Timeline for Log-concave polynomial is a log-concave function?

7 events

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Jan 23, 2016 at 4:13	comment	added	Johnny Yin		Is the transformed function is log-concave with respect to $y$？
Jan 23, 2016 at 4:01	comment	added	Johnny Yin		A good counter example. The log-concave condition is not sufficient. The original problem acctually is from a probability computation. I attempt to make the polynomial become a log-concave function by using the transform $x=1-\exp(y)$. The sequence $a_k$ is a decreasing and log-concave sequence. More than that, $0<a_k<1$ and $0<x<1$. After the transform, does the polynomial become a log-concave function? I can prove it is log-concave when $a_k=a^k, 0<a<1$. Is it still log-concave in other cases?
Jan 22, 2016 at 19:50	comment	added	Richard Stanley		A stronger condition than log-concave is having only real zeros. Such polynomials are log-concave functions.
Jan 22, 2016 at 14:49	comment	added	David E Speyer		This doesn't seem to be true. The sequence $(1,1,1)$ is log concave, but $$\frac{d^2}{(dx)^2} \log (1+x+x^2) = \frac{1-2x-2x^2}{(1+x+x^2)^2}$$ which is positive for $0 \leq x < \frac{\sqrt{3}-1}{2}$. Did I misunderstand something?
Jan 22, 2016 at 14:35	history	edited	GH from MO	CC BY-SA 3.0	added tag and changed the word "prove" to "support"
Jan 22, 2016 at 12:24	review	First posts
Jan 22, 2016 at 12:26
Jan 22, 2016 at 12:21	history	asked	Johnny Yin	CC BY-SA 3.0