1,
b
n
= 2 +

b
n−1
− 2

1 +

b
n−1
. Calculate
∞

√
a
n−1
,
so a
n
= 2
2
−n
. Then
N

n=1
b
n
2
n
=
N

n=1
(a
n
− 1)
2
2
n
=
N


− (a
N
− 1)2
N+1
= 2 − 2
2
2
−N
− 1
2
−N
.
Put x = 2
−N
. Then x → 0 as N → ∞ and so
∞

n=1
b
n
2
N
= lim
N→∞

2 − 2
2
2
−N
− 1

| = 1, where the
complex numbers α
1
, α
2
, . . . , α
n
may coincide.
We have

P (z) ≡ 2zP

(z) − nP (z) = (z + α
1
)(z − α
2
) . . . (z − α
n
) +
+(z −α
1
)(z + α
2
) . . . (z − α
n
) + ··· + (z − α
1
)(z − α
2
) . . . (z + α

n

k=1
|z|
2
− 1
|z −α
k
|
2
. From |z| = 1
it follows that Re

P (z)
P (z)
= 0. Hence

P (z) = 0 implies |z| = 1.
7
Problem 4. (15 points)
a) Prove that for every ε > 0 there is a positive integer n and real
numbers λ
1
, . . . , λ
n
such that
max
x∈[−1,1]



f(x) −
n

k=1
µ
k
x
2k+1





< ε.
Recall that f is odd means that f (x) = −f (−x) for all x ∈ [−1, 1].
Solution. a) Let n be such that (1 − ε
2
)
n
≤ ε. Then |x(1 − x
2
)
n
| < ε
for every x ∈ [−1, 1]. Thus one can set λ
k
= (−1)
k+1

n

, such
that
max
x∈[−1,1]
|f(x) − p(x)| <
ε
2
.
Set q(x) =
1
2
{p(x) − p(−x)}. Then
f(x) − q(x) =
1
2
{f(x) − p(x)} −
1
2
{f(−x) − p(−x)}
and
(1) max
|x|≤1
|f(x) −q(x)| ≤
1
2
max
|x|≤1
|f(x) −p(x)|+
1
2

0
= 0 then (1) proves b). If b
0
= 0 then one applies a) with
ε
2|b
0
|
instead
of ε to get
(2) max
|x|≤1





b
0
x −
n

k=1
b
0
λ
k
x
2k+1


| < 1, has positive as well as negative values in the period [0, 2π).
b) Prove that the function
F (x) =
100

n=1
cos (n
3
2
x)
has at least 40 zeros in the interval (0, 1000).
Solution. a) Let us consider the integral

2π
0
f(x)(1 ± cos x)dx = π(a
0
± 1).
The assumption that f (x) ≥ 0 implies a
0
≥ 1. Similarly, if f(x) ≤ 0 then
a
0
≤ −1. In both cases we have a contradiction with the hypothesis of the
problem.
b) We shall prove that for each integer N and for each real number h ≥ 24
and each real number y the function
F
N
(x) =

(x) does not change sign in (y, y + h) then we have
|I
2
| ≤

y+h
y
|F
N
(x)|dx =






y+h
y
F
N
(x)dx





= |I
1
|.
Hence, it is enough to prove that




N

n=1

y+h
y
cos (xn
3
2
)dx





≤ 2
N

n=1
1
n
3
2
< 2

1 +


1
2
N

n=2

y+h
y

cos

x(n
3
2
− 1)

+ cos

x(n
3
2
+ 1)

dx
=
1
2
h + ∆,
where
|∆| ≤

2
− 1
.
10
We use that n
3
2
− 1 ≥
2
3
n
3
2
for n ≥ 3 and we get
|∆| ≤
1
2
+
2
2
3
2
− 1
+ 3
N

n=3
1
n
3

proof is completed.
Problem 6. (20 points)
Suppose that {f
n
}
∞
n=1
is a sequence of continuous functions on the inter-
val [0, 1] such that

1
0
f
m
(x)f
n
(x)dx =

1 if n = m
0 if n = m
and
sup{|f
n
(x)| : x ∈ [0, 1] and n = 1, 2, . . .} < +∞.
Show that there exists no subsequence {f
n
k
} of {f
n
} such that lim

k→∞
f(x) for every x ∈ [0, 1].
Fix m ∈ N. From Lebesgue’s theorem we have
0 =

1
0
f
m
(x)f
n
k
(x)dx −→
k→∞

1
0
f
m
(x)f(x)dx.
Hence

1
0
f
m
(x)f(x)dx = 0 for every m ∈ N, which implies f (x) = 0 almost
everywhere. Using once more Lebesgue’s theorem we get
1 =
