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| In [[mathematics]], '''Descartes' rule of signs''', first described by [[René Descartes]] in his work ''[[La Géométrie]]'', is a technique for determining the [[s:fr:Page:Descartes La Géométrie.djvu/84|number of positive]] or negative real roots of a [[polynomial]].
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| The rule gives an upper bound number of positive or negative roots of a polynomial. However, it is not a complete criterion, and so, does not provide the exact number of positive or negative roots.
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| ==Descartes' rule of signs==
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| ===Positive roots===
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| The rule states that if the terms of a single-variable [[polynomial]] with [[real number|real]] [[coefficient]]s are ordered by descending variable exponent, then the number of positive [[Root of a function|roots]] of the polynomial is either equal to the number of sign differences between consecutive nonzero coefficients, or is less than it by an even number. [[Multiple root]]s of the same value are counted separately.
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| ===Negative root===
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| As a [[corollary]] of the rule, the number of negative roots is the number of sign changes after multiplying the coefficients of odd-power terms by −1, or fewer than it by a multiple of 2. This procedure is equivalent to substituting the negation of the variable for the variable itself.
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| For example, to find the number of negative roots of <math>f(x)=ax^3+bx^2+cx+d</math>, we equivalently ask how many positive roots there are for <math>-x</math> in <math>f(-x)=a(-x)^3+b(-x)^2+c(-x)+d = -ax^3+bx^2-cx+d \equiv g(x). </math> Using Descartes' rule of signs on <math>g(x)</math> gives the number of positive roots <math>x_i</math> of ''g'', and since <math>g(x) = f(-x)</math> it gives the number of positive roots <math>(-x_i)</math> of ''f'', which is the same as the number of negative roots <math>x_i</math> of ''f''.
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| ===Example===
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| The polynomial
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| :<math>f(x)=+x^3 + x^2 - x - 1 \,</math>
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| has one sign change between the second and third terms (the sequence of pairs of successive signs is ++, +−, −−). Therefore it has exactly one positive root. Note that the leading sign needs to be considered although in this particular example it does not affect the answer.
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| To find the number of negative roots, change the signs of the coefficients of the terms with odd exponents, i.e., apply Descartes' rule of signs to the polynomial <math>f(-x)</math>, to obtain a second polynomial
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| :<math>f(-x)=-x^3 + x^2 + x - 1 \,</math>
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| This polynomial has two sign changes (the sequence of pairs of successive signs is −+, ++, +−), meaning that this second polynomial has two or zero positive roots; thus the original polynomial has two or zero negative roots.
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| In fact, the factorization of the first polynomial is
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| :<math>f(x)=(x + 1)^{2}(x - 1), \,</math>
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| so the roots are −1 (twice) and 1.
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| The factorization of the second polynomial is
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| :<math>f(-x)=-(x - 1)^{2}(x + 1), \,</math>
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| So here, the roots are 1 (twice) and −1, the negation of the roots of the original polynomial.
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| ===Complex roots===
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| Any ''n''<sup>th</sup> degree polynomial has exactly ''n'' roots. So if ''f''(''x'') is a polynomial which does not have a root at 0 (which can be determined by inspection) then the <u>minimum</u> number of complex roots is equal to
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| :<math>n-(p+q),\,</math>
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| where ''p'' denotes the maximum number of positive roots, ''q'' denotes the maximum number of negative roots (both of which can be found using Descartes' rule of signs), and ''n'' denotes the degree of the equation. A simple example is the polynomial
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| :<math>f(x) = x^3-1\, ,</math> | |
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| which has one sign change, so the maximum number of positive real roots is 1. From
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| :<math>f(-x) = -x^3-1\, ,</math>
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| we can tell that the polynomial has no negative real roots. So the minimum number of complex roots is
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| :<math>3 - (1+0) = 2 \, .</math>
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| Since complex roots of a polynomial with real coefficients must occur in conjugate pairs, we can see that ''x''<sup>3</sup> - 1 has exactly 2 complex roots and 1 real (and positive) root.
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| ==Special case== | |
| The subtraction of only multiples of 2 from the maximal number of positive roots occurs because the polynomial may have complex roots, which always come in pairs since the rule applies to polynomials whose coefficients are real. Thus if the polynomial is known to have all real roots, this rule allows one to find the exact number of positive and negative roots. Since it is easy to determine the multiplicity of zero as a root, the sign of all roots can be determined in this case.
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| ==Generalizations==
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| If the real polynomial ''P'' has ''k'' real positive roots counted with multiplicity, then for every ''a'' > 0 there are at least ''k'' changes of sign in the sequence of coefficients of the Taylor series of the function ''e''<sup>''ax''</sup>''P''(''x'').<ref>Vladimir P. Kostov, ''A mapping defined by the Schur-Szegő composition'', Comptes Rendus Acad. Bulg. Sci. tome 63, No. 7, 2010, 943 - 952.</ref>
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| In the 1970s [[Askold Georgevich Khovanskiǐ]] developed the theory of ''[[fewnomial]]s'' that generalises Descartes' rule.<ref>A. G. Khovanskii, ''Fewnomials'', Princeton University Press (1991) ISBN 0-8218-4547-0.</ref> The rule of signs can be thought of as stating that the number of real roots of a polynomial is dependent on the polynomial's complexity, and that this complexity is proportional to the number of monomials it has, not its degree. Khovanskiǐ showed that this holds true not just for polynomials but for algebraic combinations of many transcendental functions, the so-called [[Pfaffian function]]s.
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| ==See also==
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| *[[Sturm's theorem]]
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| *[[Rational root theorem]]
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| *[[Polynomial function theorems for zeros]]
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| *[[Properties of polynomial roots]]
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| *[[Gauss–Lucas theorem]]
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| ==Notes==
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| {{reflist}}
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| ==External links==
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| *[http://www.cut-the-knot.org/fta/ROS2.shtml Descartes’ Rule of Signs] — Proof of the Rule
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| {{PlanetMath attribution|id=5997|title=Descartes' rule of signs}}
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| *[http://www.purplemath.com/modules/drofsign.htm Descartes’ Rule of Signs] — Basic explanation
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| {{DEFAULTSORT:Descartes' Rule Of Signs}}
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| [[Category:Polynomials]]
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| [[Category:Theorems in algebra]]
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