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







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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

















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







share|cite|improve this question
























  • 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















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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 16 at 13:13

























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




















  • 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












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.






share|cite|improve this answer





















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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.






    share|cite|improve this answer

























      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.






      share|cite|improve this answer























        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        user153012

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