how do you prove that the big theta of a series is its leading term?

if f(x) = (An) x^n + (An-1) x^(n-1) +...+ (A1)x + (A0) how can you prove f(x) is big theta(x^n).

I've thought about it and one could do it by proving that f(x) big O(x^n) and x^n big O(f(x)). I've figured out the proof for the fo开发者_StackOverflow中文版rmer (using triangle inequality) but could not understand how to do the latter.

Alternatively one could prove f(x) is big omega (x^n).

I've gotten stuck on this question and any hints or clues you could give me would greatly help.

Consider |An x^n + A(n-1) x^(n-1) + ... |/|x^n| as x -> oo.

The expression gets very close to |An| and if An is not zero, then for sufficiently large x, the expression will be at least |An|/2.

You could prove that it is both big O(x^n) and big Omega(x^n).

To prove f(x) is O(x^n), observe that for x >= 1, each 0 <= x^0, x^2, ... x^n <= x^n.

Hence, f(x) <= (n+1) * max(A_0 ... A_n) * x^n

But (n+1) * max(A_0 ... A_n) is a constant with respect to x, so we have our bound[*]

To prove x^n is O(f(x)) is actually quite difficult, since it isn't true unless A_n != 0. But if A_n != 0, required to prove:

x^n is O(An x^n + ... + A0 x^0)

By some theorems about limits that I can't be bothered to state, that's true iff

(1/An) x^n is O(x^n + ... + (A0/An) x^0)

which is true iff

(1/An) x^n - ... - (A0/An) x^0 is O(x^n) [**]

But now the LHS is a polynomial of the form which we just proved is O(x^n) in the first part. QED.

In practice, though, what you actually do is prove some lemmas about the big-O complexity of the sum of two functions with known big-O complexities. Then you just observe that all terms on both sides are O(x^n), and you can ignore the rest.

[*] That's a fudge, actually, since what matters is the comparison of the absolute value of the function. But for large enough x, f(x) has the same sign as A_n, so if that's negative we just do a similar inequality the other way around.

I don't think you really need any use of the triangle inequality to "escape" the abs, because polynomial functions necessarily are monotonic outside a certain range (that is, they only have finitely many points of inflection), and when considering big-O limits we only care about what happens outside a certain range.

[**] Another fudge, really I should have written the limit constant M on the RHS, and included that when taking terms across to the LHS. OK, so this is only a sketch of a proof.