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

Home > Just Math > Calculus
Weird function challenge (Posted on 2006-08-15) Difficulty: 4 of 5
Find a function f:R->R (R the set of real numbers), such that

1. f has a discontinuity in every rational number, but is continous everywhere else, and
2. f is monotonic: x<y → f(x)<f(y)

Note: Textbooks frequently present examples of functions that meet only the first condition; requiring monotonicity makes for a slightly more challenging problem.

See The Solution Submitted by JLo    
Rating: 4.3000 (10 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): Nice solution!!! Now, you Want to try this | Comment 28 of 33 |
(In reply to re: Nice solution!!! Now, you Want to try this by Ken Haley)

Ken,
actually I was asking you to construct a function that has prescribed jumps (i.e. left limit minus right limit) p^-2 at the rationals. The solution I had in mind can achieve this easily. Now posted as requested.

That makes me wonder, did you try to figure out exactly which jumps your function has (The one with domain ]0,1[)? If you order all the jumps from large to small, you must end up with a converging series, I am just curious what the series looks like.

  Posted by JLo on 2006-08-25 11:31:07

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 (11)
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