polynomial equation of odd degree


Theorem.

The equation

a0xn+a1xn-1++an-1x+an=0 (1)

with odd degree n and real coefficients ai (a00) has at least one real root x.

Proof.  Denote by f(x) the left hand side of (1).  We can write

f(x)=a0xn[1+g(x)]

where  g(x):=a1x++an-1xn-1+anxn.  But we have  lim|x|g(x)=0  because

lim|x|aixi=0

for all  i=1,,n.  Thus there exists an  M>0  such that

|g(x)|<1for|x|M.

Accordingly  1+g(±M)>0  and

signf(±M)=(signa0)(sign(±M))n1=(signa0)(±1)

since n is odd.  Therefore the real polynomial function f has opposite signs in the end pointsPlanetmathPlanetmath of the intervalMathworldPlanetmath[-M,M].  Thus the continuity of f guarantees, according to Bolzano’s theorem, at least one zero x of f in that interval.  So (1) has at least one real root x.

Title polynomial equation of odd degree
Canonical name PolynomialEquationOfOddDegree
Date of creation 2013-03-22 15:39:19
Last modified on 2013-03-22 15:39:19
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 7
Author pahio (2872)
Entry type Theorem
Classification msc 26A15
Classification msc 26A09
Classification msc 12D10
Classification msc 26C05
Related topic AlgebraicEquation
Related topic ExampleOfSolvingACubicEquation