  Obzornik mat. ﬁz. 55 (2008) 2
Vesti
      p x
1
 x
k
 A
1
 A
k
 2×2 p(x
1
,...,x
k
) = det
 
k
X
i=1
x
i
A
i
!
.
 k k
 2 p   |p(z)| ≤ 2 z  p f:R→R  c > 0  cf   f  f C  f: C →
C  f(x) ≤ f(y) x ≤ y p∈ C f(p) = p Obzornik mat. ﬁz. 55 (2008) 2
Štirinajsto mednarodno tekmovanje študentov matematike
        p(x) = ax
2
 a A
1
=

1 0
0 a

 p(x,y) = ax
2
+by
2
+cxy p(x,y) =




x by
−y ax+cy




,
 A
1
=

1 0
0 a

 A
2
=

0 b
−1 c

 k ≥ 3 p(x
1
,...,x
k
) =
k
P
i=1
x
2
i
 x
1
 x
k
 Obzornik mat. ﬁz. 55 (2008) 2
Vesti
p(x
1
,...,x
k
) = det

k
P
i=1
x
i
A
i

  2× 2 y
1
 y
k
  y
1
A
1
+ ... + y
k
A
k
 
0
0

 det(y
1
A
1
+ ... + y
k
A
k
) = 0 p(y
1
,...,y
k
) =
y
2
1
+...+y
2
k
6= 0   1+z p(z) = a
n
z
n
+...+a
0
  p  z p −p   a
0
> 0 q(z) = a
1
z +...+a
n−1
z
n−1
 q = 0  w
0
 w
n−1
 a
n
w
k
n
=
|a
n
| w
k
=
(
e
2kπi/n
, a
n
> 0;
e
(2k+1)πi/n
, a
n
< 0.
 n−1
X
k=0
q(w
k
) =
n−1
X
k=0
q

w
0
e
2kπi/n

=
n−1
X
j=1
a
j
w
j
0
n−1
X
k=0

e
2jπi/n

k
= 0.
  
e
2jπi/n

k
 n p w
k
 1
n
n−1
X
k=0
p(w
k
) =
1
n
n−1
X
k=0
(a
0
+q(w
k
)+a
n
w
n
k
) = a
0
+|a
n
|,
 2≥
1
n
n−1
X
k=0
|p(w
k
)|≥





1
n
n−1
X
k=0
p(w
k
)





= a
0
+|a
n
|≥ 2.
 a
0
=|a
n
| = 1 |p(w
k
)| =|2 +q(w
k
)| = 2 k q(w
k
)  |z− (−2)| = 2 Obzornik mat. ﬁz. 55 (2008) 2
Štirinajsto mednarodno tekmovanje študentov matematike
 q(w
k
) = 0 k q n− 1 n q = 0
 p(z) = a
0
+a
n
z
n
    |a
0
|
2
+···+|a
n
|
2
=
1
2π
2π
Z
0
|p(e
it
)|
2
dt≤
1
2π
2π
Z
0
4dt = 4.
  ±1  ce
x
= e
x+logc
  f(x) = e
x
  f(x)6= x x∈ C [a,b] C C a,b ∈ C  f C f(a) > a
 f(b) < b p = sup{x∈ C; f(x) > x}.
 C  f f(p) ≥ p f(p) > p x∈ C p f(x) < x f(f(p)) < f(p) f   Obzornik mat. ﬁz. 55 (2008) 2