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| John Schutkeker |
 Posted: Fri Feb 18, 2005 10:46 am |
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A problem I've been working on has led me to the following three equations
and three unknowns, and I don't see a path forward to their solution, even
though the equations are very tidy. The set is x+y=a, xy+z=b and xz=c.
Does anybody have any suggestions as to how to proceed with this? TIA. |
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| James Dolan |
| Posted: Fri Feb 18, 2005 10:46 am |
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in article <xns96016db1a14f0lkajehoriuasldfjknak@151.164.30.44>,
john schutkeker <jschutkeker@sbcglobal.net.nospam> wrote:
|A problem I've been working on has led me to the following three
|equations and three unknowns, and I don't see a path forward to their
|solution, even though the equations are very tidy. The set is x+y=a,
|xy+z=b and xz=c. Does anybody have any suggestions as to how to
|proceed with this? TIA.
does anyone find it suspicious or suggestive that this system of
equations reduces to the generic monic cubic polynomial, with the
coefficients even having (almost) their (somewhat) standard names?
--
[e-mail address jdolan@math.ucr.edu] |
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| Rod |
| Posted: Fri Feb 18, 2005 10:58 am |
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use 3) to replace z in 2). use 1) to replace y in new 2)
gives you a cubic in x which can be solved.
"John Schutkeker" <jschutkeker@sbcglobal.net.nospam> wrote in message
news:Xns96016DB1A14F0lkajehoriuasldfjknak@151.164.30.44...
[quote:d9c43cdbdf]
A problem I've been working on has led me to the following three equations
and three unknowns, and I don't see a path forward to their solution, even
though the equations are very tidy. The set is x+y=a, xy+z=b and xz=c.
Does anybody have any suggestions as to how to proceed with this? TIA.[/quote:d9c43cdbdf] |
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| Matt Gutting |
| Posted: Fri Feb 18, 2005 11:01 am |
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John Schutkeker wrote:
[quote:be0f8dff1f]A problem I've been working on has led me to the following three equations
and three unknowns, and I don't see a path forward to their solution, even
though the equations are very tidy. The set is x+y=a, xy+z=b and xz=c.
Does anybody have any suggestions as to how to proceed with this? TIA.
[/quote:be0f8dff1f]
Solve the first equation for z (assume for the moment x is nonzero): z=c/x
Substitute this into the second equation:
xy + (c/x) = b
Solve the first equation for y: y=a-x
Substitute this into the second equation:
x(a-x) + (c/x) = b
If you do some rearranging on this, you get the cubic equation
x^3 - ax^2 + bx - c = 0
which you can then solve by formula.
Of course, if x = 0, then y = a (by the first equation) and z = b. But that can
only hold if b=c=0.
Matt |
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| SU(2 |
| Posted: Fri Feb 18, 2005 11:03 am |
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※ 引述《jschutkeker@sbcglobal.net.nospam (John Schutkeker)》之銘言：
[quote:2c6752014e]A problem I've been working on has led me to the following three equations
and three unknowns, and I don't see a path forward to their solution, even
though the equations are very tidy. The set is x+y=a, xy+z=b and xz=c.
Does anybody have any suggestions as to how to proceed with this? TIA.
eliminate x and y get a cubic equation of variable z.[/quote:2c6752014e]
then try to solve the equation?
--
[1;32m※ Origin:[0m[1;33m 交大應數資訊站[0m[1m <bbs.math.nctu.edu.tw>[0m
[1;31m◆ From :[0m[1;36m laidback.Dorm8.NCTU.edu.tw [0m |
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| David Kastrup |
| Posted: Fri Feb 18, 2005 11:05 am |
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John Schutkeker <jschutkeker@sbcglobal.net.nospam> writes:
[quote:40b837c8f5]A problem I've been working on has led me to the following three
equations and three unknowns, and I don't see a path forward to
their solution, even though the equations are very tidy. The set is
x+y=a, xy+z=b and xz=c. Does anybody have any suggestions as to how
to proceed with this?
[/quote:40b837c8f5]
Simple enough. y=a-x from the first equation, so the second equation
becomes z = b-x(a-x) and the third equation
x^3 - a x^2 + b x - c = 0
This is the general form of a third degree equation.
So the three solutions will be, like, of the form
2:3 cos((arccos(3:2 ((9 a b - 2 a^3) / 27 - c) sqrt(3) / (b - a^2 / 3)
sqrt(a^2 / 3 - b)) - 2 pi k) / 3) sqrt(3) sqrt(a^2 / 3 - b) + a / 3
with k being 0, 1, 2 for the three different solutions.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum |
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| Randall |
| Posted: Fri Feb 18, 2005 11:06 am |
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[quote:83ec7679d3]A problem I've been working on has led me to the following three equations
and three unknowns, and I don't see a path forward to their solution, even
though the equations are very tidy. The set is x+y=a, xy+z=b and xz=c.
Does anybody have any suggestions as to how to proceed with this? TIA.
[/quote:83ec7679d3]
y = a-x
z = b - x*(a-x)
x(b - x*(a-x)) - c = 0
or
x^3 - a*x^2 + b*x - c = 0
you'll have to use Cardano's formula to solve this. |
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| John Schutkeker |
| Posted: Fri Feb 18, 2005 2:52 pm |
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jdolan@math-cl-n03.math.ucr.edu (James Dolan) wrote in news:cv5b1h$53n$1
@glue.ucr.edu:
[quote:35267fad50]in article <xns96016db1a14f0lkajehoriuasldfjknak@151.164.30.44>,
john schutkeker <jschutkeker@sbcglobal.net.nospam> wrote:
|A problem I've been working on has led me to the following three
|equations and three unknowns, and I don't see a path forward to their
|solution, even though the equations are very tidy. The set is x+y=a,
|xy+z=b and xz=c. Does anybody have any suggestions as to how to
|proceed with this? TIA.
does anyone find it suspicious or suggestive that this system of
equations reduces to the generic monic cubic polynomial, with the
coefficients even having (almost) their (somewhat) standard names?
[/quote:35267fad50]
And we have a winner! Those equations are derived by assuming the
solution to a cubic equation can be factored into form (x-a)(x^2+bx+c)
=x^3+Ax^2+Bx+C, and then grouping the coefficients of the polynomials of
the right and left hand sides.
I was hoping that somebody would simply stuff it into Mathematica and
see what she says, but I apparently underestimated you guys. Clearly,
Mathematica will also just regurgitate Cardano's result, at best.
I tried using his formulae some years ago, and I got sucked into the
black hole of not knowing which branch of the MORE THAN ONE FRICKIN'
SQUARE ROOT TO CHOOSE!!! <:-0
My idea was to find a simple solution for the single real root of the
cubic, and a quadratic formula with a single discriminator, to get the
other roots, whether real or complex. But I see now that my idea will
not survive this discussion.
This begs the question of who has ever successfully used Cardano's
formula, by hand, and how can I get in contact with them. I'm not a big
fan of sucking up whatever mess Mathematica might expectorate, and
accepting it as gospel. |
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| G. A. Edgar |
| Posted: Fri Feb 18, 2005 4:14 pm |
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Uspensky, THEORY OF EQUATIONS.
Published before computers were common, intended for use by hand.
There is a chapter on cubic and quartic solutions.
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/ |
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| John Schutkeker |
| Posted: Sat Feb 19, 2005 3:26 pm |
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"G. A. Edgar" <edgar@math.ohio-state.edu.invalid> wrote in
news:180220051614591323%edgar@math.ohio-state.edu.invalid:
[quote:116d6aac28]Uspensky, THEORY OF EQUATIONS.
Published before computers were common, intended for use by hand.
There is a chapter on cubic and quartic solutions.
[/quote:116d6aac28]
I looked at it, and it was underwhelming. I'm betting that the solution to
this problem may not be finished, and I will stick to that bet until
somebody produces a person or paper where Cardan's formula is successfully
used.
So my original challenge to the group remains. Who has come across
published work where Cardan's method was used to generate an actual
solution of meaningful importance? |
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| Dick Tjaden |
| Posted: Tue Feb 22, 2005 2:33 pm |
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On Sat, 19 Feb 2005 20:26:30 GMT, John Schutkeker
<jschutkeker@sbcglobal.net.nospam> wrote:
[quote:052f95ba48]"G. A. Edgar" <edgar@math.ohio-state.edu.invalid> wrote in
news:180220051614591323%edgar@math.ohio-state.edu.invalid:
Uspensky, THEORY OF EQUATIONS.
Published before computers were common, intended for use by hand.
There is a chapter on cubic and quartic solutions.
I looked at it, and it was underwhelming. I'm betting that the solution to
this problem may not be finished, and I will stick to that bet until
somebody produces a person or paper where Cardan's formula is successfully
used.
So my original challenge to the group remains. Who has come across
published work where Cardan's method was used to generate an actual
solution of meaningful importance?[/quote:052f95ba48] |
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