LaGrange Error for a Taylor PolynomialDate: 05/28/2000 at 10:47:29 From: Garfield Subject: LaGrange error for Taylor polynomial What is the LaGrange error for a Taylor polynomial? Thanks, Garfi Date: 05/29/2000 at 08:02:53 From: Doctor Jerry Subject: Re: LaGrange error for Taylor polynomial Hi Garfield, Given appropriate assumptions, f(x) = f(a) + f'(a)(x-a)/1! + ... + f^{(n)}(a)(x-a)^n/n! + f^{(n+1)}(c)(x-a)^{n+1}/(n+1)! The last term is the LaGrange remainder. The point c is not known in general but it does lie between x and a. I don't know the kind of question about the error term you have in mind, but it might be something like this: For f(x) = e^x and -1 <= x <= 1, find a Taylor polynomial that is within 0.001 of f(x) throughout [-1,1]. We take a = 0 because it is in the middle of the interval of interest; we know that f^{(n+1)}(c) = e^c, regardless of n. For other functions, this part may be more difficult. We want to be sure that the difference between f(x) and the Taylor polynomial: f(a) + f'(a)(x-a)/1! + ... + f^{(n)}(a)(x-a)^n/n! is less than 0.001 for all x in [-1,1]. The absolute value of this difference is |f^{(n+1)}(c)(x-a)^{n+1}/(n+1)!| = e^c |x^{n+1}|/(n+1)! Now we try to simplify this as much as possible. For example, because we know that c is between x and a = 0 and we know that e^x is an increasing function, we can replace e^c by e^1 and know that we will overestimate the possible error, that is, |f^{(n+1)}(c)(x-a)^{n+1}/(n+1)!| = e^c |x^{n+1}|/(n+1)! <= e|x^{n+1}|/(n+1)! Also, we know that x is between -1 and 1. So, |f^{(n+1)}(c)(x-a)^{n+1}/(n+1)!| < e/(n+1)! We choose n so that this is less than 0.001. Do this experimentally, by trying various values of n. For n = 6, e/(n+1)! = e/7! = 0.00053... So n = 6 would be good enough. Maybe n = 5 would work. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ |
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