On 11/1/07, Stavros Macrakis <macrakis at alum.mit.edu> wrote:
> tlimit (which uses taylor) also works, if you're willing to answer an
> unnecessary question:
>
> tlimit(1/e,e,inf)
>
> asks the sign of
> a^(k/(k-1)+1)*k^2-2*a^(k/(k-1)+1)*k+a^(k/(k-1)+1)
> and then gives a reply. Of course, it would have been nicer if Maxima could
> recognize that that expression is positive, which it can if you factor it as
> Barton shows below.
It seems that it does not work. I used the following code:
assume(k>1/2,k<1,a>0,b>a)$
e : (1/(b-a))*integrate((1-k)*((2*t-a)/k)^(k/(k-1)),t,a,b)$
tlimit(1/e,b,inf);
and no value to the limit is returned.
Paul
> On 10/31/07, Barton Willis <willisb at unk.edu> wrote:
> > Try something like this:
> >
> > (%i1) assume(k>1/2,k<1,a>0,b>a)$
> > (%i2) e :
> (1/(b-a))*integrate((1-k)*((2*t-a)/k)^(k/(k-1)),t,a,b)$
> >
> > (%i3) factor(taylor(e,b,inf,0));
> > (%o3)
> (b^(k/(k-1))*(k-1)*2^(k/(k-1)))/(k^(k/(k-1))*(2*k-1))
> >
> > (%i4) sign(%);
> > (%o4) neg
> >
> > BW
> >
> > -----maxima-bounces at math.utexas.edu wrote: -----
> >
> > >To: maxima at math.utexas.edu
> > >From: "Paul Smith" <phhs80 at gmail.com>
> > >Sent by: maxima-bounces at math.utexas.edu
> > >Date: 10/31/2007 05:53PM
> > >Subject: Re: [Maxima] Learning that limit is 0 from above
> > >
> > >On 10/31/07, Stavros Macrakis <macrakis at alum.mit.edu> wrote:
> > >> > (%i1) f(x) := exp(-x);
> > >> > (%o1) f(x) := exp(- x)
> > >> > (%i2) limit(f(x),x,inf);
> > >> > (%o2) 0
> > >> > (%i3)
> > >> >
> > >> > In this example, the convergence to zero is from above (0^+). Is it
> > >> > possible to have maxima tell us it?
> > >>
> > >> Yes, compute limit(1/f(x),x,0,plus). If the result is inf, then the
> > >> original converges from above; if minf, from below.
> > >>
> > >> By the way, Maxima is generally better at working with expressions than
> > >> functions, so I would have written the above as
> > >>
> > >> f: exp(-x)$
> > >> limit(f,x,inf);
> > >> limit(1/f,x,inf);
> > >>
> > >> Hope this helps,
> > >
> > >However, your technique not always work. See this code:
> > >
> > >assume(k>1/2,k<1,a>0,b>a);
> > >f(b):=
> (1/(b-a))*integrate((1-k)*((2*t-a)/k)^(k/(k-1)),t,a,b)$
> > >limit(f(b),b,inf);
> > >limit(1/f(b),b,inf);
> > >
> > >The first limit is zero, but the second one cannot be calculated by
> > >maxima. Any further ideas?
> > >
> > >Paul
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