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is the statement if f(x)=3x³+2x²+5x+1 then f(x) is o(x²) correct?
No. Why did you think it was?
Original post by DFranklin
No. Why did you think it was?


Because 2<3

hol up is then o(x^4) correct? cus if you divide by x^4 then the bottom of f(x)/x^4 goes to 0 quicker than f(x)
If you're concerned about the behaviour for large X, then yes, any power of x greater than 3 will work.

If your concern is what happens when x is small, that's a different story.
Original post by DFranklin
If you're concerned about the behaviour for large X, then yes, any power of x greater than 3 will work.

If your concern is what happens when x is small, that's a different story.

ok so just to clear up both these statements are correct?
if f(x)=3x³+2x²+5x+1 then f(x) is o(x^4)
if f(x)=3x³+2x²+5x+1 then f(x) is O(x³)
Original post by will'o'wisp2
ok so just to clear up both these statements are correct?
if f(x)=3x³+2x²+5x+1 then f(x) is o(x^4)
if f(x)=3x³+2x²+5x+1 then f(x) is O(x³)
Repeating myself: if you are talking about the behaviour for large x then these are correct.

Note that it would also correct to say that f(x) is o(x^pi) or o(x^27) or O(x^4) (to throw out a few random examples).
Original post by DFranklin
Repeating myself: if you are talking about the behaviour for large x then these are correct.

Note that it would also correct to say that f(x) is o(x^pi) or o(x^27) or O(x^4) (to throw out a few random examples).


Got it and cleared up that my statements are right as
Unparseable latex formula:

x\Rightarrow \infity



So then i can't use O(x2)O(x^2) for the above examples where the largest power is 3?
Original post by will'o'wisp2
Got it and cleared up that my statements are right as
Unparseable latex formula:

x\Rightarrow \infity



So then i can't use O(x2)O(x^2) for the above examples where the largest power is 3?
No.
Original post by DFranklin
No.


Great, thanks.

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