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 Apr10 comment Norm bound for the Jordan basis matrix Thank you for the comment. In this example, the bound is $O(\epsilon^{-1})$. Bounds such as this, although generally not useful, are indeed useful in my case (where even something that is exponential in $n$, or in the minimal entry, will do). Apr9 asked Norm bound for the Jordan basis matrix Nov8 comment Lower bound on a polynomial far from its zeros Yes, you are right. @AntonioVargas, can you make it into an answer so I can accept? Thanks. Nov8 comment Lower bound on a polynomial far from its zeros I thought I would be able to find something finer, but I might be wrong. Nov8 revised Lower bound on a polynomial far from its zeros Clarification Nov8 comment Lower bound on a polynomial far from its zeros I still want $a \in [-1,1]$, I added it. Thanks. Nov8 awarded Promoter Nov7 awarded Curious Nov6 asked Lower bound on a polynomial far from its zeros Oct8 awarded Commentator Oct8 comment Implications of zero elemntary symmetric polynomials over a finite field Thank you both very much, a shame that I missed that... Oct8 accepted Implications of zero elemntary symmetric polynomials over a finite field Oct8 asked Implications of zero elemntary symmetric polynomials over a finite field Jun17 accepted A sharper bound for $\|\cos(kA)\|_{\infty}$ for symmetric stochastic matrices Jun15 answered A sharper bound for $\|\cos(kA)\|_{\infty}$ for symmetric stochastic matrices May19 revised A sharper bound for $\|\cos(kA)\|_{\infty}$ for symmetric stochastic matrices deleted 272 characters in body May17 awarded Nice Question May17 comment A sharper bound for $\|\cos(kA)\|_{\infty}$ for symmetric stochastic matrices @user1551, it only happens for larger $n$ and very large $k$. Although, for $k$ which is relatively small to $n$, it does not exceed $3$ even for large $n$. It might have something to do with the accuracy of the "funm" function rather than to the result itself. I have no intuition as to why it should not be uniformly bounded. May16 comment A sharper bound for $\|\cos(kA)\|_{\infty}$ for symmetric stochastic matrices @StephenMontgomery-Smith, thank you for your comment. Trying to derive an analytic expression, even though the diagonalization is explicitly known, was not successful. However, I did try random symmtric stochastic circulant matrices in Matalb, and for $k = O(\log n)$ the results were indeed small. However, for larger $k$ (which I am not interested), the results did exceed $3$. Any other insight would be helpful, as I am trying to show that under some (maybe more) constraints I can bound the infinity norm of $\frac{1}{T}\sum_{k=1}^{T}\cos(kA)$ by a constant, for $T = O(\log n)$. May13 comment A sharper bound for $\|\cos(kA)\|_{\infty}$ for symmetric stochastic matrices @user1551, in case $A$ is the $2 \times 2$ matrix $[a,1-a;1-a, a]$, the infinity norm is given by $\max\{|\cos(k)|,|\cos((2a-1)k)|\}$. And indeed, as you stated, the maximum is obtained for $a=0.5$, which corresponds to your assumption. Unfortunately, I had no further insight. Thank you.