- #1

Izzhov

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**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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- Thread starter Izzhov
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- #1

Izzhov

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- #2

HallsofIvy

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http://pirate.shu.edu/projects/reals/cont/fp_weier.html

- #3

JonF

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

f(x) = 0 if x is irrational

but differentiable no where

- #4

Crosson

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Perhaps Izzhov would enjoy the text 'Counterexamples in Analysis".

- #5

Gib Z

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- #6

jostpuur

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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.

- #7

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.

- #8

Gib Z

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JonF, you might want to reconsider if 0 is a rational number >.<

- #9

tehno

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:uhh:

http://pirate.shu.edu/projects/reals/cont/fp_weier.html

Don't scare children with such examples...

- #10

Gib Z

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[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

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In what sense is that a function?

- #12

tehno

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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 beautiful "monster" Gib_Z,but far from being function.

(http://en.wikipedia.org/wiki/Mandelbrot_set) [Broken].

:rofl: :rofl: :zzz:

It is a beautiful "monster" Gib_Z,but far from being function.

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- #13

Gib Z

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- #14

jostpuur

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A continuous set? 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.i know its not a function, but I thought it was continuous everywhere and differentiable nowhere

- #15

DeadWolfe

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