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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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