Pavithran Iyer
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 Jul 2 awarded Curious Apr 16 answered Polynomial identity for a sum Apr 15 comment Polynomial identity for a sum Thanks for the prompt attempt. To me it doesn't seem correct, because $f(x)^j$ will have all possible products of $j-$coefficients of f(x). Whereas, the polynomial $r(x)$ in my question does not have this structure. Apr 15 asked Polynomial identity for a sum Mar 24 comment Trace of a product of a recursive series of non-full rank matrices According to en.wikipedia.org/wiki/Hadamard_product_(matrices) we can say: $\text{Tr}(P_n) = \sum_{i,j}[P_0]_{i,j}[P_{(n-1)}]_{i,j}$, i.e, an element-wise product of $P_0$ and $P_{(n-1)}$, but I am interested if it can be some function of the rank of $H$ or $n$ ? Mar 24 comment Trace of a product of a recursive series of non-full rank matrices Addition: $\text{Tr}(P_n) = \text{Tr}(P^T_0 P_{n-1})$ by cyclic property of trace. Mar 24 asked Trace of a product of a recursive series of non-full rank matrices Mar 7 asked Lower bound on diophantine system of inequalities with all but one non-linear constraint Jan 28 comment Unknown result in probability theory relating CDF of any density to the CDF of normal distribution I meant that, in the above case, $\Phi(x)$ is the CDF of a normal distribution with mean 0 and variance 1. I was asking how the mean and variance values depend on the particular CDF function $A(z)$. Jan 28 asked Unknown result in probability theory relating CDF of any density to the CDF of normal distribution Jan 26 awarded Commentator Jan 26 comment Upper Bounding an exponential integral with complex terms But that is true for the case where there is no $i$ (the first integral), but it still converges. Is there a contour method to solve this integral. I'm aware that for functions which have simple poles, a contour integral method can be used to compute the integral. Jan 26 revised Upper Bounding an exponential integral with complex terms added 7 characters in body Jan 26 asked Upper Bounding an exponential integral with complex terms Dec 12 comment Upperbound on the number of Isolated zeros of a bivariate polynomial However, this pdf arxiv.org/pdf/1102.5391.pdf seems quite helpful. Dec 12 comment Upperbound on the number of Isolated zeros of a bivariate polynomial @GerryMyerson Sorry, I didn't get why the function must have a minimum or a maximum at a zero ? I thought it could be increasing/decreasing or the zero can be an inflection point. But maybe I am missing the main point here. Dec 11 comment Isolation of zeros in the case of univariate analytic functions expressed as a bivariate function. Suppose L(v,p(v)) is a bivariate polynomial of degree n in variables {v,p(v)} , we know the number of zero curves of it are bounded by a quadratic function of its degree, say M(n). Can we use that information and claim to bound the number of zeros of L(v) also by M(n) ? Dec 11 comment Isolation of zeros in the case of univariate analytic functions expressed as a bivariate function. @RobertIsrael What I did was first to consider all the zeros of L(v,p) : these will, in general, be curves on the v-p plane. Then I consider the points at which the curve p=f(v) intersects these zero curves. These intersecting points would be zeros of L(v,p(v)). However, can it happen that p=f(v) itself is one of the zero curves ? (In which case the zeros of L(v,p(v)) would not be isolated.) Sorry if I seem to be confusing. Dec 11 awarded Tumbleweed Dec 11 comment Upperbound on the number of Isolated zeros of a bivariate polynomial @GerryMyerson I looked at Harnack's Theorem for bivariate polynomials. It seems helpful, bounding the number of zero curves of the bivariate polynomial. Thanks for the reference, the paper seems heavy for me, but from what I understood, in it they are discussing solutions of a polynomial system, whereas I am looking for the number of zeros of a single bivariate polynomial. So, I was wondering it there are infact the same. Thanks again for the reference.