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On infinite products; notational questions

  1. Jul 11, 2005 #1
    I'm not certain this is the right forum, so let me know if its not, but I was wondering:

    I'm reading a book on special functions and there are infinite products involved. In showing they converge, a capital O crops up and is not defined. Can anyone help me? I don't know how to paste in math formulas using latex (something I would like to learn if anyone can help me with that), otherwise I would. As is I have posted a picture I did with MathType.

    What is the O mean? Perhaps that answer will be enough for me to answer the next on my own, but in case not: how do they arive at the second line's equality?

    Thanks for any help you can offer.

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    Last edited: Jul 11, 2005
  2. jcsd
  3. Jul 11, 2005 #2


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    O in this context means Order. Since it's a "Big O" (as opposed to "little o"), it represents that the cropped terms are at most in the order of -3 (i.e. somewhat small). Formally, f(n) = O(g(n)) means that "the rate of growth of f(n) is no more than a constant times the rate of growth of g(n)." See http://mathforum.org/library/drmath/view/54574.html. In your case, n = -3.
  4. Jul 12, 2005 #3
    Enuma has as such detailed out what big-Oh denotes.

    To top it off with a small example,
    consider this polynomial,
    f(n) = 1+n+n^2+n^3+n^4 (for all n>=1)

    Now if i can find two constants n_0 and c, such that
    f(n) <= c*g(n) for all n>=n_0
    then we say that
    f(n) = O(g(n))
    (**This is what Enuma states as, "the rate of growth of f(n) is no more than a constant times the rate of growth of g(n)." **)

    Consider g(n) = n^4,
    and now consider the constants n_0 = 1 and c = 5, you will notice that,
    f(n) <= 5*g(n) for all n>=1 (Try to prove this if you wish)
    hence f(n) = O(g(n))
    or (1+n+n^2+n^3+n^4) = O(n^4)

    Now moving a bit further, consider this polynomial,
    f(n) = n^5 + (1+n+n^2+n^3+n^4) (for all n>=1)
    I can always replace the term in bracket with,
    f(n) = n^5 + O(n^4)

    Such a replacement is done when we are only concerned with the growth rate of the term and not the actual term itself.

    -- AI
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