Is There a 'Discriminant' for a Quartic Equation?Date: 01/12/2005 at 16:57:52 From: Romain Subject: Quartic equation Hi, My question is about quartic equations. I know how to find the roots but I would like to know if it is possible to predict the kinds of roots only with the coefficients. For example, in a quadratic equation ax^2 + bx + c = 0, you know if b^2 - 4ac > 0, there are two real roots. In a cubic equation ax^3 + bx^2 + cx + d = 0, you know if Yn^2 > h^2, there is one real root and two complex ones, etc. Does something similar exist for the quartic equation? Thanks, and congratulations on your site. Date: 01/12/2005 at 18:00:43 From: Doctor Vogler Subject: Re: Quartic equation Hi Romain, Thanks for writing to Dr. Math. Yes, it does. The "types" of roots of a polynomial has a lot to do with a thing called the "Galois group" of a polynomial (or of the splitting field of the polynomial). This is a concept from abstract algebra. If you have a basic understanding of abstract algebra, then I would recommend a book by Dummit and Foote (called Abstract Algebra) which discusses this very subject in great detail. First of all, every polynomial has a discriminant, not just quadratics. The discriminant is the first place you look to classify the types of roots a polynomial has. The discriminant is defined as the product of the squares of all of the differences of two roots of the equation. (Some definitions also multiply by a power of the leading coefficient in order to make them polynomials in the coefficients even when the polynomial isn't monic. This makes little other difference.) Of course, if you don't know the roots, that makes it hard to compute. But, in fact, there are efficient algorithms to get a formula, for each degree n, that takes the coefficients of a degree-n polynomial and gives you the discriminant, and the formula will always be a rational polynomial in the coefficients of your polynomial. See also the MathWorld page on polynomial discriminants: http://mathworld.wolfram.com/PolynomialDiscriminant.html This page also gives the discriminants for degree 2, 3, and 4 polynomials. Now let's consider. If there is a double root (a repeated root) of your polynomial, then the discriminant will be zero. (And, likewise, a zero discriminant means that there is a repeated root.) Now suppose your coefficients are all real numbers, and there happen to be four real roots (p, q, r, s). Then the discriminant is [(p-q)(p-r)(p-s)(q-r)(q-s)(r-s)]^2 > 0. But what if you have two real roots and two complex roots (p, q, r+si, r-si)? Then the discriminant is (p-q)^2 [(p-r-si)(p-r+si)]^2 [(q-r-si)(q-r+si)]^2 (2si)^2 = (p-q)^2 ((p-r)^2 + s^2)^2 ((q-r)^2 + s^2)^2 * (-4s^2) < 0. Similarly, if you have four complex roots (p+qi, p-qi, r+si, r-si), then the discriminant will be positive. So you may conclude that if the discriminant is negative, then you have two real roots and two complex roots. You have to do a little more work to distinguish between four real roots and four complex roots, but there are various ways of doing that once you know that the discriminant is positive. I'd have to work on coming up with one (probably consulting Dummit and Foote in the process), so I'll stop here for now and see if this is the kind of thing you want. I hope this is helpful. 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: 01/12/2005 at 22:12:04 From: Romain Subject: Quartic equation Thank you--this is exactly the subject I am interested in. But what I am really interested to know is if all the roots are complex or not. Is there a simple way to determine that? Date: 01/13/2005 at 11:12:45 From: Doctor Vogler Subject: Re: Quartic equation Hi Romain, After consulting the text I mentioned, I remembered how resolvents work. A resolvent is a helping polynomial. For example, one helpful resolvent for the polynomial (x - p)(x - q)(x - r)(x - s) is the polynomial (x - (p + q)(r + s))(x - (p + r)(q + s))(x - (p + s)(q + r)). It turns out that as long as the new polynomial (just like the original polynomial) is unchanged by permuting the four original roots (p, q, r, s), you can solve for the coefficients of the new polynomial using only the coefficients (not the roots) of the original polynomial. (For info on how to do this, search for "symmetric polynomials".) For example, if the original polynomial is irreducible (over the rational numbers) then just as the discriminant tells you some things about the roots, how (or whether) the resolvent factors can tell you other things about the roots. So my hope was to come up with a cubic resolvent polynomial which would have three real roots if the original quartic had four real roots but which would have only one real root (and two complex) if the original quartic had four complex roots. Then you could use your other formula (for the discriminant of the cubic) to distinguish between the two cases. But I was having trouble finding such a resolvent. Maybe I just wasn't creative enough, but the ones I came up with generally had three real roots in both of those two cases (though not always in the other case that we can already tell apart using the discriminant). Then I had a different idea. You see, we're really dealing with real numbers instead of rational numbers, since you don't care about the form the roots take, or their degree, but only whether they are real or complex. Well, every real quartic polynomial can factor into two quadratic factors. So let's see how it factors. We want to find t, u, v, and w in terms of a, b, c, and d, so that (x^2 + tx + u)(x^2 + vx + w) = x^4 + ax^3 + bx^2 + cx + d. Actually, things work out nicer when we force a = 0 by making the change-of-variables y = x + a/4 which changes your equation to y^4 + ky^2 + my + n where k = b - 3a^2/8 m = c - ab/2 + a^3/8 n = d - ac/4 + ba^2/16 - 3a^4/256. Now we solve (y^2 + ty + u)(y^2 + vy + w) = y^4 + ky^2 + my + n which gives us four equations (since these must be the same polynomial and not just equal for some y), namely t + v = 0 u + w + tv = k tw + uv = m uw = n. We can solve these like so: v = -t w = k - u - tv = k - u + t^2 Substitute into the third equation and solve for q: u = (t^3 + kt - m)/(2t). Then we substitute u and w into the last equation, finally ending up with t^6 + (2k)t^4 + (k^2 - 4n)t^2 - m^2 = 0. That means that t^2 is a root of the following resolvent cubic: z^3 + (2k)z^2 + (k^2 - 4n)z - m^2. (This is where the change-of-variables paid off. If we hadn't forced a=0, then we would have a sixth-degree polynomial with six messy coefficients instead of three.) So now let's think for a moment about how this applies to my other method. It would appear that I found precisely the resolvent that I was looking for. So that makes me wonder what form its roots take, in terms of the roots of the original quartic. With a little work, I determined that the creative thought that I didn't have was to use the three roots (p - q - r + s)^2 (p - q + r - s)^2 (p + q - r - s)^2 (where p, q, r, and s are the roots of the original quartic) because those roots (divided by 4) are the roots of the resolvent I ended up with. Perhaps the reason I didn't come up with these three roots is that it is not immediately apparent that permuting p, q, r, and s only permutes those three expressions. Perhaps the reason is that I just wasn't creative enough. In fact, since each term above is unchanged when you add the same number to every root, you get the same formula if you let p, q, r, and s be the roots of y^4 + ky^2 + my + n. Now, continuing on, my first (incorrect) inclination was to say that if we have two pairs of complex roots, then there is only one way to choose our two quadratics (well, two if you count switching them), but if we have four real roots, then there are three ways (or six) to choose them. (In fact, there is also only one/two ways if there are two real roots and two complex.) So that means we only need to check if the resolvent cubic z^3 + (2k)z^2 + (k^2 - 4n)z - m^2 has one real root or three, which we can do by checking the discriminant, as you described. But it turned out that this cubic has exactly the same discriminant as the original quartic. This surprised me, until I realized that in the case of two pairs of complex roots, mis-matching the pairs results in a value of t which is pure imaginary, meaning that t^2 is a negative real root of the cubic polynomial. (In the two-real/two-complex case, you'll get one real root and two complex roots of the resolvent cubic.) But all is not lost! We just need to be able to tell whether our resolvent cubic g(z) = z^3 + (2k)z^2 + (k^2 - 4n)z - m^2 has three positive real roots, or one positive and two negative real roots. So let's look at our cubic. Its derivative will be zero when 3z^2 + (4k)z + (k^2 - 4n) = 0. and we can solve this for z using the quadratic formula. If there are no real roots, then that means our cubic has one real root and no local maximum or minimum (which only happens in the two-real/two-complex case). If there are two real roots, then the smaller one is the z value of the local maximum of the cubic, and the larger one is the z value of the local minimum of the cubic. A useful exercise which is well worth doing is to check that the maximum value g(z) is positive and the minimum value g(z) is negative exactly when the discriminant of the cubic is positive (i.e. when the cubic has three real roots). In our case, the important thing to check is that the z value of the local maximum (which is between the smaller two roots) is negative. 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. You'll find that the z value of the local maximum is z = -(2/3)k - (1/3)sqrt(k^2 + 12n) which is negative if and only if -(2/3)k - (1/3)sqrt(k^2 + 12n) < 0 -(2/3)k < (1/3)sqrt(k^2 + 12n) -2k < sqrt(k^2 + 12n) k > 0 or 4k^2 < k^2 + 12n k > 0 or k^2 < 4n. I'll leave it up to you to rewrite these variables in terms of the coefficients of your original polynomial. That's a straight-forward but messy task, since you'll end up with some rather large expressions. If you got lost in all of this, or need more help coming up with the formulas, or have any others questions about this, please write back, and I will try to help. But before I go, I should like to point out one other completely different but very nice method to decide the same thing. It's called Sturm's Theorem. See http://en.wikipedia.org/wiki/Sturm's_theorem for a very simple algorithm to count the number of distinct real roots of a polynomial. You start with the polynomial f0(x) = f(x) and its derivative f1(x) = f'(x). Then you do polynomial division to get the next polynomial f0(x)/f1(x) = quotient(x) + remainder(x)/f1(x) or f0(x) = quotient(x)*f1(x) + remainder(x) where the remainder polynomial has to have smaller degree than the divisor f1(x) (otherwise you chose the quotient incorrectly). You call the remainder -f2(x), so that f0(x) = (polynomial)*f1(x) - f2(x) and then you repeat f0(x) = (polynomial)*f1(x) - f2(x) f1(x) = (polynomial)*f2(x) - f3(x) f2(x) = (polynomial)*f3(x) - f4(x) f3(x) = (polynomial)*f4(x) - f5(x) and so on, until you get a constant. Then you can stop. Now to find out the number of distinct real roots between a and b, you evaluate all of the polynomials f0, f1, f2, etc. at a and count the number of times the sign changes. Then you do the same at b and subtract the second number from the first. The result is the number of real roots between a and b. To get the total number of real roots, evaluate at -infinity and +infinity. Evaluating at infinity means to take the signs of the leading coefficients of the polynomials. Evaluating at -infinity means to take the signs of the leading coefficients of the polynomials times (-1)^degree. For example, the polynomial x^6 + 2x^4 + 2x^3 + 2x^2 + 4 yields the sequence f0 = x^6 + 2x^4 + 2x^3 + 2x^2 + 4 f1 = 6x^5 + 8x^3 + 6x^2 + 4x f2 = (-2/3)x^4 - x^3 - (4/3)x^2 - 4 f3 = (-19/2)x^3 - 24x^2 + 32x - 54 f4 = (1916/361)x^2 - (2200/361)x + 2848/361 f5 = (-1385518/229441)x + 484462/229441 f6 = -8517767684/1329404521 Evaluating at -infinity, the signs go f0 = + f1 = - f2 = - f3 = + f4 = + f5 = + f6 = - and we see three sign changes. Evaluating at +infinity, the signs go f0 = + f1 = + f2 = - f3 = - f4 = + f5 = - f6 = - and again we see three sign changes. Subtracting, 3 - 3 = 0, and so there are no real roots. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/