Big-Theta, Big-O, Big-Omega in $Bbb{R}^n$.
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Is the following statement true?
Let $ g : Bbb{R}^n to Bbb{R}^n $
$$ Theta_{(g)} = Omega_{(g)} cap O_{(g)} $$
asymptotics computer-science computational-complexity
asked Oct 25 at 4:00
Carlos Uribe
62
add a comment |
up vote
1
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favorite
Is the following statement true?
Let $ g : Bbb{R}^n to Bbb{R}^n $
$$ Theta_{(g)} = Omega_{(g)} cap O_{(g)} $$
asymptotics computer-science computational-complexity
asked Oct 25 at 4:00
Carlos Uribe
62
Shouldn't it be an intersection on the right hand side?
– kodlu
Oct 25 at 8:19
Yes, you are right kodlu. Thank you for the suggestion.
– Carlos Uribe
Nov 16 at 13:14
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Is the following statement true?
Let $ g : Bbb{R}^n to Bbb{R}^n $
$$ Theta_{(g)} = Omega_{(g)} cap O_{(g)} $$
asymptotics computer-science computational-complexity
asked Oct 25 at 4:00
Carlos Uribe
62
Is the following statement true?
Let $ g : Bbb{R}^n to Bbb{R}^n $
$$ Theta_{(g)} = Omega_{(g)} cap O_{(g)} $$
asymptotics computer-science computational-complexity
asymptotics computer-science computational-complexity
asked Oct 25 at 4:00
Carlos Uribe
62
asked Oct 25 at 4:00
Carlos Uribe
62
edited Nov 16 at 13:13
asked Oct 25 at 4:00
Carlos Uribe
62
asked Oct 25 at 4:00
Carlos Uribe
62
asked Oct 25 at 4:00
Carlos Uribe
62
62
Shouldn't it be an intersection on the right hand side?
– kodlu
Oct 25 at 8:19
Yes, you are right kodlu. Thank you for the suggestion.
– Carlos Uribe
Nov 16 at 13:14
add a comment |
Shouldn't it be an intersection on the right hand side?
– kodlu
Oct 25 at 8:19
Yes, you are right kodlu. Thank you for the suggestion.
– Carlos Uribe
Nov 16 at 13:14
Shouldn't it be an intersection on the right hand side?
– kodlu
Oct 25 at 8:19
Shouldn't it be an intersection on the right hand side?
– kodlu
Oct 25 at 8:19
Yes, you are right kodlu. Thank you for the suggestion.
– Carlos Uribe
Nov 16 at 13:14
Yes, you are right kodlu. Thank you for the suggestion.
– Carlos Uribe
Nov 16 at 13:14
add a comment |
1 Answer
1
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0
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The notation $f(n)=O(g(n))$ means that $|f|$ is bounded above by $g$ asymptotically. Formally,
$$
exists k>0quadexists n_0 in mathbb{N}quadforall ninmathbb{N}, n>n_0:quad|f(n)|leq kcdot g(n).
$$
Note that this implies that $f$ is also bounded above by $g$ asymptotically.
The notation $f(n)=Omega(g(n))$ means that $f$ is bounded below by $g$ asymptotically.
Formally,
$$
exists k>0quadexists n_0inmathbb{N}quadforall ninmathbb{N},n>n_0:quad f(n)geq kcdot g(n).
$$
The notation $f(n)=Theta(g(n))$ means that $f$ is bounded both above and below by $g$ asymptotically. Formally,
$$
exists k_1>0quadexists k_2>0quadexists n_0inmathbb{N}quadforall ninmathbb{N}, n>n_{0}:quad k_1cdot g(n)leq f(n)leq k_2cdot g(n).
$$
Note that if $f(n)=Theta(g(n))$, then both $f(n)=O(g(n))$ and $f(n)=Omega(g(n))$ hold, which means that
$$
Theta(g(n)) = O(g(n)) cap Omega(g(n)),
$$
by definition.
answered Nov 19 at 12:54
user153012
6,21822277
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The notation $f(n)=O(g(n))$ means that $|f|$ is bounded above by $g$ asymptotically. Formally,
$$
exists k>0quadexists n_0 in mathbb{N}quadforall ninmathbb{N}, n>n_0:quad|f(n)|leq kcdot g(n).
$$
Note that this implies that $f$ is also bounded above by $g$ asymptotically.
The notation $f(n)=Omega(g(n))$ means that $f$ is bounded below by $g$ asymptotically.
Formally,
$$
exists k>0quadexists n_0inmathbb{N}quadforall ninmathbb{N},n>n_0:quad f(n)geq kcdot g(n).
$$
The notation $f(n)=Theta(g(n))$ means that $f$ is bounded both above and below by $g$ asymptotically. Formally,
$$
exists k_1>0quadexists k_2>0quadexists n_0inmathbb{N}quadforall ninmathbb{N}, n>n_{0}:quad k_1cdot g(n)leq f(n)leq k_2cdot g(n).
$$
Note that if $f(n)=Theta(g(n))$, then both $f(n)=O(g(n))$ and $f(n)=Omega(g(n))$ hold, which means that
$$
Theta(g(n)) = O(g(n)) cap Omega(g(n)),
$$
by definition.
answered Nov 19 at 12:54
user153012
6,21822277
add a comment |
up vote
0
down vote
The notation $f(n)=O(g(n))$ means that $|f|$ is bounded above by $g$ asymptotically. Formally,
$$
exists k>0quadexists n_0 in mathbb{N}quadforall ninmathbb{N}, n>n_0:quad|f(n)|leq kcdot g(n).
$$
Note that this implies that $f$ is also bounded above by $g$ asymptotically.
The notation $f(n)=Omega(g(n))$ means that $f$ is bounded below by $g$ asymptotically.
Formally,
$$
exists k>0quadexists n_0inmathbb{N}quadforall ninmathbb{N},n>n_0:quad f(n)geq kcdot g(n).
$$
The notation $f(n)=Theta(g(n))$ means that $f$ is bounded both above and below by $g$ asymptotically. Formally,
$$
exists k_1>0quadexists k_2>0quadexists n_0inmathbb{N}quadforall ninmathbb{N}, n>n_{0}:quad k_1cdot g(n)leq f(n)leq k_2cdot g(n).
$$
Note that if $f(n)=Theta(g(n))$, then both $f(n)=O(g(n))$ and $f(n)=Omega(g(n))$ hold, which means that
$$
Theta(g(n)) = O(g(n)) cap Omega(g(n)),
$$
by definition.
answered Nov 19 at 12:54
user153012
6,21822277
add a comment |
up vote
0
down vote
up vote
0
down vote
The notation $f(n)=O(g(n))$ means that $|f|$ is bounded above by $g$ asymptotically. Formally,
$$
exists k>0quadexists n_0 in mathbb{N}quadforall ninmathbb{N}, n>n_0:quad|f(n)|leq kcdot g(n).
$$
Note that this implies that $f$ is also bounded above by $g$ asymptotically.
The notation $f(n)=Omega(g(n))$ means that $f$ is bounded below by $g$ asymptotically.
Formally,
$$
exists k>0quadexists n_0inmathbb{N}quadforall ninmathbb{N},n>n_0:quad f(n)geq kcdot g(n).
$$
The notation $f(n)=Theta(g(n))$ means that $f$ is bounded both above and below by $g$ asymptotically. Formally,
$$
exists k_1>0quadexists k_2>0quadexists n_0inmathbb{N}quadforall ninmathbb{N}, n>n_{0}:quad k_1cdot g(n)leq f(n)leq k_2cdot g(n).
$$
Note that if $f(n)=Theta(g(n))$, then both $f(n)=O(g(n))$ and $f(n)=Omega(g(n))$ hold, which means that
$$
Theta(g(n)) = O(g(n)) cap Omega(g(n)),
$$
by definition.
answered Nov 19 at 12:54
user153012
6,21822277
The notation $f(n)=O(g(n))$ means that $|f|$ is bounded above by $g$ asymptotically. Formally,
$$
exists k>0quadexists n_0 in mathbb{N}quadforall ninmathbb{N}, n>n_0:quad|f(n)|leq kcdot g(n).
$$
Note that this implies that $f$ is also bounded above by $g$ asymptotically.
The notation $f(n)=Omega(g(n))$ means that $f$ is bounded below by $g$ asymptotically.
Formally,
$$
exists k>0quadexists n_0inmathbb{N}quadforall ninmathbb{N},n>n_0:quad f(n)geq kcdot g(n).
$$
The notation $f(n)=Theta(g(n))$ means that $f$ is bounded both above and below by $g$ asymptotically. Formally,
$$
exists k_1>0quadexists k_2>0quadexists n_0inmathbb{N}quadforall ninmathbb{N}, n>n_{0}:quad k_1cdot g(n)leq f(n)leq k_2cdot g(n).
$$
Note that if $f(n)=Theta(g(n))$, then both $f(n)=O(g(n))$ and $f(n)=Omega(g(n))$ hold, which means that
$$
Theta(g(n)) = O(g(n)) cap Omega(g(n)),
$$
by definition.
answered Nov 19 at 12:54
user153012
6,21822277
answered Nov 19 at 12:54
user153012
6,21822277
answered Nov 19 at 12:54
user153012
6,21822277
answered Nov 19 at 12:54
user153012
6,21822277
6,21822277
add a comment |
add a comment |
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Shouldn't it be an intersection on the right hand side?
– kodlu
Oct 25 at 8:19
Yes, you are right kodlu. Thank you for the suggestion.
– Carlos Uribe
Nov 16 at 13:14