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Math Help - Maclaurin expansion

  1. #1
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    Maclaurin expansion

    A question says use Maclaurin's theorem to expand \cos^2{x} upto the term containing x^6. Hence or otherwise, express x+\sin^2{x} as a power series upto a term containing x^6.

    f(x) = f(0)+f'(0)x + \frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+ \cdots+\dfrac{f^{(r)}(0)}{r!}x^r


    f(x) = \cos^2{x} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;  f(0) = 1
    f'(x) = -\sin{2x} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;  \;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;  f'(0) = 0
    f''(x) = -2\cos{2x} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;  \;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;  f''(0) = -2
    f'''(x) = 4\sin{2x} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;  \;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;  f'''(0) = 0
    f''''(x) = 8\cos{2x} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;  \;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;  f''''(0) = 8
    f'''''(x) = -16\sin{2x} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;  \;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;  f'''''(0) = 0
    f''''''(x) = -32\cos{2x} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;   \;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;  f''''''(0) = -32

    \therefore \;\;\; \cos^2{x} = 1-\dfrac{2}{2!}x^2+\dfrac{8}{4!}x^4-\dfrac{32}{6!}x^6+\cdots

    Since \sin^2{x} = 1-\cos^2{x},
    \sin^2{x} = 1-\left(1-\dfrac{2}{2!}x^2+\dfrac{8}{4!}x^4-\dfrac{32}{6!}x^6+\cdots\right) = 1-1+\dfrac{2}{2!}x^2-\dfrac{8}{4!}x^4+\dfrac{32}{6!}x^6-\cdots = \dfrac{2}{2!}x^2-\dfrac{8}{4!}x^4+\dfrac{32}{6!}x^6-\cdots

    And therefore x+\sin^2{x} = x+\dfrac{2}{2!}x^2-\dfrac{8}{4!}x^4+\dfrac{32}{6!}x^6-\cdots

    Could someone confirm please, if this is right?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Hibachi View Post
    A question says use Maclaurin's theorem to expand \cos^2{x} upto the term containing x^6. Hence or otherwise, express x+\sin^2{x} as a power series upto a term containing x^6.

    f(x) = f(0)+f'(0)x + \frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+ \cdots+\dfrac{f^{(r)}(0)}{r!}x^r


    f(x) = \cos^2{x} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;  f(0) = 1
    f'(x) = -\sin{2x} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;  \;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;  f'(0) = 0
    f''(x) = -2\cos{2x} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;  \;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;  f''(0) = -2
    f'''(x) = 4\sin{2x} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;  \;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;  f'''(0) = 0
    f''''(x) = 8\cos{2x} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;  \;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;  f''''(0) = 8
    f'''''(x) = -16\sin{2x} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;  \;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;  f'''''(0) = 0
    f''''''(x) = -32\cos{2x} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;   \;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;  f''''''(0) = -32

    \therefore \;\;\; \cos^2{x} = 1-\dfrac{2}{2!}x^2+\dfrac{8}{4!}x^4-\dfrac{32}{6!}x^6+\cdots

    Since \sin^2{x} = 1-\cos^2{x},
    \sin^2{x} = 1-\left(1-\dfrac{2}{2!}x^2+\dfrac{8}{4!}x^4-\dfrac{32}{6!}x^6+\cdots\right) = 1-1+\dfrac{2}{2!}x^2-\dfrac{8}{4!}x^4+\dfrac{32}{6!}x^6-\cdots = \dfrac{2}{2!}x^2-\dfrac{8}{4!}x^4+\dfrac{32}{6!}x^6-\cdots

    And therefore x+\sin^2{x} = x+\dfrac{2}{2!}x^2-\dfrac{8}{4!}x^4+\dfrac{32}{6!}x^6-\cdots

    Could someone confirm please, if this is right?
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    Marvelous. I didn't about that site otherwise I would have just taken the answers from there and took them to the teacher without bothering to work them out. (Joking, of course).
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