Re: SF: Two square, mystery celebration

From: Tim Peters (tim.one_at_comcast.net)
Date: 04/22/05 

Date: Fri, 22 Apr 2005 12:08:33 -0400

Qutoing the whole thing in full here, for the benefit of future prosecutors.

[JSH]
>>> Some posters, mostly from sci.math, are going on and on about
>>> supposedly showing that the SFT doesn't work,

[Rick Decker]
>> Oh geez! Those of us who have looked at it agree that
>> SFT works. The problem is that it is equivalent to several pages
>> of proof that say that any rational number can be expressed as
>> the product of two rationals.

>>> but the problem is, you don't even have to check their math to know
>>> they're wrong because of the two square roots in the SFT.
>>>
>>> The first one which is the easy one is
>>>
>>> Az = Ax(Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2Ax - 2M^2)
>>>
>>> and it's easy as you have
>>>
>>> Ax = +/- (k_1 + k_2) + 2j^2
>>>
>>> where k_1 k_2 = - j^2 T, so you have that square root no problem.
>>>
>>> But notice, k_1 and k_2 are factors of Tj^2, as that's VERY
>>> important.
>>>
>>> That one is the easy square root.

>> Right. Az = ((k_i + j^2)^2 - M^2 j^2) / (k_i - T), for i = 1, 2

>>> It's the second one that's problematic:
>>>
>>> b_2 f_1 = (-(Az - 2M^2)+/- sqrt((Az - 2M^2)^2 - 4TM^2))/2
>>>
>>> as it requires you have factors of TM^2 to evaluate.

>> No it doesn't. b_2 f_1 = -k_i + T or TM^2 / (-k_i + T), for i = 1, 2

[JSH]
> So now you're back to that claim.
>
> The equation
>
> a_1 z + b_1 = f_1
>
> comes from the derivation.
>
> Your claim is equivalent to asserting a_1 = 1.

>>> But a sci.math poster claims to have evaluated and didn't use
>>> factors of TM^2, so what do you know?
>>>
>>> That poster is wrong.

>> That'd be me. The simplification above is correct. I've posted it
>> once and I'll be happy to post it again. Perhaps you missed the
>> first one.
>>
>> Maybe you haven't been keeping up with current events, but, C. Bond
>> has provided us with a specific numerical example with M = 15, j = 17.
>> If you check, you'll see that the example is consistent with what
>> I've written above.
>>
>> For goodness sake, James, these examples are easy to verify by hand--
>> you don't even need your "lost" Mathematica key. Try it and see why
>> the case M = 15, j = 17, k_1 = 136 is special. Put up or shut up,
>> laddy boy.

> Well, you may be correct, but before I'm done, you'll explain in
> detail, like you're supposed to do, versus making taunting posts and
> behaving childishly.
>
> You can taunt as much as you wish Decker, but you're not doing yourself
> or anyone else any favors.
>
> You're just showing your true personality, the person you really are
> behind the play-acting of civility.
>
> At least I don't play at being civil. I say I think you're lying scum
> who can't be trusted but must be forced to explain out carefully and
> objectively, as you won't without pressure like this post.
>
> Sure, some of you may believe that you can act anyway you want in
> response to me, as childishly as you want, and refuse to elaborate when
> requested, but I'll get it out of you anyway, one way or another.
>
> So put all your cards on the table now.
>
> If not, if you wish to play social games, then this can last for
> months.
>
> As make no mistake, I'll let it last that long, drawing you out like I
> just did in post after post after post to which you weird people feel
> compelled to reply.
>
>
> James Harris

Rick already explained in detail. He even politely offered to post it again
for you, and gave you a ready-made excuse for requesting that he do so:

    I've posted it once and I'll be happy to post it again. Perhaps
    you missed the first one.

Part of me hopes he does (incorporating the sign correction he discovered
later), because I checked the first half of it at the time, and that much
was wonderfully easy to follow -- a model for clear exposition of a long,
messy derivation.

But most of me hopes he ignores you. You don't deserve more favors from
Rick, and your reply speaks for itself on that point more damningly than I
could.

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