On infinite products; notational questions

  • #1

Main Question or Discussion Point

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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  • #2
EnumaElish
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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.
 
  • #3
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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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