# Prove the zeros of a polynomial all lie in an annulus.

I am working on a problem. It has two parts:

(a) Let $$c_{0}>c_{1}>\cdots c_{n}>0$$. Show that the polynomial $$P(z):=c_{0}+c_{1}z+\cdots+c_{n}z^{n}$$ has no zeros inside the closed unit disc.

(b) Show that the zeros of polynomial $$P_{n}(z):=1+\frac{z}{2}+\frac{z^{2}}{3}+\cdots+\frac{z^{n}}{n+1}$$ all lie in an annulus $$\{1<|z|<1+\delta_{n}\}$$ where $$\delta_{n}\rightarrow 0$$ as $$n\rightarrow\infty$$.

I've proved part (a), but I am stuck in part (b). I think the proof of part (b) may be similar to part (a), so I state part (a) above and give my proof below, then I will give my attempt for part (b).

Part (a):

Suppose there exists $$z_{0}\in\mathbb{C}$$ such that $$P(z_{0})=0$$ and $$|z_{0}|<1$$.

Since $$P(z_{0})=0$$, we also have $$(1-z_{0})P(z_{0})=0,$$ where $$LHS=c_{0}+(c_{1}-c_{0})z_{0}+(c_{2}-c_{1})z_{0}^{2}+\cdots+ (c_{n}-c_{n-1})z_{0}^{n}-c_{n}z_{0}^{n+1}.$$

Thus, we have $$c_{0}=(c_{0}-c_{1})z_{0}+(c_{1}-c_{2})z_{0}^{2}+\cdots (c_{n-1}-c_{n})z_{0}^{n}+c_{n}z_{0}^{n+1}.$$

Now, taking norm to both side, and recalling that $$c_{0}>c_{1}>\cdots>c_{n}>0$$ and $$|z_{0}|<1$$, we have \begin{align*} c_{0}& which is a contradiction.

Thus, there is no zero of $$P(z)$$ that is inside the closed unit disc.

Part (b):

For part (b), I mimic what I've done in part (a). Let $$z\in\mathbb{C}$$ be a zero of $$P_{n}(z)$$, then $$P_{n}(z)=0$$ implies that $$(1-z)P_{n}(z)=0.$$

Thus, we have $$\Big(1-\dfrac{z^{n+1}}{n+1}\Big)-\dfrac{z}{2}-\dfrac{z^{2}}{6}-\cdots-\dfrac{z^{n}}{n(n+1)}=0.$$

Set $$Q_{n}(z):=-\dfrac{z}{2}-\dfrac{z^{2}}{6}-\cdots-\dfrac{z^{n}}{n(n+1)}.$$

Then if $$|z|<1$$, we have \begin{align*} 1&\leq|Q_{n}(z)|+\Big|\dfrac{z^{n+1}}{n+1}\Big|\\ &\leq\dfrac{|z|}{2}+\dfrac{|z|^{2}}{6}+\cdots+\dfrac{|z|^{n}}{n(n+1)}+\dfrac{|z|^{n+1}}{n+1}\\ &<\dfrac{1}{2}+\dfrac{1}{6}+\cdots+\dfrac{1}{n(n+1)}+\dfrac{1}{n+1}\\ &=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\cdots+\dfrac{1}{n}-\dfrac{1}{n+1}+\dfrac{1}{n+1}\\ &=1, \end{align*} which is a contradiction.

Thus, zeros of $$P_{n}(z)$$ must lie in $$|z|\geq 1$$.

Then, I try to get rid of $$|z|=1$$ by using the same techniques, but I found something else interesting.

If $$|z|=1$$, then by definition $$|Q_{n}(z)|\leq 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\cdots+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{n}{n+1}.$$

On the other hand, since we assume $$z$$ is a zero, we have $$|Q_{n}(z)|=\Big|1-\dfrac{z^{n+1}}{n+1}\Big|\geq \Big|1-\dfrac{1}{n+1}\Big|=\dfrac{n}{n+1}.$$

Thus, if $$|z|=1$$, we have $$\dfrac{n}{n+1}\leq |Q_{n}(z)|\leq\dfrac{n}{n+1},$$ and thus $$|Q_{n}(z)|=\dfrac{n}{n+1}.$$

This did not give me any contradiction, but an idea of $$\delta_{n}$$. By the problem itself, we can see that if $$\delta_{n}\rightarrow 0$$, then the annulus will become to a unit circle $$|z|=1$$, which is exactly our case.

Also, by my argument above, I want my $$\delta_{n}$$ to be related to $$|Q_{n}(z)|$$, so I tried to set $$1+\delta_{n}=\dfrac{n}{n+1},$$ which gives us $$\delta_{n}=-\dfrac{1}{n+1},$$ which tends to be $$0$$ as $$n\rightarrow\infty$$.

But then I don't know how to proceed, what should I do now?

Thank you!

Consider the polynomial $$R_n(z)=(z-1)P_{n}(z)=-1+\dfrac{z}{2}+\dfrac{z^{2}}{6}+\cdots+\dfrac{z^{n}}{n(n+1)}+\dfrac{z^{n+1}}{n+1}$$.

We will show that the Cauchy bound $$\rho_n=1+\gamma_n$$ of $$R_n$$, satisfies $$\gamma_n \to 0$$ as $$n \to \infty$$ hence we are done since all the roots $$w_{k,n}$$ of $$R_n$$ (hence of $$P_n$$) satisfy $$|w_{k,n}| \le \rho_n$$

If not familiar with the notion, the Cauchy bound $$\rho_P$$ of a polynomial $$P(z)=\sum_{0}^{n}{a_kz^k}, a_n \ne 0, n \ge 1$$ is the unique positive root of the polynomial $$|a_n|z^n-\sum_{0}^{n-1}|a_k|z^k$$ - taken by convention as $$0$$ if the polynomial is just $$a_nz^n$$ -and by the triangle inequality the relation $$|w_k| \le \rho_P$$ for all the roots of $$P$$ is clear.

Also it is clear that if for some $$R >0, \sum_{k=0}^{n-1}{|a_k|R^k} \le |a_n|R^n$$, then $$\rho_P \le R$$ by the unicity of the positive root.

We use the inequality: $$\rho_P \le max_{k=0,..,n-1}|n\frac{a_k}{a_n}|^{\frac{1}{n-k}}$$, which is fairly obvious since if we denote by $$R$$ the maximum on the RHS, $$|a_k| \le \frac{1}{n}|a_n|R^{n-k}, k=0,...n-1$$, so $$\sum_{k=0}^{n-1}{|a_k|R^k} \le |a_n|R^n$$, hence $$R \ge \rho_P$$ as above.

In our case the degree is $$n+1$$ and the coefficients are $$|a_0|=1, a_k=\frac{1}{k(k+1)}, 1 \le k \le n, a_{n+1}=\frac{1}{n+1}$$, so we need to prove the following two results:

1: $$((n+1)^2)^{\frac{1}{n+1}}=1+b_n, b_n \to 0$$

2: $$max_{ 1 \le k \le n}(\frac{(n+1)^2}{k(k+1)})^{\frac{1}{n+1-k}}=1+c_{n}, c_{n} \to 0$$

But using the inequality $$\log(1+x) \ge \frac{x}{2}, 0 \le x \le 1$$, we get $$\log(1+b_n) = 2\frac{\log(n+1)}{n+1} \to 0$$, hence $$b_n \le 1$$ eventually, hence $$b_n \le 4\frac{\log(n+1)}{n+1} \to 0$$, so 1: is done.

Similarly if $$k \le \frac{2n}{3}$$, $$n+1-k \ge \frac{n}{3}$$ and the same proof applies since then $$\frac{\log(\frac{(n+1)^2}{k(k+1)})}{n+1-k} \le 6\frac{\log(n+1)}{n} \to 0$$

If $$k \ge \frac{2n}{3}, \frac{\log(\frac{n+1}{k})}{n+1-k} \le \frac{n+1-k}{k(n+1-k)} \le \frac{2}{n} \to 0$$ and same for the other $$\frac{\log(\frac{n+1}{k+1})}{n+1-k} \to 0$$, so we are done too for case 2; and the problem is solved with $$\delta_n$$ the maximum of $$b_n, c_n$$ above

(we used $$\log(\frac{n+1}{k})=\log(1+\frac{n+1-k}{k}) \le \frac{n+1-k}{k}$$)

• Thank you for the brilliant solution!! Commented Aug 19, 2019 at 15:32
• you are welcome - the tricky part was actually in the answer provided by you above where the polynomial was converted to one where the leading coefficient dominates the nearby ones as the Cauchy-bound results - there are lots of similar to the one above - are the only "generic" bounds on roots in terms of coefficients, so they are useful and part of the standard kit in dealing with analytic polynomials alongside Bernstein inequality Commented Aug 19, 2019 at 15:48
• Yes. I didn't learn Cauchy bound or I missed this notion. Thank you so much for pointing this out and it is really useful. Commented Aug 19, 2019 at 16:58
• There is an excellent reference book by Rahman and Schmeisser called Analytic theory of polynomials etc which is fairly dense and long so not really that useful to study at first but awesome as reference in the subject Commented Aug 19, 2019 at 17:15