Prev: .TH C4.5 1 .SH NAME A guide to the verbose output of the C4.5 decision tree generator .SH DESCRIPTION This document explains the output of the program .I C4.5 when it is run with the verbosity level (option .BR v ) set to values from 1 to 3. .SH
Next: About Linear Independence of Exponential Functions defining an alternating serie
From: Herb on 9 Jun 2010 21:28
lim n->inf { ( 1 + 2 + ... + n ) / n^2 }

= lim n->inf (1/n^2) + lim n->inf (2/n^2) + ... + lim n->inf (n/
n^2)

= 0 + 0 + ... + 0

= 0
From: porky_pig_jr on 9 Jun 2010 21:41
On Jun 9, 9:28 pm, Herb <fuzzy...(a)chol.com> wrote:
>     lim n->inf  { ( 1 + 2 + ... + n ) / n^2 }
>
>  = lim n->inf  (1/n^2) + lim n->inf (2/n^2) + ... + lim n->inf (n/
> n^2)
>
>  = 0 + 0 + ... + 0
>
>  = 0

Limit of sum is sum of the limits: that's only applicable to finite
sums. Yet in this case as we're taking the limit, each term gets
smaller and smaller but we get more and more of such terms. So you
can't say beforehand what's gonna happen. Things may go to infinity,
or to 0, or balance out at some real number, like in this case. Cf.
limit of (1 + 1/n)^n as n goes to infinity. The answer certainly isn't
1, right?
From: Pol Lux on 9 Jun 2010 21:46
On Jun 9, 6:41 pm, "porky_pig...(a)my-deja.com" <porky_pig...(a)my-
deja.com> wrote:
> On Jun 9, 9:28 pm, Herb <fuzzy...(a)chol.com> wrote:
>
> >     lim n->inf  { ( 1 + 2 + ... + n ) / n^2 }
>
> >  = lim n->inf  (1/n^2) + lim n->inf (2/n^2) + ... + lim n->inf (n/
> > n^2)
>
> >  = 0 + 0 + ... + 0
>
> >  = 0
>
> Limit of sum is sum of the limits: that's only applicable to finite
> sums. Yet in this case as we're taking the limit, each term gets
> smaller and smaller but we get more and more of such terms. So you
> can't say beforehand what's gonna happen. Things may go to infinity,
> or to 0, or balance out at some real number, like in this case. Cf.
> limit of (1 + 1/n)^n as n goes to infinity. The answer certainly isn't
> 1, right?

1+2+...+n = n(n+1)/2

The limit above is lim n(n+1)/2n^2 when n->+inf, which is the 1/2.

Pollux
From: porky_pig_jr on 9 Jun 2010 21:48
On Jun 9, 9:46 pm, Pol Lux <luxp...(a)gmail.com> wrote:
> On Jun 9, 6:41 pm, "porky_pig...(a)my-deja.com" <porky_pig...(a)my-
>
>
>
> deja.com> wrote:
> > On Jun 9, 9:28 pm, Herb <fuzzy...(a)chol.com> wrote:
>
> > >     lim n->inf  { ( 1 + 2 + ... + n ) / n^2 }
>
> > >  = lim n->inf  (1/n^2) + lim n->inf (2/n^2) + ... + lim n->inf (n/
> > > n^2)
>
> > >  = 0 + 0 + ... + 0
>
> > >  = 0
>
> > Limit of sum is sum of the limits: that's only applicable to finite
> > sums. Yet in this case as we're taking the limit, each term gets
> > smaller and smaller but we get more and more of such terms. So you
> > can't say beforehand what's gonna happen. Things may go to infinity,
> > or to 0, or balance out at some real number, like in this case. Cf.
> > limit of (1 + 1/n)^n as n goes to infinity. The answer certainly isn't
> > 1, right?
>
> 1+2+...+n = n(n+1)/2
>
> The limit above is lim n(n+1)/2n^2 when n->+inf, which is the 1/2.
>
> Pollux

Are you telling me or you're telling him?

If you're telling me, well, thanks for the obvious. His question was
why doing it his way he got zero.
From: Pol Lux on 9 Jun 2010 21:52
On Jun 9, 6:48 pm, "porky_pig...(a)my-deja.com" <porky_pig...(a)my-
deja.com> wrote:
> On Jun 9, 9:46 pm, Pol Lux <luxp...(a)gmail.com> wrote:
>
>
>
>
>
> > On Jun 9, 6:41 pm, "porky_pig...(a)my-deja.com" <porky_pig...(a)my-
>
> > deja.com> wrote:
> > > On Jun 9, 9:28 pm, Herb <fuzzy...(a)chol.com> wrote:
>
> > > >     lim n->inf  { ( 1 + 2 + ... + n ) / n^2 }
>
> > > >  = lim n->inf  (1/n^2) + lim n->inf (2/n^2) + ... + lim n->inf (n/
> > > > n^2)
>
> > > >  = 0 + 0 + ... + 0
>
> > > >  = 0
>
> > > Limit of sum is sum of the limits: that's only applicable to finite
> > > sums. Yet in this case as we're taking the limit, each term gets
> > > smaller and smaller but we get more and more of such terms. So you
> > > can't say beforehand what's gonna happen. Things may go to infinity,
> > > or to 0, or balance out at some real number, like in this case. Cf.
> > > limit of (1 + 1/n)^n as n goes to infinity. The answer certainly isn't
> > > 1, right?
>
> > 1+2+...+n = n(n+1)/2
>
> > The limit above is lim n(n+1)/2n^2 when n->+inf, which is the 1/2.
>
> > Pollux
>
> Are you telling me or you're telling him?
>
> If you're telling me, well, thanks for the obvious. His question was
> why doing it his way he got zero.

Telling him I guess. I agree with your answer too.

Pollux
 |  Next  |  Last
Pages: 1 2
Prev: .TH C4.5 1 .SH NAME A guide to the verbose output of the C4.5 decision tree generator .SH DESCRIPTION This document explains the output of the program .I C4.5 when it is run with the verbosity level (option .BR v ) set to values from 1 to 3. .SH
Next: About Linear Independence of Exponential Functions defining an alternating serie