S-expression: Difference between revisions

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 === Derivation ===
-Work in progress.
+(Note for readers: Familiarity with basic algebra is advised, especially the rules (a + b)(a - b) = a<sup>2</sup> - b<sup>2</sup> and 1/(a/b) = b/a. Often multiple steps are combined into a single step for brevity, so follow carefully.)
+For ultraparticulars, we want to show that Sk/S(k+1) = ((k+2)/(k-1)) / ((k+1)/k)<sup>3</sup>:
+Sk/S(k+1) = ( k<sup>2</sup>/(k<sup>2</sup> - 1) )/( (k+1)<sup>2</sup>/((k+1)<sup>2</sup> - 1) )
+= k<sup>2</sup>/(k+1)/(k-1) * (k<sup>2</sup> + 2k)/(k+1)<sup>2</sup>
+= k<sup>2</sup>/(k+1)<sup>3</sup>/(k-1) * (k + 2)k
+= k<sup>3</sup>/(k+1)<sup>3</sup> * (k + 2)/(k - 1) = ((k+2)/(k-1)) / ((k+1)/k)<sup>3</sup>
+For semiparticulars, we want to show that Sk/S(k+2) = ((k+3)/(k-1)) / ((k+2)/k)<sup>2</sup>:
+Sk/S(k+2) = ( k<sup>2</sup>/(k<sup>2</sup> - 1) )/( (k+2)<sup>2</sup>/((k+2)<sup>2</sup> - 1) )
+= k<sup>2</sup>/(k+1)/(k-1) * (k+3)(k+1)/(k+2)<sup>2</sup>
+= k<sup>2</sup>/(k-1) * (k+3)/(k+2)<sup>2</sup>
+= (k+3)/(k-1) * k<sup>2</sup> / (k+2)<sup>2</sup> = ((k+3)/(k-1)) / ((k+2)/k)<sup>2</sup>
+For semiparticulars, we also want to show that Sk/S(k+2) is superparticular for k =/= 0 (mod 4):
+To be continued...
+To be continued...