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Old Jun25-09, 05:28 PM                  #1
tiger_striped_cat

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expansion of sqrt(x^2-C)

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:

sqrt(x^2-C)

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
James
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Old Jun25-09, 05:39 PM                  #2
mathman
 
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Re: expansion of sqrt(x^2-C)

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.
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Old Jun25-09, 06:50 PM                  #3
tiger_striped_cat

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Re: expansion of sqrt(x^2-C)

Thank you for your reply mathman. Maybe I should take one step back.

Id like to write

LaTeX Code:  \\sqrt(x_1^2-C) - \\sqrt(x_2^2-C)

in terms of LaTeX Code: \\Delta(x) = x_1 - x_2 or LaTeX Code: \\Delta(x) = x_1^2 - x_2^2

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?

james
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Old Jun25-09, 08:02 PM                  #4
g_edgar

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Re: expansion of sqrt(x^2-C)

Maybe you would like a series valid for LARGE x ...

LaTeX Code: \\sqrt{x^2-C} = x - \\frac{C}{2x} - \\frac{C^2}{8x^3} +\\dots
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Old Jun26-09, 08:02 PM                  #5
mathman
 
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Re: expansion of sqrt(x^2-C)

Originally Posted by tiger_striped_cat View Post
Thank you for your reply mathman. Maybe I should take one step back.

Id like to write

LaTeX Code:  \\sqrt(x_1^2-C) - \\sqrt(x_2^2-C)

in terms of LaTeX Code: \\Delta(x) = x_1 - x_2 or LaTeX Code: \\Delta(x) = x_1^2 - x_2^2

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?

james
For an error analysis, you might consider mutiplying by
LaTeX Code:  \\sqrt(x_1^2-C) + \\sqrt(x_2^2-C)  /LaTeX Code:  \\sqrt(x_1^2-C) + \\sqrt(x_2^2-C)

so that you will have LaTeX Code: \\Delta(x) = x_1^2 - x_2^2 as a numerator.
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