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- Path: sparky!uunet!noc.near.net!bigboote.WPI.EDU!bdsm
- From: bdsm@poincare.WPI.EDU (Billy Don McConnell)
- Newsgroups: sci.math
- Subject: 3 "Simple" Quartics...
- Date: 23 Jan 1993 02:51:23 GMT
- Organization: Worcester Polytechnic Institute
- Lines: 105
- Distribution: world
- Message-ID: <1jqbrbINN41t@bigboote.WPI.EDU>
- NNTP-Posting-Host: poincare.wpi.edu
- Originator: bdsm@poincare.WPI.EDU
-
-
- I have spent an obscene amount of time over the past 5 years or so looking for the
- solutions to the following three (all but identical) Quartic Equations:
-
- 4 3 2 2 2
- 3 H - 4 S H + ( S - 2 B + 2 C )H - A = 0
-
- 4 3 2 2 2
- 3 J - 4 S J + ( S - 2 C + 2 A )J - B = 0
-
- 4 3 2 2 2
- 3 K - 4 S K + ( S - 2 A + 2 B )K - C = 0,
-
-
- where S = X + Y + Z + R
-
- A = ( X - R ) ( Y - Z )
- Important(?) Fact:
- B = ( Y - R ) ( Z - X )
- A + B + C = 0
- C = ( Z - R ) ( X - Y )
-
- (Why X, Y, Z, and then "R"? Well, that's a long story. Let's just say that this
- problem relates to an extension of the x, y, r Triangle from Trigonometry.)
-
- Note: The use of S, A, B, and C may not be "convenient" for finding the
- Solutions. (That is, maybe the quantity "X+Y+Z+R" isn't itself very
- important.) But their use makes writing the Equations easier.
-
- Now, Quartics are fairly easily handled by a method one can find in plenty of
- references (I use the _CRC_Standard_Mathematical_Tables_ myself); that is, they
- are easily handled this way when the coefficients are constants, not complicated
- expressions in X, Y, Z, and R.
-
- Maple V (Release 2) handles the symbol manipulation pretty well, giving the
- Solutions to these Equations in just a few minutes. However, the Solutions are in
- an inCREDibly complicated form, and Maple doesn't seem to know how to simplify
- them in any _real_ sense of "simplification". (My quest these 5 years has been to
- simplify these Solutions.)
-
- I am _convinced_ that the Solutions have a *NICE* representation, but I don't know
- what it is.
-
- When I first loosed Maple on the problem, I was discouraged, because even when I
- substituted numerical values for X,Y,Z, and R, I _still_ couldn't find any *nice*
- representation of the results; H, J, and K were a messy combination of cube roots
- and square roots of pretty big numbers ("big" in a relative sense of course: using
- 2, 3, 5, and 7 for X, Y, Z, and R resulted in solutions involving numbers of up to
- 12 digits, including some prime numbers of roughly that size).
-
- However, I do know that the answers must satisfy
-
- H + J + K = X + Y + Z + R (*)
-
- (In fact, for each Solution for H there is a Solution for J and one for K such
- that (*) holds. That is, the 12 Solutions can be put into four groups, each
- with an H, J, and K, satisfying (*).)
-
- The fact that Maple's (numerical) Solutions _do_ this is actually ENcouraging,
- since H, J, and K have completely different complicated parts, and I know that
- these complicated parts have to cancel each other out. So there is some under-
- lying pattern to the solutions. There must be some weird factorization of
- the cube-and-square-root parts that eludes Maple (not surprisingly, since Maple
- doesn't seem to think in terms of "weird" factorizations) and has eluded me lo
- these many years (also not surprisingly, although I have tried every "weird"
- factorization I could think of).
-
- So I'm exposing this problem to the Net to see what becomes of it. Any and all
- insights are welcome (special properties of Quartics without a first-power term,
- for instance), as are questions about the origins of the problem; SOLUTIONS are
- most welcome of all, of course...or (*sigh*) proofs that *NICE* Solutions don't
- exist. (I would even like to hear from folks who decided to think about the
- problem for a while, but then gave it up; tell me what made you stop.) Please
- respond by e-mail.
-
-
-
- Note: Some relationships between X, Y, Z, and R are special to this System.
- For instance, if any two of these quantities are equal, then 0 is a
- Root of one of the Equations. That one was pretty obvious. Less obvious,
- however, is: if X + Y + Z = R, then H = Y + Z, J = X + Z, and
- K = X + Y are three Roots of the Equations (observe that H + J + K =
- 2 X + 2 Y + 2 Z = X + Y + Z + R, as is required), but the other Roots are
- messy (as far as I can tell...I have to study this case some more).
-
- Note: In the original inception of the problem, X, Y, Z, and R are squares
- of quantities (if that matters). That is, X = x-squared, Y = y-squared,
- etc. (And I'm really looking for h-squared, j-squared, and k-squared.)
- This might be helpful in viewing the problem, since for instance A, B,
- and C (in x-y-z terms) are products of Differences of Squares. (One
- might observe how, translating the previous Note into x-y-z-h-j-k terms,
- the "less obvious" relationships are _Pythagorean_ in nature.)
-
- Well, that's about it. I hope SOMEone out there give this problem a shot... I've
- been working on it alone for long enough.
-
-
- Thanks a bunch!
-
- (Oh, and to all the folks who have responded to my search for History of Math
- Programs, thanks a bunch for THAT, too! I've got some good leads to follow up
- on. My apologies for not thanking all respondants personally.)
-
- Don Mc Connell
- bdsm@fermat.wpi.edu
-