Does Recasting affect Quartic Discriminant?
Mt Gargas
ax^4 + 4bx^3 + 6cx^2 + 4dx + e = 0 ........(1)
and then by the substitution of (y-b/a) for x recast the Standard Form
as
y^4 + Fy^2 + Gy + S =0 .............(2)
giving what i think is called the reduced quartic
where
F = 6(ac-b^2)/a^2
G = 4(da^2 - 3abc + 2b^3)/a^3
S = (ea^3 -4bda^2 +6acb^2 -3b^4)/a^4
Assuming that to be the case (and indeed maybe it is not) my questions
are
:-
What is the discriminant of equation(2) and is that discriminant equal
in
magnitude and sign to the discriminant of equation(1)?
What is the reason for that being/not-being the case
Pad
AP
(1)/a and (2) have the same discri because differences between two
roots are the same
the discri of (2) is
4(p^2/3+4r)^3-27(2p^3/27-8pr/3+q^2)^2
with p=F q=G r=S
Chip Eastham
But in (2) the leading coefficient has been factored
out, and that does affect the discriminant, right?
The leading coefficient appears to a power in front
of the product of differences of roots IIRC.
regards, chip
Mt Gargas
>
> - Show quoted text -
> the discri of (2) is
> 4(p^2/3+4r)^3-27(2p^3/27-8pr/3+q^2)^2
Thank you for replies
--Where AP might I get a source for 4(p^2/3+4r)^3-27(2p^3/27-8pr/
3+q^2)^2?
--Chip, Can the 4(p^2/3+4r)^3-27(2p^3/27-8pr/3+q^2)^2 still be
regarded as correct for the equation(2) discriminant despite the 'the
leading coefficient' having 'been factored out'?
AP
I have said (1)/a : x^4+4bx^3/a+... and (2)
not (1) and (2)
Ken Pledger
<6adc9859-ec38-4c02...@k9g2000yqi.googlegroups.com>,
Mt Gargas <mtga...@gmail.com> wrote:
> > ....
> > > <mtgar...@gmail.com> wrote:
> > > >If we take the Standard Form of the Quartic equation as:-
> > > >ax^4 + 4bx^3 + 6cx^2 + 4dx + e = 0 ........(1)
> ....
> --Where AP might I get a source for 4(p^2/3+4r)^3-27(2p^3/27-8pr/
> 3+q^2)^2? ....
Ah, you need a really old-fashioned book such as S. Barnard & J. M.
Child, "Higher Algebra" (1936 and reprints). Chapter XII "Cubic and
Biquadratic Equations" has plenty of information for you.
Ken Pledger.
Mt Gargas
> not (1) and (2)
Apologies AP I thought that was a typo error.
Any websites where I might see how 4(p^2/3+4r)^3-27(2p^3/27-8pr/
3+q^2)^2 is arrived at?
Thank You for replying.
Pad.
Mt Gargas
> Child, "Higher Algebra" (1936 and reprints). Chapter XII "Cubic and
> Biquadratic Equations" has plenty of information for you.
>
> Ken Pledger.
Thank you for that Ken
Any online source that is FREE where I might see how
4(p^2/3+4r)^3-27(2p^3/27-8pr/3+q^2)^2
is arrived at?
Thank you for your reply
Pad
achille
There is a reason why the discriminant of quartic polynomial
f(x) = x^4 + p x^2 + q x + r
has the very simple form:
4(p^2/3+4r)^3 - 27(2p^3/27-8pr/3+q^2)^2
For every quartic polynomial, there is a resolvent cubic
polynomial associated to it. For our f(x), it is given by
g(x) = x^3 - 2px^2 + (p^2 - 4r)x + q^2
and these two polynomials have exactly the same discriminant!
Recall the classical way to solve f(x) = 0 is to first factor
f(x) as (x^2 + kx + l)(x^2 - kx + m). Equating coefficients
of like terms gives:
l + m - k^2 = p
k(m-l) = q
lm = r
Eliminating l and m gives
-g(-k^2) = k^6 + 2 p k^4 + (p^2-4r) k^2 - q^2 = 0
Let a1, a2, a3, a4 be the roots of f(x) and suppose
x^2 + kx + l = (x-a1)(x-a2)
x^2 - kx + m = (x-a3)(x-a4)
we see (a1+a2)*(a3+a4) = -k^2 and hence is a root of g(x).
Since the indexing of the roots is arbitrary, one find
g(x) = (x - u1)(x - u2)(x - u3)
where
u1 = (a1+a2)*(a3+a4)
u2 = (a1+a3)*(a2+a4)
u3 = (a1+a4)*(a2+a3)
Notice u1 - u2 = -(a1-a4)*(a2-a3)
u1 - u3 = -(a1-a3)*(a2-a4)
u2 - u3 = -(a1-a2)*(a3-a4), we get
discriminant of g(x)
= ((u1-u2)(u1-u3)(u2-u3))^2
= ((a1-a4)(a2-a3)(a1-a3)(a2-a4)(a1-a2)(a3-a4))^2
= discriminant of f(x).
Since the discriminant of a polynomial depends only on
differences of its roots. g(x) and g(x+2p/3) has same
discriminant. Now g(x+2p/3) can be rewritten as
x^3 + (-p^2/3-4r) x + (2p^3/27-8pr/3+q^2)
Using the fact the discriminant for a cubic polynomial of
the form x^3 + Ix + J is -4I^3 - 27J^2. One immediately
fine the discriminant of g(x) and hence f(x) is given by
4(p^2/3+4r)^3 - 27(2p^3/27-8pr/3+q^2)^2
Mt Gargas
Great stuff Achille
Thank you very much for going to the trouble of providing a very clear
explanation that is most helpful.
Sincerely
Pad
Chip Eastham
Indeed you did, sorry I overlooked that!
--chip
Mt Gargas
I'm still missing something here.
I can 'follow' up to and including the following line:-
u1 = (a1+a2)*(a3+a4)
I was then expecting something like (a1+a2+a3+a4) = 0
but am a lost to see where:-
u2 = (a1+a3)*(a2+a4)
u3 = (a1+a4)*(a2+a3) came from
Likewise I can not see where (x+2p/3) comes from
or how from that we get:-
x^3 + (-p^2/3-4r) x + (2p^3/27-8pr/3+q^2)
The remainder I can follow but not these critical parts in between
Given the clear succinct presentation in the previous reply I suppose
I feel a bit embarrassed confessing my inability to 'see' what is
probably apparent to most reading this in the sci.math newsgroup.
Can anyone help please?
Pad
achille
>
> I'm still missing something here.
>
> I can 'follow' up to and including the following line:-
> u1 = (a1+a2)*(a3+a4)
>
> I was then expecting something like (a1+a2+a3+a4) = 0
> but am a lost to see where:-
> u2 = (a1+a3)*(a2+a4)
> u3 = (a1+a4)*(a2+a3) came from
>
Given 4 roots a1, a2, a3, a4 for f(x) = 0, there
are 6 ways to factorize f(x) as product of pair
of quadratics. 3 of them is given by:
f(x) = (x^2 + k*x + l) * (x^2 - k*x + m)
= ((x-a1)*(x-a2)) * ((x-a3)*(x-a4))
OR ((x-a1)*(x-a3)) * ((x-a2)*(x-a4))
OR ((x-a1)*(x-a4)) * ((x-a2)*(x-a3))
(The remain 3 can be obtained by the transformation
k -> -k, l <-> m.) This give you 3 possible values
of -k^2:
(a1+a2)*(a3+a4) OR (a1+a3)*(a2+a4) OR (a1+a4)*(a2+a3).
and all of them have to be roots of g(x).
> Likewise I can not see where (x+2p/3) comes from
For any constant c, g(x + c) is a cubic polynomial in x.
The choice of c = 2p/3 makes the x^2 coefficient in g(x + c)
vanishes.
> or how from that we get:-
> x^3 + (-p^2/3-4r) x + (2p^3/27-8pr/3+q^2)
Get a piece of paper or use a computer algebra system
to expand out g(x+2p/3) explictly.
Mt Gargas
Ah. I see.
Thank you for making clear what I was not able to discern myself
Gratitude to all who sci.math who have answered this and provided
solution.
Pad