Re: approximation by series expansion
- From: strolgen <stephane.frion@xxxxxxxxx>
- Date: Thu, 16 Aug 2007 23:35:29 -0700
On Aug 16, 2:54 pm, Robert Israel
<isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
rembremading <rembremad...@xxxxxxx> writes:
Hi everybody!
I try to approximate the term 1/(a-x)
ar 1/(a-cos(theta)) for a>1.
ar? Is that a typo for "as"?
OK, just take cos(theta) = x. Or is there some other relation
between x, theta and a that you haven't told us about?
In order to approximate it I want to expand it in a power series.
Is there a rule of thumb which kind of expansion will give me the
best convergence in the region x=[0,1]?
It is easy to get good convergence for moderate a. However, for a close to
1
(e.g. 1.01 the convergence is quite bad)
Is there a reasonable way to make the expansion be exact at x=0 and x=1 at
the same time?
I'm very unclear on exactly what it is you're looking for. Please be
more explicit. What is "it"? A series in powers of what?
Convergent where? If you're looking at 1/(a-x), what do you do
about a=x?
--
Robert Israel isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
I think he said a> 1 then a = x cannot exist ( due to x =
cos(theta ) ).
Otherwise, I'm even more confused than Robert .
SF
.
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- approximation by series expansion
- From: rembremading
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- From: Robert Israel
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