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Prove Power Series Arc Tan

  1. May 5, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove.


    2. Relevant equations
    arctan(x) = x - [itex]\displaystyle \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9}[/itex]
    for -1 < x < 1.

    3. The attempt at a solution
    [itex]\displaystyle \sum^{∞}_{n=1} \frac{x^{n+2}}{n+2}[/itex]
     
  2. jcsd
  3. May 5, 2013 #2

    tiny-tim

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    what methods have you been taught for finding the taylor series?
     
  4. May 5, 2013 #3
    we skipped taylor and mclaurin series as we did not have time.
     
  5. May 5, 2013 #4

    tiny-tim

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    then why have you been set this question?? :confused:

    ok, here's a "cheat" idea …

    doesn't that series look a bit like an integral? :wink:
     
  6. May 5, 2013 #5

    Zondrina

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    Wait. is the question to prove that arctan is equal to its power series OR to find the series for arctan?

    If it's to prove it, then you may want to consider what's happening the remainder term.
     
  7. May 5, 2013 #6

    LCKurtz

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    How did you get that answer?
     
  8. May 5, 2013 #7
    yes, when our teach show used she used integrals in part of the problem, but i've forgotten now.
     
  9. May 5, 2013 #8

    FeDeX_LaTeX

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    What happens when you differentiate that series?
     
  10. May 5, 2013 #9

    LCKurtz

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    I will ask again: How did you get the answer you presented in the OP?
     
  11. May 6, 2013 #10

    tiny-tim

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    (just got up :zzz:)
    ok, call the series A(x),

    then (as FeDeX_LaTeX :smile: said) find dA/dx …

    and then integrate! :wink:
     
  12. May 6, 2013 #11
    tiny-tim: so differentiate each term, then go back and take the integral of each? where does that get me?
     
  13. May 6, 2013 #12

    Office_Shredder

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    The derivative of arctan(x) is 1/(1+x2). Can you get that Taylor series? Then try integrating both sides (remember to add an arbitrary constant which you have to deal with!)
     
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