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Find Taylor Series

  1. Aug 19, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the Taylor series about the point x = 0 for the function [tex] \frac{1}{3-2x^3} [/tex]

    2. Relevant equations

    3. The attempt at a solution

    [tex] \frac{1}{3 - 2x^3} = \frac{1}{3(1 - \frac{2x^3}{3})} [/tex]. Let [tex] u = \frac{2x^3}{3} [/tex]. Then [tex] \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} + ...) [/tex].

    I learned that if a function can be expanded in a power series with the point x = 0 in its interval of convergence, then that power series expansion is also the expansion for the Taylor series, so this seems correct to me. Any mistakes?
    Last edited: Aug 19, 2009
  2. jcsd
  3. Aug 19, 2009 #2
  4. Aug 19, 2009 #3


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    Homework Helper

    Sure, looks ok... where does it converge?

    Your answer to my last question says that the above equality holds for [tex]|x|<C[/tex]. Try to find a series expansion for the same function that converges for [tex]|x|>C[/tex].
  5. Aug 19, 2009 #4
    Well it converges for [tex] |u| = |\frac{2x^3}{3}| < 1 [/tex]. How would I find an expansion that converges for u > 1?
  6. Aug 19, 2009 #5
    You won't.
  7. Sep 4, 2009 #6


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    Homework Helper

    Note: this implies that the series expansion given in the OP converges for [itex]|x|<\sqrt[3]{{\scriptstyle \frac{3}{2}}}[/itex].

    I think you mean, "How do I find an expansion for [itex]|x|> \sqrt[3]{{\scriptstyle \frac{3}{2}}}[/itex]?"

    Easy, do pretty much the same thing...

    [tex] \frac{1}{3 - 2x^3} = -\frac{1}{2x^3}\cdot\frac{1}{1 - \frac{3}{2x^3}} [/tex]​

    Now let [itex] u = \frac{3}{2x^3}[/itex] ... you, finish.
    Last edited: Sep 4, 2009
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