How Precise Should My McLaurin Series Be?

[Dafydd]

Homework Statement

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.

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

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

Homework Equations

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

The Attempt at a Solution

...?

 

[Hurkyl]

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)?

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.

In this application you had in your post, you note that

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

is undetermined, so the first-order approximation is inadequate.