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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 agoFavorite 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

f(1/(2*5^n))=1/2^(n+1)

and

f(1/5^n)=1/2^n

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