Here's the essential idea: since f(0) = 0 and f(1)=1, if we define y = inf { x>= 0: f(x) > 0}, we have y in [0,1). Since f is infinitely differentiable, it can't hold that f(x) = 0 for all x <= y. So f must be decreasing somewhere before y, and thus its derivative is negative. On Fri, Dec 14, 2018 at 10:26 AM Mike Stay <metaweta@gmail.com> wrote:
On Thu, Dec 13, 2018 at 8:44 PM Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
(Apologies for the double-quoting; either I never got the original or it somehow got spam-binned by my mail program.)
On Tue, Dec 4, 2018 at 5:37 PM Dan Asimov <dasimov@earthlink.net> wrote: >>>> These are the questions from the recent Putnam Exam that seemed most>> interesting to me. I've changed some typography, and a word or two, for>> clarity. (I haven't tried to solve any yet. If anyone posts a solution,>> maybe include a spoiler warning for that problem number?)...>> A5 Let f : R → R be an infinitely differentiable function satisfying>> f(0) = 0, f(1) = 1, and f(x) ≥ 0 for all x ∈ R. Show that there exist>> a positive integer n and a real number x such that the nth derivative>> of f is negative when evaluated at x:>>>> f^(n)(x) < 0.
Surely this can't be right! Why can't we take f(x) = x^2?
n=1, x=-1. -- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike https://reperiendi.wordpress.com
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