All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
A Most Unusual Evaluation (Posted on 2006-01-04) Difficulty: 5 of 5
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.

See The Solution Submitted by K Sengupta    
Rating: 3.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution | Comment 1 of 3

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)) 


  Posted by Gamer on 2006-01-05 00:28:48
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information