Let F be an increasing real function defined for all real X, where 0<=X<=1 such that:

(i) F (X/8) = F(X)/7 and
(ii) F(1-X) = 1 – F(X)

For all whole numbers M and N greater than zero, determine:
F ( 1/ ((8^M)* ( 8^N + 1)) ) in terms of M and N.

I added more explanation in this post, as well as the two proofs of the new rule i and using powers with rule i. (This part was added in the Edit, I accidentally left in the note from the previous post)

Rule i can be rewritten as 7*F(X)=F(8X)

Proof: If F(X/8) = F(X)/7, then F(U/8) = F(U)/7, with U having the same bounds as X. Let U=8X, and substitute, then multiply both sides by 7 to get 7*F(X)=F(8X).

and it also can applied any number of times, which means it applies to the powers in the goal expresssion. (ie F(64) = F(8)/7 = F(1)/49)

Proof by induction that F(X/8^N) = F(X)/7^N

F(X/8^1) = F(X)/7^1 by rule i

F(X/8^N) = F(X)/7^N
F(8X/8^(N+1)) = 7*F(X)/7^(N+1)
F(U/8^(N+1)) = F(U)/7^(N+1) by rule i again. (let U=8X again)

This means F((1/(8^N+1))/(8^M)))=F(1/(8^N+1))/7^M

By the second rule, F(1/(8^N+1))=1-F(8^N/(8^N+1))
By the first rule (using powers) the right side is equivalent to 1 - F(1/(8^N+1))*(7^N), (let X=8^N in that case) and that is similar to the left side. (Both have F(1/(8^N+1)) which is what we want.) 
Add F(1/(8^N+1))*(7^N) to both sides to get F(1/(8^N+1))+F(1/(8^N+1))*7^N = 1 or factored as F(1/(8^N+1))/(1 + 7^N) and divide by 1+7^N to get F(1/(8^N+1))=1/(7^N+1)

Substitute this equation into the right side of F((1/(8^N+1))/(8^M)))=F(1/(8^N+1))/7^M to get F((1/(8^N+1))/(8^M))=1/(7^N+1)/7^M

So F ( 1/ ((8^M)* ( 8^N + 1)) ) = 1/((7^M)* ( 7^N + 1)) 

Edited on January 5, 2006, 12:47 am

Posted by Gamer on 2006-01-05 00:45:21