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Problem 1
Functional Equation with f(2n+1)=2f(n) and f(2n)=2f(n)+1
Functional Equation with f(2n+1)=2f(n) and f(2n)=2f(n)+1
Source: India Postals Set 3
November 7, 2015
algebra
functional equation
Problem Statement
Let
f
:
N
∪
{
0
}
→
N
∪
{
0
}
f:\mathbb{N} \cup \{0\} \to \mathbb{N} \cup \{0\}
f
:
N
∪
{
0
}
→
N
∪
{
0
}
be defined by
f
(
0
)
=
0
f(0)=0
f
(
0
)
=
0
,
f
(
2
n
+
1
)
=
2
f
(
n
)
f(2n+1)=2f(n)
f
(
2
n
+
1
)
=
2
f
(
n
)
for
n
≥
0
n \ge 0
n
≥
0
and
f
(
2
n
)
=
2
f
(
n
)
+
1
f(2n)=2f(n)+1
f
(
2
n
)
=
2
f
(
n
)
+
1
for
n
≥
1
n \ge 1
n
≥
1
If
g
(
n
)
=
f
(
f
(
n
)
)
g(n)=f(f(n))
g
(
n
)
=
f
(
f
(
n
))
, prove that
g
(
n
−
g
(
n
)
)
=
0
g(n-g(n))=0
g
(
n
−
g
(
n
))
=
0
for all
n
≥
0
n \ge 0
n
≥
0
.
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