Numerical analisys close numbers question

  • Thread starter nhrock3
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  • #1
nhrock3
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[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?
 

Answers and Replies

  • #2
36,338
8,295
[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?

If x is very close to 0, you're subtracting 1 from a number that is very close to 1.
 
  • #3
nhrock3
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thanks :)
 
  • #4
nhrock3
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why multiplying and dividing by [tex]y=\sqrt{x^2+1}+1[/tex]
 
  • #5
36,338
8,295
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]
 
  • #6
nhrock3
415
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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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