Example of a bounded, increasing, discontinuous function

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The discussion revolves around defining a function f:ℝ->ℝ that is increasing, bounded, and discontinuous at every integer. Participants suggest using piecewise definitions to achieve the desired properties, with ideas like utilizing arctan and linear functions on intervals between integers. There is a focus on ensuring that the function remains bounded above and below while maintaining jump discontinuities at integers. Suggestions include defining different linear functions for each interval [n,n+1] that lie between arctan(n) and arctan(n+1). The conversation emphasizes the importance of clarity in the definitions of "increasing" and "bounded" for the function's construction.
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Homework Statement


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

Homework Equations





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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k3k3 said:
I cannot get it to be bounded with jump discontinuities.

Jump discontinuities don't have to be of the same size.
 
k3k3 said:

Homework Statement


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

Homework Equations





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.

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.
 
How about arctan(x) if x is in (n, n+1) and arctan(x)+1/x if x is an integer?
 
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.)
 
k3k3 said:
How about arctan(x) if x is in (n, n+1) and arctan(x)+1/x if x is an integer?
Is that increasing?
 
k3k3 said:
How about arctan(x) if x is in (n, n+1) and arctan(x)+1/x if x is an integer?

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?
 
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.
 
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Zondrina said:
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.

By bounded I mean it is bounded above and below, yeah.


Dick said:
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?

I can't think of a linear function that would scale with the arctan. How about 1/x when x isn't an integer?
 
  • #10
k3k3 said:
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?

Sorry I meant Integer(x) in my last post, i edited it.
 
  • #11
Maybe you can use the floor/ceiling function somehow and try to make the next jump up smaller.
 
  • #12
happysauce said:
Maybe you can use the floor/ceiling function somehow and try to make the next jump up smaller.

I've thought about that, but I have had no luck in doing so.
 
  • #13
k3k3 said:
I can't think of a linear function that would scale with the arctan. How about 1/x when x isn't an integer?

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.
 
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  • #14
Dick said:
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.

So I just need a function whose values are between -pi/2 to pi/2?
 
  • #15
k3k3 said:
So I just need a function whose values are between -pi/2 to pi/2?

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?
 
  • #16
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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