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(Spivak) - a function with strange behaviour.

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Poll closed Apr 19, 2004.
  1. Yes

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  1. Apr 12, 2004 #1
    1) Find a function, [tex]f(x)[/tex] which is discontinuous at [tex]1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4} ...[/tex], but continuous at any other points.

    Solution (I have come across, probably wrong and a half):
    f(x) = { 1 for all real x; 0 for 1/x where x is natural numbers.

    Can any one tell me the answer to this?
    Last edited: Apr 12, 2004
  2. jcsd
  3. Apr 12, 2004 #2
    The function you have founded have indeed that property, there is an infinite set of functions that can do the job...
  4. Apr 12, 2004 #3
    Excellent! Thank you. :biggrin:
  5. Apr 13, 2004 #4
    Does the function need to be defined everywhere? If not, you can construct an elegant solution as follows:


    This function is equal to 1 except at the points [tex]\frac{1}{n}[/tex], where it is undefined.
  6. Apr 13, 2004 #5

    matt grime

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    Part of the definition of a function is its domain, and it needs to be defined on its domain.

    If you mean, say, is 1/x a function from R to R? No: you've not defined what it is at zero, until you do it is at best a function from R\{0} to R.

    This is a big problem that is not taught properly when it first arises and causes many unnecessary problems.

    The one you gave has the nice property that one may define f at the points in question so that it is 1, and is continuous at all those points.
  7. Apr 16, 2004 #6


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    see attachment for a family of solutions

    Attached Files:

  8. Apr 17, 2004 #7
    Thanks for all the help :)
  9. Apr 17, 2004 #8
    Kioria, I just want to add that the function you gave is also discontinuous at x=0, not only at x=1/n. But you can change it like this to kill that bug:

    f(x) = { 0 for all real x; x for x=1/n , n any natural number
  10. Apr 17, 2004 #9

    matt grime

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    If you're going to get picky then the original definition doesn't define a function.
  11. Jun 19, 2004 #10
    Interesting. Good stuff.
  12. Jun 19, 2004 #11
    The Gauss Transformation is a function that is like that. It looks like this:

    G(x) = 1/x + [ 1/x ] for x in the interval (0,1]
    G(x) = 0 @ x=0

    [] means floor, aka least integer function.
    Last edited: Jun 19, 2004
  13. Jun 25, 2004 #12
    Another suggestion

    [tex]\Gamma( \frac {-1}{x} ), for x > 0[/tex]
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