Is it possible to expand the radical sqrt(x^2-C) for error analysis?

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Discussion Overview

The discussion revolves around the possibility of expanding the expression sqrt(x^2-C) for the purpose of conducting error analysis. Participants explore various methods of approximating this radical expression, considering different approaches and expansions.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant expresses difficulty in finding an analytical expansion for sqrt(x^2-C) and questions the lack of a series approximation for this function.
  • Another participant notes that sqrt(x) is not analytic at x=0, suggesting that a MacLaurin series cannot be used, but proposes an expansion for sqrt(C-x^2) that converges for x^2 < C, assuming C > 0.
  • A participant suggests a series expansion for large x, providing a specific expansion: sqrt(x^2-C) = x - (C/(2x)) - (C^2/(8x^3)) + ...
  • One participant seeks to express the difference between two radicals, sqrt(x_1^2-C) - sqrt(x_2^2-C), in terms of the difference in x values, Delta(x), and expresses uncertainty about the feasibility of this approach.
  • Another participant recommends a method for error analysis involving multiplying by the conjugate of the radicals to facilitate the expression in terms of Delta(x).

Areas of Agreement / Disagreement

Participants present multiple approaches and ideas regarding the expansion of the radical, indicating that there is no consensus on a single method or solution. The discussion remains unresolved with differing perspectives on the feasibility of the expansions.

Contextual Notes

Limitations include the dependence on the condition that C > 0 and the specific range of x values for which the proposed expansions are valid. The discussion also highlights the challenge of finding a suitable expansion at x=0.

tiger_striped_cat
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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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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.
 
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?

james
 
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]
 
tiger_striped_cat said:
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?

james

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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