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MathGroup Archive 2007

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Re: Normal for Limit : Example

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74447] Re: Normal for Limit : Example
  • From: "Sebastian Meznaric" <meznaric at gmail.com>
  • Date: Wed, 21 Mar 2007 02:57:11 -0500 (EST)
  • References: <etld77$o39$1@smc.vnet.net>

On Mar 19, 12:13 am, Mr Ajit Sen <senr... at yahoo.co.uk> wrote:
> Dear Sebastian,
>
>   Here is an example to illustrate what I meant:
>
>   f=(2x-5)/ (x-2)
>
>   Limit[f,x -> 3]  ---> 1
>                              Both agree here.
>   Normal[f+O[x,3]] ---> 1
>
>   However, at the point of discontinuity x = 2 (which
> I
>   referred to loosely as a "pole" : I find the whole
>   thing redolent of Laurent Series),
>
>   Limit[f,x -> 2]  ---> - Infinity   [Correct]
>
>   Normal[f+ O[x,2]]  ---> 2 - 1/(-2+x) [ = f ]
>
>   Now, I've always been using Normal to get rid of the
>   O[ ] terms in Series, and I found Andrzej's
>   alternative use of Normal rather neat, although I
>   have no idea how it works. Trace doesn't help me
> much
>   here.  So, the question is whether Normal can be
> used
>   at a point of discontinuity.
>
>   BTW, my query was prompted by Eric Smith's posts on
>   Limit.
>
>   Best Regards.
>
>   Ajit Sen.
>
> ___________________________________________________________

Oh, I see what you mean. Well all Normal does is that it truncates the
higher order terms from your expression. So by using O you are just
saying to Mathematica that your expression is an approximation of a
certain order. What you can try instead is this: Normal[f+O[x,2]/.x-
>2]  this returns -Infinity.

Why this happens? Here is a short explanation. By adding O[x,3] you're
telling Normal that you want a series of order 0 (ie constant plus
terms of order 1) about the point x=3. So to accommodate, Normal will
substitute x=3 to f. But if there is a singularity (like at x=2) there
will be residual terms of order < 0 and so Normal will retain those
terms. In either case terms higher than the order you specified with O
will be truncated. You can play around with this by say trying O[f+O[x,
3]^2] which will give you a series of order 1. In your case you will
get -2+x.

So this should tell you can use Normal at a pole by substituting the
values before you evaluate Normal, like Normal[f+O[x,2]/.x->2]. Norma[f
+O[x,2]] won't work in general.



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