Skip to main content
24 events
when toggle format what by license comment
S Feb 17, 2021 at 21:04 history bounty ended CommunityBot
S Feb 17, 2021 at 21:04 history notice removed CommunityBot
Feb 17, 2021 at 7:17 vote accept Guy Fsone
Feb 18, 2021 at 11:39
Feb 14, 2021 at 14:18 answer added Martin Väth timeline score: 1
Feb 12, 2021 at 22:35 answer added Guy Fsone timeline score: 0
Feb 10, 2021 at 22:27 history edited Guy Fsone CC BY-SA 4.0
added 1 character in body
Feb 10, 2021 at 22:22 answer added Martin Väth timeline score: 1
S Feb 9, 2021 at 19:19 history bounty started Guy Fsone
S Feb 9, 2021 at 19:19 history notice added Guy Fsone Draw attention
Feb 9, 2021 at 19:18 comment added Guy Fsone The main problem is to find an operator F such that the equation under consideration is equivalent to find $v_k$ such that $F(v_k)= v_k$ and on which one can apply the Brouwer fixed-point theorem
Feb 8, 2021 at 14:34 comment added Guy Fsone As far I know from the mentioned paper... I don't think so..the only growth assumption are given in the first paragraph. But if you are able to tackle this with additional assumptions let me know. I am opened to any suggestion
Feb 8, 2021 at 14:30 history edited Guy Fsone CC BY-SA 4.0
added 27 characters in body
Feb 8, 2021 at 12:35 history edited Guy Fsone CC BY-SA 4.0
added 69 characters in body
Feb 8, 2021 at 10:05 comment added Pietro Majer Is there anything else known on $\zeta$, e.g. growth assumptions?
Feb 8, 2021 at 8:57 comment added Guy Fsone @PietroMajer You are right. it is just an abuse of notation. the whole equation is projected in $\mathcal V_k$
Feb 8, 2021 at 4:07 comment added Pietro Majer How can you reduce to the last equation? Shouldn't you have the projection of $\zeta(v_k)v_k$ on ${\mathcal V}_k$ in it?
Feb 7, 2021 at 22:34 comment added Guy Fsone @MartinVäth I have changed the notations.. sorry for the confusion
Feb 7, 2021 at 22:18 history edited Guy Fsone CC BY-SA 4.0
deleted 4 characters in body
Feb 7, 2021 at 22:01 history edited Guy Fsone CC BY-SA 4.0
added 16 characters in body
Feb 7, 2021 at 22:00 comment added Guy Fsone @MartinVäth $v_k\cdot b_k= \sum v_{k,i}\lambda_i$ is just the standard scalar product.
Feb 7, 2021 at 17:27 comment added Martin Väth Is $\varphi=\phi$ and does $\phi(v_k)$ mean the composition then? And what is the product $v_k\cdot b_k$ of a function with a vector?
Feb 7, 2021 at 15:57 history edited Guy Fsone CC BY-SA 4.0
deleted 4 characters in body
Feb 7, 2021 at 15:29 history edited LSpice CC BY-SA 4.0
Proofreading; name of article; PDF -> abs
Feb 7, 2021 at 15:22 history asked Guy Fsone CC BY-SA 4.0