Results 1 to 4 of 4

Math Help - numerical analisys close numbers question..

  1. #1
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    numerical analisys close numbers question..

    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?
    Last edited by transgalactic; March 29th 2010 at 09:23 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,963
    Thanks
    1631
    Quote Originally Posted by transgalactic View Post
    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?
    I presume this means for x close to 0. For example, if x= 0.001, \sqrt{x^2+ 1}- 1= \sqrt{1.000001}- 1= 1.0000004999998750000624999609375- 1 = 0.0000004999998750000624999609375. If you are rounding to, say, 6 decimal places, That would be just 0. As for multiplying and dividing by \sqrt{x^2+ 1}+ 1 the result, of course, will be the same but I presume that this is talking about it being some intermediate result in a further calculation- where the " x^2" you would get in the numerator will be canceled.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    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?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Mar 2008
    Posts
    934
    Thanks
    33
    Awards
    1
    Quote Originally Posted by transgalactic View Post
    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?
    I think you have already figured out that
    y=\frac{x^2}{\sqrt{x^2+1}+1}

    The point of the exercise is that in this form the value can be computed without the catastrophic loss of significant figures that occurs in the original computation when x is small due to the subtraction of two almost-equal numbers.

    Or, to put it another way, No Subtraction!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. unitary,close to it seflf defintion question
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: July 26th 2011, 03:16 AM
  2. complement, open, close question
    Posted in the Differential Geometry Forum
    Replies: 13
    Last Post: November 9th 2010, 06:38 AM
  3. problem in understandind sultion in numerical analisys..
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: June 25th 2010, 03:18 AM
  4. numerical analisys question..
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: May 29th 2010, 03:31 AM
  5. numerical analisys question..
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: March 27th 2010, 11:47 PM

Search Tags


/mathhelpforum @mathhelpforum