Example of a bounded, increasing, discontinuous function

In summary: 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?If you don't like like the explicit piecewise approach, what does arctan(floor(x)) look like? How could you modify...If you don't like the explicit piecewise approach, what does arctan(floor(x)) look like? How could you modify it to make it increasing?
  • #1
k3k3
78
0

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.
 
Physics news on Phys.org
  • #2
k3k3 said:
I cannot get it to be bounded with jump discontinuities.

Jump discontinuities don't have to be of the same size.
 
  • #3
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.
 
  • #4
How about arctan(x) if x is in (n, n+1) and arctan(x)+1/x if x is an integer?
 
  • #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.)
 
  • #6
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?
 
  • #7
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?
 
  • #8
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:
  • #9
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.
 
Last edited:
  • #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?
 

What is an example of a bounded, increasing, discontinuous function?

An example of a bounded, increasing, discontinuous function is the step function. It is bounded because it has a finite range, increasing because it always moves upwards, and discontinuous because it has abrupt changes in value.

Why is the step function a good example of a bounded, increasing, discontinuous function?

The step function is a good example because it satisfies all three criteria of being bounded, increasing, and discontinuous. It has a finite range, always moves upwards, and has abrupt changes in value, making it a clear and simple representation of these characteristics.

What is the difference between a bounded and unbounded function?

A bounded function has a finite range, meaning its values are limited within a certain range. On the other hand, an unbounded function has an infinite range, meaning its values can continue to increase or decrease without limit.

Can a function be both increasing and decreasing?

No, a function cannot be both increasing and decreasing at the same time. An increasing function always moves upwards, while a decreasing function always moves downwards. However, a function can have different intervals where it is increasing or decreasing, but it cannot be both simultaneously.

Why is it important to understand the characteristics of a function?

Understanding the characteristics of a function, such as being bounded, increasing, or discontinuous, can help us analyze and make predictions about its behavior. It also allows us to identify and classify different types of functions, which can aid in solving mathematical problems and real-world applications.

Similar threads

Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
2
Replies
39
Views
4K
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
987
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
850
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
Back
Top