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Oct 9, 2021 at 15:25 comment added LSpice I edited to fix some typos and link to @‍abx's comment and the answer I think you wanted. Since "the first answer" is not well defined over time, I looked for one that seemed to fit your description, and the accepted answer does, so I linked to that. I apologise if it was not correct. In the meantime, I took @user347489's suggestion and moved the disclaimer to the front.
Oct 9, 2021 at 15:23 history edited LSpice CC BY-SA 4.0
Proofreading; link to comment and answer; moved edit
Aug 22, 2021 at 7:34 comment added user347489 It'd be nice if the edit was a disclaimer before the main text instead of at the very end.
Aug 21, 2021 at 20:44 history edited Libli CC BY-SA 4.0
Edit added following comments below tbe answer.
Aug 18, 2021 at 16:10 comment added Libli @abx : right, stupid mistake. Thanks for the comment. It seems I need a fourth hypersurface to get rid off the residual points of intersection. I will edit.
Aug 16, 2021 at 7:40 comment added abx Why is it sufficient to prove that your general surface in $|\mathscr{I}(d_2)|$ doesn't contain $\mathscr{C}_i$? It could intersect $\mathscr{C}_i$ in a point not contained in $\mathscr{C}$.
Aug 16, 2021 at 6:48 history edited Libli CC BY-SA 4.0
Fix some indexes typos and spelling mistakes
Aug 15, 2021 at 8:30 history answered Libli CC BY-SA 4.0