# Expansion of sqrt(x^2-C)

1. Jun 25, 2009

### 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.

James

2. Jun 25, 2009

### 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.

3. Jun 25, 2009

### 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

4. Jun 25, 2009

### g_edgar

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

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

5. Jun 26, 2009

### mathman

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.