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# Big theta notation Watch

1. 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).

.

so I think I need to show:

where the 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.
2. *Sigma
3. So I solved the above inequalities giving me:

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...
4. bump.
5. (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.
6. (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.
7. (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.
8. (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: . As n tends to infinity...
9. (Original post by maths learner)
makes sense. So the limit is then k/2. So we are left with: . 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.
10. (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...
11. (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...
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.
12. (Original post by ghostwalker)
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.
13. (Original post by maths learner)
Ahh, got it. Thanks very much for your help.

And ta for the rep.

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