Dirichlet-multinomial distribution

2014-01-15T23:21:40Z

107.0.83.66: insert missing w_{dn} from an example/* Multiple Dirichlet priors with the same hyperprior, with dependent children */

In [[mathematics]], the '''complex conjugate root theorem''' states that if ''P'' is a [[polynomial]] in one variable with [[Real numbers|real]] [[coefficients]], and ''a'' + ''bi'' is a [[root of a function|root]] of ''P'' with ''a'' and ''b'' real numbers, then its [[complex conjugate]] ''a'' − ''bi'' is also a root of ''P''.<ref>{{cite book|title=Maynooth Mathematical Olympiad Manual|author=Anthony G. O'Farell and Gary McGuire|pages=104|chapter=Complex numbers, 8.4.2 Complex roots of real polynomials|year=2002|publisher=Logic Press |isbn=0954426908}} Preview available at [http://books.google.com/books?q=Maynooth+Mathematical+Olympiad+Manual&ots=xQ0hpAQkpc&sa=X&oi=print&ct=title Google books]</ref>

It follows from this (and the [[fundamental theorem of algebra]]), that if the degree of a real polynomial is odd, it must have at least one real root.<ref name=Jeffrey>{{cite book|title=Complex Analysis and Applications|author=Alan Jeffrey|chapter=Analytic Functions|pages=22–23|year=2005|publisher=CRC Press|id=ISBN 158488553X}}</ref> That fact can also be proven by using the [[intermediate value theorem]].

== Examples and consequences ==
* The polynomial ''x''<sup>2</sup> + 1 = 0 has roots ±''i''.
* Any real square [[matrix(mathematics)|matrix]] of odd degree has at least one real [[eigenvalue]]. For example, if the matrix is [[orthogonal matrix| orthogonal]], then 1 or −1 is an eigenvalue.
* The polynomial
::<math> x^3 - 7x^2 + 41x - 87\, </math>
:has roots
::<math> 3,\, 2 + 5i,\, 2 - 5i,</math>
:and thus can be factored as
::<math> (x - 3)(x - 2 - 5i)(x - 2 + 5i).\,</math>
:In computing the product of the last two factors, the imaginary parts cancel, and we get
::<math> (x - 3)(x^2 - 4x + 29).\,</math>
:The non-real factors come in pairs which when multiplied give quadratic polynomials with real coefficients. Since every polynomial with complex coefficients can be factored into 1st-degree factors (that is one way of stating the [[fundamental theorem of algebra]]), it follows that every polynomial with real coefficients can be factored into factors of degree no higher than 2: just 1st-degree and quadratic factors.

=== Corollary on odd-degree polynomials ===
It follows from the present theorem and the [[fundamental theorem of algebra]] that if the degree of a real polynomial is odd, it must have at least one real root.<ref name="Jeffrey"/>

This can be proved as follows.
*Since non-real complex roots come in conjugate pairs, there are an even number of them;
*But a polynomial of odd degree has an odd number of roots;
*Therefore some of them must be real.

This requires some care in the presence of [[multiple root]]s; but a complex root and its conjugate do have the same [[Multiplicity (mathematics)|multiplicity]] (and this [[lemma (mathematics)|lemma]] is not hard to prove). It can also be worked around by considering only [[irreducible polynomial]]s; any real polynomial of odd degree must have an irreducible factor of odd degree, which (having no multiple roots) must have a real root by the reasoning above.

This corollary can also be proved directly by using the [[intermediate value theorem]].

== Simple proof ==
One proof of the theorem is as follows:<ref name=Jeffrey />

Consider the polynomial

: <math>P(z) = a_0 + a_1z + a_2z^2 + \cdots + a_nz^n</math>

where all ''a''<sub>''r''</sub> are real. Suppose some complex number ''ζ'' is a root of ''P'', that is ''P''(''ζ'') = 0. It needs to be shown that
: <math>P(\overline{\zeta}) = 0</math>

as well.

If ''P''(''ζ'') = 0, then

: <math>a_0 + a_1\zeta + a_2\zeta^2 + \cdots + a_n\zeta^n = 0\,</math>

which can be put as

: <math>\sum_{r=0}^n a_r\zeta^r = 0.</math>

Now
: <math>P(\overline{\zeta}) = \sum_{r=0}^n a_r\left(\overline{\zeta}\right)^r</math>

and given the [[Complex_conjugation#Properties|properties of complex conjugation]],

: <math>\sum_{r=0}^n a_r\left(\overline{\zeta}\right)^r = \sum_{r=0}^n a_r \overline{\zeta^r} = \sum_{r=0}^n \overline{a_r\zeta^r} = \overline {\sum_{r=0}^n a_r\zeta^r}.</math>

Since,
: <math>\overline {\sum_{r=0}^n a_r\zeta^r} = \overline{0}</math>

it follows that

: <math>\sum_{r=0}^n a_r\left(\overline{\zeta}\right)^r = \overline{0} = 0.</math>

That is,

: <math>P(\overline{\zeta}) = a_0 + a_1\overline{\zeta} + a_2\left(\overline{\zeta}\right)^2 + \cdots + a_n\left(\overline{\zeta}\right)^n = 0.</math>

== Notes ==
<div class="references-small"><references /></div>

[[Category:Theorems in complex analysis]]
[[Category:Polynomials]]
[[Category:Theorems in algebra]]
[[Category:Articles containing proofs]]

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Dirichlet-multinomial distribution