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From: precarion on 9 May 2007 14:55
Hi! :)

Can somebody help me on this little problem?:

Check (prove or show a counterexample) if the following assumptions are enought to make a statement that the set of all arguments for each f(x)=0 is a dense set:
(1) f: R->R differentiable,
(2) For all x in R: f'(x) < f(x).

Chris
From: Daniel Mayost on 9 May 2007 19:09
In article <8543525.1178751383729.JavaMail.jakarta(a)nitrogen.mathforum.org>,
precarion <precarion(a)yahoo.ca> wrote:
>Hi! :)
>
>Can somebody help me on this little problem?:
>
>Check (prove or show a counterexample) if the following assumptions are enought to make a statement that the set of all arguments for each f(x)=0 is a dense set:
>(1) f: R->R differentiable,
>(2) For all x in R: f'(x) < f(x).
>
>Chris

Counterexample: f(x)=exp(x/2), which has no zeroes.

--
Daniel Mayost

From: precarion on 9 May 2007 15:20
On May 9, 2007 [7:09 PM] Daniel Mayost wrote:
[BEGIN:QUOTE]
<8543525.1178751383729.JavaMail.jakarta(a)nitrogen.mathforum.org>,
precarion <precarion(a)yahoo.ca> wrote:
>Hi! :)
>
>Can somebody help me on this little problem?:
>
>Check (prove or show a counterexample) if the following assumptions are enought to make a statement that the set of all arguments for each f(x)=0 is a dense set:
>(1) f: R->R differentiable,
>(2) For all x in R: f'(x) < f(x).
>
>Chris

Counterexample: f(x)=exp(x/2), which has no zeroes.

--
Daniel Mayost
[END:QUOTE]

And what about the situation in which we have one more assumention:

(3) f has at least one zero. ?

Chris
From: precarion on 9 May 2007 15:41
Besides, an empty set is a dense set (closure of an empty set is an empty set; if you prefere another statement of density - we do have: for every 2 different elements x_1, x_2 from an empty set there exists 3rd element x* such that x_1 < x* < x_2 - it is true because it's equal to: (x_1 in [Empty set], x_2 in [Empty set], x_1 different from x_2) => [Exists] x* in [Empty set] such that x_1 < x* < x_2, which in equal to 0 => 0, which on the other hand is a true statement!)

So your counterexample is not good enought.

Chris
From: Daniel Mayost on 9 May 2007 20:01
In article <30266884.1178754134567.JavaMail.jakarta(a)nitrogen.mathforum.org>,
precarion <precarion(a)yahoo.ca> wrote:
>Besides, an empty set is a dense set (closure of an empty set is an empty set; if you prefere another statement of density - we do have: for every 2 different elements x_1, x_2 from an empty set there exists 3rd element x* such that x_1 < x* < x_2 - it is true because it's equal to: (x_1 in [Empty set], x_2 in [Empty set], x_1 different from x_2) => [Exists] x* in [Empty set] such that x_1 < x* < x_2, which in equal to 0 => 0, which on the other hand is a true statement!)
>
>So your counterexample is not good enought.
>
>Chris

That's not the correct definition of a dense set.

--
Daniel Mayost

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