14. Given $x \in \mathbf{R}, a$ is a constant, and $f(x+a)=\frac{1+f(x)}{1-f(x)}$, is $f(x)$ a periodic function? If so, find a period; if not, explain why.
$a=\frac{\pi}{4}$, and the smallest positive period of the function $y=\tan x$ is $T=\pi=4 a$, therefore, we can conjecture that $f(x)$ is a function with a period of $4 a$.
From $f(x+a)=\frac{1+f(x)}{1-f(x)}$, we can get $f(x+2 a)=f[(x+a)+a]=\frac{1+f(x+u)}{1-f(x+a)}=-\frac{1}{f(x)}$.
Thus, $f(x+4 a)=f[(x+2 a)+2 a]=-\frac{1}{f(x+2 a)}=f(x)$. Therefore, $f(x)$ is a periodic function, and $4 a$ is one of its periods.

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