19. The function $f(x)$ defined on the set of real numbers $\mathbf{R}$ satisfies $f(x+1)=\frac{1+f(x+3)}{1-f(x+3)}$, then the value of $f(1) \cdot f(2) \cdots$ $f(2008)+2009$ is $\qquad$.
19. Fill 2010. Reason: From the given, we have $f(x)=\frac{1+f(x+2)}{1-f(x+2)}$, and also $f(x-1)=\frac{1+f(x+1)}{1-f(x+1)}$, from which we can derive $f(x+1)=\frac{f(x-1)-1}{f(x-1)+1}$.
Thus, $\frac{1+f(x+3)}{1-f(x+3)}=\frac{f(x-1)-1}{f(x-1)+1}$, which means $f(x-1) \cdot f(x+3)=-1$.
Therefore, we get $f(x+3) \cdot f(x+7)=-1$, which implies $f(x-1)=f(x+7)$.
By the conclusion 6(1) of this chapter, we know that $f(x)$ is a periodic function, and 8 is one of its positive periods. Since $f(1) \cdot f(5)=f(2) \cdot f(6)=$ $f(4) \cdot f(8)=1$, it follows that $f(1) \cdot f(2) \cdot f(3) \cdots f(8)=1$. Also, $2008=251 \cdot 8$, so $f(1) \cdot f(2) \cdots f(2008)+$ $2009=1+2009=2010$