How Precise Should My McLaurin Series Be?

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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.

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

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

Homework Equations


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

The Attempt at a Solution



...?
 
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Dafydd said:
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.
 
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