Re: approximation by series expansion
- From: Robert Israel <israel@xxxxxxxxxxx>
- Date: Fri, 17 Aug 2007 01:07:15 -0700
On Aug 16, 2:00 pm, rembremading <rembremad...@xxxxxxx> wrote:
Hi everybody!
I try to approximate the term 1/(a-x)
ar 1/(a-cos(theta)) for a>1.
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?
Thank You,
Andreas
OK, maybe I can guess what you mean. The standard series is
1/(a-x) = 1/a + x/a^2 + x^2/a^3 + ... = sum_{j=1}^infty x^(j-1)/a^j
which converges whenever |a| > |x|. Yes, it does converge
very slowly when x = 1 and a is near 1 (but that shouldn't
be unexpected, because the result must be large then).
This is exact for x = 0, i.e. all terms except the first are 0.
Now if you want something similar that is "exact" for both
x=0 and x=1, perhaps you could use this slight rearrangement:
1/(a-x) = 1/(a-1) + (x-1)/(a^2-a)) + (x^2-x)/(a^3-a^2)
+ (x^3-x^2)/(a^4-a^3) + ...
= 1/(a-1) + sum_{j=1}^infty (x^j-x^(j-1))/(a^(j+1)-a^j)
Note that for x=1 all terms except the first are 0, while for x=0 all
except the first and second are 0. Convergence is basically the same
as for the first series, since this is just writing
1/(a-x) = 1/(a-1) + ((x-1)/(a-1))/(a-x) and using the first
series for the 1/(a-x) on the right.
Alternatively, if you want good polynomial (in x) approximations for 1/
(a-x) as a function of x on [0,1], you might try a Chebyshev series
for that interval. In fact, what this does essentially is
to change variables to x = (1+cos(t))/2, t in [0,Pi], and
then take a Fourier cosine series.
1/(a-x) = 2/(2 a - 1 - cos(t))
= c_0/2 + sum_{j=1}^infty c_j cos(j t)
= c_0/2 + sum_{j=1}^infty c_j T_j(cos(t))
= c_0/2 + sum_{j=1}^infty c_j T_j(2x-1)
where T_j is the j'th Chebyshev polynomial of the first kind,
and c_j = 1/pi int_{-pi}^pi 2/(2 a - 1 - cos(t)) cos(j t) dt.
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
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