Mixing
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 $O(n^3)$.
asymptotics
edited Jan 15 at 7:44
Raphael♦
57.6k24140317
asked Jan 15 at 0:01
TigerHixTigerHix
134
$endgroup$
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 $O(n^3)$.
asymptotics
edited Jan 15 at 7:44
Raphael♦
57.6k24140317
asked Jan 15 at 0:01
TigerHixTigerHix
134
$endgroup$
1
$begingroup$
Please show in the question how "to see the (tightest) Big-$O$ is $O(n^3)$".
$endgroup$
– Apass.Jack
Jan 15 at 0:05
$begingroup$
@Apass.Jack Updated the question.
$endgroup$
– TigerHix
Jan 15 at 2:01
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 $O(n^3)$.
asymptotics
edited Jan 15 at 7:44
Raphael♦
57.6k24140317
asked Jan 15 at 0:01
TigerHixTigerHix
134
$endgroup$
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 $O(n^3)$.
asymptotics
asymptotics
edited Jan 15 at 7:44
Raphael♦
57.6k24140317
asked Jan 15 at 0:01
TigerHixTigerHix
134
edited Jan 15 at 7:44
Raphael♦
57.6k24140317
asked Jan 15 at 0:01
TigerHixTigerHix
134
edited Jan 15 at 7:44
Raphael♦
57.6k24140317
edited Jan 15 at 7:44
Raphael♦
57.6k24140317
edited Jan 15 at 7:44
Raphael♦
57.6k24140317
57.6k24140317
asked Jan 15 at 0:01
TigerHixTigerHix
134
asked Jan 15 at 0:01
TigerHixTigerHix
134
asked Jan 15 at 0:01
TigerHixTigerHix
134
134
1
$begingroup$
Please show in the question how "to see the (tightest) Big-$O$ is $O(n^3)$".
$endgroup$
– Apass.Jack
Jan 15 at 0:05
$begingroup$
@Apass.Jack Updated the question.
$endgroup$
– TigerHix
Jan 15 at 2:01
add a comment |
1
$begingroup$
Please show in the question how "to see the (tightest) Big-$O$ is $O(n^3)$".
$endgroup$
– Apass.Jack
Jan 15 at 0:05
$begingroup$
@Apass.Jack Updated the question.
$endgroup$
– TigerHix
Jan 15 at 2:01
1
1
$begingroup$
Please show in the question how "to see the (tightest) Big-$O$ is $O(n^3)$".
$endgroup$
– Apass.Jack
Jan 15 at 0:05
$begingroup$
Please show in the question how "to see the (tightest) Big-$O$ is $O(n^3)$".
$endgroup$
– Apass.Jack
Jan 15 at 0:05
$begingroup$
@Apass.Jack Updated the question.
$endgroup$
– TigerHix
Jan 15 at 2:01
$begingroup$
@Apass.Jack Updated the question.
$endgroup$
– TigerHix
Jan 15 at 2:01
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 Jan 15 at 2:13
Apass.JackApass.Jack
10.8k1939
$endgroup$
1
$begingroup$
@TigerHix The fact that $O$ and $Omega$ can be different, btw, is why $Theta$ exists.
$endgroup$
– Draconis
Jan 15 at 4:27
1
$begingroup$
@Draconis That's a bit of a non-sequitur. There are situations where there may be a "nice" $Theta$-class (unlike for $f$ in the answer) but we can't prove tight bounds. Then, we need $O$ and $Omega$ to express that what we can prove.
$endgroup$
– Raphael♦
Jan 15 at 7:46
$begingroup$
@Raphael Sorry, what I'm saying is that if $O$ and $Omega$ were always the same, there'd be no reason to have $Theta$ as its own thing.
$endgroup$
– Draconis
Jan 15 at 16:02
$begingroup$
@Draconis I understood that, and I disagree. (If anything, it'd be a reason to have only $Theta$; arguably, $Theta$ is redundant.)
$endgroup$
– Raphael♦
Jan 15 at 17:07
add a comment |
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1 Answer
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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 Jan 15 at 2:13
Apass.JackApass.Jack
10.8k1939
$endgroup$
1
$begingroup$
@TigerHix The fact that $O$ and $Omega$ can be different, btw, is why $Theta$ exists.
$endgroup$
– Draconis
Jan 15 at 4:27
1
$begingroup$
@Draconis That's a bit of a non-sequitur. There are situations where there may be a "nice" $Theta$-class (unlike for $f$ in the answer) but we can't prove tight bounds. Then, we need $O$ and $Omega$ to express that what we can prove.
$endgroup$
– Raphael♦
Jan 15 at 7:46
$begingroup$
@Raphael Sorry, what I'm saying is that if $O$ and $Omega$ were always the same, there'd be no reason to have $Theta$ as its own thing.
$endgroup$
– Draconis
Jan 15 at 16:02
$begingroup$
@Draconis I understood that, and I disagree. (If anything, it'd be a reason to have only $Theta$; arguably, $Theta$ is redundant.)
$endgroup$
– Raphael♦
Jan 15 at 17:07
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 Jan 15 at 2:13
Apass.JackApass.Jack
10.8k1939
$endgroup$
1
$begingroup$
@TigerHix The fact that $O$ and $Omega$ can be different, btw, is why $Theta$ exists.
$endgroup$
– Draconis
Jan 15 at 4:27
1
$begingroup$
@Draconis That's a bit of a non-sequitur. There are situations where there may be a "nice" $Theta$-class (unlike for $f$ in the answer) but we can't prove tight bounds. Then, we need $O$ and $Omega$ to express that what we can prove.
$endgroup$
– Raphael♦
Jan 15 at 7:46
$begingroup$
@Raphael Sorry, what I'm saying is that if $O$ and $Omega$ were always the same, there'd be no reason to have $Theta$ as its own thing.
$endgroup$
– Draconis
Jan 15 at 16:02
$begingroup$
@Draconis I understood that, and I disagree. (If anything, it'd be a reason to have only $Theta$; arguably, $Theta$ is redundant.)
$endgroup$
– Raphael♦
Jan 15 at 17:07
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 Jan 15 at 2:13
Apass.JackApass.Jack
10.8k1939
$endgroup$
$$begin{align}
sum_{x=1}^{2n}(x+x^2) gt sum_{x=1}^{2n}x^2 gt sum_{x=n+1}^{2n}x^2gtsum_{x=n+1}^{2n}n^2= n^3
end{align}
$$
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 Jan 15 at 2:13
Apass.JackApass.Jack
10.8k1939
edited Jan 15 at 3:41
answered Jan 15 at 2:13
Apass.JackApass.Jack
10.8k1939
answered Jan 15 at 2:13
Apass.JackApass.Jack
10.8k1939
answered Jan 15 at 2:13
Apass.JackApass.Jack
10.8k1939
10.8k1939
1
$begingroup$
@TigerHix The fact that $O$ and $Omega$ can be different, btw, is why $Theta$ exists.
$endgroup$
– Draconis
Jan 15 at 4:27
1
$begingroup$
@Draconis That's a bit of a non-sequitur. There are situations where there may be a "nice" $Theta$-class (unlike for $f$ in the answer) but we can't prove tight bounds. Then, we need $O$ and $Omega$ to express that what we can prove.
$endgroup$
– Raphael♦
Jan 15 at 7:46
$begingroup$
@Raphael Sorry, what I'm saying is that if $O$ and $Omega$ were always the same, there'd be no reason to have $Theta$ as its own thing.
$endgroup$
– Draconis
Jan 15 at 16:02
$begingroup$
@Draconis I understood that, and I disagree. (If anything, it'd be a reason to have only $Theta$; arguably, $Theta$ is redundant.)
$endgroup$
– Raphael♦
Jan 15 at 17:07
add a comment |
1
$begingroup$
@TigerHix The fact that $O$ and $Omega$ can be different, btw, is why $Theta$ exists.
$endgroup$
– Draconis
Jan 15 at 4:27
1
$begingroup$
@Draconis That's a bit of a non-sequitur. There are situations where there may be a "nice" $Theta$-class (unlike for $f$ in the answer) but we can't prove tight bounds. Then, we need $O$ and $Omega$ to express that what we can prove.
$endgroup$
– Raphael♦
Jan 15 at 7:46
$begingroup$
@Raphael Sorry, what I'm saying is that if $O$ and $Omega$ were always the same, there'd be no reason to have $Theta$ as its own thing.
$endgroup$
– Draconis
Jan 15 at 16:02
$begingroup$
@Draconis I understood that, and I disagree. (If anything, it'd be a reason to have only $Theta$; arguably, $Theta$ is redundant.)
$endgroup$
– Raphael♦
Jan 15 at 17:07
1
1
$begingroup$
@TigerHix The fact that $O$ and $Omega$ can be different, btw, is why $Theta$ exists.
$endgroup$
– Draconis
Jan 15 at 4:27
$begingroup$
@TigerHix The fact that $O$ and $Omega$ can be different, btw, is why $Theta$ exists.
$endgroup$
– Draconis
Jan 15 at 4:27
1
1
$begingroup$
@Draconis That's a bit of a non-sequitur. There are situations where there may be a "nice" $Theta$-class (unlike for $f$ in the answer) but we can't prove tight bounds. Then, we need $O$ and $Omega$ to express that what we can prove.
$endgroup$
– Raphael♦
Jan 15 at 7:46
$begingroup$
@Draconis That's a bit of a non-sequitur. There are situations where there may be a "nice" $Theta$-class (unlike for $f$ in the answer) but we can't prove tight bounds. Then, we need $O$ and $Omega$ to express that what we can prove.
$endgroup$
– Raphael♦
Jan 15 at 7:46
$begingroup$
@Raphael Sorry, what I'm saying is that if $O$ and $Omega$ were always the same, there'd be no reason to have $Theta$ as its own thing.
$endgroup$
– Draconis
Jan 15 at 16:02
$begingroup$
@Raphael Sorry, what I'm saying is that if $O$ and $Omega$ were always the same, there'd be no reason to have $Theta$ as its own thing.
$endgroup$
– Draconis
Jan 15 at 16:02
$begingroup$
@Draconis I understood that, and I disagree. (If anything, it'd be a reason to have only $Theta$; arguably, $Theta$ is redundant.)
$endgroup$
– Raphael♦
Jan 15 at 17:07
$begingroup$
@Draconis I understood that, and I disagree. (If anything, it'd be a reason to have only $Theta$; arguably, $Theta$ is redundant.)
$endgroup$
– Raphael♦
Jan 15 at 17:07
add a comment |
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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
Jan 15 at 0:05
$begingroup$
@Apass.Jack Updated the question.
$endgroup$
– TigerHix
Jan 15 at 2:01