问题: 函数
设函数y=f(x)满足f(1)=4,f(2)=1,f(3)= 3, f(4)=5, f(5)=2,数列{xn}满足x0=5且对任意自然数均有x（n+1）=f(x),则f(x(2008))=？ (2008下标)
解答:
对任意的自然数n，都有 x(n+1) = f(xn)
且 x0 = 5
所以 x1 = f(x0) = f(5) = 2
x2 = f(x1) = f(2) = 1
x3 = f(x2) = f(1) = 4
x4 = f(x3) = f(4) = 5 = x0
故 x5 = ................ = x1
x6 = ................ = x2
x7 = ................ = x3
x8 = ................ = x4
.............................
由周期性，可得 x2008 = x0 = 5
故 f(x2008) = f(5) = 2
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