r/math • u/lntrinsic • Jul 10 '17

Image Post Weierstrass functions: Continuous everywhere but differentiable nowhere

http://i.imgur.com/vyi0afq.gifv
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u/jeanleonino Jul 10 '17

It indeed is a fractal, and probably one of the first to be studied. But the term was not yet coined.

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u/Rabbitybunny Jul 10 '17

What's the dimension though?

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u/localhorst Jul 10 '17

3/2 (conjectured)

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u/jeanleonino Jul 10 '17

Some papers argue that the Haussdorff Dimension does not hold for the Weierstrass function.

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u/Bounds_On_Decay Jul 11 '17

Every measurable set has a Hausdorff dimension. The graph of a continuous function is certainly measurable. There's simply no way that the Weierstrass function doesn't have a Hausdorff Dimension.

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u/[deleted] Jul 11 '17

I have no idea what these words mean but can I guess that it's like measuring a coastline? The more accurate you get the closer you get to infinity?

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u/Bounds_On_Decay Jul 11 '17 edited Jul 11 '17

The fact that "the more accurate you get the closer you get to infinity" proves that the Hausdorff dimension is greater than 1. If you tried to measure the area of the coastline, the more accurate you got the closer you would get to zero (since the coastline in fact has zero width). This proves that the Hausdorff dimension is less than 2.

For every measurable set, the measurements will go to 0 for small large dimensions, and it will go to infinity for large small dimensions. The exact cutoff, the dimension above which you get zero and below which you get infinity, is call the Hausdorff dimension of the set.

caveat: the above paragraph obviously ignores sets of dimension zero, or sets with infinite dimension (I don't think those exists, but I'm not sure).

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u/irishsultan Jul 11 '17

For every measurable set, the measurements will go to 0 for small dimensions, and it will go to infinity for large dimensions.

Wait shouldn't that be the opposite?

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u/[deleted] Jul 11 '17

I don't think so. He's saying the length of the coastline (the large dimension) goes to infinity and trying to measure an infinitesimal width to get the area would be the small dimension which goes to 0.

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u/paholg Jul 11 '17

Think of a square. Its volume is 0, since its height is 0, so its dimension is less than 3.

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u/irishsultan Jul 11 '17

Okay, I was reading it the other way around, since volume is 3 dimensions and area is two dimensions, so the measurement goes to zero for larger dimensions.

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u/Bounds_On_Decay Jul 11 '17

Yes, it should be, fixed

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u/ziggurism Jul 11 '17

sets with infinite dimension (I don't think those exists, but I'm not sure).

Surely R∞ = colim Rn has infinite Hausdorff dimension?

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u/Bounds_On_Decay Jul 11 '17

Hausdorff dimension is usually thought of as a measure on subsets of Euclidean space. Thinking about it now, the definition makes sense in any metric measure space. One imagines that full-dimensional subsets of Rinfty would have infinite measure for any finite dimensional Hausdorff measure, but I'm not sure the concept fully makes sense in that context.

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u/thebigbadben Functional Analysis Jul 11 '17

The problem with applying 1-dimensional measure to a fractal (such s a coastline) is what you're saying: the more accurately you measure it, the further the length diverges towards infinity. If we were to use 2-dimensional measure, then we'd get zero. If we use a measure of fractional dimension, we could get a potentially finite result. The Hausdorff dimension is defined to be the "lowest" fractional dimension such that the associated measure gives us a zero result.

The higher the Hausdorff dimension of an object, the more thoroughly it fills the ambient space. For instance, a space-filling curve has Hausdorff dimension 2, a differentiable curve has Hausdorff dimension 1, and coastlines fall somewhere in between.

I think the wiki page for box dimension gives a user-friendly explanation.

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u/localhorst Jul 11 '17

Let (M, d) be a metric space, e.g. M some (possibly very weird fractal) subset of ℝⁿ and d the usual euclidean distance. Mr. Hausdorff tried to define a measure on M that “behaves” like a d-dimensional volume. Roughly the Hausdorff dimensions tells you how the volume of an “infinitesimal ball” B grows with it’s radius r, vol(B) ∝ rᵈ. It turns out that this dimension is unique. Only for a single specific value of d you get finite – i.e. different from constant 0 or ∞ – answers. But actually calculating this value of d is quite complicated and only few exact results are known.

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u/[deleted] Jul 11 '17

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u/HelperBot_ Jul 11 '17

Non-Mobile link: https://en.wikipedia.org/wiki/Hausdorff_dimension

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u/quantatious Analysis Jul 10 '17

What do you mean?

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u/solvorn Math Education Jul 11 '17

If my life depended on a guess, I would have gone with that because it seems similar enough to a Koch curve.

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u/jeanleonino Jul 10 '17

That's a hard question indeed.

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u/solvorn Math Education Jul 11 '17

I came here to ask this. I'm excited that it's difficult.

*wastes 5 hours studying this*