S-expression: Difference between revisions

@@ Line 217: / Line 217: @@
 = (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):
+For semiparticulars, we also want to show that Sk/S(k+2) is superparticular for all but the case of S(4n-1)/S(4n+1) which is odd-particular:
-To be continued...
+Sk/S(k+2) = (k+3)/(k-1) * k<sup>2</sup> / (k+2)<sup>2</sup>
+= (k<sup>3</sup> + 3k<sup>2</sup>)/( (k-1)(k<sup>2</sup> + 4k + 4) )
+= (k<sup>3</sup> + 3k<sup>2</sup>)/( k<sup>3</sup> + 4k<sup>2</sup> + 4k - k<sup>2</sup> - 4k - 4 )
+= (k<sup>3</sup> + 3k<sup>2</sup>)/( k<sup>3</sup> + 3k<sup>2</sup> - 4 )
+Note that when k = 2n, everything in the numerator and denominator is divisible by 4 because the only instances of k have it raised to a power of 2 or greater meaning there will be a factor of (2n)<sup>2</sup> = 4n<sup>2</sup>, therefore Sk/S(k+2) is superparticular when k is even.
+When k = 4n+1, we have to do some work to show it is superparticular:
 To be continued...