(Spivak) - a function with strange behaviour.

In summary, there are multiple possible solutions for a function that is discontinuous at 1, 1/2, 1/3, 1/4, ... but continuous at all other points. Some examples include a function that is undefined at these points, a function that is equal to 1 at these points but otherwise continuous, and a function that involves the floor or Gamma function.

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  • #1
kioria
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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 anyone tell me the answer to this?
 
Last edited:
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  • #2
The function you have founded have indeed that property, there is an infinite set of functions that can do the job...
 
  • #3
ReyChiquito said:
The function you have founded have indeed that property, there is an infinite set of functions that can do the job...

Excellent! Thank you. :biggrin:
 
  • #4
Does the function need to be defined everywhere? If not, you can construct an elegant solution as follows:

[tex]f(x)=\frac{(x-1)(x-1/2)(x-1/3)...}{(x-1)(x-1/2)(x-1/3)...}[/tex]

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

see attachment for a family of solutions
 

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  • #7
Thanks for all the help :)
 
  • #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
 
  • #9
If you're going to get picky then the original definition doesn't define a function.
 
  • #10
matt grime said:
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.

Interesting. Good stuff.
 
  • #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.
 
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  • #12
Another suggestion

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

1. What is a (Spivak) function with strange behaviour?

A (Spivak) function with strange behaviour is a mathematical function that has unexpected or unusual properties, making it difficult to predict its behavior.

2. How is a (Spivak) function with strange behaviour different from a regular function?

A (Spivak) function with strange behaviour differs from a regular function in that it may have discontinuities, non-differentiable points, or other unusual features that are not typically seen in regular functions.

3. What causes a (Spivak) function to exhibit strange behaviour?

The strange behaviour of a (Spivak) function can be caused by a variety of factors, such as the function being non-analytic or having complicated algebraic properties.

4. Are there any real-world applications for (Spivak) functions with strange behaviour?

Yes, (Spivak) functions with strange behaviour have applications in fields such as chaos theory, fractals, and dynamical systems. They can also be used to model real-world phenomena with unpredictable behavior, such as weather patterns or stock market fluctuations.

5. Is there a way to predict the behaviour of a (Spivak) function with strange behaviour?

No, unfortunately, there is no general method for predicting the behaviour of a (Spivak) function with strange behaviour. Each function must be analyzed individually and may require advanced mathematical techniques to understand its behavior.

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