Functional Equation Solutions 3
Let $ f: \mathbb{R}\rightarrow \mathbb{R} $ be a function satisfying $ f(x)f(y)=f(x-y) $. Find all possible values of $ f(2017) $. Enter all the possible values, separated by commas.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
Setting $ x = y = 0, $ we get
\[f(0)^2 = f(0),\]so $ f(0) = 0 $ or $ f(0) = 1 $.
Suppose $ f(0) = 0 $. Setting $ y = 0, $ we get
\[f(x) f(0) = f(x),\]so $ f(x) = 0 $ for all $ x $. Note that this function works, and in particular, $ f(2017) = 0 $.
Now suppose $ f(0) = 1 $. Setting $ x = 0, $ we get
\[f(0) f(y) = f(-y),\]so $ f(-y) = f(y) $ for all $ y $.
Replacing $ y $ with $ -y $, we get
\[f(x) f(-y) = f(x + y).\]Then $ f(x + y) = f(x) f(-y) = f(x) f(y) = f(x - y) $ for all $ x $ and $ y $. Setting $ x = y = \frac{a}{2}, $ we get
\[f(a) = f(0) = 1\]for all $ a $. Note that this function works, and in particular, $ f(2017) = 1 $.
Therefore, the possible values of $ f(2017) $ are $ \boxed{0,1} $.
Freeflows