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To prove there exist $M>0$ and $a_0>0$ such that for $|z|>a_0$, $$\left|\frac{p_n(z)}{q_m(z)}\right|\leq \frac{M}{|z|^{m-n}}$$ where $p_n$ and $q_m$ are the polynomials of degree $n$ and $m$ respectively with $n<m$.

I encountered this question in this post.

According to the hint, it is enough to show that for large enough $R$ and for every $|z|>R$, we have $$\left|\frac{z-z_1}{z-z_2}\right|<M$$ In fact, $$\left|\frac{z-z_1}{z-z_2}\right|<\frac{|z-z_2|+|z_2-z_1|}{|z-z_2|}=1+\frac{|z_2-z_1|}{|z-z_2|}<M.$$

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Let $p_n(z)=\sum_{k=0}^{n}a_kz^k$, and $q_n(z)=\sum_{k=0}^{m}b_kz^k$ with $a_n\not=0$ and $b_m\not=0$. Note that $$\lim_{r\to +\infty }\sum_{k=0}^{m-1}\frac{|b_k|}{r^{m-k}}=\lim_{r\to +\infty }\sum_{k=0}^{n-1}\frac{|a_k|}{r^{m-k}}=0.$$ Then, for $|z|$ is "sufficiently large", $$\left|\frac{p_n(z)z^{m-n}}{q_m(z)}\right|\leq \frac{|a_n|+\sum_{k=0}^{n-1}\frac{|a_k|}{|z|^{m-k}}}{|b_m|-\sum_{k=0}^{m-1}\frac{|b_k|}{|z|^{m-k}}}\leq 2\frac{|a_n|}{|b_m|}.$$ Can you take it from here?

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  • $\begingroup$ Thank you! I update the post according to your hint. $\endgroup$ – Aolong Li Sep 28 '17 at 6:56
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    $\begingroup$ @LeoGardon OK, I was editing my answer. I hope that everything is clear now. $\endgroup$ – Robert Z Sep 28 '17 at 7:05
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This concludes from Growth Lemma says for polynomial $p(z)=a_0+\cdots+a_nz^n$ of degree $n$ and $a_n\neq0$ then for enough large $|z|$ we have $$\dfrac{|a_n|}{2}|z|^n\leq|p(z)|\leq\dfrac{3|a_n|}{2}|z|^n$$ apply this for $p$ and $q$ and find desire bound. you may find a version of Growth Lemma here.

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  • $\begingroup$ Could you point to a proof of this lemma? $\endgroup$ – CuriousKid7 Sep 28 '17 at 9:17
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    $\begingroup$ @CuriousKid7 I added it's link. $\endgroup$ – Nosrati Sep 28 '17 at 10:04

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