1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Discontinuous functions

  1. Jan 23, 2010 #1
    1. The problem statement, all variables and given/known data

    I have to find f: R [tex]\rightarrow[/tex] R which is discontinuous at the points of the set {1/n : n a positive integer}[tex]\cup[/tex] {0} but continuous everywhere else.

    Also find g: R [tex]\rightarrow[/tex] R which is discontinuous at the points of the set {1/n : n a positive integer}but continuous everywhere else.

    2. Relevant equations



    3. The attempt at a solution

    Could I define f as f(x) = 1/(integer(x) -1) for x [tex]\in[/tex] [0,1) (where integer means round up to next integer), f(x) = x otherwise.

    Similarly for g, can I say g(x) = 1/(integer(x) -1) for x [tex]\in[/tex] (0,1) (where integer means round up to next integer), g(x) = x otherwise.

    I'm not too sure about these functions but cannot think of any more 'normal' ones that would satisfy the criteria.
     
  2. jcsd
  3. Jan 24, 2010 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    If you are looking for a 'normal' function, you probably want f(x) = 1/g(x), where g(x) has a zero at every integer n and at 0. In that case, I can think of a suitable g(x) off the top of my head.

    By the way, note that restricting the domain to [0, 1) or (0, 1) does not define a function on R.
     
  4. Jan 24, 2010 #3
    Can I say f(x) = 1/g(x) where:
    g(x) = 0 if x is in (0,1)
    g(x) = 1 otherwise
     
  5. Jan 24, 2010 #4
    Well no, since then you are having it be discontinuous on (0,1). Perhaps try 1/g, where g is a step function with an appropriate argument so that you get 1/x back when x is of the form 1/n, but the left and right limits as you approach an x of this form are different fractions altogether.
     
  6. Jan 24, 2010 #5
    I think I may be confused about what 'at the points of the set' means, does it just mean the end points, I thought it meant all points in the set?

    So for {1/n: n a positive integer} U {0} could I say:
    f(x) = -1, x<0
    f(x) = -1/2, x=0
    f(x) = 1/x, 0<x<1
    f(x) = 2, x>1

    For {1/n: n a positive integer} I could change this to:
    f(x) = -1, x[tex]\leq[/tex]0
    f(x) = 1/x, 0<x<1
    f(x) = 2, x>1
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook