Prove whether $f(n)$ is $O$, $o$, $Omega$, $omega$ or $Theta$ of $g(n)$. The 2019 Stack Overflow Developer Survey Results Are Inpredicting runtime of $mathcalO(n log(n))$ algorithm, one “input size to runtime” pair is givenHow to prove that $lim_xto infty x/2^x = 0$Using Big-O to analyze an algorithm's effectivenessConverting a CFG to Chomsky Normal FormRecurrence Relation - $T(n) = 4T(n/2) + n^2log n$ - Solve using the tree methodProve whether $f(n)$ is $O$, $o$, $Omega$, $omega$ or $Theta$ of $g(n)$ for given $f$ and $g$Determining whether $g(n)$ is $O(f(n))$Prove that among any set of 34 different positive integers that are at most 99, there is always a pair of numbers that differ by at most 2.Proving $log^n(n)=omega(n^log(n)) $Is this solution correct? Prove whether $f(n)$ is $O$, $o$, $Omega$, $omega$ or $Theta$ of $g(n)$
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Prove whether $f(n)$ is $O$, $o$, $Omega$, $omega$ or $Theta$ of $g(n)$.
The 2019 Stack Overflow Developer Survey Results Are Inpredicting runtime of $mathcalO(n log(n))$ algorithm, one “input size to runtime” pair is givenHow to prove that $lim_xto infty x/2^x = 0$Using Big-O to analyze an algorithm's effectivenessConverting a CFG to Chomsky Normal FormRecurrence Relation - $T(n) = 4T(n/2) + n^2log n$ - Solve using the tree methodProve whether $f(n)$ is $O$, $o$, $Omega$, $omega$ or $Theta$ of $g(n)$ for given $f$ and $g$Determining whether $g(n)$ is $O(f(n))$Prove that among any set of 34 different positive integers that are at most 99, there is always a pair of numbers that differ by at most 2.Proving $log^n(n)=omega(n^log(n)) $Is this solution correct? Prove whether $f(n)$ is $O$, $o$, $Omega$, $omega$ or $Theta$ of $g(n)$
$begingroup$
$f(n) = e^n ln(n)$,
$g(n) = 2^n log(n)$
log can be assumed to be base $2$.
Alright so I put this in the form $f(n) / g(n)$ and then used L'Hôpital's rule giving me
$$ frac dfrace^nn +ln(n)e^n dfrac2^nnln(2) +n2^n-1log(n) $$
and I'm pretty sure if I did this right that it's still of the form infinity/infinity and will continue to be even if I differentiate again.
I think the solution has to do something with getting the exponent of $n2^n-1$ to 0 by differentiating over and over. But I'm not sure how to get a solution from this.
discrete-mathematics asymptotics computer-science
asked Mar 24 at 5:33
BrownieBrownie
3327
$endgroup$
add a comment |
$begingroup$
$f(n) = e^n ln(n)$,
$g(n) = 2^n log(n)$
log can be assumed to be base $2$.
Alright so I put this in the form $f(n) / g(n)$ and then used L'Hôpital's rule giving me
$$ frac dfrace^nn +ln(n)e^n dfrac2^nnln(2) +n2^n-1log(n) $$
and I'm pretty sure if I did this right that it's still of the form infinity/infinity and will continue to be even if I differentiate again.
I think the solution has to do something with getting the exponent of $n2^n-1$ to 0 by differentiating over and over. But I'm not sure how to get a solution from this.
discrete-mathematics asymptotics computer-science
asked Mar 24 at 5:33
BrownieBrownie
3327
$endgroup$
add a comment |
$begingroup$
$f(n) = e^n ln(n)$,
$g(n) = 2^n log(n)$
log can be assumed to be base $2$.
Alright so I put this in the form $f(n) / g(n)$ and then used L'Hôpital's rule giving me
$$ frac dfrace^nn +ln(n)e^n dfrac2^nnln(2) +n2^n-1log(n) $$
and I'm pretty sure if I did this right that it's still of the form infinity/infinity and will continue to be even if I differentiate again.
I think the solution has to do something with getting the exponent of $n2^n-1$ to 0 by differentiating over and over. But I'm not sure how to get a solution from this.
discrete-mathematics asymptotics computer-science
asked Mar 24 at 5:33
BrownieBrownie
3327
$endgroup$
$f(n) = e^n ln(n)$,
$g(n) = 2^n log(n)$
log can be assumed to be base $2$.
Alright so I put this in the form $f(n) / g(n)$ and then used L'Hôpital's rule giving me
$$ frac dfrace^nn +ln(n)e^n dfrac2^nnln(2) +n2^n-1log(n) $$
and I'm pretty sure if I did this right that it's still of the form infinity/infinity and will continue to be even if I differentiate again.
I think the solution has to do something with getting the exponent of $n2^n-1$ to 0 by differentiating over and over. But I'm not sure how to get a solution from this.
discrete-mathematics asymptotics computer-science
discrete-mathematics asymptotics computer-science
asked Mar 24 at 5:33
BrownieBrownie
3327
asked Mar 24 at 5:33
BrownieBrownie
3327
edited Mar 24 at 5:53
Brownie
asked Mar 24 at 5:33
BrownieBrownie
3327
asked Mar 24 at 5:33
BrownieBrownie
3327
asked Mar 24 at 5:33
BrownieBrownie
3327
3327
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We can compute
$$lim_n frace^nln n2^nlog n = lim_n bigg[left(frace2right)^n cdot fracln 2 ln nln nbigg] = infty.$$
Edit: To deal with the logarithms, you can use the identity
$$log_b x = fracln xln b.$$
answered Mar 24 at 5:56
Gary MoonGary Moon
921127
$endgroup$
$begingroup$
So I get that you split e^n and 2^n off separately. I'm not sure what math you did on ln(n) /log(n) though. Did you differentiate?
$endgroup$
– Brownie
Mar 24 at 6:08
1
$begingroup$
@Brownie I added an edit to (hopefully) address your question.
$endgroup$
– Gary Moon
Mar 24 at 6:12
add a comment |
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1 Answer
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1 Answer
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$begingroup$
We can compute
$$lim_n frace^nln n2^nlog n = lim_n bigg[left(frace2right)^n cdot fracln 2 ln nln nbigg] = infty.$$
Edit: To deal with the logarithms, you can use the identity
$$log_b x = fracln xln b.$$
answered Mar 24 at 5:56
Gary MoonGary Moon
921127
$endgroup$
$begingroup$
So I get that you split e^n and 2^n off separately. I'm not sure what math you did on ln(n) /log(n) though. Did you differentiate?
$endgroup$
– Brownie
Mar 24 at 6:08
1
$begingroup$
@Brownie I added an edit to (hopefully) address your question.
$endgroup$
– Gary Moon
Mar 24 at 6:12
add a comment |
$begingroup$
We can compute
$$lim_n frace^nln n2^nlog n = lim_n bigg[left(frace2right)^n cdot fracln 2 ln nln nbigg] = infty.$$
Edit: To deal with the logarithms, you can use the identity
$$log_b x = fracln xln b.$$
answered Mar 24 at 5:56
Gary MoonGary Moon
921127
$endgroup$
$begingroup$
So I get that you split e^n and 2^n off separately. I'm not sure what math you did on ln(n) /log(n) though. Did you differentiate?
$endgroup$
– Brownie
Mar 24 at 6:08
1
$begingroup$
@Brownie I added an edit to (hopefully) address your question.
$endgroup$
– Gary Moon
Mar 24 at 6:12
add a comment |
$begingroup$
We can compute
$$lim_n frace^nln n2^nlog n = lim_n bigg[left(frace2right)^n cdot fracln 2 ln nln nbigg] = infty.$$
Edit: To deal with the logarithms, you can use the identity
$$log_b x = fracln xln b.$$
answered Mar 24 at 5:56
Gary MoonGary Moon
921127
$endgroup$
We can compute
$$lim_n frace^nln n2^nlog n = lim_n bigg[left(frace2right)^n cdot fracln 2 ln nln nbigg] = infty.$$
Edit: To deal with the logarithms, you can use the identity
$$log_b x = fracln xln b.$$
answered Mar 24 at 5:56
Gary MoonGary Moon
921127
edited Mar 24 at 6:17
answered Mar 24 at 5:56
Gary MoonGary Moon
921127
answered Mar 24 at 5:56
Gary MoonGary Moon
921127
answered Mar 24 at 5:56
Gary MoonGary Moon
921127
921127
$begingroup$
So I get that you split e^n and 2^n off separately. I'm not sure what math you did on ln(n) /log(n) though. Did you differentiate?
$endgroup$
– Brownie
Mar 24 at 6:08
1
$begingroup$
@Brownie I added an edit to (hopefully) address your question.
$endgroup$
– Gary Moon
Mar 24 at 6:12
add a comment |
$begingroup$
So I get that you split e^n and 2^n off separately. I'm not sure what math you did on ln(n) /log(n) though. Did you differentiate?
$endgroup$
– Brownie
Mar 24 at 6:08
1
$begingroup$
@Brownie I added an edit to (hopefully) address your question.
$endgroup$
– Gary Moon
Mar 24 at 6:12
$begingroup$
So I get that you split e^n and 2^n off separately. I'm not sure what math you did on ln(n) /log(n) though. Did you differentiate?
$endgroup$
– Brownie
Mar 24 at 6:08
$begingroup$
So I get that you split e^n and 2^n off separately. I'm not sure what math you did on ln(n) /log(n) though. Did you differentiate?
$endgroup$
– Brownie
Mar 24 at 6:08
1
1
$begingroup$
@Brownie I added an edit to (hopefully) address your question.
$endgroup$
– Gary Moon
Mar 24 at 6:12
$begingroup$
@Brownie I added an edit to (hopefully) address your question.
$endgroup$
– Gary Moon
Mar 24 at 6:12
add a comment |
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