Gauss-Lucas theorem

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In complex analysis, the Gauss-Lucas theorem gives a geometrical relation between the roots of a polynomial P and the roots of its derivative P'. The set of roots of a real or complex polynomial is a set of points in the complex plane. The theorem states that the roots of P' all lie within the convex hull of the roots of P, that is the smallest convex polygon containing the roots of P. When P has a single root then this convex hull is a single point and when the roots lie on a line then the convex hull is a segment of this line. The Gauss-Lucas Theorem, named after Karl Friedrich Gauss and Édouard Lucas is similar in spirit to Rolle's Theorem.

1 Formal statement
2 Special cases
3 Proof
4 External links

[edit] Formal statement

If P is a (nonconstant) polynomial with complex coefficients, all zeros of P ' belong to the convex hull of the set of zeros of P.

[edit] Special cases

It is easy to see that if P(x) = ax² + bx + c is a second degree polynomial, the zero of P '(x) = 2ax + b is the average of the roots of P. In that case, the convex hull is the line segment with the two roots as endpoints and it is clear that the average of the roots is the middle point of the segment.

In addition, if a polynomial of degree n of real coefficients has n distinct real zeros $x_1<x_2<\cdots <x_n\,$ , we see, using Rolle's theorem, that the zeros of the derivative polynomial are in the interval $[x_1,x_n]\,$ which is the convex hull of the set of roots.

[edit] Proof

Over the complex numbers, P is a product of prime factors

$P(z)= \alpha \prod_{i=1}^n (z-a_i)$

where the complex numbers $a_1, a_2, \ldots, a_n$ are the – not necessary distinct – zeros of the polynomial $P$ , the complex number $α$ is the leading coefficient of $P$ and $n$ is the degree of $P$ . Let $z$ be any complex number for which $P(z) \neq 0$ . Then we have for the Logarithmic derivative

$\frac{P^\prime(z)}{P(z)}= \sum_{i=1}^n \frac{1}{z-a_i}.$

In particular, if $z$ is a zero of $P'$ and still $P(z) \neq 0$ , then

$\sum_{i=1}^n \frac{1}{z-a_i}=0.\$

$\ \sum_{i=1}^n \frac{\overline{z}-\overline{a_i} } {\vert z-a_i\vert^2}=0.$

This may also be written as

$\left(\sum_{i=1}^n \frac{1}{\vert z-a_i\vert^2}\right)\overline{z}= \sum_{i=1}^n \frac{1}{\vert z-a_i\vert^2}\overline{a_i}.$

Taking their conjugates, we see that z is a weighted sum with positive coefficients that sum to one, or the barycenter, of the complex numbers a_i (with different mass assigned on each root).

If P(z) = P'(z) = 0, then z = 1·z + 0·a_i, and is still a convex combination of the roots of P.