Linked by Thom Holwerda on Sat 7th Apr 2007 20:58 UTC, submitted by rx182
Windows Paul Thurrot writes about Windows XP SP3: "If you were looking for any glimpse into the mind of Microsoft, this is it: the company has completely abandoned Windows XP, and it has absolutely no plans to ever ship an XP SP3. My guess is that Microsoft will do what it did with the final Windows 2000 Service Pack: claim years later that it's no longer needed and just ship a final security patch roll-up. This is the worst kiss-off to any Microsoft product I've ever seen, and you'd think the company would show a little more respect to its best-selling OS of all time."
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RE[5]: so ?
by h times nue equals e on Sun 8th Apr 2007 08:23 UTC in reply to "RE[4]: so ?"
h times nue equals e
Member since:

Especially when a 'straight line' is defined as 'the shortest path between 2 points'. So No you're wrong.

You are limiting this definition to euclidean geometry, are you ?

Reply Parent Score: 2

RE[6]: so ?
by stestagg on Sun 8th Apr 2007 08:47 in reply to "RE[5]: so ?"
stestagg Member since:

No. If you were to look, from a distance, at a straight line (can be analogized by a beam of light) near a gravity well, it will apear curved yet still can be defined as a straight line.

Reply Parent Score: 2

RE[7]: so ?
by h times nue equals e on Sun 8th Apr 2007 09:10 in reply to "RE[6]: so ?"
h times nue equals e Member since:

As this is already getting to far off-topic, I would suggest we discuss this further in private if necessary.

It may very well be, that I have worded my argument too sloppy, and it is definitly possible, that your knowledge in this field surpasses mine, so feel free to correct me, as I'm very eager to learn if I have made any stupid mistakes.

Say I limit my set of objects to a curved subspace of lets say the R^N with an euclidean metric (prime example: a closed spherical surface R^(N-1) ), then (again to my knowledge) the straight line definition from the R^(N) is no longer the shortest path within the curved space that connects two points. It is in this example the shorter of the two pieces of the grand circle, that goes through both points.

I have approached this whole thing from the variational calculus point of view, where one tries to minimise the
functional of the action on the curved manifolds that result from the (holonomic) constraints, that one has to obey that I'm used to from classical mechanics.

Feel free to correct me here if I have made any obvious mistakes, otherwise I would ask you to contact me at mfschwin [at] gmx [dot] at
to discuss this further. Thanks in advance!


Edited 2007-04-08 09:14

Reply Parent Score: 2