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    Hi,

    I'm completely new to big theta notation, but have an idea of what it is/doing. I'm trying to show the following is equal to big theta(n^2).

     4!+k\sum\limits_{z=5}^n z .

    so I think I need to show:  k_{1}|n^2|\leq4!+k\sum\limits_{z  =5}^n z\leq k_{2}|n^2|

    where the  k_{1} k_{2} are constants. But I'm not really sure what to do next. Like I said I'm totally new to this. Any help would be appreciated.
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    *Sigma
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    So I solved the above inequalities giving me:

     k_{1} \leq \frac {4!+k\sum\limits_{z  =5}^n z)}{n^2} \leq k_{2}

    But then all I can tell from this is that as n tends to infinity the fraction tends to 0...

    Also if I set n=5 then the summation just becomes from z=5. Given that z starts at 5 in the first place, I think the minimum value that n can be is 5...
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    bump.
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    (Original post by maths learner)
    bump.
    Not familiar with the big theta notation, but from a quick google, you're interested in the behaviour as n tends to infinity.

    Also, I presume you're familiar with the sum of integers formula.

    The fraction doesn't tend to zero.
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    (Original post by ghostwalker)
    Not familiar with the big theta notation, but from a quick google, you're interested in the behaviour as n tends to infinity.

    Also, I presume you're familiar with the sum of integers formula.

    The fraction doesn't tend to zero.
    Hi, yeah its basically bounding it from above and below. Yeah i do, but obviously it will need to be adapted since we start at 5 and not 1... Bit confused why it doesn't tend to 0, I'll have another look.

    Opps. Just noticed it will tend to 1/2 won't it? The ratio of the powers of n.
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    (Original post by maths learner)
    Opps. Just noticed it will tend to 1/2 won't it? The ratio of the powers of n.
    The ratio of the n^2, yes. And the factor "k" too.

    Don't know how familiar you are with limits, but what I'd do is divide everything through by the n^2 and then let n tend to infinity, leaving just the k/2.
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    (Original post by ghostwalker)
    The ratio of the n^2, yes. And the factor "k" too.

    Don't know how familiar you are with limits, but what I'd do is divide everything through by the n^2 and then let n tend to infinity, leaving just the k/2.
    makes sense. So the limit is then k/2. So we are left with:  k_{1} \leq \frac{k}{2} \leq k_{2} . As n tends to infinity...
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    (Original post by maths learner)
    makes sense. So the limit is then k/2. So we are left with:  k_{1} \leq \frac{k}{2} \leq k_{2} . As n tends to infinity...
    Yep. And you're done - the limit is k/2. There's no necessity to find values of k1,k2, unless you've been asked to.
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    (Original post by ghostwalker)
    Yep. And you're done - the limit is k/2. There's no necessity to find values of k1,k2, unless you've been asked to.
    Thanks. But then I don't understand how that shows that its growing at a rate of n^2. Which is what big theta shows...
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    (Original post by maths learner)
    Thanks. But then I don't understand how that shows that its growing at a rate of n^2. Which is what big theta shows...
    \Theta shows/says (as does this working) that it is growing at a rate proportional to n^2, and in this case the constant of proportionality is k/2. And it's constant for any given k.
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    (Original post by ghostwalker)
    \Theta shows/says (as does this working) that it is growing at a rate proportional to n^2, and in this case the constant of proportionality is k/2. And it's constant for any given k.
    Ahh, got it. Thanks very much for your help.
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    (Original post by maths learner)
    Ahh, got it. Thanks very much for your help.
    :cool:

    And ta for the rep.
 
 
 
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