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Oct 25, 2012 at 11:15 answer added Thierry Zell timeline score: 2
Oct 16, 2012 at 17:00 answer added Gerhard Paseman timeline score: 0
Oct 16, 2012 at 15:01 vote accept Yair Carmon
Oct 16, 2012 at 13:14 answer added Emil Jeřábek timeline score: 6
Oct 16, 2012 at 9:28 answer added Markus Schweighofer timeline score: 2
Oct 15, 2012 at 21:35 comment added Rabee Tourky A very general class (of continuous functions) is continuous piecewise increasing or decreasing functions with finite components. If it has an odd number of affine components and its first piece is increasing, then the last piece is increasing.
Oct 15, 2012 at 21:10 comment added Gerald Edgar @Yir: true, my class is too general
Oct 15, 2012 at 19:47 answer added Aaron Meyerowitz timeline score: 0
Oct 15, 2012 at 19:18 comment added Yair Carmon But $cos(x)$ is complex-analytic and has infinitely many zeros...
Oct 15, 2012 at 16:49 comment added Gerald Edgar A general class: complex-analytic (in a neighborhood of $\mathbb R^+$).
Oct 15, 2012 at 15:04 comment added Yair Carmon Gerald - while I don't see what's not "elementary" in the function you mentioned, I'd love to hear ideas for other classes of functions for which the finite number of zeros property is provable.
Oct 15, 2012 at 14:39 comment added Igor Khavkine I meant that $\log(x+1)$ is not eventually concave. Though perhaps I'm misunderstanding something about the motivation part of your question... ah, it seems that for you $f$ is some specific function that you didn't define, rather any function.
Oct 15, 2012 at 14:09 answer added André Schlichting timeline score: 0
Oct 15, 2012 at 13:50 comment added Lierre Can you show us the explicit formula you have for $f$ ? Do you have a differential equation ?
Oct 15, 2012 at 13:01 comment added Gerald Edgar You need something better than the vague definition of "elementary function" in Wikipedia. For example (on $\mathbb R$) is $\sqrt{x^2}-x = |x|-x$ supposed to be called "elementary"? I don't think so, but you can't see that from the Wikipedia definition.
Oct 15, 2012 at 12:45 history edited Yair Carmon CC BY-SA 3.0
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Oct 15, 2012 at 11:04 comment added Yair Carmon Why so? It seems to me $log(1+x)$ equals 0 only once...
Oct 15, 2012 at 10:53 comment added Igor Khavkine Isn't $\log(x+1)$ a counterexample?
Oct 15, 2012 at 10:35 history asked Yair Carmon CC BY-SA 3.0