big O with different ordersBig-O Notation and AsymptoticsAsymptotic dominance for sum of roots.How to find the sum of Big-Oh's?Question about Big-O notationHow to rigorously simplify an expression with Big-Oh Composed within another Big-Oh.Big O notation - proofBig oh proof for a(n) using big oh hierarcyBig O Notation Review QuestionsBig Theta with Negative Coefficient ProblemProve/disprove if f1(n) = O(g1(n)) and f2(n) = O(g2(n)), then (f1(n))/(f2(n)) = O((g1(n))/(g2(n)))
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big O with different orders
Big-O Notation and AsymptoticsAsymptotic dominance for sum of roots.How to find the sum of Big-Oh's?Question about Big-O notationHow to rigorously simplify an expression with Big-Oh Composed within another Big-Oh.Big O notation - proofBig oh proof for a(n) using big oh hierarcyBig O Notation Review QuestionsBig Theta with Negative Coefficient ProblemProve/disprove if f1(n) = O(g1(n)) and f2(n) = O(g2(n)), then (f1(n))/(f2(n)) = O((g1(n))/(g2(n)))
$begingroup$
I'm trying to show the following and I'm not quite sure how:
$f_1(n) = n^n$ is $O(f_2(n) = 3^2^n)$ and $f_2(n) = 3^2^n = O(f_3(n) = 2^3^n) $
does anyone have an idea? calculating the limits is an option it's just that I don't get too far that way either.
thanks ahead.
asymptotics
asked Mar 29 at 1:21
lidor718lidor718
125
$endgroup$
add a comment |
$begingroup$
I'm trying to show the following and I'm not quite sure how:
$f_1(n) = n^n$ is $O(f_2(n) = 3^2^n)$ and $f_2(n) = 3^2^n = O(f_3(n) = 2^3^n) $
does anyone have an idea? calculating the limits is an option it's just that I don't get too far that way either.
thanks ahead.
asymptotics
asked Mar 29 at 1:21
lidor718lidor718
125
$endgroup$
add a comment |
$begingroup$
I'm trying to show the following and I'm not quite sure how:
$f_1(n) = n^n$ is $O(f_2(n) = 3^2^n)$ and $f_2(n) = 3^2^n = O(f_3(n) = 2^3^n) $
does anyone have an idea? calculating the limits is an option it's just that I don't get too far that way either.
thanks ahead.
asymptotics
asked Mar 29 at 1:21
lidor718lidor718
125
$endgroup$
I'm trying to show the following and I'm not quite sure how:
$f_1(n) = n^n$ is $O(f_2(n) = 3^2^n)$ and $f_2(n) = 3^2^n = O(f_3(n) = 2^3^n) $
does anyone have an idea? calculating the limits is an option it's just that I don't get too far that way either.
thanks ahead.
asymptotics
asymptotics
asked Mar 29 at 1:21
lidor718lidor718
125
asked Mar 29 at 1:21
lidor718lidor718
125
asked Mar 29 at 1:21
lidor718lidor718
125
asked Mar 29 at 1:21
lidor718lidor718
125
asked Mar 29 at 1:21
lidor718lidor718
125
125
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For the first,
it is enough to show that
$dfracn^n2^3^n
lt c$
for some $c > 0$
for all large enough $n$.
Since
$dfracn^n2^3^n
to 0$,
this is more than enough.
For the second,
when there are complicated exponents,
I take logs.
This makes
$f_2(n) = 3^2^n = O(f_3(n) = 2^3^n)
$
become
$ln f_2(n)
=2^n ln 3
$
and
$ln f_3(n)
=3^n ln 2
$
and clearly
$ln f_3(n)
$
is a lot larger.
answered Mar 29 at 1:51
marty cohenmarty cohen
74.9k549130
$endgroup$
$begingroup$
How can you show the first limit?
$endgroup$
– lidor718
Mar 29 at 1:54
$begingroup$
Take logs. Note that if $ln f(n)/ln g(n) to 0$ and $f(n), g(n) to infty$ then $f(n)/g(n) to 0$.
$endgroup$
– marty cohen
Mar 29 at 2:02
$begingroup$
Thanks for your help
$endgroup$
– lidor718
Mar 29 at 10:29
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For the first,
it is enough to show that
$dfracn^n2^3^n
lt c$
for some $c > 0$
for all large enough $n$.
Since
$dfracn^n2^3^n
to 0$,
this is more than enough.
For the second,
when there are complicated exponents,
I take logs.
This makes
$f_2(n) = 3^2^n = O(f_3(n) = 2^3^n)
$
become
$ln f_2(n)
=2^n ln 3
$
and
$ln f_3(n)
=3^n ln 2
$
and clearly
$ln f_3(n)
$
is a lot larger.
answered Mar 29 at 1:51
marty cohenmarty cohen
74.9k549130
$endgroup$
$begingroup$
How can you show the first limit?
$endgroup$
– lidor718
Mar 29 at 1:54
$begingroup$
Take logs. Note that if $ln f(n)/ln g(n) to 0$ and $f(n), g(n) to infty$ then $f(n)/g(n) to 0$.
$endgroup$
– marty cohen
Mar 29 at 2:02
$begingroup$
Thanks for your help
$endgroup$
– lidor718
Mar 29 at 10:29
add a comment |
$begingroup$
For the first,
it is enough to show that
$dfracn^n2^3^n
lt c$
for some $c > 0$
for all large enough $n$.
Since
$dfracn^n2^3^n
to 0$,
this is more than enough.
For the second,
when there are complicated exponents,
I take logs.
This makes
$f_2(n) = 3^2^n = O(f_3(n) = 2^3^n)
$
become
$ln f_2(n)
=2^n ln 3
$
and
$ln f_3(n)
=3^n ln 2
$
and clearly
$ln f_3(n)
$
is a lot larger.
answered Mar 29 at 1:51
marty cohenmarty cohen
74.9k549130
$endgroup$
$begingroup$
How can you show the first limit?
$endgroup$
– lidor718
Mar 29 at 1:54
$begingroup$
Take logs. Note that if $ln f(n)/ln g(n) to 0$ and $f(n), g(n) to infty$ then $f(n)/g(n) to 0$.
$endgroup$
– marty cohen
Mar 29 at 2:02
$begingroup$
Thanks for your help
$endgroup$
– lidor718
Mar 29 at 10:29
add a comment |
$begingroup$
For the first,
it is enough to show that
$dfracn^n2^3^n
lt c$
for some $c > 0$
for all large enough $n$.
Since
$dfracn^n2^3^n
to 0$,
this is more than enough.
For the second,
when there are complicated exponents,
I take logs.
This makes
$f_2(n) = 3^2^n = O(f_3(n) = 2^3^n)
$
become
$ln f_2(n)
=2^n ln 3
$
and
$ln f_3(n)
=3^n ln 2
$
and clearly
$ln f_3(n)
$
is a lot larger.
answered Mar 29 at 1:51
marty cohenmarty cohen
74.9k549130
$endgroup$
For the first,
it is enough to show that
$dfracn^n2^3^n
lt c$
for some $c > 0$
for all large enough $n$.
Since
$dfracn^n2^3^n
to 0$,
this is more than enough.
For the second,
when there are complicated exponents,
I take logs.
This makes
$f_2(n) = 3^2^n = O(f_3(n) = 2^3^n)
$
become
$ln f_2(n)
=2^n ln 3
$
and
$ln f_3(n)
=3^n ln 2
$
and clearly
$ln f_3(n)
$
is a lot larger.
answered Mar 29 at 1:51
marty cohenmarty cohen
74.9k549130
answered Mar 29 at 1:51
marty cohenmarty cohen
74.9k549130
answered Mar 29 at 1:51
marty cohenmarty cohen
74.9k549130
answered Mar 29 at 1:51
marty cohenmarty cohen
74.9k549130
74.9k549130
$begingroup$
How can you show the first limit?
$endgroup$
– lidor718
Mar 29 at 1:54
$begingroup$
Take logs. Note that if $ln f(n)/ln g(n) to 0$ and $f(n), g(n) to infty$ then $f(n)/g(n) to 0$.
$endgroup$
– marty cohen
Mar 29 at 2:02
$begingroup$
Thanks for your help
$endgroup$
– lidor718
Mar 29 at 10:29
add a comment |
$begingroup$
How can you show the first limit?
$endgroup$
– lidor718
Mar 29 at 1:54
$begingroup$
Take logs. Note that if $ln f(n)/ln g(n) to 0$ and $f(n), g(n) to infty$ then $f(n)/g(n) to 0$.
$endgroup$
– marty cohen
Mar 29 at 2:02
$begingroup$
Thanks for your help
$endgroup$
– lidor718
Mar 29 at 10:29
$begingroup$
How can you show the first limit?
$endgroup$
– lidor718
Mar 29 at 1:54
$begingroup$
How can you show the first limit?
$endgroup$
– lidor718
Mar 29 at 1:54
$begingroup$
Take logs. Note that if $ln f(n)/ln g(n) to 0$ and $f(n), g(n) to infty$ then $f(n)/g(n) to 0$.
$endgroup$
– marty cohen
Mar 29 at 2:02
$begingroup$
Take logs. Note that if $ln f(n)/ln g(n) to 0$ and $f(n), g(n) to infty$ then $f(n)/g(n) to 0$.
$endgroup$
– marty cohen
Mar 29 at 2:02
$begingroup$
Thanks for your help
$endgroup$
– lidor718
Mar 29 at 10:29
$begingroup$
Thanks for your help
$endgroup$
– lidor718
Mar 29 at 10:29
add a comment |
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