balls in 2D

[gusl]

The L₁ ball in 2D is shaped like a diamond (L₁ is also known as the Manhattan norm). The L_∞ ball is shaped like a square (L_∞ is also known as the supremum norm). They are similar, i.e. have same shape. The L₂ ball is shaped like a circle.

Hypothesis: For all n in the interval (1,2), there is m>2 such that the m-ball and the n-ball are similar.

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In case you need the mathematical background:
The L_n ball is the set of points whose L_n norm is < 1.
If we call our coordinates x and y, then the L_n norm is defined as (|x|ⁿ + |y|ⁿ)^1/n (for n=-1, we get the formula for resistance in a parallel circuit)

Comments

[rdore.livejournal.com]

you should post this on mathoverflow.

[gustavolacerda.livejournal.com]

Thanks. Which tags should I use? Algebraic geometry?

[rdore.livejournal.com]

Definitely not algebraic geometry. I would probably just use something like linear algebra or functional analysis, though they are not great fits. You can feel free to include in your question a comment that you weren't sure how to tag it.

I agree with en_ki's hunch that the answer is probably no, but I haven't thought up a counterexample.

[gustavolacerda.livejournal.com]

I can think of a probable counterexample: just pick a random number in (1,2).
If it is false, I imagine the counterexamples will be the whole set.
But I haven't thought up an argument.

[stepleton.livejournal.com]

Wouldn't the high school math approach to this be to come up with equations for the appropriate ball boundaries and then demonstrate somehow that there's no way (i.e. through scaling and picking the right corresponding exponent) for the curves to be equal?

[gustavolacerda.livejournal.com]

sure!

[gustavolacerda.livejournal.com]

done!

[en-ki.livejournal.com]

Nickel gets a dollar that's not true. If it's true, nickel gets a dollar the dimensions are Hölder conjugates.

[the-locster.livejournal.com]

Also I was trying to figure out recently where the point of minimum squared error is in L1 space. For L2 it's the coordinate-wise mean, for L1 you often see the coord-wise median used to minimize error (e.g. where to locate fire station in Manhattan), but what of you wish to minimize error^2 ?

[the-locster.livejournal.com]

.. it came up while writing:
http://sites.google.com/site/sharpneat/speciation/speciation-by-k-means-clustering

[gustavolacerda.livejournal.com]

minimizing error^2 is equivalent to minimizing error.

Similarly, people choose to maximize log-likelihood, rather than likelihood, because it gives the same result.

[the-locster.livejournal.com]

Except I don't think they're equivalent in L1 space. To minimize error (E) in L1 space we take the component wise median, rather than the mean. Thus if we take the component-wise mean then we haven't minimized E and therefore (I think) E^2.

Consider the line AB and central point C. Any point (position of C) on AB minimizes E because L1 distance is AC + CB. But mean E^2 = (AC^2 + CB^2)/2. Hence in that trivial case you *can* minimize E^2 by taking the component-wise mean of the coords A and B. But this does not hold when all points are no longer on a straight line.