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Big O notation question

  1. Aug 4, 2014 #1
    I am trying to reconcile the following statement:

    " ||u||<=eps*||f|| means ||u||=O(eps) ("||u|| is order eps")... "

    with the limit definition of "big O"; considering it's not clear that ||u|| here even depends on eps:

    " lim as eps goes to 0 of ||u||/eps, by definition of "big O", should equal some constant "; is it necessarily ||f||?

    I understand that, for example, sin(x)=O(x) because lim as x goes to 0 of sin(x)/x is bounded.

    So to casually say something is "order epsilon", if it doesn't depend on epsilon, I am not sure how to reconcile that with the definition of "big O" given by the limit one should be able to take, as above.
     
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  3. Aug 4, 2014 #2

    Simon Bridge

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  4. Aug 4, 2014 #3
    I paraphrased for the sake of drawing attention to my (pedantic) question;
    The original statement, from a book on finite element, was:
    ||u-u_s||<=eps*||u-u_s||_E <=eps^2*||f|| (the middle norm is energy norm)

    the following text then goes:
    "The point of course is that ||u-u_s||_E (energy norm) is of order eps, whereas ||u-u_s|| is of order eps^2."

    So maybe you can explain based on this context then?

    And sorry for lack of "tex" skills.
     
  5. Aug 4, 2014 #4

    Simon Bridge

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    It looks like the author is not using regular big-O notation and/or conventions here.
    Please provide a citation.

    Possibly the author just refers to the order in "epsilon".
    What is epsilon supposed to stand for?
     
    Last edited: Aug 4, 2014
  6. Aug 4, 2014 #5
    Well, that was sort of what I figured..but anyway, Brenner, Mathematical Theory of Finite Element, p.6 bottom.

    Thanks
     
  7. Aug 4, 2014 #6

    Simon Bridge

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    I don't know that one - the best way to be sure is to look for the same lesson in another text and compare.

    If I said that |y| = x|k| while |z|=x^2|k| then you could say that y = O(x) while z=O(x^2).
    I'm thinking that the dependent variable in the inequalities above is epsilon.
     
  8. Aug 4, 2014 #7
    Thanks; do you believe (as in, is it plausible) that the author means: ||u-u_s||/||f||<=eps means that thing less than eps is "of order epsilon"?
     
  9. Aug 4, 2014 #8

    Simon Bridge

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    Really need the context.
    But, especially if ||f|| is just a number, it would be fair to say that ||u-u_s||=O(eps)
    i.e. it seems one could expand the expressions as ##\sum_n a_n\epsilon^n ## ... whatever epsilon is supposed to represent. You are the one with the book in front of you.
     
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