Assume the n distinct roots of f(x) = 0 are \(\left\{ r_1, \dots r_n \right\} \subset \mathbb{C}\). This set has no intrinsic order, and so we need to talk about it as if the indices are abstract. Let \(z_j = \sum_{a} r_a^j, \quad z_{j,k} = \sum_{a \neq b} r_a^j r_b^k, \quad z_{j, k,\ell} = \sum_{a \neq b, a \neq c, b \neq c} r_a^j r_b^k r_c^{\ell}, \) and so forth... Then \(z_{k,j} = z_{j,k}, \; z_a z_b = z_{a + b} + z_{a,b}, \; z_a z_{j, k, \ell, m} = z_{j+a, k, \ell, m} + z_{j, k+a, \ell, m} + z_{j, k, \ell+a, m} + z_{j, k, \ell, m+a} + z_{j, k, \ell, m, a}, \) and \(z_{a,b} z_{j,k} = z_{j + a, k+b} + z_{j + b, k+a} + z_{j + a, k, b} + z_{j, k + a, b} + z_{j + b, k, a} + z_{j, k + b, a} + z_{j,k,a,b} \) Then \( z_1 z_{k} = z_{k+1}, \; z_1 z_{k,1} = z_{k+1,1} + z_{k,2} + z_{k,1,1}, \; z_{1,1} z_{k,1} = 2 z_{k+1,2} + 2 z_{k+1,1,1} + 2z_{k,2,1} + z_{k,1,1,1} \). Then \(z_1^2 = z_2 + z_{1,1}, \; z_1 z_2 = z_3 + z_{2,1}, \; z_1 z_{1,1} = 2 z_{2,1} + z_{1,1,1}, \; z_1^3 = z_3 + 3 z_{2,1} + z_{1,1,1}\). And \( z_1^4 = \left( z_2 + z_{1,1} \right)^2 = z_2^2 + 2 z_2 z_{1,1} + z_{1,1}^2 = z_4 + z_{2,2} + 4 z_{3,1} + 2 z_{2,1,1} + 2 z_{2,2} + 4 z_{2,1,1} + z_{1,1,1,1} = z_4 + 4 z_{3,1} + 3 z_{2,2} + 6 z_{2,1,1} + z_{1,1,1,1}\).
Assembling a multiplication table: \( \begin{array}{c|cc} \times & z_1 & z_{1,1} & z_2 & z_{1,1,1} & z_{2,1} & z_3 & z_{1,1,1,1} & z_{2,1,1} & z_{2,2} & z_{3,1} & z_4 \\ \hline \\ z_1 & z_2 + z_{1,1} \\ z_{1,1} & 2 z_{2,1} + z_{1,1,1} & 2 z_{2,2} + 4 z_{2,1,1} + z_{1,1,1,1} \\ z_2 & z_3 + z_{2,1} & 2 z_{3,1} + z_{2,1,1} \\ z_{1,1,1} & 3 z_{2,1,1} + z_{1,1,1,1} & \\ z_{2,1} & z_{3,1} + z_{2,2} + z_{2,1,1} & 2 z_{3,2} + 2 z_{3,1,1} + 2z_{2,2,1} + z_{2,1,1,1} \\ z_3 & z_4 + z_{3,1} & 2 z_{4,1} + z_{3,1,1} \\ z_{1,1,1,1} & 4 z_{2,1,1,1} + z_{1,1,1,1,1} & \\ z_{2,1,1} & z_{3,1,1} + 2 z_{2,2,1} + z_{2,1,1,1} & \\ z_{2,2} & 2 z_{3,2} + z_{2,2,1} & \\ z_{3,1} & z_{4,1} + z_{3,2} + z_{3,2,1} & 2 z_{4,2} + 2 z_{4,1,1} + 2z_{3,2,1} + z_{3,1,1,1} \\ z_4 & z_5 + z_{4,1} & 2 z_{5,1} + z_{4,1,1} \\ z_{1,1,1,1,1} & 5 z_{2,1,1,1,1} + z_{1,1,1,1,1,1} & \\ z_{2,1,1,1} & z_{3,1,1,1} + 3 z_{2,2,1,1} + z_{2,1,1,1,1} & \\ z_{2,2,1} & 2 z_{3,2,1} + z_{2,2,2} + z_{2,2,1,1} & \\ z_{3,1,1} & z_{4,1,1} + 2 z_{3,2,1} + z_{3,1,1,1} & \\ z_{3,2} & z_{4,2} + z_{3,3} + z_{3,2,1} & \\ z_{4,1} & z_{5,1} + z_{4,2} + z_{4,1,1} &2 z_{5,2} + 2 z_{5,1,1} + 2z_{4,2,1} + z_{4,1,1,1} \\ z_5 & z_6 + z_{5,1} & 2 z_{6,1} + z_{5,1,1} \end{array} \) Woops, ran out of time. Hope to add to this table later.
Reformated to fit on at least my screen. \( \begin{array}{c|cc} \times & z_{1} & z_{1,1} & z_{2} \\ \hline \\ z_{1} & z_{2} + z_{1,1} & 2 z_{2,1} + z_{1,1,1} & z_{3} + z_{2,1} \\ z_{1,1} & 2 z_{2,1} + z_{1,1,1} & 2 z_{2,2} + 4 z_{2,1,1} + z_{1,1,1,1} & 2 z_{3,1} + z_{2,1,1} \\ z_{2} & z_{3} + z_{2,1} & 2 z_{3,1} + z_{2,1,1} & z_{4} + z_{2,2} \\ z_{1,1,1} & 3 z_{2,1,1} + z_{1,1,1,1} & 6 z_{2,2,1} + 6 z_{2,1,1,1} + z_{1,1,1,1,1} & 3 z_{3,1,1} + z_{2,1,1,1} \\ z_{2,1} & z_{3,1} + z_{2,2} + z_{2,1,1} & 2 z_{3,2} + 2 z_{3,1,1} + 2 z_{2,2,1} + z_{2,1,1,1} & z_{4,1} + z_{3,2} + z_{2,2,1} \\ z_{3} & z_{4} + z_{3,1} & 2 z_{4,1} + z_{3,1,1} & z_{5} + z_{3,2} \\ z_{1,1,1,1} & 4 z_{2,1,1,1} + z_{1,1,1,1,1} & 12 z_{2,2,1,1} + 8 z_{2,1,1,1,1} + z_{1,1,1,1,1,1} & 4 z_{3,1,1,1} + z_{2,1,1,1,1} \\ z_{2,1,1} & z_{3,1,1} + 2 z_{2,2,1} + z_{2,1,1,1} & 4 z_{3,2,1} + 2 z_{3,1,1,1} + 2 z_{2,2,2} + 4 z_{2,2,1,1} + z_{2,1,1,1,1} & z_{4,1,1} + 2 z_{3,2,1} + z_{2,2,1,1} \\ z_{2,2} & 2 z_{3,2} + z_{2,2,1} & 2 z_{3,3} + 4 z_{3,2,1} + z_{2,2,1,1} & 2 z_{4,2} + z_{2,2,2} \\ z_{3,1} & z_{4,1} + z_{3,2} + z_{3,1,1} & 2 z_{4,2} + 2 z_{4,1,1} + 2 z_{3,2,1} + z_{3,1,1,1} & z_{5,1} + z_{3,3} + z_{3,2,1} \\ z_{4} & z_{5} + z_{4,1} & 2 z_{5,1} + z_{4,1,1} & z_{6} + z_{4,2} \\ z_{1,1,1,1,1} & 5 z_{2,1,1,1,1} + z_{1,1,1,1,1,1} & & \\ z_{2,1,1,1} & z_{3,1,1,1} + 3 z_{2,2,1,1} + z_{2,1,1,1,1} & & \\ z_{2,2,1} & 2 z_{3,2,1} + z_{2,2,2} + z_{2,2,1,1} & & \\ z_{3,1,1} & z_{4,1,1} + 2 z_{3,2,1} + z_{3,1,1,1} & & \\ z_{3,2} & z_{4,2} + z_{3,3} + z_{3,2,1} & & \\ z_{4,1} & z_{5,1} + z_{4,2} + z_{4,1,1} & & \\ z_{5} & z_{6} + z_{5,1} & & \\ \hline \times & z_{1,1,1} & z_{2,1} & z_{3} \\ \hline \\ z_{1} & 3 z_{2,1,1} + z_{1,1,1,1} & z_{3,1} + z_{2,2} + z_{2,1,1} & z_{4} + z_{3,1} \\ z_{1,1} & 6 z_{2,2,1} + 6 z_{2,1,1,1} + z_{1,1,1,1,1} & 2 z_{3,2} + 2 z_{3,1,1} + 2 z_{2,2,1} + z_{2,1,1,1} & 2 z_{4,1} + z_{3,1,1} \\ z_{2} & 3 z_{3,1,1} + z_{2,1,1,1} & z_{4,1} + z_{3,2} + z_{2,2,1} & z_{5} + z_{3,2} \\ z_{1,1,1} & 6 z_{2,2,2} + 18 z_{2,2,1,1} + 9 z_{2,1,1,1,1} + z_{1,1,1,1,1,1} & 6 z_{3,2,1} + 3 z_{3,1,1,1} + 3 z_{2,2,1,1} + z_{2,1,1,1,1} & 3 z_{4,1,1} + z_{3,1,1,1} \\ z_{2,1} & 6 z_{3,2,1} + 3 z_{3,1,1,1} + 3 z_{2,2,1,1} + z_{2,1,1,1,1} & z_{4,2} + z_{4,1,1} + z_{3,3} + 2 z_{3,2,1} + z_{2,2,2} + z_{2,2,1,1} & z_{5,1} + z_{4,2} + z_{3,2,1} \\ z_{3} & 3 z_{4,1,1} + z_{3,1,1,1} & z_{5,1} + z_{4,2} + z_{3,2,1} & z_{6} + z_{3,3} \\ \hline \times & z_{1,1,1,1} & z_{2,1,1} & z_{2,2} \\ \hline \\ z_{1} & 4 z_{2,1,1,1} + z_{1,1,1,1,1} & z_{3,1,1} + 2 z_{2,2,1} + z_{2,1,1,1} & 2 z_{3,2} + z_{2,2,1} \\ z_{1,1} & 12 z_{2,2,1,1} + 8 z_{2,1,1,1,1} + z_{1,1,1,1,1,1} & 4 z_{3,2,1} + 2 z_{3,1,1,1} + 2 z_{2,2,2} + 4 z_{2,2,1,1} + z_{2,1,1,1,1} & 2 z_{3,3} + 4 z_{3,2,1} + z_{2,2,1,1} \\ z_{2} & 4 z_{3,1,1,1} + z_{2,1,1,1,1} & z_{4,1,1} + 2 z_{3,2,1} + z_{2,2,1,1} & 2 z_{4,2} + z_{2,2,2} \\ \hline \times & z_{3,1} & z_{4} \\ \hline \\ z_{1} & z_{4,1} + z_{3,2} + z_{3,1,1} & z_{5} + z_{4,1} \\ z_{1,1} & 2 z_{4,2} + 2 z_{4,1,1} + 2 z_{3,2,1} + z_{3,1,1,1} & 2 z_{5,1} + z_{4,1,1} \\ z_{2} & z_{5,1} + z_{3,3} + z_{3,2,1} & z_{6} + z_{4,2} \end{array} \)
Now if all the roots are distinct, then we have polynomials with coefficients in these forms, which is just some elaborate multiplication of polynomials. \(\begin {eqnarray} x - r_1 & = & x - z_1 ; & \quad n = 1 \\ (x - r_1)(x - r_2) & = & x^2 - z_1 x + \frac{1}{2!} z_{1,1} ; & \quad n = 2 \\ (x - r_1)(x - r_2)(x - r_3) & = & x^3 - z_1 x^2 + \frac{1}{2!} z_{1,1} x - \frac{1}{3!} z_{1,1,1} ; & \quad n = 3 \\ (x - r_1)(x - r_2)(x - r_3)(x - r_4) & = & x^4 - z_1 x^3 + \frac{1}{2!} z_{1,1} x^2 - \frac{1}{3!} z_{1,1,1} x + \frac{1}{4!} z_{1,1,1,1} ; & \quad n = 4 \end{eqnarray} \) "Solving" a polynomial equation is very much just the reverse of this multiplication. For the typical quadratic equation, where \(z_1^2 \neq 2 z_{1,1}\) we have: \( \left\{ r_1, r_2 \right\} = \left\{ \frac{1}{2} z_1 \pm \frac{1}{2} \sqrt{ z_1^2 - 2 z_{1,1} } \right\} = \left\{ \frac{1}{2} z_1 \pm \frac{1}{2} \sqrt{ z_2 - z_{1,1} } \right\} = \left\{ \frac{r_1 + r_2 }{2} \pm \frac{1}{2} \sqrt{ r_1^2 + r_2^2 - r_1 r_2 - r_2 r_1 } \right\} = \left\{ \frac{r_1 + r_2 }{2} \pm \frac{1}{2} \sqrt{ (r_1 - r_2)^2 } \right\}\). So undoing the polynomial coefficients which are symmetric sums of products of roots proceeds simply from combining the coefficients and relating that to the multivalued root functions. It's because two different numbers square to the same number that we know \(\pm \sqrt{ x }\) connects us to two answers. So how much harder could this get? [Ironic orchestral hit]
I wonder if anyone other than students given an assignment try to solve 3rd/4th order polynomials analytically. The analytic solution for 3rd/4th order polynomials is time consuming & there is no analytic solution for 5th & higher order polynomials. Successive approximation methods are easy for any polynomial & might be easier for a 2nd order polynomial, although I tend to use the formula learned in grade school.
Even in the case of quadratic equation, \(x^2 + a x + b\), the standard prescription of \(\frac{-a \pm \sqrt{a^2 - 4 b}}{2}\) is not always the best solution for the finite precision of IEEE floating point numbers, in that when \(a\) is close to either \(\pm \sqrt{a^2 - 4 b}\) you can lose significant digits. I'm not suggesting that there is a lot of utility to analytic solutions to polynomials of degree < 5. But the connection between coefficients and roots is worthy of study and can result in pretty visualizations. http://math.ucr.edu/home/baez/roots/
R Penner: I think you made a typo or your memory is suspect. The correct formula is BTW: Your formatting is better than mine. What tools do you use here at SciForums?
No typo; have a closer look at the equation in the first line. This illustrates that mathematicians are free to label coefficients as they please.
Arfa Bran: Sorry about that, chief! You are correct. I am so used to the more common notation for quadratics that I scanned your Post & did not notice that it was not the normal quadratic with 3 coefficients. BTW: Are there other Posters who admit to making errors?
I believe you are speaking of Integrity and Intellectual Honesty, features slated for Web 3.0. Please Register or Log in to view the hidden image!
I believe there is a resemblance between rpenner's "indexed roots" and the elementary symmetric polynomials, which I suspect is more than superficial. I didn't get that close to the subject, but did look at latin squares.
