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Sergei
  • Member for 10 years, 8 months
  • Last seen more than a month ago
  • Voronezh, Russia
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Inverse Hadamard determinant inequality
@Gerry Myerson - thank you. May somebody help with a paper in Bulletin des Sciences Mathematiques? A first paper-not an easy reading...
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A third degree surface and a touching sphere
@Vladimir Zolotov - pochtaname(c)gmail.com
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A third degree surface and a touching sphere
@Lev Borisov - I solved it by Lagrange. The aim of the question is to find geometrical solution using convexity of related surface.
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Inverse Hadamard determinant inequality
Do you know a reference to Ostrowski's paper? Is not a first proposed inequality a consequence?
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Inverse Hadamard determinant inequality
Yes, it is exactly what I wanted, user35593. Thank you very much!
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A third degree surface and a touching sphere
Lev Borisov - I proved the underlying inequality of 3rd degree as a lemma for some result in joint paper. But the proof is not nice or short. So I am interesting to find another one using geometry and convexity.
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A third degree surface and a touching sphere
Somebody of the usual set "put on hold" may prove it? Will be thankful...
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Complex proof of $B(a,b)=\Gamma(a)\Gamma(b)/\Gamma(a+b)$
We may take an integral using a technique based on Slater's Theorem (cf. O.I.Marichev's book). In fact it will use in an inderect way contour integrals and residues.
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When does $\nabla\times(\nabla\times F)=0$ imply $\nabla \times F=0$
Is not the question we consider then equivalent that $grad g =O$? More: $rot(rot F)=O => rot F =grad g $ for some g. Question: when this $rot F=O$ ? Answer: only if g=const.
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When does $\nabla\times(\nabla\times F)=0$ imply $\nabla \times F=0$
Is not the question we consider then equivalent that $grad g =O$?
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When does $\nabla\times(\nabla\times F)=0$ imply $\nabla \times F=0$
The main problem and the pity of this site is that many participants are too fast and not objective with their "ON HOLD" decision and demonstrating "WANT TO BE CLEVER THEN OTHERS" position. Say for not simply connected domains this is an interesting question, I do not know is it solved or not.
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