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Integration of taylor series

  1. Mar 30, 2013 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    A problem from advanced calculus by Taylor :

    http://gyazo.com/5d52ea79420c8998a668fab0010857cf

    2. Relevant equations

    ##sin(x) = \sum_{n=0}^{∞} (-1)^n \frac{x^{2n+1}}{(2n+1)!}##

    ##sin(3x) = \sum_{n=0}^{∞} (-1)^n \frac{3^{2n+1}x^{2n+1}}{(2n+1)!}##

    3. The attempt at a solution

    I'm not quite seeing what the question is trying to ask me here. It wants me to find a series for the integral of f(x)?

    I wrote out the sin(px) in terms of their power series and then integrated f(x) from 0 to π/2. So I have :

    ##\int_{0}^{π/2} f(x) dx = \frac{\sum_{n=0}^{∞} \frac{(-1)^n 3^{2n+1} (π/2)^{2n+2}}{(2n+1)!(2n+2)}}{1 * 2} + \frac{\sum_{n=0}^{∞} \frac{(-1)^n 5^{2n+1} (π/2)^{2n+2}}{(2n+1)!(2n+2)}}{3 * 4} + ......##

    I don't see where to go from here or what I'm even supposed to be doing with this monster.
     
  2. jcsd
  3. Mar 30, 2013 #2
    What makes you think you are required to mess with power series? Can you integrate the series term by term? Why?
     
  4. Mar 30, 2013 #3

    Zondrina

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    Well I don't HAVE to use the power series I suppose.

    Yes I can integrate the series term by term. Hm, what if I write :

    ##f(x) = \sum_{n=1}^{∞} \frac{sin((2n+1)x)}{not sure} = \sum f_n(x)##

    Having a bit of trouble getting the denominator on that one. If I can show that the series is uniformly convergent on ##[0, π/2]## and each ##f_n(x)## is continuous on the interval, then I can integrate the terms of the series term by term which will give me the series I'm looking for.

    ##\int_{0}^{π/2} f(x) dx = \int_{0}^{π/2} \frac{sin(3x)}{1*2} + \int_{0}^{π/2} \frac{sin(5x)}{3*4} + \int_{0}^{π/2} \frac{sin(7x)}{5*6} + ...##
     
  5. Mar 30, 2013 #4
    Does (2n - 1)(2n) work for the denominator?
     
  6. Mar 30, 2013 #5

    Zondrina

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    I was just thinking (2n+1)(2n), but (2n-1) works since it goes oddeven oddeven. I was actually getting shafted by (2n+1) screwing everything.

    So the series would be uniformly bounded by ##M_n = \frac{1}{n^2}##. Since ##\sum M_n## converges by p-comparison, we know ##\sum f_n(x)## will uniformly converge by the M-Test.

    Since each term in the series is also continuous I integrate :

    ##\int_{0}^{π/2} f(x) dx = \sum_{n=1}^{∞} \int_{0}^{π/2} \frac{(2n+1)x}{4n^2 - 2n} = \sum_{n=1}^{∞} \frac{(2n+1)x^2}{8n^2 - 4n}##

    and... there's my series :eek:!
     
  7. Mar 30, 2013 #6
    I do not understand your last step. You should be integrating ## \frac {\sin (2n + 1)x} {(2n - 1)(2n)} ##, no?
     
  8. Mar 30, 2013 #7

    Zondrina

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    Oh whoops, I jumped a little too quickly and forgot the sin.

    ##\int_{0}^{π/2} f(x) dx = \sum_{n=1}^{∞} \int_{0}^{π/2} \frac{sin((2n+1)x)}{4n^2 - 2n}dx##

    Then using ##u = (2n+1)x##, I get ##\frac{1}{2n+1}du = dx## and so on and so forth.

    I'm sure I got this now :). Thanks voko.
     
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