MathDB
Problems
Contests
National and Regional Contests
Switzerland Contests
Switzerland Team Selection Test
1998 Switzerland Team Selection Test
1
1
Part of
1998 Switzerland Team Selection Test
Problems
(1)
f(x)- f(y) = f(x)f(1/y)- f(y)f(1/x), periodic
Source: Switzerland - Swiss TST 1998 p1
2/19/2020
A function
f
:
R
−
{
0
}
→
R
f : R -\{0\} \to R
f
:
R
−
{
0
}
→
R
has the following properties: (i)
f
(
x
)
−
f
(
y
)
=
f
(
x
)
f
(
1
y
)
−
f
(
y
)
f
(
1
x
)
f(x)- f(y) = f(x)f\left(\frac{1}{y}\right)- f(y)f\left(\frac{1}{x}\right)
f
(
x
)
−
f
(
y
)
=
f
(
x
)
f
(
y
1
)
−
f
(
y
)
f
(
x
1
)
for all
x
,
y
≠
0
x,y \ne 0
x
,
y
=
0
, (ii)
f
f
f
takes the value
1
2
\frac12
2
1
at least once. Determine
f
(
−
1
)
f(-1)
f
(
−
1
)
. Prove that
f
f
f
is a periodic function
functional
periodic
function
algebra