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S Apr 9, 2019 at 15:01 history bounty ended CommunityBot
S Apr 9, 2019 at 15:01 history notice removed CommunityBot
S Apr 2, 2019 at 14:27 history edited R.P.
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Apr 2, 2019 at 14:04 review Suggested edits
S Apr 2, 2019 at 14:27
Apr 2, 2019 at 14:03 answer added Alex Ravsky timeline score: 0
S Apr 2, 2019 at 3:56 history suggested CommunityBot CC BY-SA 4.0
\operatorname{} has context-dependent spacing an in some other respects does not always have the same effect as \mbox{}. Also some other minor edits.
Apr 2, 2019 at 1:40 review Suggested edits
S Apr 2, 2019 at 3:56
S Apr 1, 2019 at 13:53 history bounty started j.s.
S Apr 1, 2019 at 13:53 history notice added j.s. Canonical answer required
Apr 25, 2017 at 9:18 answer added Rodrigo de Azevedo timeline score: 1
S Apr 23, 2017 at 19:33 history edited Taras Banakh CC BY-SA 3.0
Improved the appearance of {i,j}
S Apr 23, 2017 at 19:33 history suggested Rodrigo de Azevedo CC BY-SA 3.0
Improved the question
Apr 23, 2017 at 19:11 review Suggested edits
S Apr 23, 2017 at 19:33
Apr 23, 2017 at 12:04 history edited Tom Goodwillie CC BY-SA 3.0
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Apr 23, 2017 at 11:49 history edited Tom Goodwillie CC BY-SA 3.0
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Apr 23, 2017 at 10:06 history edited j.s. CC BY-SA 3.0
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Apr 23, 2017 at 1:26 history edited j.s. CC BY-SA 3.0
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Apr 22, 2017 at 20:36 history edited j.s. CC BY-SA 3.0
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Apr 22, 2017 at 19:21 comment added j.s. @Rasberry: No, entries of $\nabla{f}$ indexed by all two-element subsets of ${\{1,...,n}\}$.
Apr 22, 2017 at 19:07 comment added Thomas Rasberry Apologies for running too far off-topic, but is $\mathbb{R}^{\binom{n}{2}}$ the same as $\mathbb{R}^{n \times n}?$
Apr 22, 2017 at 18:56 history edited j.s. CC BY-SA 3.0
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Apr 22, 2017 at 17:43 history edited j.s. CC BY-SA 3.0
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Apr 22, 2017 at 15:52 history edited j.s. CC BY-SA 3.0
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Apr 22, 2017 at 13:08 comment added j.s. @Poloni. I have checked it for many more than 10.000 random functions.
Apr 22, 2017 at 12:42 comment added Federico Poloni This is the kind of claims that I start believing in only after I have done at least 10.000 random experiments without finding any counterexample...
Apr 22, 2017 at 12:42 comment added j.s. @Firsching: Yes, you are right. I forgut mention that the angle between $f$ and $g$ is at most $\pi/2$. I edit question.
Apr 22, 2017 at 12:40 history edited j.s. CC BY-SA 3.0
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Apr 22, 2017 at 12:35 comment added Moritz Firsching what if you take $f=(1,-1)$ and $g=(-1,1)$? Isn't then $\nabla f = \nabla g = (2)$, and their distance 0, while $dist(f,g)=2\sqrt{2}$?
Apr 22, 2017 at 11:56 history edited j.s. CC BY-SA 3.0
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Apr 22, 2017 at 11:51 comment added j.s. @Beenakker: That is Euclidean distance or squared Euclidean distance (i.e. sum of squared differences of all elements).
Apr 22, 2017 at 11:44 comment added Carlo Beenakker how is is the "distance" of $\nabla f$ and $\nabla g$ defined? sum of difference of all elements squared?
Apr 22, 2017 at 11:12 history edited j.s.
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Apr 22, 2017 at 11:01 history asked j.s. CC BY-SA 3.0