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f(1/2008) =?

Given: f(x) is a real function on [0,1]. f(0) = 0, f(x) + f(1-x) = 1, f(x/5) = f(x)/2, f(x1) ≤ f(x2) if x1 ≤ x2.

1 Answer

  • 1 decade ago
    Favorite Answer

    f(0)=0 => f(1)=1 => f(1/5)=1/2

    f(x)+f(1-x)=1 => f(1/2)=1/2

    Now, by induction on the identity f(x/5)=f(x)/2, we can establish that




    Since 1/5^5 ≤ 1/2008 ≤ 1/(2*5^4),

    1/2^5 = f(1/5^5) ≤ f(1/2008) ≤ f(1/(2*5^4)) = 1/2^5

    Consequently, f(1/2008)=1/32

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