Cubic EquationsDate: 01/19/98 at 00:17:43 From: COLIN Subject: Polynomials I know how to solve a polynomial like this: ax^2 + bx + c = 0 but how would you solve (for example) a problem like this: 3x^3 + x^2 + 15x + 27 = 0 or x^2673 + x + 3265782635529 = 0 Date: 01/19/98 at 07:10:56 From: Doctor Mitteldorf Subject: Re: Polynomials Dear Colin - Good question! It turns out a lot of the math that they teach you in school is problems that we know how to solve. There are a lot more problems that we don't know how to solve, and they get less mention, even though some of them are just as important or more important than the other kind. Polynomials with an x^2 as the highest power are called quadratic equations, and there is a formula for solving them, which you know. Polynomials with an x^3 as the highest power are called cubic equations, and there is also a formula for solving them, but it is so complicated that it is rarely used. There's even a fantastically complicated formula for "quartic" equations which have an x^4 in them - that formula is almost useless. Beyond x^4, there are tricks that work sometimes, but there is no general formula. What people do is called a "numerical solution". This sounds less satisfactory than a formula, but in practice it tells you everything you need to know. A numerical solution is a method of finding better and better approximations to the solution, good to more and more decimal places. Most numerical solutions start with a guess, and then use that guess to make another guess that's closer, and use the closer guess to get closer still. There's no limit to how close you can get this way, and in that sense a numerical solution is a complete and satisfactory solution to the problem. Here's an example of a numerical solution to the equation you asked about, 3x^3 + x^2 + 15x + 27 = 0 First, solve for x in a trivial way: x = - (3x^3 + x^2 + 27) / 15 Now make a guess. I'm going to make a fairly intelligent guess to start, and say x = -1. Let all the x's on the right side of the equation be -1, and solve for x on the left. In other words, find - (3x^3 + x^2 + 27) / 15 where x = -1. Let that be our new x. It comes out to -1.6667. Now let -1.6667 be our new x, and do the same thing again. The next x comes out to -1.05926. If you do this on a calculator or a computer, you'll see that each time you go through this process, you get a little closer to the right answer. But it takes a long, long time, because you keep "stepping past" the right answer, and coming going back and forth, back and forth. A trick that almost always works in this situation is to improve your method by AVERAGING the last two answers. In other words, if you start with x = -1 and your formula comes out -1.66667, average the two to give -1.33333 as your next guess instead of just using -1.66667. This will get you closer much faster, and you'll soon home in on the answer -1.390852. You can adapting this method to your other example x^2673 + x + 3265782635529 = 0 and see what you get. You will probably have to think about sensitivity: this formula changes awfully fast if you change x by just a bit. -Doctor Mitteldorf, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 01/22/98 at 00:04:12 From: COLIN Subject: Polynomials Can any one tell me the cubic formula - a formula to solve polynomials like this: aX^3 + bX^2 + cX + D = 0? I know it is very complex. I would also like to know if there is a way to make up a formula, like say you get a polynomial where the highest power is 4 and you don't have it memorized, so could you go though a procedure to figure it out no matter what the highest power is? Date: 01/22/98 at 08:53:56 From: Doctor Anthony Subject: Re: Polynomials The general cubic ax^3 + 3bx^2 + 3cx + d = 0 is changed to a simpler cubic without a term in x^2 by the substitution y = ax+b (or x = (y-b)/a) to get y^3 + 3Hy + G = 0 where H = ac-b^2 G = a^2.d - 3abc + 2b^3 We now make use of the identity shown below, where w is a complex cube root of unity x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x+wy+w^2.z)(x+w^2.y +wz) This is easily proved if you know how to factorize determinants, but if not, you can prove it the long way by multiplying out the factors, and remembering that w^3 = 1 The cubic is now expressed as y^3 - 3pqy + p^3 + q^3 = 0 with G = p^3 + q^3 H = -pq So y^3 -3pqy + p^3+q^3 = (y+p+q)(y+wp+w^2.q)(y+w^2p+wq) = 0 and the three roots will be: (1) y = -p - q (2) y = -wp - w^2q (3) y = -w^2p - wq So all that is required is that we have the values of p and q from the two equations p^3 + q^3 = G and pq = -H Here, think in terms of p^3 and q^3 rather than p and q, and consider instead the two equations p^3 + q^3 = G and p^3.q^3 = -H^3 Now we can see that p^3 and q^3 are the roots of the quadratic t^2 - Gt - H^3 = 0 and their cube roots must be chosen so that pq = -H The roots of this quadratic are (1/2)[G +-sqrt(G^2 + 4H^3)] If G^2 + 4H^3 > 0 p and q are real and distinct If G^2 + 4H^3 = 0 p and q are real and equal. The corresponding roots of the cubic are -2p, p, p. If G^2 + 4H^3 < 0 p and q are complex. The cubic then, ironically, has 3 real roots. In this situation it is better to use a trigonometrical method. If you wish to see the trig method, write back. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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