What is the Big-Ω of the following function?
$begingroup$
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
asked 4 hours ago
TigerHixTigerHix
134
New contributor
TigerHix is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
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
asked 4 hours ago
TigerHixTigerHix
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.
$endgroup$
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
add a comment |
$begingroup$
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
asked 4 hours ago
TigerHixTigerHix
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.
$endgroup$
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
asked 4 hours ago
TigerHixTigerHix
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.
asked 4 hours ago
TigerHixTigerHix
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.
edited 2 hours ago
TigerHix
asked 4 hours ago
TigerHixTigerHix
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.
asked 4 hours ago
TigerHixTigerHix
134
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$begin{align}
sum_{x=1}^{2n}(x+x^2) gt sum_{x=1}^{2n}x^2 gt sum_{x=n+1}^{2n}x^2gtsum_{x=n+1}^{2n}n^2= n^3
end{align}
$$
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 $$sum_{x=1}^{m}(x+x^2)=frac{m(m+1)(m+2)}{3}$$
answered 2 hours ago
Apass.JackApass.Jack
8,1121633
$endgroup$
$begingroup$
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
22 mins ago
add a comment |
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$begingroup$
$$begin{align}
sum_{x=1}^{2n}(x+x^2) gt sum_{x=1}^{2n}x^2 gt sum_{x=n+1}^{2n}x^2gtsum_{x=n+1}^{2n}n^2= n^3
end{align}
$$
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 $$sum_{x=1}^{m}(x+x^2)=frac{m(m+1)(m+2)}{3}$$
answered 2 hours ago
Apass.JackApass.Jack
8,1121633
$endgroup$
$begingroup$
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
22 mins ago
add a comment |
$begingroup$
$$begin{align}
sum_{x=1}^{2n}(x+x^2) gt sum_{x=1}^{2n}x^2 gt sum_{x=n+1}^{2n}x^2gtsum_{x=n+1}^{2n}n^2= n^3
end{align}
$$
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 $$sum_{x=1}^{m}(x+x^2)=frac{m(m+1)(m+2)}{3}$$
answered 2 hours ago
Apass.JackApass.Jack
8,1121633
$endgroup$
$begingroup$
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
22 mins ago
add a comment |
$begingroup$
$$begin{align}
sum_{x=1}^{2n}(x+x^2) gt sum_{x=1}^{2n}x^2 gt sum_{x=n+1}^{2n}x^2gtsum_{x=n+1}^{2n}n^2= n^3
end{align}
$$
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 $$sum_{x=1}^{m}(x+x^2)=frac{m(m+1)(m+2)}{3}$$
answered 2 hours ago
Apass.JackApass.Jack
8,1121633
$endgroup$
$$begin{align}
sum_{x=1}^{2n}(x+x^2) gt sum_{x=1}^{2n}x^2 gt sum_{x=n+1}^{2n}x^2gtsum_{x=n+1}^{2n}n^2= n^3
end{align}
$$
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 $$sum_{x=1}^{m}(x+x^2)=frac{m(m+1)(m+2)}{3}$$
answered 2 hours ago
Apass.JackApass.Jack
8,1121633
edited 1 hour ago
answered 2 hours ago
Apass.JackApass.Jack
8,1121633
answered 2 hours ago
Apass.JackApass.Jack
8,1121633
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
22 mins ago
add a comment |
$begingroup$
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
22 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
22 mins ago
$begingroup$
@TigerHix The fact that $O$ and $Omega$ can be different, btw, is why $Theta$ exists.
$endgroup$
– Draconis
22 mins ago
add a comment |
TigerHix is a new contributor. Be nice, and check out our Code of Conduct.
TigerHix is a new contributor. Be nice, and check out our Code of Conduct.
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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