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Maclaurin Series

  1. Oct 10, 2009 #1
    1. The problem statement, all variables and given/known data

    How can I calculate it for [tex]\frac{1}{1+cos^2(x)}[/tex] by using the fact that [tex]\frac{1}{1+x^2} = 1 - x^2 + x^4 - ...[/tex]?


    2. Relevant equations

    Given in the problem.

    3. The attempt at a solution

    I tried letting u = cos(x), then

    [tex]\frac{1}{1+cos^2(x)} = \frac{1}{1+u^2} = 1 - u^2 + u^4 - ... = 1 - cos^2(x) + cos^4(x) - ...[/tex]

    But I don't think this is right because the first term should be 0.5, not 1... and I don't see how a -0.5 might pop out of this series of cos terms....and even if it somehow does, I think this question is not meant to be that difficult...

    The answer I am trying to get is [tex]\frac{1}{2} + \frac{1}{4}x^2 + ...[/tex]

    Any ideas?

    Thanks
     
  2. jcsd
  3. Oct 10, 2009 #2

    Dick

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    Homework Helper

    Use cos(x)=1-x^2/2!+x^4/4!-... to express the first few terms of the series for 1+cos(x)^2. Then, sure, use your geometric series. Feel free to drop terms that you know won't be important.
     
  4. Oct 10, 2009 #3
    I don't follow, could you spell it out a bit more?

    So [tex]cos^2(x) = 1-x^2+\frac{1}{3}x^4-...[/tex]

    Now what?

    Thanks
     
  5. Oct 10, 2009 #4

    Dick

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    Alright. So now you've got 1/(1+1-x^2+x^4/3+...). That's 1/(2-x^2/+x^4/3...). Write it as (1/2)*1/(1-(x^2/2-x^4/6...)). Now use 1/(1-a)=1+a+a^2+... with a=x^2/2-x^4/6+... If you only want a quadratic approximation then you didn't need to keep the x^4 stuff, right?
     
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