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I'm very interested in finding a way or hint for the construction of the Weierstrass function which is everywhere continuous but nowhere differentiable - let's call this (ECND). My most humble example is the iteration.

$y_0 = |1-x|$

$y_n = \frac{1}{2^n}|1-2^n y_{n-1}| \text{ ; for } n > 0$

The vertical lines are absolute values.

Iterating produces an serrated function over $(0,2)$ which seems to disappear. However, since I'm guessing this iteration keeps the length constant, this can't be a problem (i.e the function is still there and each iteration gives increases the serrations to twice the previous ones.). I'm not claiming it is ECND, but it is an idea.

Do you have any info on how the original function was produced? What inspired the definition? Is the cosine function the only appropriate option or are there other functions that can satisfy the ECND condition?

I leave some iterations here:

enter image description here

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OK. This is quite the visual proof that the length is constant. I inverted the functions upside down. The iteration is

$y_0 = 1-|1-x|$

$y_n = \frac{1}{2^n}\left(1-|1-2^n y_{n-1}|\right) \text{ ; for } n > 0$

enter image description here

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You do have a problem, the iterations will converge to the zero function. What you want to do is look at the sums of the iterations. Do this in such a way that the infinite sum converges (take the amplitude of the $i$th iteration to be small enough, as you already have). Try graphing $y_0$, $y_0+y_1$, $y_0+y_1+y_2$, etc... I believe this is exactly how the original nowhere differentiable, continuous function was produced; although, the particular form of the iterates was different (they still had the qualitative behaviour of yours). –  David Mitra Feb 8 '12 at 23:10
    
@DavidMitra Ok. Look at the new image I uploaded. It's quite "fractalish" isn't it? –  Pedro Tamaroff Feb 8 '12 at 23:15

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up vote 2 down vote accepted

As I mentioned above, just considering your iterates $y_i$ will lead you to the 0 function, which is differentiable. What you want to do is consider the sums of the iterates, $y_0+y_1$, $y_0+y_1+y_2$, etc...

Some care needs to be taken so that the infinite sum converges. You are ok, since the sum of the amplitudes of the iterates converges.

With $y_0=1-|1-x|$. And $y_n=2^{-n}(1-|1-2^ny_{n-1}|)$ (click on image, then on "view image" for the full size picture).

enter image description here

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With what soft did you do that? –  Pedro Tamaroff Feb 9 '12 at 0:48
    
@Peter jsxgraph.org –  David Mitra Feb 9 '12 at 0:51
    
The batrachion... interesting. :) –  J. M. Feb 9 '12 at 0:54
    
@J.M. Could you add some more JM? If its in an answer better. –  Pedro Tamaroff Feb 9 '12 at 1:02
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Actually, this curve is exactly the Blancmange curve. –  Willie Wong Feb 9 '12 at 1:12

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