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National and Regional Contests
Switzerland Contests
Switzerland - Final Round
2013 Switzerland - Final Round
4
4
Part of
2013 Switzerland - Final Round
Problems
(1)
f ( x / (y + 1)) = 1 - xf(x + y)
Source: Switzerland - 2013 Swiss MO Final Round p4
12/30/2022
Find all functions
f
:
R
>
0
→
R
>
0
f : R_{>0} \to R_{>0}
f
:
R
>
0
→
R
>
0
with the following property:
f
(
x
y
+
1
)
=
1
−
x
f
(
x
+
y
)
f \left( \frac{x}{y + 1}\right) = 1 - xf(x + y)
f
(
y
+
1
x
)
=
1
−
x
f
(
x
+
y
)
for all
x
>
y
>
0
x > y > 0
x
>
y
>
0
.
algebra
functional
functional equation