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

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SUMMARY

This discussion addresses the challenge of computing high precision square roots in the context of special relativity, specifically for evaluating the gamma factor. The user encounters limitations with standard calculators and Mathematica's default settings, which round small differences to 1. A solution is provided through Mathematica's SetPrecision function, allowing for accurate floating-point calculations. Additionally, the approximation (1+x)^(1/2) ≈ 1 + x/2 is mentioned as a useful method for manual calculations.

PREREQUISITES
  • Understanding of special relativity concepts, particularly the gamma factor.
  • Familiarity with Mathematica, specifically its floating-point precision functions.
  • Basic knowledge of mathematical approximations and Taylor series.
  • Experience with high precision calculations in computational tools.
NEXT STEPS
  • Research the use of Mathematica's SetPrecision for high precision calculations.
  • Explore the Taylor series expansion for functions near 1, specifically for square roots.
  • Learn about numerical methods for computing square roots with high precision.
  • Investigate other computational tools that support high precision arithmetic, such as Python's mpmath library.
USEFUL FOR

Students and educators in physics, particularly those studying special relativity, as well as anyone needing to perform high precision calculations in mathematical software.

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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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