Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question about big O

  1. Nov 20, 2009 #1
    hi,

    I am reading Prime Obsession and john talks about Big o. He goes if Function A is trapped within + and - of function B then F(A) is Big O of F(B)

    So, y=Sin x is Big O of Y = 1 right?

    He also goes that if F(A) is Big O of F(B) then so is any multiple of F(A).

    so y=2sinx is also big O of Y=1? Even though is breaks the limit infinite amount of times?

    Any clarification appreciated. Thanks!
     
  2. jcsd
  3. Nov 20, 2009 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    Yes, so the definition is not: "A is trapped between + and - the value of B", but "A is trapped between + and - a multiple of the value of B".

    Technically speaking, we say that
    [tex]f(x) = \mathcal O(g(x)) \qquad (x \to \infty)[/tex]
    if there exists some number M > 0 and some value x0 such that
    [tex]x > x_0 \implies |f(x)| < M |g(x)|[/tex]

    If you have f(x) = c sin(x), then you can easily show that f(x) = O(c), where c is the constant function which takes value c everywhere.
    Note that even g(x) = c sin(x) / xn is O(c) (for all n > 1), because for x > 1 we have |g(x)| < |f(x)| < c.

    Analogously to the definition of limit as x goes to infinity vs. limit as x goes to some finite number a, we can also define
    [tex]f(x) = \mathcal O(g(x)) \qquad (x \to a)[/tex]
    if there exists some number M > 0 and some value [itex]\delta > 0[/itex] such that
    [tex]|x - a| < \delta \implies |f(x)| < M |g(x)|[/tex]
     
  4. Nov 20, 2009 #3
    oh yeh, forgot about that. Thanks! :D
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook