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Math Help - Maclaurin Series

  1. #1
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    Maclaurin Series

    Compute the Maclaurin series (i.e., the Taylor series about 0) of

    f(x)=x^2 + arcsin(x)

    up to and including terms of order two. Then: evaluate (a-zero)^2 + (a-one)^2 + (a-two)^2.

    Hint: d/dx arcsin(x) = 1/sqrt(1 - x^2)

    I'm getting 9/4 by just computing the Series to the second derivative and then using the rest of the formula. It's incorrect.

    Also, can someone explain what the bolded part.

    Thanks!
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  2. #2
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    Re: Maclaurin Series

    f(x) = x^2 + \arcsin(x) ... f(0) = 0

    f'(x) = 2x + \frac{1}{\sqrt{1-x^2}} ... f'(0) = 1

    f''(x) = 2 - \frac{x}{(1-x^2)^{3/2}} ... f''(0) = 2

    P_2(x) = f(0) + f'(0) \cdot x + \frac{f''(0) \cdot x^2}{2!}

    P_2(x) = x + x^2

    a_0 = 0

    a_1 = x

    a_2 = x^2
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    Re: Maclaurin Series

    Um, I think you might have accidentally misinterpreted the question. The answers are one of the following:

    (A)0 (B)1 (C)2 (D) 9/4 (E)3 (F) 13/4 (G)4 (H)5

    and I'm not close to any of them lol.
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  4. #4
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    Re: Maclaurin Series

    Quote Originally Posted by bhaktir View Post
    Um, I think you might have accidentally misinterpreted the question. The answers are one of the following:

    (A)0 (B)1 (C)2 (D) 9/4 (E)3 (F) 13/4 (G)4 (H)5

    and I'm not close to any of them lol.
    if a_0 , a_1 , and a_2 are what I think they are, then the answer is (H) 5
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  5. #5
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    Re: Maclaurin Series

    Quote Originally Posted by bhaktir View Post
    Compute the Maclaurin series (i.e., the Taylor series about 0) of

    f(x)=x^2 + arcsin(x)

    up to and including terms of order two. Then: evaluate (a-zero)^2 + (a-one)^2 + (a-two)^2.
    This is meaningless because "a-zero", "a-one", and "a-two" have not been defined. What does the problem really say?

    Hint: d/dx arcsin(x) = 1/sqrt(1 - x^2)
    If g(x)= arcsin(x), then g(0)= 0. As given, g'= 1/sqrt(1- x^2), so g'(0)= 1. That is the same as g'= (1- x^2)^(1/2) so g''= (1/2)(1- x^2)^(-1/2)(-2x)= -x(1- x^2)^(-1/2) and g''(0)= 0.

    Since you are asked only for the MacLaurin series up to "order 2", that is sufficient. The MacLaurin series for arcsin, up to degree 2, is 0+ x+ 0 so the MacLaurin series, up to degree 2, for x^2+ arcsin(x) isx+ x^2. I still have no idea how to answer the given questio because I do not know what "a-zero", "a_one", and "a_two" mean. I might guess that they are the coefficents of x^0, x^1, and x^2, respectively, in which case "[itex]a_0^2+ a_1^2+ a_2^2= 0+ 1+ 1= 2[/itex]", but I don't know that.

    I'm getting 9/4 by just computing the Series to the second derivative and then using the rest of the formula. It's incorrect.

    Also, can someone explain what the bolded part.

    Thanks!
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