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Expansion of sqrt(x^2-C)

  1. Jun 25, 2009 #1
    I'm trying to do some rough error analysis and I came into a problem I can't do. I want to do a quick expansion of the radical:


    I'm sure I can do a substitution of the x^2 or even (x^2-C), but nowhere is there an expansion listed for sqrt(x). I don't know why one couldn't expand this analytically to get an series approximation of this function.

    Thanks for your help
  2. jcsd
  3. Jun 25, 2009 #2


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    sqrt(x) is not analytic at x=0, so you can't get a MacLaurin series. However you can get an expansion for sqrt(C-x^2) which will coverge for x^2 < C (assuming C >0). You can use the binomial. To get what you want, multiply by i.
  4. Jun 25, 2009 #3
    Thank you for your reply mathman. Maybe I should take one step back.

    Id like to write

    [tex] \sqrt(x_1^2-C) - \sqrt(x_2^2-C) [/tex]

    in terms of [tex]\Delta(x) = x_1 - x_2[/tex] or [tex]\Delta(x) = x_1^2 - x_2^2[/tex]

    which is why I was hoping to do an expansion of the radicals. Even if it meant doing an expansion at x=0.001, or even doing this at only the first order, but I'm beginning to think that this isn't possible. What do you think?

  5. Jun 25, 2009 #4
    Maybe you would like a series valid for LARGE x ...

    [tex]\sqrt{x^2-C} = x - \frac{C}{2x} - \frac{C^2}{8x^3} +\dots[/tex]
  6. Jun 26, 2009 #5


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    For an error analysis, you might consider mutiplying by
    [tex] \sqrt(x_1^2-C) + \sqrt(x_2^2-C) [/tex]/[tex] \sqrt(x_1^2-C) + \sqrt(x_2^2-C) [/tex]

    so that you will have [tex]\Delta(x) = x_1^2 - x_2^2[/tex] as a numerator.
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