Re: SF: Generalized SFT's
From: Tim Peters (tim.one_at_comcast.net)
Date: 04/24/05
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Date: Sat, 23 Apr 2005 18:32:39 -0400
[JSH]
>> Abstracted Surrogate Factoring Theorem:
>>
>> Given non-zero integer A and B, let
>>
>> f_1 f_2 = A^2 (A^2 - B^2)
>>
>> then
>>
>> f_1 = (-(z - 2A^2)+ sqrt((z - 2A^2)^2 - 4A^2(A^2 - B^2)))/2
>>
>> and
>>
>> f_2 = (-(z - 2A^2) - sqrt((z - 2A^2)^2 - 4A^2(A^2 - B^2)))/2
[Tim Smith]
> Let K = z-2A^2,
> L = A^2(A^2-B^2)
>
> Then what you are saying is:
>
> f_1 f_2 = L
> f_1 = -K + sqrt(K^2-4L)/2
> f_2 = -K - sqrt(K^2-4L)/2
Sorry, nope, you misread the scope to which "/2" applies; i.e.,
(-(z - 2A^2)+ sqrt((z - 2A^2)^2 - 4A^2(A^2 - B^2)))/2
^--------^ ^--------^ ^---------^
^------------------------------^
^-------------------------------------------------^
> However, multiplying f_1 by f_2 gives:
>
> f_1 f_2 = K^2 - (K^2-4L)/4 = 3K^2/4 + L
>
> This is only equal to L when K = 0.
Given the mireading, yes. Given what James actually wrote, and using your
substitutions:
[so sue me for dropping the clumsy underscores]
f1 = (-K + sqrt(K^2-4L))/2
f2 = (-K - sqrt(K^2-4L))/2
Then multiplying f1 by f2 gives (K^2-(K^2-4L))/4 = 4L/4 = L, as claimed.
> Thus, you have produced the following factorization of L:
>
> L = sqrt(-L) * sqrt(-L).
>
> We are not impressed.
That I don't dispute <wink>.
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