Indeed, I tried to find a counterexample to the problem in the form of: f(x) := 0 (x <= 0); f(x) := exp(1 - 1/x) (otherwise); but its second derivative is annoyingly negative somewhere. Best wishes, Adam P. Goucher
Sent: Saturday, December 15, 2018 at 12:01 AM From: "Gareth McCaughan" <gareth.mccaughan@pobox.com> To: math-fun@mailman.xmission.com Subject: Re: [math-fun] Some questions from the 2018 Putnam Exam
On 14/12/2018 16:11, Victor Miller wrote:
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.
At the risk of once again showing myself to be a moron, that doesn't sound right. Suppose g(x) = 0 for x<=0 and g(x) = exp(-1/x^2) for x>0. Then g is infinitely differentiable but has the property you say f can't have on account of being infinitely differentiable.
(I am not claiming that g or anything like it is a counterexample to the theorem the question is asking for a proof of. It's easy to find places where, say, its second derivative is zero. But it's a counterexample to the I-think-not-a-theorem that says that an infinitely differentiable function can't be 0 for all x <= some x0 and strictly increasing thereafter.)
-- g
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