In contrast to my last post...

[pbrane]

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 liya_rowan a nice and practical 0'th birthday present like a pack of tushie wipes or a portable high chair

Comments

[pbrane]

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)).

[mmcirvin]

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.

[mmcirvin]

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.

[pbrane]

why can't sin^(N-2)[theta] ever be negative?

[pbrane]

Mmm, yesyes. This makes sense, and the 1/N scaling jives with my unit cube calculation.

[mmcirvin]

The answer should certainly go to zero in the large N limit. More dimensions, less chance the vectors will be almost parallel.

[pbrane]

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'".

[pbrane]

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?

[mmcirvin]

Theta is the polar angle, goes from zero to pi. Sin theta is positive over that range.

[pbrane]

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...