IS/ssps3

For any \(n\ge 2\), consider the stability region \(S_n:=\{\nu\in \mathbb{C} : \left| \frac{n \nu^{(n-1)^2}}{2 n-1}+\frac{(n-1) \nu^{n^2}}{2 n-1}\right| \le 1 \}\). By introducing polar coordinates \(\nu=\rho e^{i \phi}\), we rewrite \(S_n\) as \[ \{ (\rho,\phi)\in [0,+\infty)\times [0,2\pi) : \mu_n(\rho,\phi)\le 0 \}, \] where

\[ \mu_n(\rho,\phi):=-1+\frac{\rho ^{(n-1)^2} }{2 n-1}\sqrt{(n-1)^2 \rho ^{4 n-2}+2n (n-1)\rho ^{2 n-1} \cos ((2 n-1 )\phi )+n^2}. \]

Claim:

\(\bullet\) \(S_n\) contains the closed unit disk (as shown by a triangle inequality)

\(\bullet\) \(S_n\) is rotationally symmetric, so it is enough to consider \(0\le \phi\le\frac{2\pi}{2n-1}\), moreover, due to the symmetry of the cosine function, \(0\le \phi\le\frac{\pi}{2n-1}\). Let \(\widehat{S_n}\) denote this restricted set.

\(\bullet\) the farthest point of \(\widehat{S_n}\) from the origin is the endpoint of the slice corresponding to \(\phi=\frac{\pi}{2n-1}\). Steps of the proof:

\(\mu_n(\cdot,0):[0,+\infty)\to \mathbb{R}\) is strictly increasing and has a unique root at 1.

\(\mu_n(\rho,\phi)<0\) for \(0<\phi<\frac{\pi}{2n-1}\) and \(0\le \rho <1\).

\(\mu_n(2,\phi)>0\) for \(0<\phi<\frac{\pi}{2n-1}\).

For each \(0<\phi<\frac{\pi}{2n-1}\), the derivative of \(\rho\mapsto \mu_n(\rho,\phi)\) is positive for \(\rho\ge 2\).

So, for each \(0<\phi<\frac{\pi}{2n-1}\), by the intermediate value theorem, \(\mu_n(\cdot,\phi)\) has at least one root in the interval \((1,2)\), but no roots outside.

Show that there is at most one critical point of \(\mu_n(\cdot,\phi)\) in the interval \([1,2)\) (for any such \(\phi\)). (There can be at most two critical points altogether in \([0,2)\), since the sign of the derivative is determined by a quadratic polynomial in \(\lambda:=\rho^{2 n-1}\).)

This means that \(\mu_n(\cdot,\phi)\) has a unique zero \(\zeta_n(\phi)\in (1,2)\) for each \(0<\phi<\frac{\pi}{2n-1}\) (because the assumption “it has 2 roots” leads to a contradiction by Rolle’s theorem), moreover, for any fixed \(\phi\), \(\mu_n(\rho,\phi)<0\) if \(0\le \rho < \zeta_n(\phi)\), and \(\mu_n(\rho,\phi)>0\) if \(\rho > \zeta_n(\phi)\).

Notice that the derivative of \(\mu_n(\rho,\cdot)\) is negative on \(\left(0,\frac{\pi}{2n-1}\right)\) (and 0 on the boundary).

These imply that \(\zeta_n(\cdot):\left(0,\frac{\pi}{2n-1}\right)\to (1,2)\) is strictly increasing.

Finally, due to continuity, for any fixed \(0\le \rho<+\infty\), we have \(\mu_n(\rho,\phi)\to \mu_n\left(\rho,\frac{\pi}{2n-1}\right)\) as \(\phi\to\frac{\pi^-}{2n-1}\). Notice that the graph of \(\rho\mapsto \mu_n\left(\rho,\frac{\pi}{2n-1}\right)\) is non-differentiable at one point \(<\zeta_n\left(\frac{\pi}{2n-1}\right)\).

These imply that the farthest point of \(\widehat{S_n}\) from the origin has distance \(\zeta_n\left(\frac{\pi}{2n-1}\right)\), that is, when \(\mu_n(\cdot,\frac{\pi}{2n-1})=0\).

Let us introduce the notation

\[ \tilde{\nu}_n:=\max\{|\nu| : \nu\in S_n \}. \]

(For example, \(\tilde{\nu}_n=\zeta_n(\pi)=\zeta_n\left(\frac{\pi}{2n-1}\right)\).)

So it is enough to consider the “real slice” of \(S_n\), i.e., we can suppose that \(\nu\) is real. We know that \(\tilde{\nu}_n>1\). The real slice of \(S_n\) is then the interval \([-\tilde{\nu}_n,1]\).

Claim: For \(\nu\in\mathbb{R}\)

\(\bullet\) by examining the graphs, we can prove that in a neighborhood of \(-\tilde{\nu}_n\) we have that \(\left| \frac{n \nu^{(n-1)^2}}{2 n-1}+\frac{(n-1) \nu^{n^2}}{2 n-1}\right| -1\) equals \(\frac{n \nu^{(n-1)^2}}{2 n-1}+\frac{(n-1) \nu^{n^2}}{2 n-1}-1\) if \(n\) is even, and \(-\left(\frac{n \nu^{(n-1)^2}}{2 n-1}+\frac{(n-1) \nu^{n^2}}{2 n-1}\right)-1\) if \(n\) is odd. Therefore omitting the absolute value this way does not change the location of the unique root of \(\left| \frac{n \nu^{(n-1)^2}}{2 n-1}+\frac{(n-1) \nu^{n^2}}{2 n-1}\right| -1=0\) in \((-2,-1)\), this root being \(-\tilde{\nu}_n\). (We also see that omitting the absolute value this way does not introduce any new roots in \((-2,-1)\).) Moreover, we know from the above that any of these 3 functions is positive if \(-2<\nu<-\tilde{\nu}_n\), and negative if \(-\tilde{\nu}_n<\nu<-1\).

\(\bullet\) The sequence \(-\tilde{\nu}_n\) (for \(n\ge 2\)) is strictly increasing. Proof: in view of the above, it is enough to show that for any fixed \(\nu\in(-2,-1)\) (or in fact, for any fixed \(\nu<0\)) and \(k\ge 1\) \[ \frac{n \nu^{(n-1)^2}}{2 n-1}+\frac{(n-1) \nu^{n^2}}{2 n-1}-1\Big|_{n=2k}<-\left(\frac{n \nu^{(n-1)^2}}{2 n-1}+\frac{(n-1) \nu^{n^2}}{2 n-1}\right)-1\Big|_{n=2k+1}<\frac{n \nu^{(n-1)^2}}{2 n-1}+\frac{(n-1) \nu^{n^2}}{2 n-1}-1\Big|_{n=2k+2}. \] The left inequality, for example, leads to \[(4 k^2-k)\nu^{8 k}+(8 k^2-1)\nu^{4 k-1}+4 k^2+k>0 \quad (k\ge 1).\] (The right inequality is completely analogous.) We can show that this spiky 3-term polynomial is first decreasing, then increasing, and is positive for all \(k\) at its unique critical point (using a factorization), so it is positive on the real line.

So it is enough to prove estimates for even \(n\) values, say.

Claim: If \(n\ge 2\) is any even integer, then

\[1+\frac{\ln (n)}{n^2}-\frac{\ln (\ln (n))}{n^2}<\tilde{\nu}_n\,\]

further, if \(n\ge 24\) is any even integer, then

\[\tilde{\nu}_n<1+\frac{\ln (n)}{n^2}-\frac{1}{2}\frac{\ln (\ln (n))}{n^2}.\]

These bounds imply the following estimates: if \(n\ge 2\) is any even integer, then

\[\frac{99}{100}\frac{n}{\ln (n)}<\left(\tilde{\nu}_n\right)^{n^2},\]

further, if \(n\ge 24\) is any even integer, then

\[\left(\tilde{\nu}_n\right)^{n^2}<\frac{n}{\sqrt{\ln (n)}}.\]

Numerical simulations nicely show this sublinear behavior.

A heuristic argument showing the above asymptotic behavior of \(\tilde{\nu}_n\) can be given as follows. When appropriate, we are going to use the approximations \(\ln (1+x)\approx x\) and \(e^{x}\approx 1+x\) valid if \(|x|\) is small. Take and rearrange the defining equation of \(-\tilde{\nu}_n\):

\[\left|\frac{n (-\tilde{\nu}_n)^{(n-1)^2}}{2 n-1}+\frac{(n-1) (-\tilde{\nu}_n)^{n^2}}{2 n-1}\right|-1=0\]

so

\[\tilde{\nu}_n=\left(\frac{2 n-1}{|n (-\tilde{\nu}_n)^{1-2 n}+n-1|}\right)^{1/n^2}.\quad\quad\quad(*)\]

Since \(-\tilde{\nu}_n\approx -1\), we have

\[\tilde{\nu}_n\approx \left(\frac{2 n-1}{|n (-1)^{1-2 n}+n-1|}\right)^{1/n^2}=(2n-1)^{1/n^2}.\]

But then

\[\tilde{\nu}_n\approx \left(2 n-1\right)^{1/n^2}=\exp\left({\frac{\ln \left(1-\frac{1}{2 n}\right)+\ln (2 n)}{n^2}}\right)\approx \exp\left({\frac{-\frac{1}{2 n}+\ln (2 n)}{n^2}}\right)=\exp\left(-\frac{1}{2 n^3}+\frac{\ln 2 }{n^2}+\frac{\ln ( n)}{n^2}\right)\approx \exp\left(\frac{\ln ( n)}{n^2}\right)\approx 1+\frac{\ln ( n)}{n^2}.\]

Substituting this approximation into (*), we get \[ \tilde{\nu}_n\approx \left(\frac{2 n-1}{\left|n \left(-1-\frac{\ln ( n)}{n^2}\right)^{1-2 n}+n-1\right|}\right)^{1/n^2}= \left(\frac{2 n-1}{n\left(1-\left(1+\frac{\ln ( n)}{n^2}\right)^{1-2 n}-\frac{1}{n}\right)}\right)^{1/n^2}=\exp\left(\frac{\ln(2-\frac{1}{n})}{n^2}-\frac{1}{n^2}\ln\left(1-\left(1+\frac{\ln ( n)}{n^2}\right)^{1-2 n}-\frac{1}{n}\right)\right)\approx \]

\[ \exp\left(-\frac{1}{n^2}\ln\left(1-\frac{1}{n}-\left(1+\frac{\ln ( n)}{n^2}\right)^{1-2 n}\right)\right)= \exp\left(-\frac{1}{n^2}\ln\left(1-\frac{1}{n}-\exp\left((1-2n)\ln\left(1+\frac{\ln ( n)}{n^2}\right)\right)\right)\right)\approx \]

\[ \exp\left(-\frac{1}{n^2}\ln\left(1-\frac{1}{n}-\exp\left((1-2n)\frac{\ln ( n)}{n^2}\right)\right)\right)= \exp\left(-\frac{1}{n^2}\ln\left(1-\frac{1}{n}-\exp\left(\frac{\ln ( n)}{n^2}-\frac{2\ln ( n)}{n}\right)\right)\right)\approx \]

\[\exp\left(-\frac{1}{n^2}\ln\left(1-\frac{1}{n}-1-\frac{\ln ( n)}{n^2}+\frac{2\ln ( n)}{n}\right)\right)\approx \exp\left(-\frac{1}{n^2}\ln\left(\frac{2\ln ( n)}{n}\right)\right)= \exp\left(-\frac{\ln 2}{n^2}-\frac{1}{n^2}\ln\left(\frac{\ln ( n)}{n}\right)\right)\approx \]

\[ \exp\left(-\frac{1}{n^2}\ln\left(\frac{\ln ( n)}{n}\right)\right)= \exp\left(\frac{\ln(n)-\ln \ln(n)}{n^2}\right)\approx 1+\frac{\ln(n)-\ln \ln(n)}{n^2}=1+\frac{\ln(n)}{n^2}-\frac{\ln \ln(n)}{n^2}. \]

By iterating this process, we’d get finer and finer asymptotic estimates.