Truncation Error and its Bounds

[theuniverse]

Homework Statement

a) Derive a Taylor series with n+1 terms and the associated truncation error for
the function $$f(x)=(log(x)-x+1)/(x-1)^2$$
b) Construct an expression that bounds the truncation error, assuming n > 2, for a given value of x

2. The attempt at a solution
a) So I came up with the following series: (from k=2 to k=n+2) $$Ʃ((-1)^k*(x-1)^{k-1})/(k+1)$$

But I'm not really sure how to derive the truncation error. Is the error simply in the n+2 term, and all I have to do is just sub it into the Lagrange remainder equation?

b) not sure on what interval I should be doing it. As well as taking taking n+2 derivatives to figure out the bound doesn't seem right...

Thanks for your help,

 

[mighty2000]

a) To derive the truncation error, we can use the Lagrange remainder formula for Taylor series. This formula states that the truncation error, denoted by R_n(x), can be written as R_n(x) = f^(n+1)(c)(x-a)^(n+1)/(n+1)! for some c between a and x. In this case, our function f(x) = (log(x)-x+1)/(x-1)^2 has a singularity at x=1, so we will choose a = 1. Then, we need to find the (n+1)th derivative of f(x). This can be done by using the quotient rule repeatedly, or by using the general formula for the derivative of a quotient function. After some simplification, we get:

f^(n+1)(x) = (-1)^n*(n+1)!*(x-1)/x^(n+2)

Substituting this into the Lagrange remainder formula, we get:

R_n(x) = (-1)^n*(x-1)^(n+1)/(n+2)

b) To find an expression that bounds the truncation error, we can use the fact that for any n > 2, the (n+2)th derivative of f(x) will be a decreasing function for x > 1. This means that the maximum value of the (n+2)th derivative will occur at x = 1. Therefore, we can bound the truncation error as follows:

|R_n(x)| ≤ |f^(n+1)(1)|*|x-1|^(n+1)/(n+1)!

Substituting in our expression for f^(n+1)(x) from part (a), we get:

|R_n(x)| ≤ (n+1)!*|x-1|^(n+1)/(n+2)^(n+2)

This gives us an expression that bounds the truncation error for any given value of x > 1.