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1986 Poland - Second Round
1
1
Part of
1986 Poland - Second Round
Problems
(1)
2f(2x) = f(x) + x., continuous at 0
Source: Polish MO Recond Round 1986 p1
9/9/2024
Determine all functions
f
:
R
ā
R
f : \mathbb{R} \to \mathbb{R}
f
:
R
ā
R
continuous at zero and such that for every real number
x
x
x
the equality holds
2
f
(
2
x
)
=
f
(
x
)
+
x
.
2f(2x) = f(x) + x.
2
f
(
2
x
)
=
f
(
x
)
+
x
.
algebra
continuous
functional
functional equation