Proof of Roots of Odd Degree PolynomialDate: 05/10/2000 at 00:08:37 From: Anonymous Subject: Taylor Series (specific problem) Hello, I am a student in a Calculus II course and we're currently studying complex numbers (power series, Taylor series, sequences, etc.) I was given on a homework assignment the following question: 1) Prove that any real polynomial p: R->R of odd degree has a root. I don't know how to begin this proof (or finish it for that matter). It is one of the simpler problems on the assignment so if you could do this and post it so that I may use it as a guide for all other proofs, that would be fantastic. Please help! Thank you. Date: 05/10/2000 at 13:23:21 From: Doctor Rob Subject: Re: Taylor Series (specific problem) Thanks for writing to Ask Dr. Math. Assume that the leading coefficient is positive. Observe that lim p(x) as x -> +infinity is +infinity, so that from some point on in the positive direction, p(x) > 0. (Eventually the leading term will dominate.) Observe that lim p(x) as x -> -infinity is -infinity, so that from some point on in the negative direction, p(x) < 0. Now use the fact that polynomial functions are continuous, and apply the Intermediate Value Theorem, to conclude that p(x) = 0 somewhere in between (maybe more than one place, but at least one.) A similar argument works if the leading coefficient is negative. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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