问题及解答

定义函数 $f(x)$ 如下,

\[
f(x)=\begin{cases}
-x, & x < 0,\\
\frac{1}{2}f(x-f(x-1)), & x\geqslant 0.
\end{cases}
\]

求 $f(3)$.

注:  题目来源于 QQ群 数学竞赛之窗

(1)  $f(0)$

\[
f(0)=\frac{1}{2}f(0-f(-1))=\frac{1}{2}f(0-1)=\frac{1}{2}f(-1)=\frac{1}{2}.
\]

f(0) --> f(-1) --> f(-1) --> 1|2

(2)  $f(1)$

\[
f(1)=\frac{1}{2}f(1-f(0))=\frac{1}{2}f(1-\frac{1}{2})=\frac{1}{2}f(\frac{1}{2})
\]

\[
f(\frac{1}{2})=\frac{1}{2}f(\frac{1}{2}-f(-\frac{1}{2}))=\frac{1}{2}f(\frac{1}{2}-\frac{1}{2})=\frac{1}{2}f(0)=\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4},
\]

因此 $f(1)=\frac{1}{2}f(\frac{1}{2})=\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}$.

f(1) --> f(0) --> f(-1) --> f(-1) --> f(1|2) --> f(-1|2) --> f(0) --> f(-1) --> f(-1) --> 1|8

(3)  $f(2)$

\[
f(2)=\frac{1}{2}f\bigl(2-f(1)\bigr)=\frac{1}{2}f(2-\frac{1}{8})=\frac{1}{2}f(\frac{15}{8})=\cdots=\frac{1}{1024}
\]

f(2) --> f(1) --> f(0) --> f(-1) --> f(-1) --> f(1|2) --> f(-1|2) --> f(0) --> f(-1) --> f(-1) --> f(15|8) --> f(7|8) --> f(-1|8) --> f(3|4) --> f(-1|4) --> f(1|2) --> f(-1|2) --> f(0) --> f(-1) --> f(-1) --> f(29|16) --> f(13|16) --> f(-3|16) --> f(5|8) --> f(-3|8) --> f(1|4) --> f(-3|4) --> f(-1|2) --> f(7|4) --> f(3|4) --> f(-1|4) --> f(1|2) --> f(-1|2) --> f(0) --> f(-1) --> f(-1) --> f(13|8) --> f(5|8) --> f(-3|8) --> f(1|4) --> f(-3|4) --> f(-1|2) --> f(3|2) --> f(1|2) --> f(-1|2) --> f(0) --> f(-1) --> f(-1) --> f(5|4) --> f(1|4) --> f(-3|4) --> f(-1|2) --> f(1) --> f(0) --> f(-1) --> f(-1) --> f(1|2) --> f(-1|2) --> f(0) --> f(-1) --> f(-1) --> 1|1024>

Claim.  当 $n > 2$ 时

\[
f(\frac{1}{n})=\frac{1}{2}-\frac{1}{n}.
\]

Pf. 

\[
f(\frac{1}{n})=\frac{1}{2}f\bigl(\frac{1}{n}-f(\frac{1}{n}-1)\bigr)=\frac{1}{2}f\bigl(\frac{1}{n}-(1-\frac{1}{n})\bigr)=\frac{1}{2}f(\frac{2}{n}-1)
\]

由于 $n > 2$, 故 $\frac{2}{n}-1 < 0$, 因此 $f(\frac{2}{n}-1)=1-\frac{2}{n}$. 故

\[
f(\frac{1}{n})=\frac{1}{2}\cdot\Bigl(1-\frac{2}{n}\Bigr)=\frac{1}{2}-\frac{1}{n}.
\]