Theta complexity of the expression












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$begingroup$


Why is the complexity of expression is given as follows?




  1. $(1 + 2 + 3 + … + (n – 2) + (n – 1) + n)! = Theta((n^2)!)$

  2. $1 + 2 + 2^2 + … + 2^{n-2} + 2^{n-1} + 2^n = Theta(2^{n + 1})$

  3. $log(1) + log(2) + … + log(n – 1) + log(n) = Theta(n log(n))$






algorithms asymptotics







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$endgroup$












  • $begingroup$
    $$1+2+...+n=frac{ncdot(n+1)}2$$$$1+2+...+2^n=2^{n+1}-1$$$$log(1)+log(2)+...<log(n)+log(n)+...=nlog(n)$$$$log(1)+log(2)+...+log(n)geqlog(frac n2)+...+log(n)geqlog(frac n2)+log(frac n2)+...+log(frac n2)=frac n2log(frac n2)$$
    $endgroup$
    – Don Thousand
    Dec 19 '18 at 16:27












  • $begingroup$
    It's a little strange to see this presented as a "complexity." Certainly you are using Theta to describe the behavior of a function and not the space or time complexity of an algorithm.
    $endgroup$
    – Mason
    Dec 19 '18 at 16:30
















-1












$begingroup$


Why is the complexity of expression is given as follows?




  1. $(1 + 2 + 3 + … + (n – 2) + (n – 1) + n)! = Theta((n^2)!)$

  2. $1 + 2 + 2^2 + … + 2^{n-2} + 2^{n-1} + 2^n = Theta(2^{n + 1})$

  3. $log(1) + log(2) + … + log(n – 1) + log(n) = Theta(n log(n))$






algorithms asymptotics







share|cite|improve this question











$endgroup$












  • $begingroup$
    $$1+2+...+n=frac{ncdot(n+1)}2$$$$1+2+...+2^n=2^{n+1}-1$$$$log(1)+log(2)+...<log(n)+log(n)+...=nlog(n)$$$$log(1)+log(2)+...+log(n)geqlog(frac n2)+...+log(n)geqlog(frac n2)+log(frac n2)+...+log(frac n2)=frac n2log(frac n2)$$
    $endgroup$
    – Don Thousand
    Dec 19 '18 at 16:27












  • $begingroup$
    It's a little strange to see this presented as a "complexity." Certainly you are using Theta to describe the behavior of a function and not the space or time complexity of an algorithm.
    $endgroup$
    – Mason
    Dec 19 '18 at 16:30














-1












-1








-1





$begingroup$


Why is the complexity of expression is given as follows?




  1. $(1 + 2 + 3 + … + (n – 2) + (n – 1) + n)! = Theta((n^2)!)$

  2. $1 + 2 + 2^2 + … + 2^{n-2} + 2^{n-1} + 2^n = Theta(2^{n + 1})$

  3. $log(1) + log(2) + … + log(n – 1) + log(n) = Theta(n log(n))$






algorithms asymptotics







share|cite|improve this question











$endgroup$




Why is the complexity of expression is given as follows?




  1. $(1 + 2 + 3 + … + (n – 2) + (n – 1) + n)! = Theta((n^2)!)$

  2. $1 + 2 + 2^2 + … + 2^{n-2} + 2^{n-1} + 2^n = Theta(2^{n + 1})$

  3. $log(1) + log(2) + … + log(n – 1) + log(n) = Theta(n log(n))$






algorithms asymptotics




algorithms asymptotics






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edited Dec 19 '18 at 16:19









SmileyCraft

3,591517




3,591517










asked Dec 19 '18 at 16:09









JuneJune

122




122












  • $begingroup$
    $$1+2+...+n=frac{ncdot(n+1)}2$$$$1+2+...+2^n=2^{n+1}-1$$$$log(1)+log(2)+...<log(n)+log(n)+...=nlog(n)$$$$log(1)+log(2)+...+log(n)geqlog(frac n2)+...+log(n)geqlog(frac n2)+log(frac n2)+...+log(frac n2)=frac n2log(frac n2)$$
    $endgroup$
    – Don Thousand
    Dec 19 '18 at 16:27












  • $begingroup$
    It's a little strange to see this presented as a "complexity." Certainly you are using Theta to describe the behavior of a function and not the space or time complexity of an algorithm.
    $endgroup$
    – Mason
    Dec 19 '18 at 16:30


















  • $begingroup$
    $$1+2+...+n=frac{ncdot(n+1)}2$$$$1+2+...+2^n=2^{n+1}-1$$$$log(1)+log(2)+...<log(n)+log(n)+...=nlog(n)$$$$log(1)+log(2)+...+log(n)geqlog(frac n2)+...+log(n)geqlog(frac n2)+log(frac n2)+...+log(frac n2)=frac n2log(frac n2)$$
    $endgroup$
    – Don Thousand
    Dec 19 '18 at 16:27












  • $begingroup$
    It's a little strange to see this presented as a "complexity." Certainly you are using Theta to describe the behavior of a function and not the space or time complexity of an algorithm.
    $endgroup$
    – Mason
    Dec 19 '18 at 16:30
















$begingroup$
$$1+2+...+n=frac{ncdot(n+1)}2$$$$1+2+...+2^n=2^{n+1}-1$$$$log(1)+log(2)+...<log(n)+log(n)+...=nlog(n)$$$$log(1)+log(2)+...+log(n)geqlog(frac n2)+...+log(n)geqlog(frac n2)+log(frac n2)+...+log(frac n2)=frac n2log(frac n2)$$
$endgroup$
– Don Thousand
Dec 19 '18 at 16:27






$begingroup$
$$1+2+...+n=frac{ncdot(n+1)}2$$$$1+2+...+2^n=2^{n+1}-1$$$$log(1)+log(2)+...<log(n)+log(n)+...=nlog(n)$$$$log(1)+log(2)+...+log(n)geqlog(frac n2)+...+log(n)geqlog(frac n2)+log(frac n2)+...+log(frac n2)=frac n2log(frac n2)$$
$endgroup$
– Don Thousand
Dec 19 '18 at 16:27














$begingroup$
It's a little strange to see this presented as a "complexity." Certainly you are using Theta to describe the behavior of a function and not the space or time complexity of an algorithm.
$endgroup$
– Mason
Dec 19 '18 at 16:30




$begingroup$
It's a little strange to see this presented as a "complexity." Certainly you are using Theta to describe the behavior of a function and not the space or time complexity of an algorithm.
$endgroup$
– Mason
Dec 19 '18 at 16:30










1 Answer
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$begingroup$

The first one is actually false. We have $1+2+...+(n-1)+n=n(n+1)/2=Theta(n^2)$, however that does not imply that we can take a factorial on both sides. Consider the fact that $n!neqmathcal{O}((n-1)!)$, since the ratio is $n$ and can therefore be arbitrarily large.



For the second one, we use the geometric series formula $1+2+2^2+...+2^n=2^{n+1}-1$.



The third one is the most interesting one. I hope it is clear why $log(1)+log(2)+...+log(n)=mathcal{O}(nlog(n))$. For the other way around, we note that $log(1)+log(2)+...+log(n)geqlog(n/2)+log(n/2+1)+...+log(n)geq(n/2)log(n/2)$.






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    $begingroup$

    The first one is actually false. We have $1+2+...+(n-1)+n=n(n+1)/2=Theta(n^2)$, however that does not imply that we can take a factorial on both sides. Consider the fact that $n!neqmathcal{O}((n-1)!)$, since the ratio is $n$ and can therefore be arbitrarily large.



    For the second one, we use the geometric series formula $1+2+2^2+...+2^n=2^{n+1}-1$.



    The third one is the most interesting one. I hope it is clear why $log(1)+log(2)+...+log(n)=mathcal{O}(nlog(n))$. For the other way around, we note that $log(1)+log(2)+...+log(n)geqlog(n/2)+log(n/2+1)+...+log(n)geq(n/2)log(n/2)$.






    share|cite|improve this answer









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      0












      $begingroup$

      The first one is actually false. We have $1+2+...+(n-1)+n=n(n+1)/2=Theta(n^2)$, however that does not imply that we can take a factorial on both sides. Consider the fact that $n!neqmathcal{O}((n-1)!)$, since the ratio is $n$ and can therefore be arbitrarily large.



      For the second one, we use the geometric series formula $1+2+2^2+...+2^n=2^{n+1}-1$.



      The third one is the most interesting one. I hope it is clear why $log(1)+log(2)+...+log(n)=mathcal{O}(nlog(n))$. For the other way around, we note that $log(1)+log(2)+...+log(n)geqlog(n/2)+log(n/2+1)+...+log(n)geq(n/2)log(n/2)$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The first one is actually false. We have $1+2+...+(n-1)+n=n(n+1)/2=Theta(n^2)$, however that does not imply that we can take a factorial on both sides. Consider the fact that $n!neqmathcal{O}((n-1)!)$, since the ratio is $n$ and can therefore be arbitrarily large.



        For the second one, we use the geometric series formula $1+2+2^2+...+2^n=2^{n+1}-1$.



        The third one is the most interesting one. I hope it is clear why $log(1)+log(2)+...+log(n)=mathcal{O}(nlog(n))$. For the other way around, we note that $log(1)+log(2)+...+log(n)geqlog(n/2)+log(n/2+1)+...+log(n)geq(n/2)log(n/2)$.






        share|cite|improve this answer









        $endgroup$



        The first one is actually false. We have $1+2+...+(n-1)+n=n(n+1)/2=Theta(n^2)$, however that does not imply that we can take a factorial on both sides. Consider the fact that $n!neqmathcal{O}((n-1)!)$, since the ratio is $n$ and can therefore be arbitrarily large.



        For the second one, we use the geometric series formula $1+2+2^2+...+2^n=2^{n+1}-1$.



        The third one is the most interesting one. I hope it is clear why $log(1)+log(2)+...+log(n)=mathcal{O}(nlog(n))$. For the other way around, we note that $log(1)+log(2)+...+log(n)geqlog(n/2)+log(n/2+1)+...+log(n)geq(n/2)log(n/2)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 '18 at 16:29









        SmileyCraftSmileyCraft

        3,591517




        3,591517






























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