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Maclaurin series help

  1. Apr 26, 2005 #1
    i have trouble expandind f(x)= (sin2x)^2 into Maclaurin series
    for sin(x) Maclaurin series is

    [tex]\sum^{\infty}_{n=0} (-1)^n \frac{x^{2n+1}}{(2n+1)!} [/tex]

    probably the key is to change (sin2x)^2 into new shape. I found that

    (sin2x)^2=2sin(2x^2), but that coefficent 2 is bothering me, what to do?
  2. jcsd
  3. Apr 26, 2005 #2


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    That relation does not hold (the left side is always positive, the right side isn't).

    But you can use another identity:

    [tex]\sin^2 x=\frac{1}{2}(1-\cos 2x)[/tex]
  4. Apr 26, 2005 #3
    [tex]\sin^2 2x=\frac{1}{2}(1-\cos 4x)=0.5-0.5cos4x[/tex]

    and now if i apply for cosx maclaurin series expansion considering the function f(x)= -0,5cos4x i get the right answer but im puzzled, where does the coefficent 0,5 go????? i dont have to count that??
    Last edited: Apr 26, 2005
  5. Apr 26, 2005 #4


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    The constant term of the series for sin2(2x) is 0.
    The 0.5 cancels the -0.5 from the expansion of -0.5*cos(4x).
    Last edited: Apr 26, 2005
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