Numerical analisys close numbers question

  • Thread starter Thread starter nhrock3
  • Start date Start date
  • Tags Tags
    Numbers Numerical
nhrock3
Messages
403
Reaction score
0
y=\sqrt{x^2+1}-1
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 y=\sqrt{x^2+1}+1
makes it go away?
 
Physics news on Phys.org
nhrock3 said:
y=\sqrt{x^2+1}-1
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 y=\sqrt{x^2+1}+1
makes it go away?

If x is very close to 0, you're subtracting 1 from a number that is very close to 1.
 
thanks :)
 
why multiplying and dividing by y=\sqrt{x^2+1}+1
 
What do you get when you do this multiplication?
\left(\sqrt{x^2+1}-1\right)\frac{\sqrt{x^2+1}+1}{\sqrt{x^2+1}+1}
 
we get x^2 in the nominator
and a sum in the denominator
now there is some stuff about relative error
but i don't know what?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top