User dmitry karp - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T19:04:25Z http://mathoverflow.net/feeds/user/22384 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108309/a-subclass-of-log-concave-functions-satifying-a-sum-inequality A subclass of log-concave functions satifying a sum inequality Dmitry Karp 2012-09-28T03:55:27Z 2012-09-28T04:30:36Z <p>Suppose $f:[0,\infty)\to[0,\infty)$ is continuous and for all positive integers $m$ and real $x\geq0$, $\alpha,\beta>0$: $$\sum\limits_{k+n=m}[f(x+\alpha+k)f(x+\beta+n)-f(x+k)f(x+\alpha+\beta+n)]\geq0.$$ Is anything known about this class of functions? If $m=0$ we have $$f(x+\alpha)f(x+\beta)-f(x)f(x+\alpha+\beta)\geq0$$ which is equivalent to log-concavity of $f$. Similar question about functions satisfying $$\sum\limits_{k+n=m}\binom{m}{k}[f(x+\alpha+k)f(x+\beta+n)-f(x+k)f(x+\alpha+\beta+n)]\geq0.$$ Is there a reasonably simple sufficient condition on $f$ so that it satisfiesthe above inequalities?</p> http://mathoverflow.net/questions/92122/proving-non-negativity-of-a-hypergeometric-type-sum Proving non-negativity of a hypergeometric type sum Dmitry Karp 2012-03-25T02:17:44Z 2012-03-25T02:17:44Z <p>I am trying to prove the following inequality: <code>$$\sum\limits_{k=0}^{m}\frac{(a)_k(a+\mu)_{m-k}}{(c)_k(c+\mu)_{m-k}}\binom{m}{k}(m-2k+\mu)\geq{0},$$</code> where <code>$(a)_0=1$, $(a)_k=a(a+1)\cdots(a+k-1)$</code> is rising factorial and $a\geq{c}\geq{1/2}$, $\mu\geq{0}$. It is easy to prove it for $c\geq{a}>0$ by Gauss pairing but this method fails for <code>$c&lt;a$</code>. I also managed to prove it for integer $\mu$. I believe the general case can be handled once $0&lt;\mu&lt;1$ is proved. </p> <p>Any help welcome.</p> http://mathoverflow.net/questions/92122/proving-non-negativity-of-a-hypergeometric-type-sum Comment by Dmitry Karp Dmitry Karp 2012-03-26T10:02:27Z 2012-03-26T10:02:27Z Thanks to everyone who looked at the problem. I have found a proof using a combination of Gosper's algorithm to find the antidifference of <code>$(a)&#95;k(a+\mu)&#95;{m-k}/[(c)&#95;k(c+\mu)&#95;{m-k}](m-2k+\mu)$</code> and summation by parts. The resulting sum can be handled by Gauss pairing. Dmitry