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Homework Help: Numerical analisys close numbers question

  1. Mar 29, 2010 #1
    [tex]y=\sqrt{x^2+1}-1[/tex]
    how do we know that there is a subtraction of two close numbers thus making loss of significance?
    there is no close numbers there is variable X
    it could give use a close result or otherwise

    and why multiplying and dividing by [tex]y=\sqrt{x^2+1}+1[/tex]
    makes it go away?
     
  2. jcsd
  3. Mar 29, 2010 #2

    Mark44

    Staff: Mentor

    If x is very close to 0, you're subtracting 1 from a number that is very close to 1.
     
  4. Mar 29, 2010 #3
    thanks :)
     
  5. Mar 29, 2010 #4
    why multiplying and dividing by [tex]y=\sqrt{x^2+1}+1[/tex]
     
  6. Mar 29, 2010 #5

    Mark44

    Staff: Mentor

    What do you get when you do this multiplication?
    [tex]\left(\sqrt{x^2+1}-1\right)\frac{\sqrt{x^2+1}+1}{\sqrt{x^2+1}+1}[/tex]
     
  7. Mar 29, 2010 #6
    we get x^2 in the nominator
    and a sum in the denominator
    now there is some stuff about relative error
    but i dont know what?
     
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