定义在R上的函数f(x)满足当x≤0时 log2 (1-x) 当x>0时 f(x-1)-f(x-2) 则f(2014)=

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  • f(-1)=log2(1-(-1))=1

    f(0)=log2(1-0)=0

    f(1)=f(0)-f(-1)=-1

    f(2)=f(1)-f(0)=-1

    f(3)=f(2)-f(1)=0

    f(4)=f(3)-f(2)=1

    f(5)=f(4)-f(3)=1

    f(6)=f(5)-f(4)=0

    f(7)=f(6)-f(5)=-1

    f(8)=f(7)-f(6)=-1

    f(9)=f(8)-f(7)=0

    f(10)=f(9)-f(8)=1

    f(11)=f(10)-f(9)=1

    f(12)=f(11)-f(10)=0

    可见x>0时,f(x)的值以6为一个最小周期

    2014÷6=335.4

    所以f(2010)=-1

    另证明:

    f(x)=f(x-1)-f(x-2)

    f(x-1)=f(x-2)-f(x-3)

    f(x)=f(x-1)-f(x-2)=f(x-2)-f(x-3)-f(x-2)=-f(x-3)

    f(x+3)=-f(x)

    f(x+6)=f[(x+3)+3]=-f(x+3)=-[-f(x)]=f(x)

    故f(x)周期为6