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Math Help - Power series representation of this function.

  1. #1
    s3a
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    Power series representation of this function.

    I'm having trouble getting the power series representation of this function.

    (My work and the question are attached)

    Any help would be greatly appreciated!
    Thanks in advance!
    Attached Thumbnails Attached Thumbnails Power series representation of this function.-mywork.jpg  
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  2. #2
    Super Member Random Variable's Avatar
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     \arctan x = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7} + ... \  \ |x|<1

     \arctan x^{5} = x^{5} - \frac{x^{15}}{3} + \frac{x^{25}}{5} - \frac{x^{35}}{7} + ...

     2x^{2} \arctan x^{5} = 2x^{7} - \frac{2x^{17}}{3} + \frac{2x^{27}}{5} - \frac{2x^{37}}{7} + ...
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    wait. what does  c_n and  x^n represent?
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  4. #4
    s3a
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    I don't even understand how to get to the first step you wrote Is it something I am supposed to have memorized?
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  5. #5
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    Quote Originally Posted by lilaziz1 View Post
    wait. what does  c_n and  x^n represent?
    x^n, oddly enough, means the variable x to the nth power. c_n is the coefficient of x^n.
    Last edited by HallsofIvy; May 17th 2010 at 03:41 AM.
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  6. #6
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    Quote Originally Posted by s3a View Post
    I don't even understand how to get to the first step you wrote Is it something I am supposed to have memorized?
    Not necessarily memorized. The important point is that to find the Taylor's series for 2x^2 arctan(x^5) you don't have to work directly the entire thing. Start by finding the Taylor's series for arctan(x):

    arctan(0)= 0.

    (arctan(x))'= \frac{1}{x^2+ 1} (That's a very important formula- you should have it memorized!) so (arctan(0))'= 1.

    (arctan(x))"= \left(\frac{1}{x^2+1}\right)'= \frac{-2x}{(x^2+1)^2} which is 0 at x= 0

    etc.
    Last edited by HallsofIvy; May 17th 2010 at 03:42 AM.
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  7. #7
    s3a
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    Sorry, I am very confused with this stuff. Is Taylor series for arctan(x) the f^(n) (x-a)^n /n! stuff or is that something else?
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  8. #8
    s3a
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    I had what seemed like an epiphany and I did what I'm attaching but I'm stuck still .
    Attached Thumbnails Attached Thumbnails Power series representation of this function.-mywork.jpg  
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  9. #9
    Super Member Random Variable's Avatar
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     \frac{d}{dx} \arctan x = \frac{1}{1+x^{2}} = \sum_{n = 0}^{\infty} (-1)^n x^{2n}, \ |x|<1

    so  \arctan x = \int \sum^{\infty}_{n=0} (-1)^{n} x^{2n} \ dx = \sum^{\infty}_{n=0} (-1)^{n} \int x^{2n} \ dx = \sum^{\infty}_{n=0} (-1)^{n}\frac{x^{2n+1}}{2n+1}

    and  \arctan x^5 = \sum^{\infty}_{n=0} (-1)^{n}\frac{(x^5)^{2n+1}}{2n+1}= \sum^{\infty}_{n=0} (-1)^{n}\frac{x^{10n+5}}{2n+1}

    and finally  2x^{2} \arctan x^5 = 2 \sum^{\infty}_{n=0} (-1)^{n}\frac{x^{10n+7}}{2n+1}
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  10. #10
    s3a
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    Are you sure the exponent of your final form on x isn't 10n + 8 since you add 7 to (10n + 1) ?
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  11. #11
    Super Member Random Variable's Avatar
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    Quote Originally Posted by s3a View Post
    Are you sure the exponent of your final form on x isn't 10n + 8 since you add 7 to (10n + 1) ?
    I'm positive.

     x^{2} (x^{5})^{2n+1} = x^{2} x^{10n} x^{5} = x^{2 + 10n +5} = x^{10n+7}
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