LaGrange Error for a Taylor Polynomial

Date: 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/