# Talk:Square superparticular

## I prefer a much bruter method to show the semiparticulars' superparticularity

If |
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If |
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So we see for *k* = 4*n*, 4*n* + 1, and 4*n* - 2, a coefficient of 2^{2} is canceled out, whereas for the *k* = 4*n* - 1, a coefficient of 2 is canceled out. FloraC (talk) 18:31, 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 it's superparticular for those cases. Also, it took me a little while but I believe S(k)/S(k+2) = (k+3)/(k-1) * k
^{2}/(k+2)^{2}is the equation you substituted into for the four cases? (Just arranged as one single fraction.) --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. FloraC (talk) 00:55, 8 February 2022 (UTC)

## What is So supposed to stand for?

[math]So(k) = \frac{ 4k^2 + 4k + 1 }{ 4k^2 - 4k - 3 }[/math] simplifies to [math]\frac{2 k + 1}{2 k - 3}[/math] so it doesn't feel very "square". --Frostburn (talk) 12:08, 5 July 2024 (UTC)

- From Square superparticular#Abstraction: "A suggestion is to use the notation So
*k*, if this is not unambiguous, with the letter "o" standing for "odd"." - Specifically, it's "square" w.r.t. the "odd harmonic series" (as opposed to the natural harmonic series); with respect to odds it's technically only one off of the odd analogue of superparticular.
- If that still doesn't make sense to you, think of [math] {\rm So}(k) = h_k^2 h_{k-1}^{-1} h_{k+1}^{-1} [/math] as the "source" of its squareness (as h
_{k}is the analogue of the*k*th harmonic so this is the direct analogue of k^{2}/(k+1)/(k-1)). Ideally, for the analogy to be perfect, So*k*would be odd-particulars (as it'd be impossible to have them be superparticular when they are defined entirely in terms of ratios between odd numbers), but quodd-particular is thus the next best thing. - Hope that clarifies. --Godtone (talk) 22:31, 5 July 2024 (UTC)
- I should note that "S" stands for "(Shorthand for) Second-order/Square Superparticular", so by analogy and contrast:
- you can think of "So" as standing for ""(Shorthand for) Second-order/Square Odd-particular". --Godtone (talk) 13:17, 6 July 2024 (UTC)