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seen Aug 10 at 21:44

Jun
7
asked Lower bound for the probability that a certain component of a Gaussian vector dominates all others
May
4
accepted Positive roots of a polynomial
May
4
comment Positive roots of a polynomial
Neat derivation!
May
4
awarded  Nice Question
May
3
comment Positive roots of a polynomial
@Greg Martin: $n=1$ does not satisfy the condition that not all $a_i$'s are equal. For $n=2$ we get $p(x)=\frac{1}{2} \left(a_1-a_2\right){}^2 x^2-\frac{1}{2} a_1 \left(a_1-a_2\right){}^2 a_2$ and so clearly the only positive root is $x_0=\sqrt{a_1 a_2}$. For $n=3$ I don't know already...
May
3
awarded  Yearling
May
3
revised Positive roots of a polynomial
added 55 characters in body
May
3
comment Positive roots of a polynomial
@Seva some numerical evidence, I'll add this to the question
May
3
asked Positive roots of a polynomial
May
1
comment Maximizing the discrepancy in Jensen's inequality for a certain function
@Sergei: see for instance Theorem 1 here: iub.edu/~caepr/RePEc/PDF/2012/CAEPR2012-004.pdf
Apr
30
asked Maximizing the discrepancy in Jensen's inequality for a certain function
Jul
2
awarded  Curious
Jun
5
answered Property/Relations using Fourier series/transform, which give complete information about all the jump singularities of a function.
Jun
2
accepted Is the following conjecture regarding rotational iterates of collection of points on a circle true?
Jun
2
comment Is the following conjecture regarding rotational iterates of collection of points on a circle true?
@Anthony Quas: This comes from estimating condition numbers of Vandermonde matrices with nodes on the circle.
Jun
2
comment Is the following conjecture regarding rotational iterates of collection of points on a circle true?
Since your rescale everything to $[0,1)$, I think $\alpha$ should be multiplied by $\pi$. Also, could you please provide more intuition regarding the usage of the Dirichlet principle? Thanks!
Jun
1
asked Is the following conjecture regarding rotational iterates of collection of points on a circle true?
May
12
comment Inverse of matrix of generalised harmonic numbers
@ChristianRemling: Unfortunately, this is not the case. $H^{-1}$ does have negative entries (as does the inverse of the Hilbert matrix), so even for a 2-by-2 case the matrix $H^{-1}R-I$ has $O(1)$ entries.
May
11
comment Inverse of matrix of generalised harmonic numbers
@ChristianRemling: the problem is that one cannot write $R=H(I+O(1/n))$ in this case. Rather, $O(1/n)H$ stands for a matrix whose $i,j$-th entry is $O(1/n)$ times the $i,j$-th entry of $H$.
May
11
awarded  Informed