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Math Help - Use the Maclaurin series of cos(x) to state the rst three non-zero terms of the Macl

  1. #1
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    Use the Maclaurin series of cos(x) to state the rst three non-zero terms of the Macl

    (a) Use the Maclaurin series of cos(x) to state the rst three non-zero
    terms of the Maclaurin series for (1-(cos(x))/(x^2)
    x2 .
    (b) Explain how you would use your answer in (a) to estimate
    (integral from 0 to 1) (1-(cos(x))/(x^2) dx

    This is a pretty complicated problem, but I got part a)
    I got:

    first derivative: -2(1-cos(x))(x^-3) + sinx(x^-2)
    second derivative: 6(1-cos(x))(x^-4)-2(sinx)(x^-3)-2sinx(x^-3)+cosx(x^-2)
    third derivative: -24(1-cosx)(x^-5)+6(sinx)(x^-4)+6sinx(x^-4)-2cosx(x^-3)+6sinx(x^-4)-2cosx(x^-3)-2cosx(x^-3)-sinx(x^-2)

    From here, I know I am supposed to plug in (first derivative + second derivative + third derivative) in place of (1-(cos(x))/(x^2) to get the
    (integral from 0 to 1) (1-(cos(x))/(x^2) dx, but I don't know how to integrate the whole thing. Please help! Thank you
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  2. #2
    MHF Contributor

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    NO, you are NOT supposed to do that!

    Read the problem again- it said "use the MacLaurin series of cos(x)" and you did not do that. The MacLaurin series of cos(x) is \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}. The first four non-zero terms of that are 1- \frac{1}{2}x^2+ \frac{1}{24}x^4- \frac{1}{720}x^6.

    Replaced cos(x) in \frac{1- cos(x)}{x^2} and you will see why I gave the first four terms of cos(x) rather than just the first three.

    That is just a 6th degree polynomial. Integrate the polynomial between 0 and 1.
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  3. #3
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    Please clarify

    So the answer for part a) is
    1 - 0.5x^2 + (1/24)x^4 - (1/720)x^6

    you say you put the first four terms because the first term is 0, but 1 isn't 0, so I don't get what you mean.

    and for part b) I got
    4241/5040
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