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Finding Maclaurin expansion and interval of convergence

  1. Jul 23, 2017 #1
    1. The problem statement, all variables and given/known data
    Find the Maclaurin series and inverval of convergence for ##f(x) = \log (\cos x)##

    2. Relevant equations


    3. The attempt at a solution
    I used the fact that ##\log (\cos x) = \log (1+ (\cos x - 1))##, and the standard expansions for ##\cos x## and ##\log (x+1)## to get that ##\displaystyle \log (\cos x) = -\frac{x^2}{2} + \frac{x^4}{12} - \frac{x^6}{720} + O (x^8)##. How do I find the interval of convergence for this? Also, how do I know that this is the valid expansion?
     
  2. jcsd
  3. Jul 23, 2017 #2

    haruspex

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    Try working from the outside in: what is the interval for log( 1+y)?
     
  4. Jul 24, 2017 #3

    vela

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    According to Mathematica, the series is
    $$-\frac{x^2}{2}-\frac{x^4}{12}-\frac{x^6}{45}-\frac{17 x^8}{2520}-\frac{31 x^{10}}{14175}+O\left(x^{11}\right)$$
    It might be easier to find the series for ##f'(x) = -\tan x## and then integrate term by term.
     
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