Piotr Semenov
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# 18 Comments

 Dec 9 comment For $f(x) \in \mathcal{F}_{q^r}[x]$, how to find all $p(x)f(x)$ s.t. its coefficients are $\mathcal{F}_q$-linear over coefficients of $f(x)$? @JyrkiLahtonen So all the $p(x)$ that I want are only $\mathcal{F}_q[x]$? How it can be proven that there are no $p(x) \in \mathcal{F}_{q^r}[x] \setminus \mathcal{F}_q[x]$ satisfying my requirements? Dec 8 comment For $f(x) \in \mathcal{F}_{q^r}[x]$, how to find all $p(x)f(x)$ s.t. its coefficients are $\mathcal{F}_q$-linear over coefficients of $f(x)$? @JyrkiLahtonen: I mean that all the coefficients of $p(x)f(x) = \sum_i b_i \cdot x^i$ are the $\mathcal{F}_q$-linear combinations of coefficients of $f(x) = \sum_i a_i \cdot x^i$. In another words, each $b_i = \sum_j c_j \cdot a_j$, where $c_j \in \mathcal{F}_q$. Nov 26 comment Given the roots of polynomial over finite field, what is the count of its distinct nonzero coefficients? @RobertIsrael: I have no assumptions about the finite field. It can be any $GF(q^p)$ for any prime $q$ and positive $p$. Feb 7 comment How many ways to distribute $k$ indistinguishable balls over $m$ of $n$ distinguishable bins of finite capacity $l$? sure, I specified the upper limit just to see the finite summation. Thank you for your answer! Feb 6 comment How many ways to distribute $k$ indistinguishable balls over $m$ of $n$ distinguishable bins of finite capacity $l$? Just to improve your answer, the upper limit of sum is $\lfloor \frac{k}{l} \rfloor-1$. Feb 6 comment How many ways to distribute $k$ indistinguishable balls over $m$ of $n$ distinguishable bins of finite capacity $l$? Just a small typo in your answer. Add $l$ balls to first bin, one have to distribute $k-l$ balls (instead of "$m-l$ units"). Feb 6 comment How many ways to distribute $k$ indistinguishable balls over $m$ of $n$ distinguishable bins of finite capacity $l$? Can you kindly explain, why $\binom{k-jl-1}{m-1}$ and not $\binom{k-jl-1}{m-j-1}$? Feb 3 comment Number of ways to put $n$ unlabeled balls in $k$ bins with a max of $m$ balls in each bin @Brian What are the sum limits in your formula? The expression from book Enumerative Combinatorics (p. $360$) is $L(n,k,m)=\sum_{j=0}^k (-1)^j C_k^j C_{k+n-j(m+1)-1}^{k-1}$ according to your notation, where $C_k^j$ means the binomial coefficient. Unfortunately, $L(n,k,m)=0$ for any $n \leq k \cdot m$ (Mathematica' showed me that). Somewhere the mistake lives. Can you kindly help me with this problem? Oct 21 comment When is a cyclotomic polynomial over a finite field a minimal polynomial? Thanks! So I did a mistake while suggesting that cyclotomic polynomials are also irreducible in case of finite fields. Oct 21 comment When is a cyclotomic polynomial over a finite field a minimal polynomial? Please explain it for the following example. Given finite field $\mathrm{F}_q$, I have a decomposition $x^n-1=\prod_{i=0}^{m-1}f_i(x)$ over $\mathrm{F}_q$, where all $f_i(x)$ are cyclotomic. I am not sure that all $f_i(x)$ are minimal polynomials for some elements from extension $[\mathrm{F}_q^r : \mathrm{F}_q]$, where $r$ is a smallest number that $n$ divides $q^r - 1$. What is wrong in that? Jun 26 comment Finding Eigenvalues and Eigenvectors weird equations I got my answers just be definition of eigenvector: for matrix $A$ the eigenvector $v$ corresponding to eigenvalue $\lambda$ is the solution of system of linear equations $(A - \lambda \cdot I) \cdot v = 0$. In another words, dot product of matrix $(A - \lambda \cdot I)$ rows and vector $v$ vanishes. So I have got the equation $2x - y - 2z = 0$ for $[x, y, z]$ being the eigenvector we are looking for. Setting $x$ to $1$ you can got the eigenvectors $[1, 0, 1]$ and $[1, 2, 0]$ correspondingly. Jun 26 comment Finding Eigenvalues and Eigenvectors weird equations Just consider equations $2 \cdot x - y - 2 \cdot z = 0$. These are for example, $[x,y,z] = [1, 0, 1]$ and $[x,y,z] = [1,2,0]$. These ones are the basis of eigenspace corresponding to eigenvalue $1$. May 23 comment Multivariate normal distribution density function @Babla Sorry for delay. Let $Y_1, Y_2$ have the covariance matrices $A, B$ correspondingly. Let $Cov(Y_1,Y_2)=0$. So the covariance matrix of $2n$-dimensional variable $(Y_1,Y_2)$ is the just $\Sigma=\left(\begin{array} AA & 0 \\ 0 & B \end{array}\right)$. Note that $|\Sigma|=|A||B|$ and $\Sigma^{-1}=\left(\begin{array} AA^{-1} & 0 \\ 0 & B^{-1} \end{array}\right)$. Put this in density function $f_{(Y_1,Y_2)}(x)$ and it decomposes to $f_{Y_1}(x) \cdot f_{Y_2}(x)$. This is the independence of Gaussian variables $Y_1,Y_2$. Hope, it helps. Unfortunately, I cannot update my answer with this text. May 18 comment Multivariate normal distribution density function @Babla $Y_1$ and $Y_2$ are uncorrelated if and only if $Cov(Y_1, Y_2) = 0$. This means that $Y_1$ and $Y_2$ variates in ways those do not share any statistical data between. From $Cov(Y_1, Y_2) = 0$ you can not reason anything about independence of $Y_1$ and $Y_2$. It must be only the a-priori assumption. May 18 comment $0$-th moment of product of gaussian and sinc function @JFNJr See my answer. Unfortunately, Mathematica was able to evaluate only the first integral. May 18 comment Multivariate normal distribution density function @Babla I see. I have rewritten my answer according to your comment. May 18 comment Multivariate normal distribution density function @Babla $Cov(\mathbf{c} + B \cdot \mathbf{X}, \mathbf{c} + B \cdot \mathbf{X}) = B \cdot \Sigma \cdot B^\top$, where $\Sigma$ is covariance matrix of random Gaussian vector $\Sigma$. May be I do not understand your problem... May 18 comment $0$-th moment of product of gaussian and sinc function If you need only the answer, I can calculate it with help of Wolfram Mathematica.