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Math Help - Big Omega question

  1. #1
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    Big Omega question

    Show that f(x) = 21x^2 - 17x - 3 is Big omega (x^2) .

    I really dont know where to start. Please help.

    Thanks
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  2. #2
    Senior Member
    Joined
    Nov 2008
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    Hi buddy!

    Quote Originally Posted by micky_577 View Post
    Show that f(x) = 21x^2 - 17x - 3 is Big omega (x^2) .

    I really dont know where to start. Please help.
    You have to show:
    \exists\ c > 0\ : |f(x)| \ge c\cdot |g(x)|

    g(x) is in this case equal to x^2

    So,

    21x^2-17x - 3 \ge c*x^2

    This is what we want to show

    Deviding by x≤ leads to

    \frac{21x^2-17x - 3}{x^2} \ge c

    or easier

    21 - \frac{17}{x} - \frac{3}{x^2} \ge c

    Now we determine x to infty and get

    lim_{x \to \infty} [ 21 - \frac{17}{x} - \frac{3}{x^2} ] \ge c

    lim_{x \to \infty} [ 21 - 0 - 0 ] \ge c

    So the claim is correct, it is in big omega(x≤), because it is

    0 < c \le 21

    Any questions?

    Yours
    Rapha
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