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Nov
8
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.
Nov
8
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.
Nov
8
revised Lower bound on a polynomial far from its zeros
Clarification
Nov
8
comment Lower bound on a polynomial far from its zeros
I still want $a \in [-1,1]$, I added it. Thanks.
Nov
8
awarded  Promoter
Nov
7
awarded  Curious
Nov
6
asked Lower bound on a polynomial far from its zeros
Oct
8
awarded  Commentator
Oct
8
comment Implications of zero elemntary symmetric polynomials over a finite field
Thank you both very much, a shame that I missed that...
Oct
8
accepted Implications of zero elemntary symmetric polynomials over a finite field
Oct
8
asked Implications of zero elemntary symmetric polynomials over a finite field
Jun
17
accepted A sharper bound for $\|\cos(kA)\|_{\infty}$ for symmetric stochastic matrices
Jun
15
answered A sharper bound for $\|\cos(kA)\|_{\infty}$ for symmetric stochastic matrices
May
19
revised A sharper bound for $\|\cos(kA)\|_{\infty}$ for symmetric stochastic matrices
deleted 272 characters in body
May
17
awarded  Nice Question
May
17
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.
May
16
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 $T = O(\log n)$.
May
13
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.
May
8
revised A sharper bound for $\|\cos(kA)\|_{\infty}$ for symmetric stochastic matrices
alternative formulation
May
5
revised A sharper bound for $\|\cos(kA)\|_{\infty}$ for symmetric stochastic matrices
added 4 characters in body