Re: Sum of alternating series with large items
- From: Bill August <hui.song@xxxxxxxxxx>
- Date: Tue, 07 Jul 2009 13:09:20 EDT
On Tue, 07 Jul 2009 10:40:30 EDT, Bill August
<hui.song@xxxxxxxxxx>
wrote:
Hi Guys,let's assume S is finite.
I met this problem recently.
Say S is an alternating series,
S = a1 - a2 + a3 - a4 + ....
It could be finite or infinite. For simplicity,
a large number of digits), how to calculate the value
The problem is when abs(ai) is extremely large (with
of S by numerical methods?
Anyone can help? Thanks in advance.
More information is needed. If ai is a function of i
there is a
whole class of things that can be done. If ai is
simply data the
result is restricted to the accuracy of the data. I
suggest you
look up Kahan summation as a start on the finite
case. In the
infinite series case there are a host of methods for
accelerating
convergence.
Richard Harter, cri@xxxxxxxx
http://home.tiac.net/~cri, http://www.varinoma.com
If I do not see as far as others, it is because
I stand in the footprints of giants.
Thanks Richard.
My problem is little bit different with the accuracy or convergence problem. An example is that:
(a - b)^k ------------(1)
Where a and b are extremely large number and k is integer.
If we expand this item by binomial theory, then
k
----
\ k k i-k
| (-1) b a ------------(2)
/
----
i=0
Suppose a = XXXXXXXXX...XXXXXX2, and b = XXXXXXXXXX...XXXXXX1; The result (1) is 1, but how to get the same result as (1)? When a and b are extremely big? It may not be possible to represent each item by i.e. double.
.
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