Pavithran Iyer

58 reputation

bio	website
	location	Mohali, India
	age	22
visits	member for	6 months
	seen	May 14 at 16:08
stats	profile views	23

	58 reputation	bio	website		visits	member for	6 months
6 badges		location	Mohali, India		seen	May 14 at 16:08

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Mar 21	revised	Sphere containment problem inside a rational convex polytope of general dimensions. in polynomial time, define B(n,r) - the ball of radius r.
Mar 21	revised	Sphere containment problem inside a rational convex polytope of general dimensions. in polynomial time:
Mar 21	revised	Sphere containment problem inside a rational convex polytope of general dimensions. reordering of the tags according to relavance
Mar 21	asked	Sphere containment problem inside a rational convex polytope of general dimensions.
Mar 15	comment	Solving an overdetermined system of inequalities using null-space arguments I mean element-vise inequalities. Sorry, I skipped to mention it as I thought it was implicit.
Mar 15	asked	Solving an overdetermined system of inequalities using null-space arguments
Mar 7	asked	Lower bound on diophantine system of inequalities with all but one non-linear constraint
Mar 1	asked	Generating matrix for a normal distribution?
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.