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Yuval Filmus's user avatar
Yuval Filmus's user avatar
Yuval Filmus's user avatar
Yuval Filmus
  • Member for 14 years, 5 months
  • Last seen more than a week ago
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Identification of a curious function
To quote my post, "During computation of some Shapley values (details below), I encountered the following function".
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Bijective proof for a partition identity
Here is another proof of your lemma: $\prod_{i=0}^\infty \frac{1}{1+x^{2^i}} = \prod_{i=0}^\infty \frac{1-x^{2^i}}{1-x^{2^{i+1}}} = 1-x$.
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A rather curious identity on sums over triple binomial terms
This is essentially the same proof as Sean Eberhard's.
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Connection between cyclic group and exponential function
This answer can serve as a reference.
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An attempt to find expected value of clique number of special random graph
The estimates are pretty good, the error is $O(1)$ or perhaps even smaller. Asking for more is too much to hope for.
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An attempt to find expected value of clique number of special random graph
Your random graph is not special at all. It's the standard $G(n,p)$ model. You can find the answer in textbooks.
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