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Re: Normal for Limit : Example
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.