# Approximating mac/taylor series to z decimal accuracy

1. Dec 6, 2009

### frozenguy

1. The problem statement, all variables and given/known data
Any problem. Ex Use the Mac series of $$tan^{-1}\left(x\right)$$ to approximate at $$x=\frac{1}{2}$$ to three decimal place accuracy.

2. Relevant equations
$$\sum^{\infty}_{k=0}(-1)^{k}\frac{x^{2k+1}}{2k+1}=x-\frac{x^3}{3}+\frac{x^5}{5}-...$$

3. The attempt at a solution

I found out using the remainder estimation theorem that all I need is up to the fifth power. So I put x in to the expanded series to the fifth power (third term) and then I take three decimal places, rounding the third.
The books answer is four decimal places, fourth rounded and an answer that comes from using the seventh power (fourth term).

So, what am I doing wrong.. I have encountered this on several problems in the book.
Am I using my remainder estimation theorem wrong?

$$\frac{\left(\frac{1}{2}\right)^{n+1}}{n+1}<0.0005$$

Using my calculator, I find at n=5, the left side is less then 0.0005

2. Dec 6, 2009

### frozenguy

Anyone able to shed some light?

3. Dec 7, 2009

### TheoMcCloskey

frozenguy - Just look at the terms of the series. At what point will the next sucessive term yield precision beyond your target. That's your cut-off point.

Using the Remainder Estimation therom requires knowledge of the (n+1)th derivative of f(x), something that might be very messy for your function, and will not (in this case) help you solve for the desired value of n (unless you try several trial-by-error cases).

Also, please recall/review, the Remainder Therom is

$$R_{n}(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!} \cdot (x-a)^{(n+1)}$$

where $\xi=\xi(x)$ and here $a= 0$.