Talk:S-expression: Difference between revisions

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: I figured something like this would be possible but I sometimes get lost in how to simplify and group stuff in intermediate steps when I do it that way or I make mistakes simplifying/expanding so I tried to use the most intuitive approach I could think of. The observation that k=2n leads to a factor of 4 I think is a relatively intuitive explanation of why its superparticular for those cases. Also, it took me a little while but I believe S(k)/S(k+2) = (k+3)/(k-1) * k<sup>2</sup>/(k+2)<sup>2</sup> is the equation you substituted into for the four cases? (Just arranged as one single fraction.) --[[User:Godtone|Godtone]] ([[User talk:Godtone|talk]]) 23:45, 7 February 2022 (UTC)

:: Yup that's the equation I started from. We may go even more primitive if we want, such as from the definitions. It doesn't matter in the end. I consider the elementary algebraic operations an easy problem, i.e. the solution is guaranteed by a known routine. All the tricks on the other hand involve observation and intuition. I mean, it's totally reasonable for one to apply them to instantly gain the insight, but I take the liberty of assuming the readers' expectation here is a solid and sound presentation of the result. [[User:FloraC|FloraC]] ([[User talk:FloraC|talk]]) 00:55, 8 February 2022 (UTC)

Revision as of 23:45, 7 February 2022 view source Godtone (talk \| contribs) autopatrolled, emailconfirmed 2,420 edits m reply ← Older edit		Revision as of 00:55, 8 February 2022 view source FloraC (talk \| contribs) autopatrolled, Bureaucrats, Interface administrators, Sysops 24,088 edits →I prefer a much bruter method to show the semiparticulars' superparticularity: re Newer edit →
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	: I figured something like this would be possible but I sometimes get lost in how to simplify and group stuff in intermediate steps when I do it that way or I make mistakes simplifying/expanding so I tried to use the most intuitive approach I could think of. The observation that k=2n leads to a factor of 4 I think is a relatively intuitive explanation of why its superparticular for those cases. Also, it took me a little while but I believe S(k)/S(k+2) = (k+3)/(k-1) * k<sup>2</sup>/(k+2)<sup>2</sup> is the equation you substituted into for the four cases? (Just arranged as one single fraction.) --[[User:Godtone\|Godtone]] ([[User talk:Godtone\|talk]]) 23:45, 7 February 2022 (UTC)		: I figured something like this would be possible but I sometimes get lost in how to simplify and group stuff in intermediate steps when I do it that way or I make mistakes simplifying/expanding so I tried to use the most intuitive approach I could think of. The observation that k=2n leads to a factor of 4 I think is a relatively intuitive explanation of why its superparticular for those cases. Also, it took me a little while but I believe S(k)/S(k+2) = (k+3)/(k-1) * k<sup>2</sup>/(k+2)<sup>2</sup> is the equation you substituted into for the four cases? (Just arranged as one single fraction.) --[[User:Godtone\|Godtone]] ([[User talk:Godtone\|talk]]) 23:45, 7 February 2022 (UTC)

			:: Yup that's the equation I started from. We may go even more primitive if we want, such as from the definitions. It doesn't matter in the end. I consider the elementary algebraic operations an easy problem, i.e. the solution is guaranteed by a known routine. All the tricks on the other hand involve observation and intuition. I mean, it's totally reasonable for one to apply them to instantly gain the insight, but I take the liberty of assuming the readers' expectation here is a solid and sound presentation of the result. [[User:FloraC\|FloraC]] ([[User talk:FloraC\|talk]]) 00:55, 8 February 2022 (UTC)