1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook