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Big O notation

  1. Jan 24, 2008 #1
    [tex]\biggl(-\frac{x^2}2 + \frac{x^4}{24} - \frac{x^6}{720} +\mathcal{O}(x^8)\biggr)-\frac12\biggl(-\frac{x^2}2+\frac{x^4}{24}+\mathcal{O}(x^6)\biggr)^2+\frac13\biggl(-\frac{x^2}2+\mathcal{O}(x^4)\biggr)^3 + \mathcal{O}(x^8)\\ & =-\frac{x^2}2 + \frac{x^4}{24}-\frac{x^6}{720} - \frac{x^4}8 + \frac{x^6}{48} - \frac{x^6}{24} +\mathcal{O}(x^8)\\


    This is a Taylor expansion of f(x) = ln(cos(x)) . I just wonder what happened with the first three O's, especially with (O(x^6))^2 and (O(x^4))^3. Are they somehow incorporated in O(x^8)?
  2. jcsd
  3. Jan 24, 2008 #2


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    Both of those go to [itex]\mathcal{O}(x^8)[/itex] when you multiply out...
    For example:
    [tex]\left (- \frac{x^2}{2}+ \mathcal{O}(x^4}) \right)^3[/tex]
    multiplies out to:
    [tex]-\frac {x^6}{8} + 3\frac{x^4}{4} \mathcal{O}(x^4) - 3\frac{x^2}{2} \mathcal{O}{x^8} + \mathcal{O}(x^12)[/tex]
    [tex]-\frac{x^6}{8} + \left( \mathcal{O}(x^8) - \mathcal{O}(x^{10}) + \mathcal{O}(x^{12}) \right)[/tex]
    [tex]-\frac{x^6}{8} + \mathcal{O}(x^8)[/tex]
  4. Jan 24, 2008 #3
    that's exactly what i don't get... shouldn't the result of

    [tex]\mathcal{O}(x^8) - \mathcal{O}(x^{10}) + \mathcal{O}(x^{12}) \right)[/tex]

    be O(x^12) since that is the largest term?
    Last edited: Jan 24, 2008
  5. Jan 25, 2008 #4


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    Doesn't that depend on [itex]x[/itex]? (In Taylor expansions [itex]x[/itex] is 'small', so lower exponents are more important.)
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