The discussion clarifies the meaning of big O notation in the context of Taylor polynomials and limits. Specifically, it explains that O(p(x)) indicates that a function f(x) is bounded above by a constant multiple of p(x) as x approaches infinity. The conversation also highlights the distinction between big O and little o notation, with little o indicating that f(x) approaches zero faster than p(x) as x approaches zero. The example provided, exp(x) = 1 + x + x^2/4 + O(x^3), illustrates the application of big O in expressing the remainder of a Taylor series.
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Yes, it's 2! = 2 :DOffice_Shredder said:I feel a compelling need to point out it's x2/2 actually. Regardless.
If you have f(x) = o(p(x)) for some f and p, it means that as x goes to zero, f goes to zero faster than p, i.e. f/p -> 0 as x goes to zero. This is meant to indicate that near zero something is so small, it's going to vanish when you take a limit to zero.
O is the opposite. f(x) = O(p(x)) means that as x goes to infinity, f is bounded above by p, i.e. there is some constant C such that f(x) <= C*p(x) for all large x.
So in your Taylor series example, I don't agree with what's written (assuming you meant little o as you stated in your second sentence) as the remainder is
x^3/6 + ... and when you divide through by x3 you get 1/6 + x*(stuff that's at least constant) which doesn't go to zero. It IS o(x2 as when you divide through by x2 you get
x/6 + x^2/24 + ... = x*(stuff) and that goes to zero as x goes to zero