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.
Here is a concise solution with many of the boring proofs left out. For a rigorous solution, see the next post.
Rule i can be rewritten as 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)
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), and that is similar to the left side.
Add F(1/(8^N+1))*(7^N) to both sides and divide by 1+7^N to get F(1/(8^N+1))=1/(7^N+1)
This means 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))
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Posted by Gamer
on 2006-01-05 00:28:48 |