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I'm going to teach a preparation course for the complex analysis qualifying exam from my university (which basically consists of me doing some problems from past exams) and I'm trying to solve some questions from previous exams. One of the questions is the following:

Prove that $\displaystyle{\sum_{n = 0}^{2013} a_n z^n \neq 0}$ if the coefficients satisfy $a_0 > a_1 > \dots > a_{2013} > 0$ and $|z| \leq 1$.

I tried to approach the problem by using the reverse triangle inequality to try to isolate the leading term, which I thought should be the constant coefficient, as follows:

\begin{array} . \left| \sum_{n = 0}^{2013} a_n z^n \right| &= \left| a_0 + \sum_{n = 1}^{2013} a_n z^n \right| \\ &\geq |a_0| - \left| \sum_{n = 1}^{2013} a_n z^n \right|\\ &\geq |a_0| - \sum_{n = 1}^{2013} |a_n z^n|\\ &\geq a_0 - \sum_{n = 1}^{2013} a_n |z^n|\\ &\geq a_0 - \sum_{n = 1}^{2013} a_0 |z^n|\\ &= a_0\left( 1 - \sum_{n = 1}^{2013} |z^n| \right) \end{array}

Then for $|z| < 1$, we have that this last expression is equal to

$$ a_0\left( 1 - |z| \frac{|z|^{2013} - 1}{|z| - 1} \right) $$

and my hope was to prove that this was always positive, nevertheless, after failing to prove it, I plotted the expression inside the parenthesis as a function of $|z|$, and found out that it changes sign in the interval $(0, 1)$.

Question

Can my approach somehow be made to work? And if not, how can I prove that the sum $\displaystyle{\sum_{n = 0}^{2013} a_n z^n \neq 0}$ under the given conditions?

Thank you very much for any help.

Antonio Vargas
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Adrián Barquero
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    I think the 2013 in the upper limit has got nothing to do with the question. Next year, the same question can be asked with 2014 as the upper limit. So perhaps you could try to prove it by taking infinity as the upper limit. The |z|^2013 then diminishes. Regraph and see what you get...Good luck teaching that course. You are certainly smarter than me... – imranfat Jul 07 '13 at 23:53
  • @imranfat Thank you very much for your good wishes. If I consider the same problem, but with an infinite sum I basically have the same problem, if I graph the corresponding function, it changes sign in the interval $(0, 1)$. But you're right that the problem doesn't depend on the "year it is asked" =) – Adrián Barquero Jul 08 '13 at 00:04
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    This is (apparently) called the Enestrom-Kakeya theorem; see here and here. – icurays1 Jul 08 '13 at 00:11
  • @icurays1 Thanks a lot for the links. That's very nice to know. – Adrián Barquero Jul 08 '13 at 00:19

3 Answers3

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You can look at (with $N$ = 2013). $$ g(z) = zf(z) - f(z) = a_N z^{N+1} + (a_{N-1} - a_N) z^{N} + (a_{N-2} - a_{N-1}) z^{N-1} + \cdots + (a_0 - a_1) z - a_0 $$

Since $g(z) = (z-1)f(z)$, strictly inside the unit disk the zeros of $g$ are the same as the zeros of $f$. On the boundary $g$ has a zero at $z = 1$, which needs to be handled separately.

$$ |g(z)| \ge \Big| |a_0| - |a_N z^{N+1} + (a_{N-1} - a_N) z^{N} + (a_{N-2} - a_{N-1}) z^{N-1} + \cdots + (a_0 - a_1) z | \Big| $$ $$ |g(z)| \ge \Big| |a_0| - |a_N + (a_{N-1} - a_N) + (a_{N-2} - a_{N-1}) + \cdots + (a_0 - a_1) | \Big| $$ $$ |g(z)| \ge \Big| |a_0| - |a_0 | \Big| = 0 $$ Strictly inside the unit disk there is never equality. When $|z| = 1$ you get equality when $z, z^2, \cdots z^{N+1}$ are collinear, which only happens at $z=1$. But $z = 1$ is not a zero of $f$.

bryanj
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Your approach cannot work because if e.g. $a_0\to-a_0$ the result stops working, but your argument essentially depends only on $|a_0|$.

As pointed out in the comments, this result is shown elsewhere, such as at Showing that the roots of a polynomial with descending positive coefficients lie in the unit disc.

Expanding on the above, let $a_0=1,a_1=\frac 3 4,a_2=\frac 1 2$ and everything else vanish (or be $\epsilon$, whatever). Then the first inequality you write down has $$|a_0|-\left|\sum_{n=1} a_nz^n\right| \text{ which at }z=1\text{ is just } 1-\frac 5 4 - \cdots \le -\frac 1 4$$ so you have no hope of bounding it below by zero. You need to use the relationship between $a_0$ and the other $a_i$, you cannot simply split it off via a single application of the triangle inequality.

Community
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not all wrong
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  • Dear Sharkos, thank you very much for the link. I don't understand what do you mean by $a_0 \rightarrow -a_0$ though since I'm assuming that $a_0 > 0$. – Adrián Barquero Jul 08 '13 at 00:21
  • I meant your argument doesn't really use enough of the hypotheses to work - in particular it makes use of the relative sign of $a_0,a_n$ too late, in the sense that the very first inequality you write down isn't strong enough. (See updated answer.) – not all wrong Jul 08 '13 at 00:30
  • I see, I understand your point now. Thank you very much for the help. – Adrián Barquero Jul 08 '13 at 00:41
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Instead of an answer I will state a slightly stronger assertion: $\newcommand{\CC}{\mathbb{C}}$

Let $a_0\geq a_1\geq\cdots\geq a_n>0$, $n\geq 1$, and set $f(z):=\sum_{k=0}^na_kz^k~$ for $z\in\CC$. If $a_{k-1}>a_k$ for some integer $k$, $1\leq k\leq n$, that is coprime to $n+1$ (in particular if $a_0>a_1$ or $a_{n-1}>a_n$), then $f(z)\neq 0$ whenever $|z|\leq 1$.

See my answer to this duplicate question.

Here is a more precise result: $\newcommand{\set}[1]{{\{#1\}}} $$\newcommand{\suchthat}{\mid} $$\newcommand{\union}{\cup} $$\newcommand{\NN}{\mathbb{N}} $

Let $a_0\geq a_1\geq\cdots\geq a_n>0$, $n\geq 1$, and set $f(z):=\sum_{k=0}^na_kz^k~$ for $z\in\mathbb{C}$, $K:=\set{k\in\NN\suchthat\text{$1\leq k\leq n$ and $a_{k-1}>a_k$}}$, $d:=\gcd(K\union\set{n\!+\!1})$. $~$If $d=1$, then $f(z)\neq 0$ for every $z$ in the closed unit disk $D:=\set{z\in\CC\suchthat |z|\leq 1}$. $~$If $d>1$, then all the zeros of $f(z)$ that lie in $D$ are $\,\exp(l\!\cdot\!2\pi i/d)$, $1\leq l\leq d-1$, and every one of them is simple.

chizhek
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