• Support PF! Buy your school textbooks, materials and every day products Here!

Numerical analisys close numbers question

  • Thread starter nhrock3
  • Start date
  • #1
415
0
[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
33,505
5,191
[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
415
0
thanks :)
 
  • #4
415
0
why multiplying and dividing by [tex]y=\sqrt{x^2+1}+1[/tex]
 
  • #5
33,505
5,191
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
415
0
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?
 

Related Threads on Numerical analisys close numbers question

  • Last Post
Replies
2
Views
643
Replies
0
Views
689
Replies
7
Views
3K
Replies
7
Views
2K
  • Last Post
Replies
0
Views
876
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
7
Views
3K
Replies
5
Views
5K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
3
Views
1K
Top