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Math Help - taylor series

  1. #1
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    taylor series

    find the taylor series expansion of

    f(x) = ___1____
    .........(3-2 x^3)

    at the point x = 0
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  2. #2
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    Not sure if this is correct (I hope more knowledgeable users correct me if it isn’t!).
     \frac{1}{3 - 2x^3} = \frac{1}{3(1 - \frac{2x^3}{3})} . Let  u = \frac{2x^3}{3} . Then  = \frac{1}{3(1 - \frac{2x^3}{3})} = \frac{1}{3} \frac{1}{1 - u} which is the sum of a geometric series, so just use the expansion for a geometric series now.
    Last edited by JG89; August 19th 2009 at 07:41 AM.
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  3. #3
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    can you explain further please
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  4. #4
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    <br />
       \frac{1}{3(1 - \frac{2x^3}{3})} = \frac{1}{3} \frac{1}{1 - u} = \frac{1}{3} (1 + u + u^2 + u^3 + ... )= \frac{1}{3} (1 + \frac{2x^3}{3} + \frac{4x^6}{9} + \frac{8x^9}{27} + ...) <br />
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