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Math Help - Find a power series for f'(x) by starting with the convergent geometric series

  1. #1
    Junior Member moonman's Avatar
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    Find a power series for f'(x) by starting with the convergent geometric series

    a) Find a power series for f'(x) by starting with the convergent geometric series

    f(x) = ln(x^3 + 27)

    f'(x) = \frac{2x^2}{x^3 +27}

    b) Find a power series of f(x) and it's radius of convergence.
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  2. #2
    MHF Contributor

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    Re: Find a power series for f'(x) by starting with the convergent geometric series

    Do you understand what the question is asking you to do? You know that f'= \frac{3x^2}{x^3+ 27} (note the "3", not "2", in the numerator. I assume that was a typo). You should also know that a "geometric series" is a series of the form a+ ar+ ar^2+ ar^3+ ... and that the sum of such a series is \frac{a}{1- r} as long as |r|< 1 so that the series converges.

    Now look at \frac{3x^2}{x^3+ 27}. That is not quite in the form \frac{a}{1- r} but can be put in that form. Divide both numerator and denominator by 27 and you get \frac{\frac{3}{27}x^2}{\frac{x^3}{27}+ 1}= \frac{\left(\frac{x}{3}\right)^2}{1- (-\left(\frac{x}{3}\right)^3)}. Now do you see what "a" and "r" must be?
    Last edited by HallsofIvy; December 8th 2012 at 05:46 AM.
    Thanks from moonman and topsquark
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    Junior Member moonman's Avatar
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    Re: Find a power series for f'(x) by starting with the convergent geometric series

     a = \frac{x}{3} ,  r = \frac{x}{3}

    How do we go about finding the radius of convergence?
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