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Math Help - Radius of convergence

  1. #1
    Senior Member Educated's Avatar
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    Radius of convergence

    Radius of convergence-question-3.png

    I can do part a) and b) but I'm not getting how to do part c)

    So I'm guessing the part under the square root cannot be negative, and so \dfrac{64b^2}{a^4}x^2 \le 1

    Solving that I would get:

    x^2 \le \dfrac{64b^2}{a^4}

    |x| \le \dfrac{8b}{a^2}

    And so the radius of convergence is: \dfrac{8b}{a^2} ?

    And how do I show that for the series to converge, b < \dfrac{a}{4} ?
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  2. #2
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    Re: Radius of convergence

    Quote Originally Posted by Educated View Post
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    I can do part a) and b) but I'm not getting how to do part c)

    So I'm guessing the part under the square root cannot be negative, and so \dfrac{64b^2}{a^4}x^2 \le 1
    Your inequality is correct (though you may not necessarily have equality), but not the reason. The radius of convergence being one is the same as saying the expansion of \sqrt{1+z} converges whenever |z|<1. In your case, z=\frac{64b^2}{a^4}x^2 which gives you \frac{64b^2}{a^4}x^2<1

    Solving that I would get:

    x^2 \le \dfrac{64b^2}{a^4}
    I suggest you take a look at this equation again. It should be x^2<\frac{a^4}{64b^2}. Then |x|<\frac{a^2}{8b}. When x=\frac{a}{2}, we have \frac{a}{2}< \frac{a^2}{8b} (since a is positive, we drop the absolute value sign). This is true if and only if b<\frac{a}{4} simply by manipulating the inequality.
    Last edited by Gusbob; April 10th 2013 at 04:32 AM.
    Thanks from HallsofIvy and Educated
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  3. #3
    Senior Member Educated's Avatar
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    Re: Radius of convergence

    Quote Originally Posted by Gusbob View Post
    I suggest you take a look at this equation again. It should be x^2<\frac{a^4}{64b^2}. Then |x|<\frac{a^2}{8b}. When x=\frac{a}{2}, we have \frac{a}{2}< \frac{a^2}{8b} (since a is positive, we drop the absolute value sign). This is true if and only if b<\frac{a}{4} simply by manipulating the inequality.
    Oh right, thank-you!
    (I can't believe I made such a simple mistake by re-arranging incorrectly)
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