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    I have no clue what O(x) is in this, I'm sure it has something to do with limiting behaviour but I'm not too sure... Thanks for any help!

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    x^3 + x = O(x), as x \rightarrow 0
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    (Original post by Blue7195)
    I have no clue what O(x) is in this, I'm sure it has something to do with limiting behaviour but I'm not too sure... Thanks for any help!

    Show that

    x^3 + x = O(x), as x \rightarrow 0
    Big O notation describes the limiting behaviour of a function


    Was there more to the question
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    (Original post by Blue7195)
    I have no clue what O(x) is in this, I'm sure it has something to do with limiting behaviour but I'm not too sure... Thanks for any help!

    Show that

    x^3 + x = O(x), as x \rightarrow 0
    it means it is of the Order x

    as x gets very small x3 is very small so this behaves like x

    if you have a graphic package plot the cubic and the line y=x and look at the two close to zero
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    (Original post by TenOfThem)
    Big O notation describes the limiting behaviour of a function


    Was there more to the question
    Nope! It's just a show that, question.
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    (Original post by TeeEm)
    it means it is of the Order x

    as x gets very small x3 is very small so this behaves like x

    if you have a graphic package plot the cubic and the line y=x and look at the two close to zero
    Thankyou! But it is a show that question, how would you show x^3 + x --> O(x)? Or is just explaining what you said enough?
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    (Original post by Blue7195)
    Nope! It's just a show that, question.
    Ok

    As x gets very small x^3 becomes irrelevant when compared to x

    So as x tends to zero the function behaves as x


    Edit ... As indeed teeem said

    I was fooled as I expected it to be tending to infinity and mis read
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    (Original post by TeeEm)
    it means it is of the Order x

    as x gets very small x3 is very small so this behaves like x

    if you have a graphic package plot the cubic and the line y=x and look at the two close to zero
    ...
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    (Original post by Blue7195)
    Thankyou! But it is a show that question, how would you show x^3 + x --> O(x)? Or is just explaining what you said enough?
    I do not do pure maths, so I do not know the level/rigour that you are expected to use.
    To me that suffices for the level/type of maths I do.
    I am sure some purists will look at this question and if they can add more rigor I am sure they will.
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    (Original post by TeeEm)
    I do not do pure maths, so I do not know the level/rigour that you are expected to use.
    To me that suffices for the level/type of maths I do.
    I am sure some purists will look at this question and if they can add more rigor I am sure they will.
    Thankyou! I've never seen a question like it before, it's Uni level, not in the notes anywhere so maybe it was old content they threw out but just wanted to make sure! :P
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    (Original post by Blue7195)
    I have no clue what O(x) is in this, I'm sure it has something to do with limiting behaviour but I'm not too sure... Thanks for any help!

    Show that

    x^3 + x = O(x), as x \rightarrow 0
    To show formally that f(x) = O(g(x)) as x \rightarrow 0 means that you have to show that:

    \displaystyle \lim_{x \rightarrow 0} |\frac{f(x)}{g(x)}| = L > 0

    That's the definition of the original statement in terms of something you can calculate. So you need to show that:

    \displaystyle \lim_{x \rightarrow 0} |\frac{x^3+x)}{x}|

    is some +ve constant (which is pretty easy).

    In general, in order to show that X is true, you need to work from the definition of X.
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    (Original post by atsruser)
    To show formally that f(x) = O(g(x)) as x \rightarrow 0 means that you have to show that:

    \displaystyle \lim_{x \rightarrow 0} |\frac{f(x)}{g(x)}| = L > 0

    That's the definition of the original statement in terms of something you can calculate. So you need to show that:

    \displaystyle \lim_{x \rightarrow 0} |\frac{x^3+x)}{x}|

    is some +ve constant (which is pretty easy).

    In general, in order to show that X is true, you need to work from the definition of X.
    That makes sense, thanks a lot!
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    (Original post by atsruser)
    To show formally that f(x) = O(g(x)) as x \rightarrow 0 means that you have to show that:

    \displaystyle \lim_{x \rightarrow 0} |\frac{f(x)}{g(x)}| = L > 0
    Sorry, but this isn't correct.

    There are two issues.

    Firstly, the limit doesn't have to exist: e.g. sin(1/x) is O(1) as x->0.

    Secondly, even if the limit does exist, it doesn't have to be > 0. e.g. x^2 is O(x) as x->0 (*)

    Correct definition is more like: \exists M \in \mathbb{R} and \delta > 0 s.t. for 0 < x < \delta, \left|\dfrac{f(x)}{g(x)} \right| < M.

    Edit: (*) If it helps, informally f(x) = O(g(x)) means f(x) is no larger than g(x). There is also f(x) = \Theta(g(x)) which means f(x) is roughly "the same size" as g(x). In this case if the limit exists it must be > 0. This may be what you were thinking of.
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    (Original post by DFranklin)
    Sorry, but this isn't correct.

    There are two issues.

    Firstly, the limit doesn't have to exist: e.g. sin(1/x) is O(1) as x->0.
    I'll have to think about this.

    Correct definition is more like: \exists M \in \mathbb{R} and \delta > 0 s.t. for 0 < x < \delta, \left|\dfrac{f(x)}{g(x)} \right| < M.
    This seems to be correct - I was trying to express the concept "bounded by g(x)" in some sense and didn't get it quite right.

    Edit: (*) If it helps, informally f(x) = O(g(x)) means f(x) is no larger than g(x). There is also f(x) = \Theta(g(x)) which means f(x) is roughly "the same size" as g(x). In this case if the limit exists it must be > 0. This may be what you were thinking of.
    Possibly. I was recalling it incorrectly and was too lazy to check. I'd have to go and read up on this stuff properly to make a more sensible response but I don't have the time at the moment.
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    (Original post by atsruser)
    Possibly. I was recalling it incorrectly and was too lazy to check. I'd have to go and read up on this stuff properly to make a more sensible response but I don't have the time at the moment.
    I did the same thing on another thread today
 
 
 
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