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Is There a "Discriminant" for a Quartic Equation ... in Closed Form?

Date: 02/17/2012 at 21:13:24
From: Ian
Subject: quartic discriminant

Without revealing it, you alluded to a closed form for the discriminant of
the quartic at the end of a Dr. Math conservation entitled "Is There a
'Discriminant' for a Quartic Equation?" Will you help me come up with it?

According to you, the closed form involves two components: the
discriminant of the quartic and the value of the independent variable z at
the local maximum of the resolvent cubic. I assume that you want to find
the discriminant by using the equation on the Wolfram website you
indicated earlier in the archived conversation (if not, please indicate
how else). So that takes care of the discriminant's sign (as discussed
below).

Here's the closed form that I came up with:

   If the discriminant is positive but the z value of the local maximum is
   negative, then you have four complex roots.

   If the discriminant is positive but the z value of the local maximum is
   positive, then you have four real roots.

   If the discriminant is negative, then you have two real roots and two
   complex roots.

Now, in that archive, you also indicated that the positivity or negativity
of the z value of the local maximum could be determined by the facts k > 0
or k^2 < 4n (for z negative). I added k^2 > 4n (z positive). Earlier in
the entry, you wrote that k and n are equal to e:

   k = b - 3a^2/8
   n = d - ac/4 + ba^2/16 - 3a^4/256

Finally, you used the word "messy" to describe this closed form. The
results of my work, above, do not seem particularly messy. If you
substitute the expressions in terms of the original coefficients, and then
evaluate and see which inequality is true, you should have your second
condition (the first being the positivity or negativity of the
discriminant). Would that be so messy?

I do believe you, so I must have interpreted your thinking wrongly, but I
am not sure how.

Thanks.



Date: 02/21/2012 at 01:03:11
From: Doctor Vogler
Subject: Re: quartic discriminant

Hi Ian,

Thanks for writing to Dr. Math.

Whoa! Blast from the past. Let's see what I wrote five years ago ...

   http://mathforum.org/library/drmath/view/68049.html 

Well, yes, you seem to have correctly understood the process for
determining how many real roots you have. I suppose the question of
whether or not it is "messy" is a matter of opinion. You need not consider
it to be messy if you don't want to. And you might be able to simplify the
final result of the (possibly) messy task to the point where you get a
very tidy answer.

If you have any questions about this or need more help, please write back
and show me what you have been able to do, and I will try to offer further
suggestions.

- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 02/21/2012 at 07:53:05
From: Ian
Subject: quartic discriminant

When I substitute the expressions in terms of the original coefficients
for k and n into the final inequalities k > 0 or k^2 < 4n (the ones you
gave for z negative) and k^2 > 4n (the one I assumed for z positive), it
does indeed seem that there should be more to the second condition (z
component) for the closed form. After all, from the beginning of that
archive to the end, a lot of intermediate steps came into play:
substitutions, resolvent cubic, etc. It seems like maybe there should be
some nested expressions, comparison of original variables to those that
were substituted, etc. -- or maybe not.

You said it seems that I have understood the process. I am not sure if you
mean that my math is exactly correct. If not, can you please let me know
what exactly the closed form is and how to derive it from where you left
off in that entry?

I really cannot express my gratitude enough. Thanks.



Date: 02/22/2012 at 13:42:50
From: Doctor Vogler
Subject: Re: quartic discriminant

Hi Ian,

Thanks for writing to Dr. Math. 

Well, you already know that

   k = b - 3*a^2/8

And we can calculate that

   4*n - k^2 = -a*c - 3/16*a^4 + b*a^2 - b^2 + 4*d.

I already said that the z value of the local maximum is negative exactly
when k > 0 or 4*n - k^2 > 0, so that means that the z value of the local
maximum is negative when either ...

   8*b > 3*a^2

... or

   b*a^2 + 4*d > a*c + 3/16*a^4 + b^2.

You can get the formula for the discriminant from MathWorld or Wikipedia,
and if the discriminant is positive and (at least) one of the above two
inequalities is satisfied, then you have four complex roots.

If the discriminant is positive and neither of the above two inequalities
is satisfied, then you have four real roots.

If the discriminant is negative, then you have two real roots and two
complex roots.

You should try this for a few quartic polynomials, because it's certainly
possible that I made a mistake somewhere in all of this.

- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 02/28/2012 at 20:22:03
From: Ian
Subject: Thank you (quartic discriminant)

Thank you for answering my questions. 

I tried the discriminant-resolvent cubic method for classifying roots on
several quartics; and for the cases I tried, it worked.

I really cannot imagine how people become as creative in thinking as the
doctors at Dr. Math. Your response was very helpful.



Date: 03/06/2012 at 15:45:28
From: Doctor Vogler
Subject: Re: Thank you (quartic discriminant)

Hi Ian,

I should also point out that, in the article you referenced, I mentioned
that you can answer this question using Sturm's Theorem. For example, we
had already reduced the original problem to the equation

   p0 = y^4 + ky^2 + my + n

Then we can create a Sturm sequence with

   p1 = p0'
      = 4y^3 + 2ky + m
      
   p2 = -p0 + p1*(x/4) 
      = (-2kx^2 - 3mx - 4n)/4
      
   p3 = -p1 - p2*4*(2kx - 3m)/k^2 
      = -(Rx + S)/k^2
      
   p4 = D*k^2/4R^2

Here, D is the discriminant:

    D = 16nk^4 - 4k^3m^2 - 27m^4 + 144knm^2 - 128(kn)^2 + 256n^3.

Also,

    R = 2k^3 - 8kn + 9m^2
    S = (k^2 + 12n)m

This is a valid Sturm sequence as long as k and R are nonzero. If k is
nonzero but R = 0, then you can use

   p1 = p0'
      = 4y^3 + 2ky + m
     
   p2 = -p0 + p1*(x/4) 
      = (-2kx^2 - 3mx - 4n)/4
     
   p3 = -p1 - p2*4*(2kx - 3m)/k^2 
      = -S/k^2 
      = -m(8k^3 + 27m^2)/2k^3

In this case, D can be simplified to

    D  = m^2(8k^3 + 27m^2)^2/2k^3

And if k = 0, then you can use

   p0 = y^4 + my + n

   p1 = p0' 
      = 4y^3 + m

   p2 = -p0 + p1*(x/4) 
      = (-3mx - 4n)/4

   p3 = D/27m^3

In each case, you can check (by looking at the leading coefficients) that
Sturm's Theorem tells you that

   If D < 0, then there are 2 real roots.
   If D > 0, k < 0, and R < 0, then there are 4 real roots.
   If D > 0, but either k >= 0 or R >= 0, then there are 0 real roots.
   If D = 0, then there is a repeated root.

So this gives you another way to classify your quartic polynomial.

In fact, in the cases where D = 0, you can classify them, too. For example,

   If D = 0 but R is nonzero, then there is a double root.
   
   If D = 0, R = 0, and k = 0, then m = n = 0, too; and there is a
   quadruple root at y = 0.
   
   If D = 0, R = 0, k is nonzero, but m = 0, then you have the square of a
   quadratic, (y^2 + k/2)^2.
   
   If D = 0, R = 0, but k and m are nonzero, then you have a triple root.

And so forth.

- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 03/29/2012 at 12:49:05
From: Ian
Subject: quartic discriminant

Thank you greatly for the part you added about Sturm's Theorem. 

However, there are a few things in this most recent addition that have
left me stumped.

I understand that you start with p0 as the original equation in y. p1 is
the derivative of p0. But from what I understand from the original Dr.
Math article that I referenced -- the one that first mentions Sturm's
Theorem -- p2 is the opposite of the remainder of the division of p0 by
p1. Therefore, I found

   p2 = -((1/2)ky^2 + (3/4)my + n)

You wrote 

   p2 = -p0 + p1*(x/4).
   
I used these substitutions to see if the resulting expression was equal to
the expression I obtained for p2:

   p0 = y^4 + ky^2 + my + n
   p1 = 4y^3 + 2ky + m

But I could not understand the switch to variable x from y, since a
substitution in the original article from x to y included an "a."
Furthermore, p1 and p0 are expressions with third and fourth powers, so I
did not see how p2 would come out. I found this expression to have a
highest power of 2, and we've performed only additions and subtractions.

If you could write me to tell me how you got your p2, I would be very
grateful.



Date: 03/29/2012 at 16:06:53
From: Doctor Vogler
Subject: Re: quartic discriminant

Hi Ian,

I'm sorry. I got my x's and y's mixed up. You can change all instances of
x in my previous answer to y's.

Does that clear everything up?

- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 03/31/2012 at 21:49:01
From: Ian
Subject: quartic discriminant

I understand everything now that I mentioned in my most recent note,
though there is something about which I am still confused.

In each of the three cases with Sturm's Theorem (neither R nor k zero, R
zero, k zero), I cannot see how you determined the expression you found
for D in each case.

In case 3 (k zero), I can see that you found 27m^4 + 192n^3 to be the
discriminant, although not explicitly shown. But in that case, as in the
other two, I cannot see what originally led you to choose that specific
expression. 

Could you please let me know your reasoning?

Thank you.



Date: 04/02/2012 at 13:05:48
From: Doctor Vogler
Subject: Re: quartic discriminant

Hi Ian,

I don't understand. You asked "how I determined the expression I found for
D." Are you asking how I knew the formula for the discriminant? I admit
that I used a computer program to help with some of the mundane
computations of algebraic formulas; this also helps to avoid silly
mistakes like accidentally dropping terms or switching signs. But the
formula for discriminant can be found on any of several descriptions of
discriminants (such as the Wikipedia page, or MathWorld, or a text book)
or by doing any of a variety of computations that determine a resultant,
such as taking the resultant of the polynomial and its derivative, or
computing the determinant of the appropriate Sylvester matrix. I just took
the formula that you get, which I wrote down as

   D = 16nk^4 - 4k^3m^2 - 27m^4 + 144knm^2 - 128(kn)^2 + 256n^3,

For the other cases, you just simplify. For example, if k = 0, then you
just substitute 0 for k in the above expression and get

   D = -27m^4 + 256n^3.

For the R = 0 case, you can't just do a simple substitution, since

   R = 2k^3 - 8kn + 9m^2

But since I've already given the solution as ...

  R = 0      implies      D = m^2(8k^3 + 27m^2)^2/2k^3,

... then you can just check that ...

  D - m^2(8k^3 + 27m^2)^2/2k^3

... is R times a polynomial.

The way I computed this expression for D is by looking for an
easy-to-solve-for variable in this equation:

  2k^3 - 8kn + 9m^2 = 0

This led me to

  n = (2k^3 + 9m^2)/(8k)

Then I substituted this into my expression for D. Now I can write D in
terms of just k and m.

Is that what you were asking about?

- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/ 
  


Date: 04/06/2012 at 15:40:38
From: Ian
Subject: Thank you (quartic discriminant)

Yes, everything is clear to me now. Thank you
soooooooooooooooooooo.......................ooooooooooo much.

I love Dr. Math.

I love what it does. 

I love how valuable it is -- in my opinion, more than gold. Because while
gold can wrap around someone's finger or neck and gleam, Dr. Math helps to
improve the mind and mathematical abilities, which I see as priceless,
especially when they are as polished as the ones you have.

You have shared your knowledge with me; and for that ... and for your
input ... and for the extensiveness of your responses ... I am extremely
grateful -- all the way from 2005, during the first entry on the subject,
up through these last few weeks of our correspondence!
Associated Topics:
College Modern Algebra

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