bounding roots of a polynomial with Rouche's Theorem - MathOverflow

bounding roots of a polynomial with Rouche's Theorem

2010-02-24T15:24:05Z

Suppose f(z) = z^n - k [ z^(n-1) + ... + z + 1 ] where n is a positive integer and k is a real constant such that nk<1. I have shown that a root of this polynomial must satisy |z|<1, but I want a slightly better bound such as 1-k. This seems plausible from computational results but is difficult to prove. I am trying to use Rouche's theorem to do this but finding an appropriate bounding function is difficult. Is there any other result about holomorphic functions that may help?

Answer by Gabriel Benamy for bounding roots of a polynomial with Rouche's Theorem

2010-02-24T15:34:28Z

I don't know about any further boundings, but n = 3 and k = 1/4, or polynomial $4z^3 - z^2-z-1 = 0$ has a solution (1/12 + 1/12 (235 - 6 Sqrt[1473])^(1/3) + 1/12 (235 + 6 Sqrt[1473])^(1/3)), whose absolute value is ~ 0.868877, which is greater than 1-k. Other {n,k} pairs are {2,3}, {4,6}, and {5,6}.

EDIT I noticed that if the roots are multiplied by nk, then as k goes from 0 to 1/n, the largest root in absolute value (which happens to be the largest root) goes from 0 to about 1. So I suppose that the roots are bound in the range (0, 1/n).

Answer by Harald Hanche-Olsen for bounding roots of a polynomial with Rouche's Theorem

2010-02-24T21:34:48Z

Summary of the discussion: Using the triangle inequality, one sees that that $|f(z)|\ge f(|z|)$, and so the root of largest absolute value is the positive real root $z_k$. Differentiating $f(z)(z-1)$, one gets a bound:

$$z_k < \frac{1 + k}{1 + n^{-1}}.$$

When $k \rightarrow 1/n$, the largest real root approaches $1$ (by continuity, since $f(1) = 1 - nk$). Thus any bound must involve $n$.

The OP complains that he wants something better. It is pointed out that as $k \rightarrow 1/n$, the quantity $1 - z_k$ is asymptotic to

$$\frac{2(1 - kn)}{(1 + n)}.$$

The OP then complains that he wants a bound in $n$ and $k$ (which was already given). The OP askes whether the asymptotic above was found in the following way: "Are you simply using the fact that the root would occur roughly twice as far as the turning point?" No --- mathematics was used at this point.

The OP says that he simply wants an upper bound on the real part of each root. Since the real part of the real root $z_k$ is itself, this question has already been answered. The asymptotic result shows it is impossible to impove this bound significantly.

It's hard to tell if the problem with the OP's repeated questions involve English, Mathematics, or both. In either case, this has already wasted 15 minutes of my time. To paraphrase Zagier, that's the equivalent of 15 days of the OP's time. Feel free to edit this post to make it more "civil".

Answer by Josh for bounding roots of a polynomial with Rouche's Theorem

2010-02-24T23:01:21Z

I got the same bound via a different method. I looked at fixing the complex part of z and taking the partial derivative of g(z) with respect to the real part. The first turning point to the left of the root at z=1 is at 1-(1-kn)/((1+n) = (1+k)/(1+n^(-1)).

The fact that this is where the turning point lies motivates the fact that there is still some more room to push to the left of this value where a root cannot lie.

Answer by Josh for bounding roots of a polynomial with Rouche's Theorem

2010-02-25T11:37:18Z

I would like a bound in terms of k and/or n. One which would work whether or not k is close to 1/n. My method was different in the sense that I took partial derivatives which would tell you the effect of fixing different values for the complex part.

Are you simply using the fact that the root would occur roughly twice as far as the turning point? I would like a better lower bound of |1-z_k| where |z_k| is the largest root with positive real part (better than (1-kn)/(1+n) I mean).

I actually simply want to find an upper bound on the real part of any root. i.e find an upper bound for p where z=p+iq is a root and p,q are real.

However it seems more natural to consider |z|. I hope this makes more sense.