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Sep 25, 2023 at 8:29 answer added AccidentalTaylorExpansion timeline score: 2
Nov 30, 2022 at 4:02 answer added thone timeline score: 0
May 19, 2019 at 21:49 answer added user232454 timeline score: 0
Apr 13, 2017 at 12:39 history edited CommunityBot
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Mar 4, 2015 at 6:27 history edited Kuhlambo CC BY-SA 3.0
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Mar 2, 2015 at 18:34 comment added Kuhlambo I venture physicists just take the variables to be normalized? can any one at least confirm or deny that suspicion or point me to some source on this?
Mar 2, 2015 at 9:13 comment added Kuhlambo read the wikipedia definition it uses the covariance. The definition every physicist uses is very different from this. So why is that?
Mar 2, 2015 at 9:07 comment added Martin As I understand it now, it seems that you are asking more about what the covariance measures? And what "correlation" means in general? Then this might be better suited for mathematics.
Mar 2, 2015 at 5:20 answer added Geoff Ryan timeline score: 13
Mar 2, 2015 at 1:50 review Close votes
Mar 3, 2015 at 13:04
Mar 1, 2015 at 16:13 history edited Kuhlambo CC BY-SA 3.0
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Mar 1, 2015 at 16:04 comment added Kuhlambo shure. I know that the correlation is defined as covariance function normalized as in wikipedia. The normalization seemes to alwys be missing in physics. In qm the normalisation is possibly in the states, I don't know but either way i have never seen this addressed. Secondly i read in many places "It will have value 0 when the covariance is zero and value 1" but why this is i don't know. Shurely there must be a way to proove or at least understand this theoretically. The expression in the covariance $cov(X,Y)=E(XY)-E(X)E(Y)$ achives this, but how or why? I hope this is more helpful.
Mar 1, 2015 at 15:58 history edited glS CC BY-SA 3.0
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Mar 1, 2015 at 15:55 history edited Qmechanic CC BY-SA 3.0
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Mar 1, 2015 at 15:45 comment added ACuriousMind I think Lubos' answer there answers this question. Could you elaborate what dissatisfies you about it?
Mar 1, 2015 at 15:43 history asked Kuhlambo CC BY-SA 3.0