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?

  • $\begingroup$ Does anything useful happen if you apply L'Hospital's rule and differentiate under the integral sign? $\endgroup$ – Nate Eldredge Jun 22 '18 at 15:25
  • $\begingroup$ @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 Jun 22 '18 at 16:13
  • 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 Martin Jun 23 '18 at 6:54
  • 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 Jun 23 '18 at 7:56
  • 1
    $\begingroup$ numerically it seems that $f(n)=4\pi n-6\pi+O(1/\sqrt{n})$ $\endgroup$ – Henri Cohen Jun 23 '18 at 18:18

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.