f(2)=-f(0)
f(3)=[1-f(0)]/[1+f(0)]
f(4)=f(0)
所以f(x)为周期为4的周期函数
f(2014)=f(2)=-f(0)=-2010f(2)=f(1+1)=[1+f(1)]/[1-f(1)]={1+[1+f(0)]/[1-f(0)]}/{1-[1+f(0)]/[1-f(0)]}
={2/[1-f(0)]}/{-2f(0)/[1-f(0)]}=-1/f(0)
f(3)=f(2+1)=[1+f(2)]/[1-f(2)]=-(1-f(0))/(1+f(0))
f(4)=f(0)
很抱歉，中间步骤有些错误~
非常感谢指出来~
