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Power Series of f'(x)

  1. Apr 24, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the power series of f'(x), given f(x) = x2cos2(x)

    2. Relevant equations

    attachment.php?attachmentid=58196&stc=1&d=1366887854.png
    Correct me if I'm wrong

    3. The attempt at a solution

    Can I just take the derivative of the solution I got previously? If so, what's a good way to write the sequence out so I can easily make a series representation. Or is there a better approach to the problem?
     

    Attached Files:

    Last edited by a moderator: Apr 25, 2013
  2. jcsd
  3. Apr 24, 2013 #2

    LCKurtz

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    Looks good. I would multiply the ##x^2/2## into the parentheses and include it under the summation. Then differentiate it.
     
  4. Apr 24, 2013 #3
    My main problem is trying to figure out how to represent what I'm getting as a series. When I multiply in the x^2 and differentiate I get

    x + x - 8x^3 + 4x^5 - (32/45)x^7...

    I can't figure it out
     
  5. Apr 25, 2013 #4

    LCKurtz

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    When you multiply the ##\frac{x^2} 2## in you have$$
    \frac {x^2} 2 +\sum_{n=0}^\infty\frac{(-1)^n2^{2n}x^{2n+2}}{(2n)!}$$Just differentiate the first term as you did and differentiate under the sum and you will have your formula.
     
  6. Apr 25, 2013 #5
    When you say differentiate under the sum, do you mean differentiate each term in the sequence or do you mean you can actually take the derivative of the series equation? I tried googling and checking my textbook and I don't know how to take a derivative of a series in equation form, with the n's and such. If you can show me, that'd be great!
     
  7. Apr 25, 2013 #6

    HallsofIvy

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    Yes, "differentiate under the sum" means "differentiate term by term". I don't know what you mean by "take the derivative of the equation"!
     
  8. Apr 25, 2013 #7

    LCKurtz

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    If you look at each term inside the sum, it just a constant times a power of ##x##. Use the power rule.
     
  9. Apr 25, 2013 #8
    Oh okay, I got it now. Thanks for everything man!
     
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