Newton created his identities for this purpose. In today's blog, I will show how Newton's identities can be used to determine if two polynomials of degree 3 have a common root. In a future blog, I will show Euler's generalization of this idea which led to the concept of the resultant.
Consider two cubic equations:
f(x) = x3 + bx2 + cx + d
g(x) = x3 + Bx2 + Cx + D
Let r,s,t be the roots of f(x).
It is clear that f(x) and g(x) have common roots if and only if g(r)*g(s)*g(t) = 0.
Here's Newton's insight:
Theorem: Common Roots in Cubic Equations
For any two cubic equations:
f(x) = x3 + bx2 + cx + d
g(x) = x3 + Bx2 + Cx + D
There exists an equation P(b,c,d,B,C,D) which only equals 0 if and only if the two cubic equations have a common root.
Proof:
(1) Let r,s,t be the three roots for f(x). [We know this is the case from the Fundamental Theorem of Algebra, see here, and also from the general formula for cubic equations, see here]
(2) Assume that g(x) shares a root with f(x) so that either g(r)=0 or g(s)=0 or g(t)=0.
(3) So, clearly:
g(r)*g(s)*g(t) = 0
(4) This means that:
[r3 + br2 + cr + d][s3 + bs2 + cs + d][t3 + bt2 + ct + d] = 0
(5) Now we can divide up this equation into a sum of the following factors:
g(r)*g(s)*g(t) = a1*r3s3t3 + a2r3s3t2 + ... + a34d3
(6) Since all three equations are symmetric, that it we can switch any two and get the same result, it is clear that all combinations have the same coefficients.
For example, consider every term of the form: r3s2t.
r3*bs2*ct = bc*(r3s2t)
r3*cs*bt2 = bc*(r3t2s)
br2*s3*ct = bc*(s3r2t)
cr*s3*bt2 = bc*(s3t2r)
br2*cs*t3 = bc*(t3r2s)
cr*bs2*t3 = bc*(t3s2r)
(7) So, by grouping the terms by form (and I will rxsytz to refer to all combinations of this form), we have:
g(r)g(s)g(t) = r3s3t3 + b*r3s3t2 + c*r3s3t + ... + c3rst + d3
(8) It turns out that each of these terms is one of Newton's identities. Besides d3, there are 19 possible combinations. If we combine this with Newton's identities (see here), we can restate the sums rxsytz into a term that consists of the coefficients B,C,D.
For example:
(every r3s2t) = -CD
So, the term itself is:
bc*(-CD) = -bcCD
(9) Since the entire equation can be expressed in terms b,c,d,B,C,D, we have shown that there exists an equation P(b,c,d,B,C,D) that equals 0 if and only if f(x) and g(x) have a common root.
So, we get:
P = -d3 + Bcd2 - B2bd2 + B3d2 - Cc2d + 2Cbd2 - 3BCd2 - B2Ccd - C2b2d + 2C2cd + BC2bd - C3d - 3Dbcd + 3Dd2 + 2DBb2d + DBcd + seventeen more terms obtained from these by reversing the sign and interchanging B with b, C with c, and D with d, that is D3 - bCD2 + b2BD2 - ...
QED
References
- Harold M. Edwards, Galois Theory, Springer, 1984.
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