How Can I Compute High Precision Square Roots for Special Relativity Homework?

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Discussion Overview

The discussion revolves around computing high precision square roots in the context of special relativity homework, specifically focusing on evaluating the gamma factor and handling very small numbers in calculations.

Discussion Character

  • Homework-related, Mathematical reasoning

Main Points Raised

  • One participant expresses difficulty in evaluating the gamma factor due to the need to compute 1 minus a very small number (8.57*10^-13), noting that standard calculators and Mathematica return a value of 1.
  • Another participant suggests that Mathematica can handle floating point calculations with sufficient precision to avoid returning 1 for the difference, referencing a specific function for setting precision.
  • A later post reiterates the previous suggestion regarding Mathematica's capabilities, indicating that it resolves the initial problem.
  • One participant introduces a mathematical approximation, stating that (1+x)^(1/2) is approximately 1 + x/2, which may be relevant for small x.

Areas of Agreement / Disagreement

Participants generally agree that standard calculators struggle with such small differences, while there is a consensus that Mathematica can be used effectively for high precision calculations. However, there is no consensus on the best method to compute the square root by hand.

Contextual Notes

The discussion does not address potential limitations of the approximation method introduced or the specific conditions under which it may apply.

Who May Find This Useful

Students working on special relativity problems, particularly those involving high precision calculations and the gamma factor.

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Hello,

i'm having trouble evaluating my gamma factor for my special relativity homework, because I need to compute 1 minus a very small number (8.57*10^-13). My calculator treats this value as simply 1, as does Mathematica. Although I don't know much about it, and maybe there's a way to force it to consider the small number.

Are there any methods that I could do by hand?

(I am also looking for a computer method elsewhere, but I figured this might be interesting).

Thanks!
 
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SteamKing said:
Evaluating such a difference will be a problem with a standard calculator. Mathematica however should be able to handle floating point calculations to such a precision that you don't get a value of 1 for the difference.

http://reference.wolfram.com/language/ref/SetPrecision.html

Thanks, that solves the problem at hand.
 
(1+x)^(1/2) is approximately 1 + x/2.
 

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