The Student Room Group

Big theta notation

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

so I think I need to show: k1n24!+kz=5nzk2n2 k_{1}|n^2|\leq4!+k\sum\limits_{z=5}^n z\leq k_{2}|n^2|

where the k1k2 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.
*Sigma
So I solved the above inequalities giving me:

k14!+kz=5nz)n2k2 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...
(edited 9 years ago)
bump.
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.
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.
(edited 9 years ago)
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.
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: k1k2k2 k_{1} \leq \frac{k}{2} \leq k_{2} . As n tends to infinity...
Original post by maths learner
makes sense. So the limit is then k/2. So we are left with: k1k2k2 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.
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...
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...


Θ\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.
Original post by ghostwalker
Θ\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.
Original post by maths learner
Ahh, got it. Thanks very much for your help.


:cool:

And ta for the rep.

Quick Reply

Latest