expansion of sqrt(x^2-C)

[tiger_striped_cat]

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

[mathman]

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.

[tiger_striped_cat]

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

Id like to write

\sqrt(x_1^2-C) - \sqrt(x_2^2-C)

in terms of \Delta(x) = x_1 - x_2 or \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

[g_edgar]

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

\sqrt{x^2-C} = x - \frac{C}{2x} - \frac{C^2}{8x^3} +\dots

[mathman]

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

Id like to write

\sqrt(x_1^2-C) - \sqrt(x_2^2-C)

in terms of \Delta(x) = x_1 - x_2 or \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
\sqrt(x_1^2-C) + \sqrt(x_2^2-C) / \sqrt(x_1^2-C) + \sqrt(x_2^2-C)

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