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Math Help - URGENT, NEEDED BY 3:30! Power Series and error.

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    URGENT, NEEDED BY 3:30! Power Series and error.

    I was having trouble with this problem:

    Use a power series to approximate the value of the integral with an error less than 0.0001. Assume that the integrand is defined as 1 when x=0. \int_{0}^{1/2}{{arctan(x)\over{x}}}
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  2. #2
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    Quote Originally Posted by dark_knight_307 View Post
    I was having trouble with this problem:

    Use a power series to approximate the value of the integral with an error less than 0.0001. Assume that the integrand is defined as 1 when x=0. \int_{0}^{1/2}{{arctan(x)\over{x}}}

    The series for

    \tan^{-1}(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}

    \frac{\tan^{-1}(x)}{x}=\frac{1}{x} \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{2n+1}

    \int_{0}^{1/2}\frac{\tan^{-1}(x)}{x}dx=\int_{0}^{1/2} \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{2n+1}=

    \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)^2}|_{0}^{1/2}=\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{2n+1}(2n+1)^2}

    This is an alternating series so the error is always less than the absolute value of next term

    so |a_n|=\frac{(-1)^n}{2^{2n+1}(2n+1)^2}

    We need a_n < .0001

    \left| \frac{(-1)^n}{2^{2n+1}(2n+1)^2} \right|< .0001

    Solving I get n = 3.24 so we need to take the first four terms.

    Good luck
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  3. #3
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    I'm confused about how you create the MacLauren Series though, because I'm having difficulty coming up with the derivatives of f(x).
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  4. #4
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    I think using geometric series would be the way to go because finding the derivatives of arctan(x) and generalizing looks like it's going to be a pain.

    Recall the geometric series: \sum_{n=0}^{\infty} ar^{n} = \frac{a}{1-r}

    Looking at the derivative of arctan(x), we have  \frac{1}{1+x^{2}} which resembles the geometric series with a = 1, r = -x^{2}

    This gives us:
    \frac{1}{1+x^{2}} = \sum_{n=0}^{\infty}(-1)^{n}(x)^{2n} = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} + ...
    \arctan(x) = \int \sum_{n=0}^{\infty}(-1)^{n}(x)^{2n}dx

    Integrating gives you the power series for arctan(x), then the rest follows as TheEmptySet showed you.

    --

    If you need to find the maclaurin series for arctan x by finding all the derivatives, it's done here: Maclaruin Series (Scroll down a bit)

    Gets messy ...
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