Numerical analisys close numbers question

[nhrock3]

$$y=\sqrt{x^2+1}-1$$
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?

 

[Mark44]

$$y=\sqrt{x^2+1}-1$$
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.

 

[nhrock3]

thanks :)

 

[nhrock3]

why multiplying and dividing by $$y=\sqrt{x^2+1}+1$$

 

[Mark44]

What do you get when you do this multiplication?
$$\left(\sqrt{x^2+1}-1\right)\frac{\sqrt{x^2+1}+1}{\sqrt{x^2+1}+1}$$

 

[nhrock3]

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?