S-expression: Difference between revisions
m →Mathematical derivation: finished proofs (needs proofchecking) |
m →Mathematical derivation: added indentation to try to help with clearer reading as well as section off the (mostly) separate proofs |
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Sk/S(k+1) = ( k<sup>2</sup>/(k<sup>2</sup> - 1) )/( (k+1)<sup>2</sup>/((k+1)<sup>2</sup> - 1) ) | 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)/(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>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> | : = 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>: | For semiparticulars, we want to show that Sk/S(k+2) = ((k+3)/(k-1)) / ((k+2)/k)<sup>2</sup>: | ||
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Sk/S(k+2) = ( k<sup>2</sup>/(k<sup>2</sup> - 1) )/( (k+2)<sup>2</sup>/((k+2)<sup>2</sup> - 1) ) | 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-1) * (k+3)(k+1)/(k+2)<sup>2</sup> | ||
= k<sup>2</sup>/(k-1) * (k+3)/(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> | : = (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 all but the case of S(4n-1)/S(4n+1) which is odd-particular: | 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: | ||
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Sk/S(k+2) = (k+3)/(k-1) * k<sup>2</sup> / (k+2)<sup>2</sup> | 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-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> + 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 ) | : = (k<sup>3</sup> + 3k<sup>2</sup>)/( k<sup>3</sup> + 3k<sup>2</sup> - 4 ) | ||
This result will be useful, so we will refer to it as [Eq. 1]: | This result will be useful, so we will refer to it as '''[Eq. 1]''': Sk/S(k+2) = (k<sup>3</sup> + 3k<sup>2</sup>)/( k<sup>3</sup> + 3k<sup>2</sup> - 4 ) | ||
: 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. | |||
: When k = 4n+1 in [Eq. 1], we have to do some extra work to show the result is superparticular: | |||
: (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> | : (4n+1)<sup>2</sup> = (4n)<sup>2</sup> + 2*4n + 1 is also of the form 4m+1. | ||
(4n+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 set k=4n+1 in [Eq. 1] and then do the replacements and simplify to a superparticular: | ||
: 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 ) | |||
:: = ( 4m+1 + 3(4a+1) )/( 4m+1 + 3(4a+1) - 4) | |||
:: = ( 4m + 4(3a) + 4 )/( 4m + 4(3a) ) = ( m + 3a + 1 )/( m + 3a ) | |||
= ( 4m + 4(3a) + 4 )/( 4m + 4(3a) ) = ( m + 3a + 1 )/( m + 3a ) | |||
Then for the final case we want to show that S(4n-1)/S(4n+1) is odd-particular by setting k=4n-1 in [Eq. 1]: | Then for the final case we want to show that S(4n-1)/S(4n+1) is odd-particular by setting k=4n-1 in [Eq. 1]: | ||
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S(4n-1)/S(4n+1) = ( (4n-1)<sup>3</sup> + 3(4n-1)<sup>2</sup> )/( (4n-1)<sup>3</sup> + 3(4n-1)<sup>2</sup> - 4 ) | S(4n-1)/S(4n+1) = ( (4n-1)<sup>3</sup> + 3(4n-1)<sup>2</sup> )/( (4n-1)<sup>3</sup> + 3(4n-1)<sup>2</sup> - 4 ) | ||
As before we make replacements: | : As before we make replacements: | ||
: (4n-1)<sup>3</sup> = (4n)<sup>3</sup> - 3*(4n)<sup>2</sup> + 3*4n - 1 is of the form 4m-1 so will be replaced with such. | |||
: (4n+1)<sup>2</sup> = (4n)<sup>2</sup> - 2*4n + 1 is of the form 4a+1 so will be replaced with such. | |||
: Therefore: | |||
(4n+1) | : S(4n-1)/S(4n+1) = ( 4m-1 + 3(4a+1) )/( 4m-1 + 3(4a+1) - 4 ) | ||
:: = ( 4m-1 + 4(3a) + 3 )/( 4m-1 + 4(3a) + 3 - 4 ) | |||
= ( 4m | :: = ( 4m + 4(3a) + 2 )/( 4m + 4(3a) - 2 ) | ||
:: = ( 2m + 2(3a) + 1 )/( 2m + 2(3a) - 1 ) | |||
...which is of the form (x + 1)/(x - 1) meaning it is odd-particular. | :: ...which is of the form (x + 1)/(x - 1) meaning it is odd-particular. | ||
In conclusion: Sk/S(k+2) is superparticular for all integer 2 < k =/= 3 (mod 4) and is odd-particular when k = 3 (mod 4). | In conclusion: Sk/S(k+2) is superparticular for all integer 2 < k =/= 3 (mod 4) and is odd-particular when k = 3 (mod 4). | ||