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Math Help - maclaurin series for cos 3x

  1. #1
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    maclaurin series for cos 3x

    okay so i took the first 7 derivatives and evaluated them at zero and got:

    1,0,-9,0,81,0,-729,0,...

    so, if you remove the zero terms, you know that the series is alternating.. and it looks like there's a (-9)^n term in there

    what i'm confused about is how to find the nth derivative at zero, and how to just isolate the nonzero terms of the series..and then i'm not sure how to plug it in and continue from there
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  2. #2
    MHF Contributor matheagle's Avatar
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    Get the series for \cos \theta, which must be in your textbook, then let \theta =3x.
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    really? because my calc teacher told us not to do that >.< i thought maybe it was because the derivatives were different?
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  4. #4
    MHF Contributor matheagle's Avatar
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    YES, this works.
    But if you want to do it from scratch, look at ther derivation of cosine at x and apply the chain rule repeatedly, when you take all of those derivatives.
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    did i not use the chain rule appropriately?

    i don't WANT to do it from scratch, i just, think i remember him saying not to use the definition for cosine..

    but, again, i don't really understand what i'm doing when it comes to these anyway
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  6. #6
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    Quote Originally Posted by buttonbear View Post
    did i not use the chain rule appropriately?

    i don't WANT to do it from scratch, i just, think i remember him saying not to use the definition for cosine..

    but, again, i don't really understand what i'm doing when it comes to these anyway
    Assume that \cos{X} = a + bX + cX^2 + dX^3 + \dots

    Then \cos{0} = a \implies a = 1

    So \cos{X} = 1 + bX + cX^2 + dX^3 + \dots

    Now to evaluate b, take the derivative of both sides.

    -\sin{X} = b + 2cX + 3dX^2 + 4eX^3 + \dots

    So -\sin{0} = b \implies b = 0

    Now to find c, take the second derivative.

    -\cos{x} = 2c + 6dX + 12eX^2 + 20fX^3 + \dots

    So -\cos{0} = 2c \implies c = -\frac{1}{2} = -\frac{1}{2!}


    Taking further derivatives yields

    \sin{X} = 6d + 24eX + 60fX^2 + \dots

    \sin{0} = 6d \implies d = 0


    \cos{X} = 24e + 120fX + \dots

    \cos{0} = 24e \implies e = \frac{1}{24} = \frac{1}{4!}


    -\sin{X} = 120f + \dots

    -\sin{0} = 120f \implies f = 0


    If you continue on like this you should find that

    \cos{X} = 1 - \frac{1}{2!}X^2 + \frac{1}{4!}X^4 - \frac{1}{6!}X^6 + \frac{1}{8!}X^8 + \dots

    So \cos{(3x)} = 1 - \frac{1}{2!}(3x)^2 + \frac{1}{4!}(3x)^4 - \frac{1}{6!}(3x)^6 + \frac{1}{8!}(3x)^8 + \dots.


    I'll let you do the expanding and simplifying.
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