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Finding the power series of a square root

  1. Dec 17, 2013 #1
    1. The problem statement, all variables and given/known data
    Find a power series for [itex]f(x) = \frac{1}{\sqrt{4+x^{2}}}[/itex], at [itex]x=0[/itex].

    2. The attempt at a solution
    I have looked up the Taylor series of [itex]\frac{1}{\sqrt{4+x^{2}}}[/itex], but I don't find any similarity with a power serie like [itex]\sum_{n\geq 0} a_{n} x^{n}[/itex]

    I don't know how to start, anyone can lend me a hand?
  2. jcsd
  3. Dec 17, 2013 #2

    Simon Bridge

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    Start with the definition of a power series.
    Look at what happens when you differentiate the power series.
    (Or just look up "power series")
  4. Dec 17, 2013 #3
    I know that the differentiation of a power series is [itex]\sum n a_{n} x^{n-1}[/itex], so if by differentiating [itex]f(x) = \frac{1}{\sqrt{4+x^{2}}}[/itex] I got rid of the square root, I would know how to find a power series: integrating the result found by the power series of the differentiation. But I think it can't be applied here since any differentiation leads to a square root.
  5. Dec 17, 2013 #4


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    Why not use the binomial expansion?
  6. Dec 18, 2013 #5

    Simon Bridge

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    You have ##f(x)=\sum_{n=0}^\infty a_nx^n## it remains only for you to find out what the individual ##a_n##'s are.

    How may you do that?

    Write out the first three or so terms explicitly.
    What is f(x=0) in the power series? What is f(x=0) for the actual function?

    You have ##f'(x)=\sum_{n=0}^\infty na_nx^n## ... good. What is f'(0) in the series? What is f'(0) for the actual function?

    See the pattern?


    Aside 1:
    Considering that this is homework - I'd expect that you have been asked specifically to find a power series expansion rather than some other approximation method. I would have expected that you have had some lessons that include the power series expansion and how to go about finding the coefficients. Have you reviewed your coursework and notes?

    Or maybe look it up?
    ... if f(x) is analytic then the coefficients have a simple form.

    Aside 2:
    I'm puzzled that you don't find resemblance with the Taylor series - since the Taylor series is a power series.

    In the Maclaurin series (a Taylor series about x=0) the terms are $$f(x)=\sum_{n=0}^\infty a_nx^n:a_n=\frac{f^{(n)}}{n!} $$ ... which is exactly the form you are looking for.

    The suggestions above are supposed to help you realize this.
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