User talk:Fly by Night/Archive Sep 10

The following content exists solely as an archive.

PLEASE DO NOT MODIFY IT IN ANY WAY.

Hello Fly by Night. re our discussion in the math ref desk I suggest you study the article 1 + 2 + 4 + 8 + ..., and search the archives of the math desk for 1 + 2 + 4 + 8 ... It is an old discussion and the result −1 is not 'absurd'. Kind regards. Bo Jacoby (talk) 06:31, 4 September 2010 (UTC).[reply]

Thanks for your suggestion. I have already stated on the maths reference desk why I find the conclusion to be absurd, and I have no wish to investigate this murky backwater any further. Thanks again for your suggested reading. — Fly by Night (talk) 22:02, 4 September 2010 (UTC)[reply]

Having read the 1 + 2 + 4 + 8 + ... article, as you suggested, I noticed that it contains the following: "...it is absurd that a series of positive terms could have a negative value." I also searched the archive, see here for example. Some selected quotes are:

• "It's probably more natural to say that the sum is either infinite or undefined..."
• "In the real numbers, with the usual definition of summation, the sum does not converge and thus you cannot assign a real value to it."
• "You can assign those values [e.g. s = –1], but it's not particularly meaningful to do so when working in the standard real numbers."
• "...contrary to Bo's assertion, you will definitely get into all kinds of problems..."

— Fly by Night (talk) 15:07, 6 September 2010 (UTC)[reply]

Have a look at Real projective plane#Homogeneous coordinates which explains why in projective space the set of lines in the plane give the real projective plane. If we are working in real space we need to exclude lines through infinity which basically means we remove a curve from the projective plane hence giving a mobius strip.--Salix (talk): 21:47, 8 September 2010 (UTC)[reply]

Thanks Salix! I kind of worked that out today actually. The unoriented lines are given by ax + by + c = 0 which corresponds to the projective point (a:b:c). We want lines, so we have to disallow a = b = 0 which corresponds to removing the projective point (0:0:c) = (0:0:1) from the projective plane. I used the model of the projective plane that's a unit circle centred at the origin, with (±1,0) identified and the upper and lower arcs identified with opposite orientations. Deleting the points (±1,0), which are really just a point in the projective plane, gives two little open arcs around (±1,0). These open arcs can be stretched so the whole set-up gives the square { (x,y) ∈ R² : −1 < x < 1 and −1 ≤ y ≤ 1} with (x,1) identified with (−x,−1). This is exactly the open Möbius band!

The space or oriented lines is exactly the open cylinder S¹ × R. You choose a point of the circle to give the line's slope, orienting the line anti-clockwise when thought of as tangent to the unit circle. You then choose a perpendicular distance from the origin, allowing negative distances. So the line corresponding to (θ,r) is the same line, but with opposite orientation, as the one corresponding to (θ + π,−r). I'll take a look at the article that you suggested too. Thanks again Salix! — Fly by Night (talk) 19:18, 9 September 2010 (UTC)[reply]

The above content exists solely as an archive.

PLEASE DO NOT MODIFY IT IN ANY WAY.