Does such a function even exist?

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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?
 

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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
 
1,222
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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".
 
Gib Z
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OP: Try searching up fractals. The Weierstrass function is a good example, and somewhat the first one found.
 
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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.
 
arildno
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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.
 
Gib Z
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JonF, you might want to reconsider if 0 is a rational number >.<
 
Gib Z
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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.
 
HallsofIvy
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In what sense is that a function?
 
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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.
 
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Gib Z
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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.
 
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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.
 
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I would imagine that if there was a smooth parameterization of the boundary then it would have to have Hausdorff dimension 1.
 

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