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eugene
science forum Guru
Joined: 24 Nov 2005
Posts: 331
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 Posted: Fri Jun 23, 2006 10:46 pm Post subject:
Re: fractional part of the function at integer points+concave
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| Quote: | In article
33398815.1151085054168.JavaMail.jakarta@nitrogen.math
forum.org>,
eugene <jane1806@mail.ru> wrote:
In article
15558845.1151055387707.JavaMail.jakarta@nitrogen.math
forum.org>,
eugene <jane1806@mail.ru> wrote:
Let f be a continious concave function from (0,
+infty) -> R such that lim_{x -> infty} f(x) =
infty
and f(x) = o(x) when x -> +infty.
How can i prove that sup_{n -naturals} {f(n)} =
1,
where
{f(n)} is fractional part of f(n).
Hint: show that f(n+1) - f(n) tends to 0 as n
tends
to infinity.
Thanks. I had the same idea, surely if we prove it,
using the fact that f
tends to infitiy everything becomes clear about the
problem.
But, i'm not sure about my proof of that
lim_{n -> infy} (f(n+1)-f(n)) = 0.
show f(n+1)-f(n) is a positive decreasing sequence.
It thus has a
limit. If the limit were positive, then the o(x)
condition
couldn't hold.
|
Ah, thanks, it seems i've got it.
First, since f is concave g(x) = (f(x)-f(n))/(x-n) is decsreasing for any fixed natural n , so
g(n+1) >= lim_{x -> infty} g(x) = 0 due to that o(x) condition.
so we have that f(n+1) > f(n).
And to this end, we may notice that since f is concave
1/2(f(n)+ f(n+2)) <= f((n+n+2)/2) = f(n+1), so
f(n+1) - f(n) >= f(n+2) - f(n+1) which means that the sequence f(n+1) - f(n) is decreasing.
Is it ok?
Thanks |
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The World Wide Wade
science forum Guru
Joined: 24 Mar 2005
Posts: 790
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| Posted: Fri Jun 23, 2006 6:36 pm Post subject:
Re: fractional part of the function at integer points+concave
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In article
<33398815.1151085054168.JavaMail.jakarta@nitrogen.mathforum.org>,
eugene <jane1806@mail.ru> wrote:
| Quote: | In article
15558845.1151055387707.JavaMail.jakarta@nitrogen.math
forum.org>,
eugene <jane1806@mail.ru> wrote:
Let f be a continious concave function from (0,
+infty) -> R such that lim_{x -> infty} f(x) = infty
and f(x) = o(x) when x -> +infty.
How can i prove that sup_{n -naturals} {f(n)} = 1,
where
{f(n)} is fractional part of f(n).
Hint: show that f(n+1) - f(n) tends to 0 as n tends
to infinity.
Thanks. I had the same idea, surely if we prove it, using the fact that f
tends to infitiy everything becomes clear about the problem.
But, i'm not sure about my proof of that
lim_{n -> infy} (f(n+1)-f(n)) = 0.
|
show f(n+1)-f(n) is a positive decreasing sequence. It thus has a
limit. If the limit were positive, then the o(x) condition
couldn't hold. |
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eugene
science forum Guru
Joined: 24 Nov 2005
Posts: 331
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| Posted: Fri Jun 23, 2006 5:50 pm Post subject:
Re: fractional part of the function at integer points+concave
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| Quote: | In article
15558845.1151055387707.JavaMail.jakarta@nitrogen.math
forum.org>,
eugene <jane1806@mail.ru> wrote:
Let f be a continious concave function from (0,
+infty) -> R such that lim_{x -> infty} f(x) = infty
and f(x) = o(x) when x -> +infty.
How can i prove that sup_{n -naturals} {f(n)} = 1,
where
{f(n)} is fractional part of f(n).
Hint: show that f(n+1) - f(n) tends to 0 as n tends
to infinity.
Thanks. I had the same idea, surely if we prove it, using the fact that f tends to infitiy everything becomes clear about the problem. |
But, i'm not sure about my proof of that
lim_{n -> infy} (f(n+1)-f(n)) = 0.
So,using the properties of f i stated here
http://mathforum.org/kb/thread.jspa?threadID=1402907&tstart=0
f'(x) where it exists is decreasing and thus it may have finite limit L when x -> infty or it may have infinite limit. In both cases using the relation
f(x) = f(a) + int_[a,x] f'(t)dt we would have a
contradiction with that f(x) = o(x), x-> infty unless
L=0, whih means that lim f'(x) = 0 when x -> infty, whihch together with
f(n+1) - f(n) = f'(ksi) , ksi \in (n,n+1) gives the
desired lim_{n ->infty} (f(n+1)-f(n)) = 0.
I think these solutoon has some flaws and i'm not sure about it.
Is it in the whole near to be ok.
Thanks
> Mike Guy |
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M.J.T. Guy
science forum addict
Joined: 03 May 2005
Posts: 62
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| Posted: Fri Jun 23, 2006 5:18 pm Post subject:
Re: fractional part of the function at integer points+concave
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In article <15558845.1151055387707.JavaMail.jakarta@nitrogen.mathforum.org>,
eugene <jane1806@mail.ru> wrote:
| Quote: | Let f be a continious concave function from (0, +infty) -> R such that lim_{x -> infty} f(x) = infty and f(x) = o(x) when x -> +infty.
How can i prove that sup_{n -naturals} {f(n)} = 1, where
{f(n)} is fractional part of f(n).
|
Hint: show that f(n+1) - f(n) tends to 0 as n tends to infinity.
Mike Guy |
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eugene
science forum Guru
Joined: 24 Nov 2005
Posts: 331
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| Posted: Fri Jun 23, 2006 9:35 am Post subject:
fractional part of the function at integer points+concave
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Let f be a continious concave function from (0, +infty) -> R such that lim_{x -> infty} f(x) = infty and f(x) = o(x) when x -> +infty.
How can i prove that sup_{n -naturals} {f(n)} = 1, where
{f(n)} is fractional part of f(n).
Any ideas would be welcome.
Thanks. |
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