Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm supposed to prove that :

For $g(x) = a_0+a_1 \cdot x+\cdots+a_n \cdot x^n$ a polynomial of degree $ n$ where $ n \ge 0$ and $a_n \ne 0$.

Prove that $\log|g(x)|$ is $O(\log(x))$.

I've been able to do it for specific polynomials, but I can't seem to prove this with a generic polynomial.

share|improve this question
add comment

1 Answer

up vote 0 down vote accepted

We can assume that $a_n=1$. Let $R$ such that if $|x|>R$ then $\left|\frac{a_0+\dots+a_{n-1}x^{n-1}}{x^n}\right|\leq 1$. Then $$\log|g(x)|=\log\left(|x|^n\left(\left|\frac{a_0+\dots+a_{n-1}x^{n-1}}{x^n}\right|\right)+1\right)\leq \log(|x|^n+1)\leq \log 2+n\log |x|$$ if $|x|>\max\{R,1\}$. This gives the wanted result, as if we assume $|x|\geq \max\{R,2\}$, we get $\log|g(x)|\leq (n+1)\log|x|$.

share|improve this answer
    
Note that the constant implied by the big-O notation depends on $n$. –  marty cohen Oct 6 '12 at 22:50
    
@martycohen Yes, for example when $g(x)=x^n$ we have to take a constant $\geq n$. –  Davide Giraudo Oct 7 '12 at 9:21
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.