jake ([info]pbrane) wrote,
@ 2007-08-02 06:35:00
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In contrast to my last post...
It's not that I don't love you guys... it's just hard to post (for some reason, my 6am brain wrote "poast" first... and I thought it was right - I must be hungry for some tost) these days. Roughly 8 weeks until impending fatherhood (if you want to get [info]liya_rowan a nice and practical 0'th birthday present like a pack of tushie wipes or a portable high chair, swing on over to her gift registry: here [if that link doesn't work, just go to myregistry.com and search for my first and last name]), whee!

In other news, to contrast with my last post - at work yesterday, the ability to get my job done depended on knowing what the average angle between two randomly chosen vectors in N-dimensional space (for large [but finite] N). It's not hard to figure out what the answer is if the vectors lie in the unit cube (it's O(log(N)/N), I think), but on the unit sphere is trickier (and as N grows, the disparity of the two answers is probably large). Anyone else happen to know the average dot product of two unit vectors in R^N, given an even distribution on S^N-1 ?

And now, I'm late - off yer butt, [info]pbrane! Dog must be walked, bus must be caught! Next week: World Diplomacy Championships in Vancouver, where my diplomacy scoring system (and rails app to calculate it) will be used to determine The World Champion BackstabberTM.

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[info]evan
2007-08-02 03:42 pm UTC (link)
For some reason it takes ~8 seconds to load pages on your rails app. I think you need to use FastCGI.

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[info]pbrane
2007-08-02 04:00 pm UTC (link)
I'm on shared hosting (dreamhost), and I've tried to make sure that I've got it using FastCGI, but for some reason it occasionally acts like it's totally not... :\

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[info]onhava
2007-08-02 04:07 pm UTC (link)
Probably I'm being dumb and/or misunderstanding the question, but: if we say the first vector is along, say, the z-axis, then we're just asking for the average z-component of a randomly chosen vector on the sphere, right? And isn't that just zero? In other words, I would think the average angle is 90 degrees, in any number of dimensions.... Do you want the average absolute value of cos(theta), instead of just the average cos(theta)?

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[info]pbrane
2007-08-02 04:15 pm UTC (link)
Do you want the average absolute value of cos(theta), instead of just the average cos(theta)?

Yes, that's what I meant (or equally useful, the average of cos^2(theta)).

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[info]mmcirvin
2007-08-02 04:48 pm UTC (link)
Integrate cos^N theta times area of unit S^(N-2) from zero to pi, then divide by area of S^(N-1).

I think.

I used to know what the area of S^N was back when I was in Sidney Coleman's field theory class. Now I don't but I think you can look it up.

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[info]mmcirvin
2007-08-02 04:53 pm UTC (link)
No, no, that can't be right.

That cos^N theta should be cos^2 theta * sin^(N-2) theta. Can't ever be negative.

Uh, maybe.

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[info]pbrane
2007-08-02 05:08 pm UTC (link)
why can't sin^(N-2)[theta] ever be negative?

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[info]onhava
2007-08-02 05:57 pm UTC (link)
Well, it's really an absolute value |sin(theta)|: at each choice of z=cos(theta) (might as well say between 0 and 1, the other half contributes equally), you have one lower-dimensional sphere of radius sqrt(1-z^2). The area (or volume, or whatever you want to call it) of that sphere scales like (1-z^2)^((N-1)/2). So it's the integral dz of z^2 (1-z^2)^((N-1)/2) from z=0 to 1, divided by the same integral without the z^2. I think.

Sound OK? I think it's the same thing [info]mmcirvin was saying....

Mathematica tells me this is 1/(N+2).

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[info]pbrane
2007-08-02 06:33 pm UTC (link)
Mmm, yesyes. This makes sense, and the 1/N scaling jives with my unit cube calculation.

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[info]mmcirvin
2007-08-02 08:09 pm UTC (link)
The answer should certainly go to zero in the large N limit. More dimensions, less chance the vectors will be almost parallel.

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[info]pbrane
2007-08-02 08:21 pm UTC (link)
Which is what the actual application deals with: I've got an algorithm which hunts for vectors, and to test it, I can feed in data which I know the answer to; then I need to see whether I'm getting close or not. Taking the (normalized) dot product gets me something, and then the question becomes "how close to 1 is 'close' when talking about the dot product of two vectors on S^N?" This discussion leads me to believe it's "anything much bigger (esp. parametrically) than 1/N is 'close'".

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[info]pbrane
2007-08-02 06:40 pm UTC (link)
But should the distribution of choices of z-coordinate of the second vector be uniform on [0,1]? This is the part I'm not sure about - do uniformly distributed points on the sphere have uniform z-coordinates on [-1,1]? Is there an obvious thing I'm seeing that shows that this is true?

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[info]onhava
2007-08-02 07:35 pm UTC (link)
Hrmm, you're right.... Looking at the one case I actually understand without thinking too hard, the 2-sphere, we compute the surface area as an integral of 2 pi times d(cos theta), or 2 pi times sin theta d(theta). So the right thing is to integrate d(theta), not dz... In other words, I was saying this one is the integral of sin(theta) d(cos(theta)) so I need to remove an extra factor of sin(theta). So what I want is z^2 (1-z^2)^((N-3)/2) in the first integral, and the same thing without z^2 in the second.

So, it's just 1/N, not 1/(N+2). Unless there's some further subtlety in the right integration measure in higher dimensions. I don't think there is though.

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[info]onhava
2007-08-02 08:56 pm UTC (link)
Ugh, I keep screwing up. Let me try again: I'm removing one power of *sine*, not of *sine^2*, so that should be neither N-1 nor N-3 in the exponent but N-2, and the answer is 1/(N+1).

In other words, what [info]mmcirvin said somewhere above. But definitely O(1/N), even if I'm still wrong.

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[info]mmcirvin
2007-08-02 08:11 pm UTC (link)
Theta is the polar angle, goes from zero to pi. Sin theta is positive over that range.

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[info]pbrane
2007-08-02 09:29 pm UTC (link)
ah, damn you and your "theta is the polar angle" ways. I was always in the phi camp for that (since that way theta can be the same angle in both 2 in 3 dimensions), and ya gots me confused!

Thanks for your help though. 1/N scaling is something I like to see. Although now that I think about it, I did the cube wrong - that case goes like O(1/sqrt(N)), not O(log(N)/N)... and that's a different scaling entirely... hmmm...

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