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Homework Help: McLaurin - when do you stop?

  1. Aug 15, 2009 #1
    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. jcsd
  3. Aug 15, 2009 #2


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    Staff Emeritus
    Science Advisor
    Gold Member

    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

    [tex]\lim_{x \rightarrow 0} \frac{O(x^2)}{O(x^2)}[/tex]

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