What is the Big-Ω of the following function?












2












$begingroup$


For the following function:



$$
sum_{n=1}^{2n}x+x^2
$$



It is easy to see the (tightest) Big-Oh is $O(n^3)$, but I am not so sure about the Big-Omega. Here is my attempt:



$$
sum_{n=1}^{2n}x+x^2
$$
$$
geq sum_{n=1}^{2n}x^2
$$



But not sure how to continue. Also, does tightest Big-Ω always equal tightest Big-Oh?



EDIT: The way I worked out the Big-Oh is shown below:



$$
sum_{n=1}^{2n}x+x^2
$$
$$
leq sum_{n=1}^{2n}x^2+x^2
$$
$$
= sum_{n=1}^{2n}2x^2
$$
$$
leq 2n ( 2(2n)^2)
$$
$$
= 8n^3
$$



Therefore the Big-Oh is $O(n^3)$.






time-complexity asymptotics runtime-analysis







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TigerHix is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 1




    $begingroup$
    Please show in the question how "to see the (tightest) Big-$O$ is $O(n^3)$".
    $endgroup$
    – Apass.Jack
    4 hours ago










  • $begingroup$
    @Apass.Jack Updated the question.
    $endgroup$
    – TigerHix
    2 hours ago
















2












$begingroup$


For the following function:



$$
sum_{n=1}^{2n}x+x^2
$$



It is easy to see the (tightest) Big-Oh is $O(n^3)$, but I am not so sure about the Big-Omega. Here is my attempt:



$$
sum_{n=1}^{2n}x+x^2
$$
$$
geq sum_{n=1}^{2n}x^2
$$



But not sure how to continue. Also, does tightest Big-Ω always equal tightest Big-Oh?



EDIT: The way I worked out the Big-Oh is shown below:



$$
sum_{n=1}^{2n}x+x^2
$$
$$
leq sum_{n=1}^{2n}x^2+x^2
$$
$$
= sum_{n=1}^{2n}2x^2
$$
$$
leq 2n ( 2(2n)^2)
$$
$$
= 8n^3
$$



Therefore the Big-Oh is $O(n^3)$.






time-complexity asymptotics runtime-analysis







share|cite|improve this question









New contributor




TigerHix is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    Please show in the question how "to see the (tightest) Big-$O$ is $O(n^3)$".
    $endgroup$
    – Apass.Jack
    4 hours ago










  • $begingroup$
    @Apass.Jack Updated the question.
    $endgroup$
    – TigerHix
    2 hours ago














2












2








2





$begingroup$


For the following function:



$$
sum_{n=1}^{2n}x+x^2
$$



It is easy to see the (tightest) Big-Oh is $O(n^3)$, but I am not so sure about the Big-Omega. Here is my attempt:



$$
sum_{n=1}^{2n}x+x^2
$$
$$
geq sum_{n=1}^{2n}x^2
$$



But not sure how to continue. Also, does tightest Big-Ω always equal tightest Big-Oh?



EDIT: The way I worked out the Big-Oh is shown below:



$$
sum_{n=1}^{2n}x+x^2
$$
$$
leq sum_{n=1}^{2n}x^2+x^2
$$
$$
= sum_{n=1}^{2n}2x^2
$$
$$
leq 2n ( 2(2n)^2)
$$
$$
= 8n^3
$$



Therefore the Big-Oh is $O(n^3)$.






time-complexity asymptotics runtime-analysis







share|cite|improve this question









New contributor




TigerHix is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




For the following function:



$$
sum_{n=1}^{2n}x+x^2
$$



It is easy to see the (tightest) Big-Oh is $O(n^3)$, but I am not so sure about the Big-Omega. Here is my attempt:



$$
sum_{n=1}^{2n}x+x^2
$$
$$
geq sum_{n=1}^{2n}x^2
$$



But not sure how to continue. Also, does tightest Big-Ω always equal tightest Big-Oh?



EDIT: The way I worked out the Big-Oh is shown below:



$$
sum_{n=1}^{2n}x+x^2
$$
$$
leq sum_{n=1}^{2n}x^2+x^2
$$
$$
= sum_{n=1}^{2n}2x^2
$$
$$
leq 2n ( 2(2n)^2)
$$
$$
= 8n^3
$$



Therefore the Big-Oh is $O(n^3)$.






time-complexity asymptotics runtime-analysis




time-complexity asymptotics runtime-analysis






share|cite|improve this question









New contributor




TigerHix is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




TigerHix is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 2 hours ago







TigerHix













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asked 4 hours ago









TigerHixTigerHix

134




134




New contributor




TigerHix is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





TigerHix is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






TigerHix is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    $begingroup$
    Please show in the question how "to see the (tightest) Big-$O$ is $O(n^3)$".
    $endgroup$
    – Apass.Jack
    4 hours ago










  • $begingroup$
    @Apass.Jack Updated the question.
    $endgroup$
    – TigerHix
    2 hours ago














  • 1




    $begingroup$
    Please show in the question how "to see the (tightest) Big-$O$ is $O(n^3)$".
    $endgroup$
    – Apass.Jack
    4 hours ago










  • $begingroup$
    @Apass.Jack Updated the question.
    $endgroup$
    – TigerHix
    2 hours ago








1




1




$begingroup$
Please show in the question how "to see the (tightest) Big-$O$ is $O(n^3)$".
$endgroup$
– Apass.Jack
4 hours ago




$begingroup$
Please show in the question how "to see the (tightest) Big-$O$ is $O(n^3)$".
$endgroup$
– Apass.Jack
4 hours ago












$begingroup$
@Apass.Jack Updated the question.
$endgroup$
– TigerHix
2 hours ago




$begingroup$
@Apass.Jack Updated the question.
$endgroup$
– TigerHix
2 hours ago










1 Answer
1






active

oldest

votes


















2












$begingroup$

$$begin{align}
sum_{x=1}^{2n}(x+x^2) gt sum_{x=1}^{2n}x^2 gt sum_{x=n+1}^{2n}x^2gtsum_{x=n+1}^{2n}n^2= n^3
end{align}
$$



So the function is $Omega(n^3)$. You have shown it is $O(n^3)$. So the function is $Theta(n^3)$.



Although $n^3$, the asymptotic lower bound ignoring a constant factor is the same as the asymptotic upper bound ignoring a constant factor in the current case, it is not always true because some functions fluctuate. For example,
$$f(n)=begin{cases}1 quad text {when }ntext{ is odd,}\ n quad text{when }ntext{ is even.}end{cases}$$ $f(n)=O(n)$ and $f(n)=Omega(1)$. Both bounds are tightest.




Exercise. Show that $$sum_{x=1}^{m}(x+x^2)=frac{m(m+1)(m+2)}{3}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for the detailed answer! Could you also answer the other question: does the tightest Big-Ω always equal the tightest Big-Oh?
    $endgroup$
    – TigerHix
    2 hours ago










  • $begingroup$
    That clears it up, thanks!
    $endgroup$
    – TigerHix
    1 hour ago






  • 1




    $begingroup$
    @TigerHix The fact that $O$ and $Omega$ can be different, btw, is why $Theta$ exists.
    $endgroup$
    – Draconis
    30 mins ago













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1 Answer
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1 Answer
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active

oldest

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2












$begingroup$

$$begin{align}
sum_{x=1}^{2n}(x+x^2) gt sum_{x=1}^{2n}x^2 gt sum_{x=n+1}^{2n}x^2gtsum_{x=n+1}^{2n}n^2= n^3
end{align}
$$



So the function is $Omega(n^3)$. You have shown it is $O(n^3)$. So the function is $Theta(n^3)$.



Although $n^3$, the asymptotic lower bound ignoring a constant factor is the same as the asymptotic upper bound ignoring a constant factor in the current case, it is not always true because some functions fluctuate. For example,
$$f(n)=begin{cases}1 quad text {when }ntext{ is odd,}\ n quad text{when }ntext{ is even.}end{cases}$$ $f(n)=O(n)$ and $f(n)=Omega(1)$. Both bounds are tightest.




Exercise. Show that $$sum_{x=1}^{m}(x+x^2)=frac{m(m+1)(m+2)}{3}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for the detailed answer! Could you also answer the other question: does the tightest Big-Ω always equal the tightest Big-Oh?
    $endgroup$
    – TigerHix
    2 hours ago










  • $begingroup$
    That clears it up, thanks!
    $endgroup$
    – TigerHix
    1 hour ago






  • 1




    $begingroup$
    @TigerHix The fact that $O$ and $Omega$ can be different, btw, is why $Theta$ exists.
    $endgroup$
    – Draconis
    30 mins ago


















2












$begingroup$

$$begin{align}
sum_{x=1}^{2n}(x+x^2) gt sum_{x=1}^{2n}x^2 gt sum_{x=n+1}^{2n}x^2gtsum_{x=n+1}^{2n}n^2= n^3
end{align}
$$



So the function is $Omega(n^3)$. You have shown it is $O(n^3)$. So the function is $Theta(n^3)$.



Although $n^3$, the asymptotic lower bound ignoring a constant factor is the same as the asymptotic upper bound ignoring a constant factor in the current case, it is not always true because some functions fluctuate. For example,
$$f(n)=begin{cases}1 quad text {when }ntext{ is odd,}\ n quad text{when }ntext{ is even.}end{cases}$$ $f(n)=O(n)$ and $f(n)=Omega(1)$. Both bounds are tightest.




Exercise. Show that $$sum_{x=1}^{m}(x+x^2)=frac{m(m+1)(m+2)}{3}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for the detailed answer! Could you also answer the other question: does the tightest Big-Ω always equal the tightest Big-Oh?
    $endgroup$
    – TigerHix
    2 hours ago










  • $begingroup$
    That clears it up, thanks!
    $endgroup$
    – TigerHix
    1 hour ago






  • 1




    $begingroup$
    @TigerHix The fact that $O$ and $Omega$ can be different, btw, is why $Theta$ exists.
    $endgroup$
    – Draconis
    30 mins ago
















2












2








2





$begingroup$

$$begin{align}
sum_{x=1}^{2n}(x+x^2) gt sum_{x=1}^{2n}x^2 gt sum_{x=n+1}^{2n}x^2gtsum_{x=n+1}^{2n}n^2= n^3
end{align}
$$



So the function is $Omega(n^3)$. You have shown it is $O(n^3)$. So the function is $Theta(n^3)$.



Although $n^3$, the asymptotic lower bound ignoring a constant factor is the same as the asymptotic upper bound ignoring a constant factor in the current case, it is not always true because some functions fluctuate. For example,
$$f(n)=begin{cases}1 quad text {when }ntext{ is odd,}\ n quad text{when }ntext{ is even.}end{cases}$$ $f(n)=O(n)$ and $f(n)=Omega(1)$. Both bounds are tightest.




Exercise. Show that $$sum_{x=1}^{m}(x+x^2)=frac{m(m+1)(m+2)}{3}$$






share|cite|improve this answer











$endgroup$



$$begin{align}
sum_{x=1}^{2n}(x+x^2) gt sum_{x=1}^{2n}x^2 gt sum_{x=n+1}^{2n}x^2gtsum_{x=n+1}^{2n}n^2= n^3
end{align}
$$



So the function is $Omega(n^3)$. You have shown it is $O(n^3)$. So the function is $Theta(n^3)$.



Although $n^3$, the asymptotic lower bound ignoring a constant factor is the same as the asymptotic upper bound ignoring a constant factor in the current case, it is not always true because some functions fluctuate. For example,
$$f(n)=begin{cases}1 quad text {when }ntext{ is odd,}\ n quad text{when }ntext{ is even.}end{cases}$$ $f(n)=O(n)$ and $f(n)=Omega(1)$. Both bounds are tightest.




Exercise. Show that $$sum_{x=1}^{m}(x+x^2)=frac{m(m+1)(m+2)}{3}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 1 hour ago

























answered 2 hours ago









Apass.JackApass.Jack

8,1121633




8,1121633












  • $begingroup$
    Thank you for the detailed answer! Could you also answer the other question: does the tightest Big-Ω always equal the tightest Big-Oh?
    $endgroup$
    – TigerHix
    2 hours ago










  • $begingroup$
    That clears it up, thanks!
    $endgroup$
    – TigerHix
    1 hour ago






  • 1




    $begingroup$
    @TigerHix The fact that $O$ and $Omega$ can be different, btw, is why $Theta$ exists.
    $endgroup$
    – Draconis
    30 mins ago




















  • $begingroup$
    Thank you for the detailed answer! Could you also answer the other question: does the tightest Big-Ω always equal the tightest Big-Oh?
    $endgroup$
    – TigerHix
    2 hours ago










  • $begingroup$
    That clears it up, thanks!
    $endgroup$
    – TigerHix
    1 hour ago






  • 1




    $begingroup$
    @TigerHix The fact that $O$ and $Omega$ can be different, btw, is why $Theta$ exists.
    $endgroup$
    – Draconis
    30 mins ago


















$begingroup$
Thank you for the detailed answer! Could you also answer the other question: does the tightest Big-Ω always equal the tightest Big-Oh?
$endgroup$
– TigerHix
2 hours ago




$begingroup$
Thank you for the detailed answer! Could you also answer the other question: does the tightest Big-Ω always equal the tightest Big-Oh?
$endgroup$
– TigerHix
2 hours ago












$begingroup$
That clears it up, thanks!
$endgroup$
– TigerHix
1 hour ago




$begingroup$
That clears it up, thanks!
$endgroup$
– TigerHix
1 hour ago




1




1




$begingroup$
@TigerHix The fact that $O$ and $Omega$ can be different, btw, is why $Theta$ exists.
$endgroup$
– Draconis
30 mins ago






$begingroup$
@TigerHix The fact that $O$ and $Omega$ can be different, btw, is why $Theta$ exists.
$endgroup$
– Draconis
30 mins ago












TigerHix is a new contributor. Be nice, and check out our Code of Conduct.










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