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Problem/Conjecture :

Let $a_i>0$ such that $a_{n}\geq a_{n-1}\geq \cdots \geq a_1\geq a_0$ then define :

$$\sum_{i=1}^{n}a_ix^{2i}+a_0=0$$

Then it seems we have all the complex roots in the unit circle or let $x_{root}=a+bi$ so :

$$a^2+b^2\leq 1$$

I made several random test with Wolfram alpha and it seems true notably as the $a_i\simeq 1$

Motivation :

See https://en.wikipedia.org/wiki/Sendov%27s_conjecture .

For an attempt see my answer here Prove magnitude of zeros of polynomial $\le\frac{\sqrt5+1}2$

On the other hand using Lagrange's bound it's true for :

$$\max\left(1,\sum_{i=0}^{n-1}\frac{a_{i}}{a_{n}}\right)\leq 1$$

But it's the simple case .

Question :

How to (dis)prove it ?(answered now)

The real case :

Let $a_i>0,a_0=-1,n\geq 3$ then define :

$$\sum_{i=1}^{n}a_ix^{i}+a_0=0\tag{I}$$

Then there exits a constant $0<C_n<1/2$ depending on the degree of the polynomial such that :

$$\prod_{i=1}^{n}a_i>C_n>0$$

Then $\exists! x_{root}\in [0,1]$ . Which is a root of $I$

In my answer I provide a lower bound near $C_3\simeq 0.039$ then next bound is lower than $$C_4\simeq 0.0079$$

Have you an asymptotics for $C_n$ ?

To concentrate the effort we can try to show :

$$\frac{C^2_{n-1}}{C_{n}C_{n-2}}\simeq \phi$$

Which is a Turan-Type estimation . Then use https://mathoverflow.net/questions/102540/product-of-fibonacci-numbers

Nota bene :

An very accurate solution (the bound $x=1$ is reached at a limit ) can be found if we consider the equality case in HM-GM considering the coefficients of the highest degree of the polynomials (numerator/denominator in the algorithm) and solving a trinomial using Lagrange reversion theorem from the geometric series (equality case).

Example :

$$\frac{x}{8}(x+1+\frac{y}{x+1})-1=0$$

Now $x\to 2x,y=11$ we have the following iteration :

$$f(x)=\frac{4}{2x+1+\frac{11}{2x+1}}$$

With five iteration with itself ($f(x)$) :

We get the coefficient :

$$\frac{3573\cdot7355}{48631411}\simeq x_{root}$$

Using a proportional argument we get in the case above a solution close to one .

Miss and Mister cassoulet char
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  • $a,b\in [-1,1]$ is not equivalent with the root $a+ib$ being in the unit circle. Take for instance $1+i\cdot 1$. – Gary Oct 24 '23 at 13:55
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    This is a duplicate of https://math.stackexchange.com/q/188039/42969, it is known as the “Eneström–Kakeya theorem.” – Martin R Oct 24 '23 at 14:00
  • “In my answer I provide a lower bound near $C_3\simeq 0.039$ ...” – Where exactly is that in your answer? It is also very unfortunate that you changed the contents of your question significantly, as it invalidates Alex's answer, which was given before that edit. – Martin R Oct 26 '23 at 09:43
  • @MartinR "$$abc\geq r$$ with the polynomial where $2\frac{\left(2r^{2}+3r+1\right)}{r\left(r^{2}+4r+2\right)}-27=0,r>0$(take the numerator) :".I change nothing just correct it and add new things .Alex can just make an edit with the theorem you found and I accept it . – Miss and Mister cassoulet char Oct 26 '23 at 13:34

2 Answers2

1

Let us simplify the problem first. For each nonnegative integer $k\le n$ put $b_k=\frac {a_k}{a_n}$. Then $1=b_n\ge\dots\ge b_0$. Put $p(y)=\sum_{k=0}^n b_kx^k$. Then the conjecture is that each root $y$ of $p(y)$ has the absolute value at most $1$. Consider cases of small degree $n$.

Suppose that $n=1$. Then $y=-b_0$ and $|y|=|-b_0|\le 1$.

Suppose that $n=2$. Then $y=-\frac{b_1}2\pm\sqrt{\frac{b_1^2}4-b_0}$. If $\frac{b_1^2}4-b_0\ge 0$ then $y$ is real and $|y|\le \frac{b_1}2+\frac{b_1}2\le 1$. If $\frac{b_1^2}4-b_0<0$ then $y$ is not real and $$|y|^2=\left(\frac{b_1}2\right)^2+\left(b_0-\frac{b_1^2}4\right)=b_0\le 1.$$

Suppose that $n=3$. To be continued...

ErikSatie part:

As quoted by user MartinR see Showing that the roots of a polynomial with descending positive coefficients lie in the unit disc.

Miss and Mister cassoulet char
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Alex Ravsky
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-1

General case viewed with an example :

Abstract :

All of it and follows need to be really justify because I assume something like $x_{root}=x_{iteration}=0$ which is obviously false in general but conducts to nice result .I underlying what which need to be justify.

Using my question here :Got a factored version of the Taylor's series?

We have using HM-GM :

$$1/27\left(a\left(x+1\right)\cdot\frac{b}{x+1}\cdot\frac{c}{\frac{1}{x+1}+1}\right)^{-1}=\frac{\left(x+2\right)}{abc\left(x+1\right)}\leq^?1$$

Iterating two time (even case is greater than odd case in a continued fraction with positive coefficient) setting in the algorithm x=0 :

$$2/27\frac{\left(2\left(abc\right)^{2}+3abc+1\right)}{abc\left(\left(abc\right)^{2}+4abc+2\right)}\leq^?1$$

Setting in the algorithm $x=0$ we have a one variable inequality in $y=abc$ which as constraint :

$$abc\geq r$$ with the polynomial where $2\frac{\left(2r^{2}+3r+1\right)}{r\left(r^{2}+4r+2\right)}-27=0,r>0$(take the numerator) :

$$P(x)=x\left(a\left(x+1\right)+\frac{b}{x+1}+\frac{c}{\frac{1}{x+1}+1}\right)-1=0$$

To finish use the trick $x^2=-1$ make $x\to x^2$ and square .

Some remark :

With this method it seems we got the "greatest" root and so applying Gauss-Lucas's theorem the roots of the derivative are also less than one which is a particular example true of the Sendov's conjecture .

As the derivative have zero as root the conjecture is true in this case .

Some other remark :

Using the trick $x\neq 0$:

$$f\left(x\right)=\left(\frac{1}{\frac{x}{a}+\frac{x}{b}+\frac{x}{c}}\right)^{3}-\frac{abc}{27x^{3}}=-\frac{1}{3}xf'\left(x\right)=\frac{1}{12}x^{2}f''\left(x\right)=\cdots=(-1)^n 2/(n+2)!x^{n}f^{(n)}(x),n\ge 1$$

We can find a power series .

Reference :

https://mathworld.wolfram.com/SendovConjecture.html

https://en.wikipedia.org/wiki/De_Branges%27s_theorem

Miss and Mister cassoulet char
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