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Power series/taylor

  1. Apr 7, 2009 #1
    Use series to approximate the definite integral I. (Give your answer correct to 3 decimal places.)

    1
    [tex]\int x*cos(x^2) dx[/tex]
    0

    the series I derived was
    inf
    [tex]\sum (-1)^n * x^4^n^+^1 / 2n![/tex]
    n=0

    how would I evaluate it?
     
  2. jcsd
  3. Apr 7, 2009 #2

    Mark44

    Staff: Mentor

    Your Taylor's series is the series for xcos(x^2), although you have the factorial in the denominator wrong - it should be (2n)!, not 2n!. Integrate your series term by term to get a new series. Then evaluate the new series at x = 1 (you won't need to evaluate it at x = 0, since all the terms in your series vanish there). When your partial sums are the same in 3 decimal places, you have your approximation.
     
  4. Apr 7, 2009 #3
    I obtained that series by integrating the series of cos(n). * x.

    I am not getting what you mean. I did integrate it term by term.
     
  5. Apr 8, 2009 #4

    Mark44

    Staff: Mentor

    Here's where you are:
    x*cos(x2) = x - x5/2! + x9/4! - x13/6! + x17/8! -+ ...

    So the antiderivative of the above is:
    [itex]\int x*cos(x^2) dx [/itex] = x2/2 - x6/(6 *2!) + x10/(10*4!) - x14/(14*6!) + x18/(18*8!) -+ ...

    I have omitted the constant of integration on the right side above, and also have omitted the general term.

    What you want is
    [tex]\int_0^1 x*cos(x^2) dx [/tex]

    so what you need to do is to evaluate the antiderivative a few lines up at x = 1. You don't need to evaluate it at x = 0, since all of the terms in the antiderivative are zero when x is zero. The terms up to the 18th power are probably enough to get you three decimal place precision in your answer. If not, include enough terms so that your partial sums are the same in the first three decimal places.
     
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