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Suppose $\large{P(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_{n-1} x^{n-1} + x^n}$

where all the $\large{a_i}$ are strictly positive.

Find a sharp (ish?) lower bound for the smallest magnitude of all the roots in terms of the coefficients.

Can this be done? Or is it impossible, apart from special cases?

The case I am after is for strictly decreasing coefficients. $$a_0 > a_1 > a_2 > \cdots > a_{n-2} > a_{n-1}$$

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  • $\begingroup$ An obvious lower bound is $0$. A sharp lower bound would (almost) give us a root of $P$ in terms of the coefficients, and it is well know that for $n>4$ there exist no general radical expression. So it seem unlikely that a practical lower bound exists. $\endgroup$ – Servaes Mar 8 '18 at 10:54
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In 1912, Kakeya has obtained following result:

The roots of the polynomial $$p(z) = a_0 + a_1 z + \cdots + a_{n-1} z^{n-1} + a_n z^n$$ with real and positive coefficients lie in the annulus $R_1 \le |z| \le R_2$ where $$R_1 = \min_{0\le j \le n-1} \frac{a_j}{a_{j+1}}\quad\text{and}\quad R_2 = \max_{0\le j \le n-1} \frac{a_j}{a_{j+1}}$$

For the problem at hand, one can take $a_n = 1$. This leads to following lower bound for the smallest magnitude of the roots: $$|z| \ge \min\left\{ \frac{a_0}{a_1}, \frac{a_1}{a_2}, \cdots, \frac{a_{n-2}}{a_{n-1}}, a_{n-1} \right\}$$ Other types of bounds are available, a good search key is the keyword "Eneström-Kakeya Theorem".

Update

I just notice there is an error in Kakeya's 1912 paper$\color{blue}{{}^{[1]}}$. It wrongly assert the inequalities are strict (i.e $R_1 < |z| < R_2$ instead of the correct version $R_1 \le |z| \le R_2$). When $R_1 < R_2$, it is possible for some roots lie on one (but not both) of the circles $|z| = R_1$ and $|z| = R_2$. For a sufficient and necessary condition for this to happen, please refer to Anderson's paper$\color{blue}{{}^{[2]}}$ below.

References

  • $\color{blue}{[1]}$ Kakeya, S., On the Limits of the Roots of an Algebraic Equation with Positive Coefficients, Tôhoku Mathematical Journal (First Series),2, 140–142 (1912–13).
    ( an online copy can be found here)

  • $\color{blue}{[2]}$ N. Anderson, E. B. Saff, and R. S. Varga, On the Eneström-Kakeya theorem and its sharpness, Linear Algebra Appl. 28 (1979), 5-16.

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Since you are working over the complex numbers, the fundamental theorem of algebra says that any (monic) polynomial splits in its roots, i.e. given $P(x)=\sum_{i=0}^na_ix^i$ (with $a_n=1$), we have that $$\sum_{i=0}^na_ix^i=\prod_{i=1}^n(x-\lambda_i).$$

Looking at the right-hand side, we see that the constant term is given by $\prod_{i=1}^n \lambda_i$. In other words $a_0=\prod_{i=1}^n (-\lambda_i)$. Thinking a bit harder we find that $a_1=(-1)^{n-1}\sum_{i}\lambda_{1}\dots \widehat{\lambda_i}\dots \lambda_n$ where $\widehat{\lambda_i}$ means that we ignore this factor in the product. From here it easy to find the general formula of $a_k$ in terms of the roots $\lambda_1, \dots , \lambda_n$.

As far as I can see, this is the only connection you will be able to find between the $a_i$'s and the $\lambda_i$'s. Perhaps imposing restrictions on the $a_i$'s allows you to conclude something for the $\lambda_i$'s as well but I really doubt it. Indeed, the above connection is what led Galois and others to investigate formula's for solution in general.

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  • $\begingroup$ I just edited to include the special case I am after. $\endgroup$ – Jack Tiger Lam Mar 8 '18 at 11:05

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