comparing Asymptotic functions $ 3^{{2}^{n}}$ and $n! times n^{3}$
$begingroup$
$f(n)= 3^{{2}^{n}}$
$g(n)=n! times n^{3}$
$text{which is asymptotically greater} ?$
My Approach-:
$$f(n)=e^{log 4^{{3}^{n}} }$$
$$g(n)=e^{log n! times n^{3}}$$
i am comparing just the power of the functions
i.e
$$log 3^{{2}^{n}}$$ and $$log n! times n^{3}$$
$$f_1(n)=log 3^{{2}^{n}}=2^{n} log 3$$
$$f_2(n)=log (n! times n^{3})=log (n!) +log (n^{3}) =n log n+3 log n$$
Again we can express $f_1(n) $ and $f_2(n)$ as
$f_1(n)=e^{log( 2^n log 3)}$ and $f_2(n)=e^{log( n log n)}$
Again comparing the powers,
$f_{11}(n)=log (2^n log 3)=n + log log 3$
$f_{12}=log( n log n)=log n + log log n $
here $f_{11}(n) >f_{12}(n) Rightarrow f_{1}(n)>f_{12}(n) Rightarrow f(n)>g(n) $
here $>$ means asymptotically greater
Am i correct?
functions algorithms asymptotics
asked Dec 1 '18 at 16:35
viratvirat
18118
$endgroup$
add a comment |
$begingroup$
$f(n)= 3^{{2}^{n}}$
$g(n)=n! times n^{3}$
$text{which is asymptotically greater} ?$
My Approach-:
$$f(n)=e^{log 4^{{3}^{n}} }$$
$$g(n)=e^{log n! times n^{3}}$$
i am comparing just the power of the functions
i.e
$$log 3^{{2}^{n}}$$ and $$log n! times n^{3}$$
$$f_1(n)=log 3^{{2}^{n}}=2^{n} log 3$$
$$f_2(n)=log (n! times n^{3})=log (n!) +log (n^{3}) =n log n+3 log n$$
Again we can express $f_1(n) $ and $f_2(n)$ as
$f_1(n)=e^{log( 2^n log 3)}$ and $f_2(n)=e^{log( n log n)}$
Again comparing the powers,
$f_{11}(n)=log (2^n log 3)=n + log log 3$
$f_{12}=log( n log n)=log n + log log n $
here $f_{11}(n) >f_{12}(n) Rightarrow f_{1}(n)>f_{12}(n) Rightarrow f(n)>g(n) $
here $>$ means asymptotically greater
Am i correct?
functions algorithms asymptotics
asked Dec 1 '18 at 16:35
viratvirat
18118
$endgroup$
$begingroup$
Final result is correct, the way you got to that result might be debatable
$endgroup$
– Jakobian
Dec 1 '18 at 16:37
add a comment |
$begingroup$
$f(n)= 3^{{2}^{n}}$
$g(n)=n! times n^{3}$
$text{which is asymptotically greater} ?$
My Approach-:
$$f(n)=e^{log 4^{{3}^{n}} }$$
$$g(n)=e^{log n! times n^{3}}$$
i am comparing just the power of the functions
i.e
$$log 3^{{2}^{n}}$$ and $$log n! times n^{3}$$
$$f_1(n)=log 3^{{2}^{n}}=2^{n} log 3$$
$$f_2(n)=log (n! times n^{3})=log (n!) +log (n^{3}) =n log n+3 log n$$
Again we can express $f_1(n) $ and $f_2(n)$ as
$f_1(n)=e^{log( 2^n log 3)}$ and $f_2(n)=e^{log( n log n)}$
Again comparing the powers,
$f_{11}(n)=log (2^n log 3)=n + log log 3$
$f_{12}=log( n log n)=log n + log log n $
here $f_{11}(n) >f_{12}(n) Rightarrow f_{1}(n)>f_{12}(n) Rightarrow f(n)>g(n) $
here $>$ means asymptotically greater
Am i correct?
functions algorithms asymptotics
asked Dec 1 '18 at 16:35
viratvirat
18118
$endgroup$
$f(n)= 3^{{2}^{n}}$
$g(n)=n! times n^{3}$
$text{which is asymptotically greater} ?$
My Approach-:
$$f(n)=e^{log 4^{{3}^{n}} }$$
$$g(n)=e^{log n! times n^{3}}$$
i am comparing just the power of the functions
i.e
$$log 3^{{2}^{n}}$$ and $$log n! times n^{3}$$
$$f_1(n)=log 3^{{2}^{n}}=2^{n} log 3$$
$$f_2(n)=log (n! times n^{3})=log (n!) +log (n^{3}) =n log n+3 log n$$
Again we can express $f_1(n) $ and $f_2(n)$ as
$f_1(n)=e^{log( 2^n log 3)}$ and $f_2(n)=e^{log( n log n)}$
Again comparing the powers,
$f_{11}(n)=log (2^n log 3)=n + log log 3$
$f_{12}=log( n log n)=log n + log log n $
here $f_{11}(n) >f_{12}(n) Rightarrow f_{1}(n)>f_{12}(n) Rightarrow f(n)>g(n) $
here $>$ means asymptotically greater
Am i correct?
functions algorithms asymptotics
functions algorithms asymptotics
asked Dec 1 '18 at 16:35
viratvirat
18118
asked Dec 1 '18 at 16:35
viratvirat
18118
asked Dec 1 '18 at 16:35
viratvirat
18118
asked Dec 1 '18 at 16:35
viratvirat
18118
asked Dec 1 '18 at 16:35
viratvirat
18118
18118
$begingroup$
Final result is correct, the way you got to that result might be debatable
$endgroup$
– Jakobian
Dec 1 '18 at 16:37
add a comment |
$begingroup$
Final result is correct, the way you got to that result might be debatable
$endgroup$
– Jakobian
Dec 1 '18 at 16:37
$begingroup$
Final result is correct, the way you got to that result might be debatable
$endgroup$
– Jakobian
Dec 1 '18 at 16:37
$begingroup$
Final result is correct, the way you got to that result might be debatable
$endgroup$
– Jakobian
Dec 1 '18 at 16:37
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
You can do it easily by applying $log$ function to $f(n)$ and $g(n)$.
$$log(f(n)) = 2^nlog(3), log(g(n)) = log(n!) + 3log(n)$$
As we know $log(n!) = Theta(nlog(n))$, we have:
$$g(n) = Theta(nlog(n)), f(n) = Theta(2^n)$$
Hence, $f(n)$ is greater than $g(n)$.
answered Dec 1 '18 at 16:51
OmGOmG
2,407722
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
You can do it easily by applying $log$ function to $f(n)$ and $g(n)$.
$$log(f(n)) = 2^nlog(3), log(g(n)) = log(n!) + 3log(n)$$
As we know $log(n!) = Theta(nlog(n))$, we have:
$$g(n) = Theta(nlog(n)), f(n) = Theta(2^n)$$
Hence, $f(n)$ is greater than $g(n)$.
answered Dec 1 '18 at 16:51
OmGOmG
2,407722
$endgroup$
add a comment |
$begingroup$
You can do it easily by applying $log$ function to $f(n)$ and $g(n)$.
$$log(f(n)) = 2^nlog(3), log(g(n)) = log(n!) + 3log(n)$$
As we know $log(n!) = Theta(nlog(n))$, we have:
$$g(n) = Theta(nlog(n)), f(n) = Theta(2^n)$$
Hence, $f(n)$ is greater than $g(n)$.
answered Dec 1 '18 at 16:51
OmGOmG
2,407722
$endgroup$
add a comment |
$begingroup$
You can do it easily by applying $log$ function to $f(n)$ and $g(n)$.
$$log(f(n)) = 2^nlog(3), log(g(n)) = log(n!) + 3log(n)$$
As we know $log(n!) = Theta(nlog(n))$, we have:
$$g(n) = Theta(nlog(n)), f(n) = Theta(2^n)$$
Hence, $f(n)$ is greater than $g(n)$.
answered Dec 1 '18 at 16:51
OmGOmG
2,407722
$endgroup$
You can do it easily by applying $log$ function to $f(n)$ and $g(n)$.
$$log(f(n)) = 2^nlog(3), log(g(n)) = log(n!) + 3log(n)$$
As we know $log(n!) = Theta(nlog(n))$, we have:
$$g(n) = Theta(nlog(n)), f(n) = Theta(2^n)$$
Hence, $f(n)$ is greater than $g(n)$.
answered Dec 1 '18 at 16:51
OmGOmG
2,407722
answered Dec 1 '18 at 16:51
OmGOmG
2,407722
answered Dec 1 '18 at 16:51
OmGOmG
2,407722
answered Dec 1 '18 at 16:51
OmGOmG
2,407722
2,407722
add a comment |
add a comment |
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$begingroup$
Final result is correct, the way you got to that result might be debatable
$endgroup$
– Jakobian
Dec 1 '18 at 16:37