# McLaurin - when do you stop?

1. Aug 15, 2009

### Dafydd

1. The problem statement, all variables and given/known data

I have a problem with McLaurin series. I never know when to stop. How do I know if O(x3) is adequate, or O(x5)?

Let's take this exam question as an example.

$$\lim_{x \to 0} \frac{(x+1)e^x -1-2x}{cosx-1}$$
$$\frac{(x+1)e^x -1-2x}{cosx-1} = \frac{(x+1)(1+x+\frac{x^2}{2!}+...) -1-2x}{1-\frac{x^2}{2!}+...-1} =$$
$$\frac{(1+2x+\frac{3x^2}{2}+O(x^3) -1-2x}{-\frac{x^2}{2}+O(x^4)} =$$
$$\frac{\frac{3x^2}{2}+O(x^3)}{-\frac{x^2}{2}+O(x^4)} = \frac{\frac{3}{2}+O(x^3)}{-\frac{1}{2}+O(x^4)} = -3$$

How do I know to stop at O(x3) for ex and at O(x4) for cosx? Any tactics?

2. Relevant equations
$$e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+ \frac{x^n}{n!}+O(x^{n+1})$$
$$cosx = 1+x+\frac{x^2}{2!}+\frac{x^4}{4!}+...+(-1)^{n-1} \frac{x^{2n}}{2n!}+O(x^{2n+2})$$
3. The attempt at a solution

...?

Last edited: Aug 15, 2009
2. Aug 15, 2009

### Hurkyl

Staff Emeritus
The application tells you.

If you're looking to approximate the value of a function, you use the Taylor remainder theorem (or similar) to compute how many terms you need.

$$\lim_{x \rightarrow 0} \frac{O(x^2)}{O(x^2)}$$