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Oct 21, 2015 at 21:25 vote accept majmun
Oct 7, 2015 at 15:03 answer added Jack LaVigne timeline score: 0
Oct 7, 2015 at 4:58 comment added majmun For your first question: yes, I guess g would have no output (I've tried having g be h, if that makes sense, but I couldn't find a way to make that work). Secondly, I think you may be correct. However, if I make the $\leq$ be $<$ then no $h(v)$ (h for no v) can be set to two different values (although different v's could give the same h(v) which I am okay with). It is also very possible that I incorrectly articulated my idea. Thirdly, thank you for your comment.
Oct 7, 2015 at 4:19 answer added march timeline score: 1
Oct 7, 2015 at 4:05 answer added m_goldberg timeline score: 2
Oct 7, 2015 at 2:57 comment added march I'm confused on some points. First, g should have no output? I.e. it's sole purpose is to set the values for h? Secondly, it seems like in the general case, h for a particular v could be set to different values, since you scan over possible x's. Either the list is such that this doesn't happen, or you want to define an h for every x. I'm not sure which it is. (Or some third option based on my misunderstanding of the problem.)
Oct 7, 2015 at 2:11 comment added bbgodfrey Welcome to Mathematica.SE! I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise.
Oct 7, 2015 at 2:06 review First posts
Oct 7, 2015 at 2:11
Oct 7, 2015 at 2:06 history asked majmun CC BY-SA 3.0