how to prove that $f(n)=n^3+nlog^2n$ = $theta(n^3)$?
i have $f(n)=n^3+nlog^2n$
and i was trieng to prove that $f(n)=n^3+nlog^2n$ = $theta(n^3)$. but i feel that i am doing it all wrong , which means i understand why the statent is true but i don't know if the way i proved it was enough..
here is what i did so far :
first i need to prove that :
$f(n)=n^3+nlog^2n$ = $O(n^3)$
We can choose constant C like this : if we devide the equation with $n^3$ then we get : $1+nlog^2n/n^3$ $<=$ $C $ but $nlog^2n/n^3$ <= 1 for every n so if we choose C to be C>=2 then this will be proved.
second i need to prove :$f(n)=n^3+nlog^2n$ = $pi(n^3)$
this i think doesn't need to be proved because if i choose any C<1 then for every n we will get : $n^3+nlog^2n$ >= $C(n^3)$
here is
algorithms asymptotics
edited Nov 20 at 18:25
Key Flex
7,52441232
asked Nov 20 at 18:18
jasmin
12
add a comment |
i have $f(n)=n^3+nlog^2n$
and i was trieng to prove that $f(n)=n^3+nlog^2n$ = $theta(n^3)$. but i feel that i am doing it all wrong , which means i understand why the statent is true but i don't know if the way i proved it was enough..
here is what i did so far :
first i need to prove that :
$f(n)=n^3+nlog^2n$ = $O(n^3)$
We can choose constant C like this : if we devide the equation with $n^3$ then we get : $1+nlog^2n/n^3$ $<=$ $C $ but $nlog^2n/n^3$ <= 1 for every n so if we choose C to be C>=2 then this will be proved.
second i need to prove :$f(n)=n^3+nlog^2n$ = $pi(n^3)$
this i think doesn't need to be proved because if i choose any C<1 then for every n we will get : $n^3+nlog^2n$ >= $C(n^3)$
here is
algorithms asymptotics
edited Nov 20 at 18:25
Key Flex
7,52441232
asked Nov 20 at 18:18
jasmin
12
What are $theta$ and pi?..... BTW the codes le and leq give $le$ while ge and geq give $ge$.
– DanielWainfleet
Nov 21 at 7:03
add a comment |
i have $f(n)=n^3+nlog^2n$
and i was trieng to prove that $f(n)=n^3+nlog^2n$ = $theta(n^3)$. but i feel that i am doing it all wrong , which means i understand why the statent is true but i don't know if the way i proved it was enough..
here is what i did so far :
first i need to prove that :
$f(n)=n^3+nlog^2n$ = $O(n^3)$
We can choose constant C like this : if we devide the equation with $n^3$ then we get : $1+nlog^2n/n^3$ $<=$ $C $ but $nlog^2n/n^3$ <= 1 for every n so if we choose C to be C>=2 then this will be proved.
second i need to prove :$f(n)=n^3+nlog^2n$ = $pi(n^3)$
this i think doesn't need to be proved because if i choose any C<1 then for every n we will get : $n^3+nlog^2n$ >= $C(n^3)$
here is
algorithms asymptotics
edited Nov 20 at 18:25
Key Flex
7,52441232
asked Nov 20 at 18:18
jasmin
12
i have $f(n)=n^3+nlog^2n$
and i was trieng to prove that $f(n)=n^3+nlog^2n$ = $theta(n^3)$. but i feel that i am doing it all wrong , which means i understand why the statent is true but i don't know if the way i proved it was enough..
here is what i did so far :
first i need to prove that :
$f(n)=n^3+nlog^2n$ = $O(n^3)$
We can choose constant C like this : if we devide the equation with $n^3$ then we get : $1+nlog^2n/n^3$ $<=$ $C $ but $nlog^2n/n^3$ <= 1 for every n so if we choose C to be C>=2 then this will be proved.
second i need to prove :$f(n)=n^3+nlog^2n$ = $pi(n^3)$
this i think doesn't need to be proved because if i choose any C<1 then for every n we will get : $n^3+nlog^2n$ >= $C(n^3)$
here is
algorithms asymptotics
algorithms asymptotics
edited Nov 20 at 18:25
Key Flex
7,52441232
asked Nov 20 at 18:18
jasmin
12
edited Nov 20 at 18:25
Key Flex
7,52441232
asked Nov 20 at 18:18
jasmin
12
edited Nov 20 at 18:25
Key Flex
7,52441232
edited Nov 20 at 18:25
Key Flex
7,52441232
edited Nov 20 at 18:25
Key Flex
7,52441232
7,52441232
asked Nov 20 at 18:18
jasmin
12
asked Nov 20 at 18:18
jasmin
12
asked Nov 20 at 18:18
jasmin
12
12
What are $theta$ and pi?..... BTW the codes le and leq give $le$ while ge and geq give $ge$.
– DanielWainfleet
Nov 21 at 7:03
add a comment |
What are $theta$ and pi?..... BTW the codes le and leq give $le$ while ge and geq give $ge$.
– DanielWainfleet
Nov 21 at 7:03
What are $theta$ and pi?..... BTW the codes le and leq give $le$ while ge and geq give $ge$.
– DanielWainfleet
Nov 21 at 7:03
What are $theta$ and pi?..... BTW the codes le and leq give $le$ while ge and geq give $ge$.
– DanielWainfleet
Nov 21 at 7:03
add a comment |
1 Answer
1
active
oldest
votes
Hint: $lim_{n to infty} frac{log n}{sqrt n} = (infty / infty) = (by L'Hospital) lim_{n to infty}frac{2sqrt n}{n} =0$
answered Nov 20 at 19:26
Vasya
3,3241515
i didn't really understand the hint .. is this a way toprove that nlog2n/n3 <= 1 ?
– jasmin
Nov 20 at 20:12
@jasmin: basically, $log n$ grows slower than $sqrt n$ so $log^2 n$ grows slower than $n$ which means $nlog^2 n/n^3< 1/n le 1$
– Vasya
Nov 20 at 20:35
To the proposer: For any positive $A,B$ we have $ lim_{xto infty}(ln x)^A/x^B=0.$
– DanielWainfleet
Nov 21 at 7:07
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
votes
Hint: $lim_{n to infty} frac{log n}{sqrt n} = (infty / infty) = (by L'Hospital) lim_{n to infty}frac{2sqrt n}{n} =0$
answered Nov 20 at 19:26
Vasya
3,3241515
i didn't really understand the hint .. is this a way toprove that nlog2n/n3 <= 1 ?
– jasmin
Nov 20 at 20:12
@jasmin: basically, $log n$ grows slower than $sqrt n$ so $log^2 n$ grows slower than $n$ which means $nlog^2 n/n^3< 1/n le 1$
– Vasya
Nov 20 at 20:35
To the proposer: For any positive $A,B$ we have $ lim_{xto infty}(ln x)^A/x^B=0.$
– DanielWainfleet
Nov 21 at 7:07
add a comment |
Hint: $lim_{n to infty} frac{log n}{sqrt n} = (infty / infty) = (by L'Hospital) lim_{n to infty}frac{2sqrt n}{n} =0$
answered Nov 20 at 19:26
Vasya
3,3241515
i didn't really understand the hint .. is this a way toprove that nlog2n/n3 <= 1 ?
– jasmin
Nov 20 at 20:12
@jasmin: basically, $log n$ grows slower than $sqrt n$ so $log^2 n$ grows slower than $n$ which means $nlog^2 n/n^3< 1/n le 1$
– Vasya
Nov 20 at 20:35
To the proposer: For any positive $A,B$ we have $ lim_{xto infty}(ln x)^A/x^B=0.$
– DanielWainfleet
Nov 21 at 7:07
add a comment |
Hint: $lim_{n to infty} frac{log n}{sqrt n} = (infty / infty) = (by L'Hospital) lim_{n to infty}frac{2sqrt n}{n} =0$
answered Nov 20 at 19:26
Vasya
3,3241515
Hint: $lim_{n to infty} frac{log n}{sqrt n} = (infty / infty) = (by L'Hospital) lim_{n to infty}frac{2sqrt n}{n} =0$
answered Nov 20 at 19:26
Vasya
3,3241515
answered Nov 20 at 19:26
Vasya
3,3241515
answered Nov 20 at 19:26
Vasya
3,3241515
answered Nov 20 at 19:26
Vasya
3,3241515
3,3241515
i didn't really understand the hint .. is this a way toprove that nlog2n/n3 <= 1 ?
– jasmin
Nov 20 at 20:12
@jasmin: basically, $log n$ grows slower than $sqrt n$ so $log^2 n$ grows slower than $n$ which means $nlog^2 n/n^3< 1/n le 1$
– Vasya
Nov 20 at 20:35
To the proposer: For any positive $A,B$ we have $ lim_{xto infty}(ln x)^A/x^B=0.$
– DanielWainfleet
Nov 21 at 7:07
add a comment |
i didn't really understand the hint .. is this a way toprove that nlog2n/n3 <= 1 ?
– jasmin
Nov 20 at 20:12
@jasmin: basically, $log n$ grows slower than $sqrt n$ so $log^2 n$ grows slower than $n$ which means $nlog^2 n/n^3< 1/n le 1$
– Vasya
Nov 20 at 20:35
To the proposer: For any positive $A,B$ we have $ lim_{xto infty}(ln x)^A/x^B=0.$
– DanielWainfleet
Nov 21 at 7:07
i didn't really understand the hint .. is this a way toprove that nlog2n/n3 <= 1 ?
– jasmin
Nov 20 at 20:12
i didn't really understand the hint .. is this a way toprove that nlog2n/n3 <= 1 ?
– jasmin
Nov 20 at 20:12
@jasmin: basically, $log n$ grows slower than $sqrt n$ so $log^2 n$ grows slower than $n$ which means $nlog^2 n/n^3< 1/n le 1$
– Vasya
Nov 20 at 20:35
@jasmin: basically, $log n$ grows slower than $sqrt n$ so $log^2 n$ grows slower than $n$ which means $nlog^2 n/n^3< 1/n le 1$
– Vasya
Nov 20 at 20:35
To the proposer: For any positive $A,B$ we have $ lim_{xto infty}(ln x)^A/x^B=0.$
– DanielWainfleet
Nov 21 at 7:07
To the proposer: For any positive $A,B$ we have $ lim_{xto infty}(ln x)^A/x^B=0.$
– DanielWainfleet
Nov 21 at 7:07
add a comment |
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What are $theta$ and pi?..... BTW the codes le and leq give $le$ while ge and geq give $ge$.
– DanielWainfleet
Nov 21 at 7:03