Does such a function even exist?

  • Thread starter Izzhov
  • Start date
  • #1
117
0
When my father was in college, one of his professors showed him a function that is continuous everywhere and differentiable nowhere. All my dad remembers is that the function was like an infinitely small sawtooth. If this function exists, how can it be defined?
 

Answers and Replies

  • #3
612
1
i think this function is continuous everywhere:

f(x) = 1/x if x is rational
f(x) = 0 if x is irrational

but differentiable no where
 
  • #4
1,250
2
The function given by JonF is continuous on the irrationals, and discontinuous on the rationals. So it is not continuous everywhere.

Perhaps Izzhov would enjoy the text 'Counterexamples in Analysis".
 
  • #5
Gib Z
Homework Helper
3,346
5
OP: Try searching up fractals. The Weierstrass function is a good example, and somewhat the first one found.
 
  • #6
2,111
18
The function given by JonF is continuous on the irrationals, and discontinuous on the rationals. So it is not continuous everywhere.
Doesn't look continuous anywhere.
 
  • #7
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,970
132
Doesn't look continuous anywhere.
Indeed you are right.
However, the function:
[tex]f(x)=\frac{1}{q}, x=\frac{p}{q}, p,q\in\mathbb{N}[/tex]
and f(x) zero elsewhere IS continuous on the irrationals, but not on the rationals.
 
  • #8
Gib Z
Homework Helper
3,346
5
JonF, you might want to reconsider if 0 is a rational number >.<
 
  • #10
Gib Z
Homework Helper
3,346
5
Would this be a simpler example?

[tex]x_n=(x_{n-1})^2 + C[/tex]
[tex]x_0=C[/tex].

I think I saw something like this in the favorite equations thread, general math section.
 
  • #11
HallsofIvy
Science Advisor
Homework Helper
41,833
956
In what sense is that a function?
 
  • #12
364
0
I think he considers Mandelbrot set a "function" .
(http://en.wikipedia.org/wiki/Mandelbrot_set) [Broken].
:rofl: :rofl: :zzz:

It is a beutiful "monster" Gib_Z,but far from being function.
 
Last edited by a moderator:
  • #13
Gib Z
Homework Helper
3,346
5
O i know its not a function, but I thought it was continuous everywhere and differentiable nowhere, its a fractal right? I know very little about this area.
 
  • #14
2,111
18
i know its not a function, but I thought it was continuous everywhere and differentiable nowhere
A continuous set? :rolleyes: I would guess, that you are talking about the boundary of the Mandelbrot set? It looks bad enough, that if you had a parametrisation for it, then it would not be differentiable. But at this point I must raise a different question: Is there parametrisation for boundary of Mandelbrot set? I have a feel that there's not, but I'm not sure.
 
  • #15
453
0
I would imagine that if there was a smooth parameterization of the boundary then it would have to have Hausdorff dimension 1.
 

Related Threads on Does such a function even exist?

Replies
9
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
11
Views
3K
  • Last Post
Replies
3
Views
948
  • Last Post
Replies
4
Views
24K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
6
Views
2K
Replies
4
Views
3K
  • Last Post
Replies
2
Views
2K
Top