Consider the polynomial $p(z)=\sum_0^na_iz^k$ where $a_n=1$ and $a_k \sim N(0,1)$, $k=0,1,2,\dotsc,n-1.$ What is the probability that 2 will be a bound of the roots of the polynomial? How can we find the asymptotic probability for the same? I did a simulation for a polynomial of this type of degree 5 and I found im almost 91% of the cases, the roots lie within $|z| \leq 2$. However,I do not know how to go about it anlytically. Also, kindly reference the relevant literature, if any, about these kinds of problems about the random polynomials of the above-mentioned type. I would be delighted and highly obliged for any help/hints/references.
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$\begingroup$ You mean 2 being an upper bound for the absolute values of the complex roots? $\endgroup$– Fedor PetrovCommented Jan 26, 2023 at 8:16
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$\begingroup$ yes ,of course! $\endgroup$– AgnostMysticCommented Jan 26, 2023 at 8:18
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$\begingroup$ i mean 2 being the bound for the absolute value of all roots $\endgroup$– AgnostMysticCommented Jan 26, 2023 at 8:22
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1$\begingroup$ If you apply the Gershgorin circle theorem to the transpose of the companion matrix of $p$ then a lower bound for this probability can be obtained: it reduces to the probability that $\mathbb{P}(1 - a_{n-1} \leq 2 \text{ and } -1 - a_{n-1} \geq -2)$, i.e. $\mathbb{P}(|a_{n-1}| \leq 1)$. However, this works out as about 60%, so far from optimal. $\endgroup$– David HughesCommented Jan 26, 2023 at 14:28
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$\begingroup$ It is the same as to ask what is the probability that the random function $1+\sum_{k=1}^\infty 2^{-k}a_k z^k$ has no root in the unit disk. The only (obvious) thing I can say is that it is some value between $0$ and $1$ and we can get some reasonable lower and upper bounds for it, but I doubt the exact value can be given by any decent expression, though I'll be happy to be proved wrong. $\endgroup$– fedjaCommented Jan 27, 2023 at 1:54
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1 Answer
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Not exactly your setting, with $a_n=1$ (and i.i.d. $a_0,\dots,a_{n-1}$?), but closely related settings were considered e.g. by Götze and Zaporozhets and Ibragimov and Zaporozhets. See also further references there.
In particular, in the paper by Götze and Zaporozhets we find the following:
Shparo and Shur [9] showed that under quite general assumptions the roots [...] concentrate near the unit circle as $n$ tends to $\infty$ with asymptotically uniform distribution of the argument.
Ibragimov and Zaporozhets [4] improved this result
Shepp and Vanderbei [8] considered real-valued standard Gaussian coefficients
Ibragimov and Zeitouni [3] extended this relation to the case of arbitrary i.i.d. coefficients from the domain of attraction of an $\alpha$-stable law
Even though your setting is a bit different, I am sure that arguments in the mentioned papers can be adapted to your situation.
answered Jan 26, 2023 at 14:05
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$\begingroup$ thank you for the references.Since my result is weaker in its claims and I do not have a strong background in random polynomials ,could you kindly suggest some approach or give me some hints $\endgroup$ Commented Jan 26, 2023 at 15:20
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$\begingroup$ @AgnostMystic : I suggest you read the proof of Theorem 1 by Götze and Zaporozhets. It is elementary and only about 3.5 pages in length, including the auxiliary lemmas. I think it should be rather easy to adapt that proof to your situation. However, such an adaptation will likely be too long for a MathOverflow answer. If you have difficulties reading that proof, please let me know. $\endgroup$ Commented Jan 26, 2023 at 15:57
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$\begingroup$ thank you @losif Pinnelis ,i will definitely go through it and try to understand it $\endgroup$ Commented Jan 27, 2023 at 11:01
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$\begingroup$ I found that the papers deal with the asymptotic cases when n is very large and my problem is about a certain number c being a bound for finite degree polynomials ,which requires to compute the probability that all roots lie within |z|≤c and I could not find any similar problem in the literature.Or may be I am wrong ,may be their technique can be adapted to this problem.I would be highly obliged for just some hints $\endgroup$ Commented Jan 27, 2023 at 16:41