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Feb 14, 2013 at 18:25 vote accept Chris Birkbeck
Feb 13, 2013 at 2:24 history edited Cam McLeman CC BY-SA 3.0
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Feb 13, 2013 at 1:11 answer added Demin Hu timeline score: 2
Feb 8, 2013 at 0:43 comment added Chris Gerig but it's not just that, $d(f'') = \bar{u}$, and you can see its relation to $N$ on the previous page.
Feb 8, 2013 at 0:33 comment added Chris Birkbeck Thats the thing I though since $d$ was given by a norm, it would be that $d(f'')=N(f'')=\sum_{g \in G} g.f''(s)$?
Feb 7, 2013 at 23:54 comment added Chris Gerig It follows from the definition of the norm $N$, and is what he calls his element "a", which is $N(f') = \sum u(t,s)$.
Feb 7, 2013 at 23:29 history asked Chris Birkbeck CC BY-SA 3.0