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Example of a bounded, increasing, discontinuous function

  1. Oct 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Define a function f:ℝ->ℝ that is increasing, bounded, and discontinuous at every integer.

    2. Relevant equations



    3. The attempt at a solution

    I've tried defining a fuction using the greatest integer function but I cannot get it to be bounded with jump discontinuities.

    Floor(x)*arctan(x) and variations like that are what I keep going toward.
     
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  3. Oct 10, 2012 #2

    George Jones

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    Jump discontinuities don't have to be of the same size.
     
  4. Oct 10, 2012 #3

    Dick

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    You might find it lot easier to describe your function piecewise on intervals [n,n+1] rather than trying to write a single formula for it.
     
  5. Oct 10, 2012 #4
    How about arctan(x) if x is in (n, n+1) and arctan(x)+1/x if x is an integer?
     
  6. Oct 10, 2012 #5
    By what definition are you using the term "increasing"?

    I've always known "increasing" to be different from "strictly increasing" but I understand some people may use the former to mean the latter.

    If you don't mean "strictly increasing", then you can define a piece-wise function much more simply than using arctan or such like. (In fact, you can even if you do want it strictly increasing.)
     
  7. Oct 10, 2012 #6

    SammyS

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    Is that increasing?
     
  8. Oct 10, 2012 #7

    Dick

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    That's not increasing when x is an integer. Your function jumps up and then jumps back down, does't it? Define f(n)=arctan(n) where n is an integer. Now you just have to describe f on the intervals (n,n+1). How about using a linear function?
     
  9. Oct 10, 2012 #8

    Zondrina

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    Define a function f:ℝ->ℝ that is increasing, bounded, and discontinuous at every integer.

    Hmm try f(x) = 1/Integer(x) maybe? When you say 'bounded' do you mean your function could be bounded below or above? Then yeah it would be easy just to define a piecewise function using intervals.

    Maybe even f(x) = Integer(x)/sinx would work too?

    Where Integer(x) means the integer part of x.
     
    Last edited: Oct 10, 2012
  10. Oct 10, 2012 #9
    By bounded I mean it is bounded above and below, yeah.


    I can't think of a linear function that would scale with the arctan. How about 1/x when x isn't an integer?
     
  11. Oct 10, 2012 #10

    Zondrina

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    Sorry I meant Integer(x) in my last post, i edited it.
     
  12. Oct 10, 2012 #11
    Maybe you can use the floor/ceiling function somehow and try to make the next jump up smaller.
     
  13. Oct 10, 2012 #12
    I've thought about that, but I have had no luck in doing so.
     
  14. Oct 10, 2012 #13

    Dick

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    Define a different linear function in each integer interval [n,n+1]. It doesn't have to do much. It's values just have to lie between arctan(n) and arctan(n+1) and it has to increase and be discontinuous at at least one endpoint.
     
    Last edited: Oct 10, 2012
  15. Oct 10, 2012 #14
    So I just need a function whose values are between -pi/2 to pi/2?
     
  16. Oct 10, 2012 #15

    Dick

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    No. I don't think you actually have been following me. Draw a graph of all of the points (n,arctan(n)) for n an integer. Join adjacent points with a line segment. Now decrease the slope of each line segment so it doesn't hit the upper point. That will give you the sort of graph you want, yes?
     
  17. Oct 11, 2012 #16

    Dick

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    If you don't like like the explicit piecewise approach, what does arctan(floor(x)) look like? How could you modify it?
     
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