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National and Regional Contests
Poland Contests
Poland - Second Round
1973 Poland - Second Round
3
3
Part of
1973 Poland - Second Round
Problems
(1)
f(x+1) = f(x) + 1, f(f(f(O))) = p
Source: Polish MO Second Round 1973 p3
9/8/2024
Let
f
:
R
→
R
f:\mathbb{R} \to \mathbb{R}
f
:
R
→
R
be an increasing function satisfying the following conditions: 1.
f
(
x
+
1
)
=
f
(
x
)
+
1
f(x+1) = f(x) + 1
f
(
x
+
1
)
=
f
(
x
)
+
1
for each
x
∈
R
x \in \mathbb{R}
x
∈
R
, 2. there exists an integer p such that
f
(
f
(
f
(
O
)
)
)
=
p
f(f(f(O))) = p
f
(
f
(
f
(
O
)))
=
p
. Prove that for every real number
x
x
x
lim
n
→
∞
x
n
n
=
p
3
.
\lim_{n\to \infty} \frac{x_n}{n} = \frac{p}{3}.
n
→
∞
lim
n
x
n
=
3
p
.
where
x
1
=
x
x_1 = x
x
1
=
x
and
x
n
=
f
(
x
n
−
1
)
x_n =f(x_{n-1})
x
n
=
f
(
x
n
−
1
)
for
n
=
2
,
3
,
…
n = 2, 3, \ldots
n
=
2
,
3
,
…
.
algebra
limit
calculus