S-expression: Difference between revisions

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= (k3 + 3k2)/( k3 + 3k2 - 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)2 = 4n2~~, ~~therefore Sk/S(k+2) is superparticular when k is even~~.

This result will be useful, so we will refer to it as [Eq. 1]:

~~When~~ k = 4n+~~1, we have to do some work to show it is superparticular:~~

'''[Eq. 1]''' Sk/S(k+2) = (k3 + 3k2)/( k3 + 3k2 - 4 )

~~First let's note (4n+1)3~~ = ~~(4n+~~1)(4n+1)(4n+1) = (4n)3 + 3*4(4n)2 ~~+ 3*(~~4n)2 + 1 is ~~of the form 4m+1~~.

Note that when k = 2n in [Eq. 1], 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)2 = 4n2, therefore Sk/S(k+2) is superparticular when k is even.

~~Next let's note (~~4n+1~~)2 = (4n)2 + 4(4n) +~~ 1 is ~~also of the form 4m+1.~~

When k = 4n+1 in [Eq. 1], we have to do some work to show the result is superparticular:

Therefore we can replace their occurrences with 4m+1 and 4a+1 respectively, without having to worry about what m and a are (as we know ~~they~~ are positive integers).

(4n+1)3 = (4n)3 + 3*(4n)2 + 3*4n + 1 is of the form 4m+1.

(4n+1)2 = (4n)2 + 2*4n + 1 is also of the form 4m+1.

Therefore we can replace their occurrences in [Eq. 1] with 4m+1 and 4a+1 respectively, without having to worry about what m and a are (as we only need to know that m and a are positive integers). Therefore to show S(4n+1)/S(4n+3) is superparticular, we plug in k=4n+1 into [Eq. 1] and then do the replacements and simplify to a superparticular, shown below:

S(4n+1)/S(4n+3) = ( (4n+1)3 + 3(4n+1)2 )/( (4n+1)3 + 3(4n+1)2 - 4 )

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 = (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.
+This result will be useful, so we will refer to it as [Eq. 1]:
-When k = 4n+1, we have to do some work to show it is superparticular:
+'''[Eq. 1]'''   Sk/S(k+2) = (k<sup>3</sup> + 3k<sup>2</sup>)/( k<sup>3</sup> + 3k<sup>2</sup> - 4 )
-First let's note (4n+1)<sup>3</sup> = (4n+1)(4n+1)(4n+1) = (4n)<sup>3</sup> + 3*4(4n)<sup>2</sup> + 3*(4n)<sup>2</sup> + 1 is of the form 4m+1.
+Note that when k = 2n in [Eq. 1], 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.
-Next let's note (4n+1)<sup>2</sup> = (4n)<sup>2</sup> + 4(4n) + 1 is also of the form 4m+1.
+When k = 4n+1 in [Eq. 1], we have to do some work to show the result is superparticular:
-Therefore we can replace their occurrences with 4m+1 and 4a+1 respectively, without having to worry about what m and a are (as we know they are positive integers).
+(4n+1)<sup>3</sup> = (4n)<sup>3</sup> + 3*(4n)<sup>2</sup> + 3*4n + 1 is of the form 4m+1.
+(4n+1)<sup>2</sup> = (4n)<sup>2</sup> + 2*4n + 1 is also of the form 4m+1.
+Therefore we can replace their occurrences in [Eq. 1] with 4m+1 and 4a+1 respectively, without having to worry about what m and a are (as we only need to know that m and a are positive integers). Therefore to show S(4n+1)/S(4n+3) is superparticular, we plug in k=4n+1 into [Eq. 1] and then do the replacements and simplify to a superparticular, shown below:
 S(4n+1)/S(4n+3) = ( (4n+1)<sup>3</sup> + 3(4n+1)<sup>2</sup> )/( (4n+1)<sup>3</sup> + 3(4n+1)<sup>2</sup> - 4 )