# Order at most and order at leastWatch

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#1
Hello,
I have to prove that 1/4n(n+1)-1/2 is order exactly n^2
I thought that it would be bounded order at most for big n^2 for all n>=1 but in this case how could it have an order at least. Is my order at most wrong? Thank you
0
1 year ago
#2
(Original post by MissMathsxo)
Hello,
I have to prove that 1/4n(n+1)-1/2 is order exactly n^2
I thought that it would be bounded order at most for big n^2 for all n>=1 but in this case how could it have an order at least. Is my order at most wrong? Thank you
Do you have an image of the question?
0
#3
(Original post by mqb2766)
Do you have an image of the question?
This is all my teacher gave me 0
1 year ago
#4
(Original post by MissMathsxo)
This is all my teacher gave me Not sure what the E(n) means and I'm presuming the O(n^2) is the big O notation.
In this context, it really just means that C(n) is a quadratic. For large n, when n doubles then C(n) will quadruple.
Simply expand C(n) into the usual n^2, n, constant terms and make the argument that the quadratic is dominant term.
0
#5
(Original post by mqb2766)
Not sure what the E(n) means and I'm presuming the O(n^2) is the big O notation.
In this context, it really just means that C(n) is a quadratic. For large n, when n doubles then C(n) will quadruple.
Simply expand C(n) into the usual n^2, n, constant terms and make the argument that the quadratic is dominant term.
Sorry, I think they're both meant I be C(n). And the big symbol means both big o and omega so I need to shown that it can be less then or equal depending on the coefficient on n^2
0
1 year ago
#6
(Original post by MissMathsxo)
Sorry, I think they're both meant I be C(n). And the big symbol means both big o and omega so I need to shown that it can be less then or equal depending on the coefficient on n^2
In the image is it
Theta(n^2)
or
O(n^2)
0
#7
(Original post by mqb2766)
In the image is it
Theta(n^2)
or
O(n^2)
Theta
0
1 year ago
#8
(Original post by MissMathsxo)
Theta
Ok, so you need to show that there is both a
* quadratic upper bound: c*n^2 for n>n0
and a
* quadratic lower bound: d*n^2 for n>n0
Have a go at doing either / both and if you have problems just post? Some examples at
http://www-cgrl.cs.mcgill.ca/~godfri...-Functions.pdf
1
#9
(Original post by mqb2766)
Ok, so you need to show that there is both a
* quadratic upper bound: c*n^2 for n>n0
and a
* quadratic lower bound: d*n^2 for n>n0
Have a go at doing either / both and if you have problems just post? Some examples at
http://www-cgrl.cs.mcgill.ca/~godfri...-Functions.pdf
Oh, I have just realised why I was confused. I thought that the values you referred to as c and d had to be integers but they don't. Thank you for your response 😊
1
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