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Math Help - Check my work on Taylor series?

  1. #1
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    Question Check my work on Taylor series?

    Find the first 4 terms (k=0,1,2,3) of the Taylor polynomial for f(x) = sin(x) centered at c=(pi/4)

    sin(x) = Sum(k-->0 to inf)[((-1)^k * x^(2k+1))/((2k+1)!)]

    [f^(k) * (c) * (x^(2k+1))] / (k!)
    ^^^^ f^(k) = kth derivative of f
    I get
    x - ((x^3)/3!) - ((x^5)/5!) - ((x^7)/7!)

    Is that correct?
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  2. #2
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    Quote Originally Posted by pinkprincess08 View Post
    Find the first 4 terms (k=0,1,2,3) of the Taylor polynomial for f(x) = sin(x) centered at c=(pi/4)

    sin(x) = Sum(k-->0 to inf)[((-1)^k * x^(2k+1))/((2k+1)!)]

    [f^(k) * (c) * (x^(2k+1))] / (k!)
    ^^^^ f^(k) = kth derivative of f
    I get
    x - ((x^3)/3!) - ((x^5)/5!) - ((x^7)/7!)

    Is that correct?
    No the signs should alternate.

    CB
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  3. #3
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    Question

    so then it's
    x - ((x^3)/3!) + ((x^5)/5!) - ((x^7/7!) ?

    I thought that was the Maclaurin series for sin(x)
    What is the difference between Taylor series of sin(x) centered at (pi/4) and Maclaurin series?
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  4. #4
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    Quote Originally Posted by pinkprincess08 View Post
    so then it's
    x - ((x^3)/3!) + ((x^5)/5!) - ((x^7/7!) ?

    I thought that was the Maclaurin series for sin(x)
    What is the difference between Taylor series of sin(x) centered at (pi/4) and Maclaurin series?
    Opps, you are right, not reading the question closly enough.

    Then no, since \sin(\pi/4)=\sqrt{2} there should be a constant term, and square roots of 2 all over the place.

    CB
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  5. #5
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    so when k = 0, i get (pi*root(2))/8
    k=1, {(pi*root(2))/8} * {x - (pi/4)}
    k=2, [-pi*root(2) * (16x^2 - 8 pi x + pi^2)] / 128
    k=3, [pi*root(2) * (64x^3 - 48 pi x^2 + 12 pi^2 x - pi^3)] / 3072

    What do I do now?
    just add those?
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  6. #6
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    Quote Originally Posted by pinkprincess08 View Post
    so when k = 0, i get (pi*root(2))/8
    k=1, {(pi*root(2))/8} * {x - (pi/4)}
    k=2, [-pi*root(2) * (16x^2 - 8 pi x + pi^2)] / 128
    k=3, [pi*root(2) * (64x^3 - 48 pi x^2 + 12 pi^2 x - pi^3)] / 3072

    What do I do now?
    just add those?
    f(x)=\sin(x)

    f'(x)=\cos(x)

    f''(x)=-\sin(x)

    f'''(x)=-\cos(x)

    so:

    f(\pi/4)=\sqrt{2}

    f'(\pi/4)=\sqrt{2}

    f''(\pi/4)=-\sqrt{2}

    f'''(\pi/4)=-\sqrt{2}

    so:

     <br />
\sin(x)=\sqrt{2}[1+ (x-\pi/4) - \frac{(x-\pi/4)^2}{2!} - \frac{(x-\pi/4)^3}{3!} + ...<br />
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  7. #7
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    i though sin (pi/4) was (root(2))/2 ..?
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  8. #8
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    Indeed it is.
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  9. #9
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    Quote Originally Posted by pinkprincess08 View Post
    i though sin (pi/4) was (root(2))/2 ..?
    Good, then you can use that to get the correct answer.

    CB
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