Re: Taylor Series
- From: hrubin@xxxxxxxxxxxxxxxxxxxx (Herman Rubin)
- Date: 10 May 2007 12:56:09 -0400
In article <4641ce6b$0$10311$426a74cc@xxxxxxxxxxxx>,
=?ISO-8859-1?Q?=B5?= <mu--@xxxxxxxxx> wrote:
Clearbluesky a crit :
exp(10^3) = (approx) 1 + 10^3 + (1/2)*10^6 + (1/6)*(10^9) +......
How many terms would i need to obtain exp(10^3) with error less than
1%?
You have Taylor-Lagrange inequality for positive x:
|exp(x)-(1+x+...+x^n/n!)| <= exp(x)x^(n+1)/(n+1)!
thus relative error is less than x^(n+1)/(n+1)!
You can try to find a correct n+1 using Stirling formula as a rough
approximation of n!
One can do better than that. For even moderate x,
the sum of the terms for n > x is approximately
exp(x)/2, so the number of terms needed to any
reasonable accuracy is larger than n, if one starts
from the beginning.
So suppose we look at the remainder for some n > x.
The series from n+1 to infinity is dominated by
the geometric series for the first two terms, which
is [x^(n+1)/(n+1)!]/(1 - x/(n+2)).
One can do something similar on the small side;
if one does this, to get a given relative error
the number of terms needed is O(sqrt(x)), but this
mean dropping the early terms as well.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@xxxxxxxxxxxxxxx Phone: (765)494-6054 FAX: (765)494-0558
.
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