A friend of mine, obtained a lower bound for the trace norm of matrices described in this question (for the special case $a_{ij} = \pm 1$). That lower bound is $ \frac{f(n)}{2\pi}$ where $$ f(n) := \int_0^\infty \log\left( \frac{(1+t)^n +(1-t)^n}{2} +n(n-1) t(1+t)^{n-2}\right)t^{- 3/2} \ \mathrm{d}t $$ Numerical computaions suggest that $$ f(n) = 4 \pi n + o(n) $$ How to justify it? Moreover, is it possible to obtain a good rate of convergence?
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asked Jun 22, 2018 at 14:57
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$\begingroup$ Does anything useful happen if you apply L'Hospital's rule and differentiate under the integral sign? $\endgroup$– Nate EldredgeCommented Jun 22, 2018 at 15:25
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$\begingroup$ @NateEldredge The function under integral divided by $n$ converges pointwise to $log(1+t)$ but it seems that there is no dominating function to use convergence theorems for integrals. $\endgroup$– Mahdi - Free PalestineCommented Jun 22, 2018 at 16:13
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4$\begingroup$ Can't you just consider the difference $f(n) - 4n\pi = f(n) - \int_0^\infty \log((1+t)^n)t^{-3/2}\,dt$, combined into a single integral that looks like $\int_0^\infty \log(g(n,t))t^{-3/2}\,dt$, where $g(n,t)$ is bounded below away from $0$ and bounded above by say $n^2$? That would show that $f(n) = 4n\pi + O(\log n)$. $\endgroup$– Greg MartinCommented Jun 23, 2018 at 6:54
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3$\begingroup$ @GregMartin In fact using this idea and and a change of variable $nt=u$ gives $f(n)=4\pi n +O(1)$. $\endgroup$– Mostafa - Free PalestineCommented Jun 23, 2018 at 7:56
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1$\begingroup$ numerically it seems that $f(n)=4\pi n-6\pi+O(1/\sqrt{n})$ $\endgroup$– Henri CohenCommented Jun 23, 2018 at 18:18
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This is an improvement of my previous post. I claim that $$4\pi n-6\pi<f(n)<4\pi n-2\pi.$$ Starting from $$f(n)-2\pi(n-1)=\int_0^\infty \log\left(\frac{1+t}{2}+\frac{1+t}{2}\left(\frac{1-t}{1+t}\right)^n+n(n-1)\frac{t}{1+t}\right)\,t^{-3/2}\,dt,$$ we see that $$f(n)-2\pi(n-1)<\int_0^\infty \log\bigl(1+n^2 t\bigr)\,t^{-3/2}\,dt=2\pi n.$$ Hence the upper bound $f(n)<4\pi n-2\pi$ follows. For the lower bound, we assume $n\geq 2$ without loss of generality, and we start from $$f(n)-2\pi(n-2)=\int_0^\infty \log\left(\frac{(1+t)^2}{2}+\frac{(1+t)^2}{2}\left(\frac{1-t}{1+t}\right)^n+n(n-1)t\right)\,t^{-3/2}\,dt.$$ Combining this with the inequality $$\frac{(1+t)^2}{2}+\frac{(1+t)^2}{2}\left(\frac{1-t}{1+t}\right)^n>1-(n-1)t,$$ which can be verified for $t<1/(n-1)$ and $t\geq 1/(n-1)$ separately, we see that $$f(n)-2\pi(n-2)>\int_0^\infty \log\bigl(1+(n-1)^2t\bigr)\,t^{-3/2}\,dt=2\pi(n-1).$$ Hence the lower bound $f(n)>4\pi n-6\pi$ follows also.
answered Jun 23, 2018 at 21:47