Sum of Powers of Roots of Polynomial Equations
Date: 01/02/2007 at 10:29:58 From: Denise Subject: Sum of powers of roots I am an Algebra II math team coach looking for a quick trick to show my students how to find the sum of powers of roots for third degree polynomial equations. For example, given a third-degree polynomial equation with roots a, b, and c, find a^3 + b^3 + c^3 or a^4 + b^4 + c^4. Must they go through Tartaglia's formula to actually find the roots or is there a simpler way using the sum and product of roots? We know how to find a^3 + b^3 and a^4 + b^4 when given the sum and product of two roots. Is there a similar approach for three roots? I have tried analyzing the expansion of (a + b + c)^3 but I'm not seeing a way to find just a^3 + b^3 + c^3 knowing the sum and product of roots. I have not tried (a + b + c)^4, yet.
Date: 01/02/2007 at 15:17:15 From: Doctor Pete Subject: Re: Sum of powers of roots Hi Denise, Let us define S[n] = a^n + b^n + c^n, where a, b, c are roots of the cubic x^3 + px^2 + qx + r. Consequently, p = -(a+b+c) q = ab + bc + ca r = -abc. Now we are interested in the relationship between S[n] and p, q, r. We have -S[n]p = (a^n + b^n + c^n)(a + b + c) = S[n+1] + a^n (b+c) + b^n (c+a) + c^n (a+b). But it is not clear how to proceed with the remaining terms. However, note S[n-1]q = (a^(n-1) + b^(n-1) + c^(n-1))(ab + bc + ca) = a^n (b+c) + b^n (c+a) + c^n (a+b) + a^(n-1) bc + b^(n-1) ca + c^(n-1) ab = -S[n]p - S[n+1] + abc S[n-2]. Equivalently, S[n+1] = -(S[n]p + S[n-1]q + S[n-2]r). This rather fascinating (and easy to remember) identity is the key to determining the value of the sum of the higher powers of the roots of a cubic. But we need to establish some initial values. Clearly S = a^0 + b^0 + c^0 = 3, S = a + b + c = -p, but we still need S. This is easy enough, since p^2 = (-(a + b + c))^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) = S + 2q, so S = p^2 - 2q. Therefore we see that S = -(Sp + Sq + Sr) = -(p^3 - 2pq - pq + 3r) = -p^3 + 3pq - 3r. In turn, we can write S = -(Sp + Sq + Sr) = -(-p^4 + 3p^2 q - 3pr + p^2 q - 2q^2 - pr) = p^4 - 4p^2 q + 4 pr + 2q^2. Of course, higher powers quickly become complicated to compute. The identity we established suggests two questions: 1) Is the identity true for negative values of n; i.e., can we compute the sum of the reciprocals of the roots of a cubic? 2) Can the identity be generalized for higher-degree polynomials? In particular, if we let r, r, ..., r[m] be the roots of a degree-m monic polynomial F[x] = c + cx + cx^2 + ... + c[m-1]x^(m-1) + x^m and S[n,m] = r^n + r^n + ... + r[m]^n, is it true that S[n+1,m] = -(S[n,m]c[m-1] + S[n-1,m]c[m-2] + ... + S[n-m+1,m]c)? If so, how do we prove it? If not, what is the correct relationship? - Doctor Pete, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum