Taylor Polynomials yet again!!?

Question

Let P(x) be the taylor polynomial of degree n be the function: f(x) = log(1-x) about a = 0. How large should n be chosed to have |f(x) - p(x)| <= 10^-4

for -1/2 < x < = 1/2 ?

I know that R(x) = (x-a)^(n+1) / (n+1)! ** f^(n+1)(Cx)

where C is between a and x

but what do I put in for Cx? and f^(n+1) ? I need to assume a large n I know, but f^(n+1) is the n+ 1 derivative of log(1-x) which is negative?

Please work this for me completely.

Answer 1

back again...you cannot find C...the theory just says that there is some x value {C} so that R(x) is given by that formula...comes from the mean value theorem ...{your f^(n+1) = - n! / (1-x)^(n+1)} ...so the easiest is to estimate |R(x)| using a bound on | f^(n+1)| for x in [-1/2,1/2]....{n! / (1/2)^(n+1)}...|R(x) | ≤ | 1/2 - 0|^(n+1) / (n+1)! * n! / (1/2)^(n+1) =  1/ [2(n+1)]...now find n so this is ≤ 1/10^4....about 5000