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Which is faster? $n^log n$ or $(log n)^n$?
Scaleless (or self-similar) function: $sin ( log x)$How to calculate $limlimits_nto infty n^frac log nn^2$?At which parameter value $c>0$ do the number of solutions of $log(1+x^2)=x^c$ change?Degree of smoothness of real functions and Fourier seriesProving exponential is growing faster than polynomialWhy is a double exponential function faster than $x!$?How can we conclude that this series converges faster?Proving $n^epsilon > log(n)$ for sufficiently large $n$For which $p>0$ does $sum_n=3^inftyfraclog(n)n^p$ converge?Little oh notation question: $x^o(1)$
$begingroup$
Which is faster? $n^log n$ or $(log n)^n$?
Should I look at the base or at the exponent when I'm confronting two exponential functions?
real-analysis
edited Mar 10 at 20:36
Dr. Mathva
2,316526
asked Mar 10 at 20:31
user649882user649882
133
$endgroup$
|
show 3 more comments
$begingroup$
Which is faster? $n^log n$ or $(log n)^n$?
Should I look at the base or at the exponent when I'm confronting two exponential functions?
real-analysis
edited Mar 10 at 20:36
Dr. Mathva
2,316526
asked Mar 10 at 20:31
user649882user649882
133
$endgroup$
2
$begingroup$
Hint: write both as $e^textsomething$.
$endgroup$
– Wojowu
Mar 10 at 20:32
$begingroup$
@Wojowu in this case $n^logn$ is faster but I checked their graphics and $n^logn$ seems to be slower
$endgroup$
– user649882
Mar 10 at 20:33
$begingroup$
@user649882 Really? $n^log n = 2^log^2 n$, how would it be "faster" than $(log n)^n = 2^nlog n$?
$endgroup$
– Clement C.
Mar 10 at 20:36
$begingroup$
$overbrace fraclog(n)^2n ^frac1nlogleft(color#C00n^log(n)right)to0$ and $overbracelog(log(n))vphantomfrac(n)^2n^frac1nlogleft(color#C00log(n)^nright)toinfty$, so it would seem that $log(n)^n$ grows faster.
$endgroup$
– robjohn♦
Mar 10 at 20:38
$begingroup$
@ClementC. to see which function is the fastest between two on a graphic like on desmos.com/calculator you have to see which function is closer to the y axis as it grows, am I wrong?
$endgroup$
– user649882
Mar 10 at 20:42
|
show 3 more comments
$begingroup$
Which is faster? $n^log n$ or $(log n)^n$?
Should I look at the base or at the exponent when I'm confronting two exponential functions?
real-analysis
edited Mar 10 at 20:36
Dr. Mathva
2,316526
asked Mar 10 at 20:31
user649882user649882
133
$endgroup$
Which is faster? $n^log n$ or $(log n)^n$?
Should I look at the base or at the exponent when I'm confronting two exponential functions?
real-analysis
real-analysis
edited Mar 10 at 20:36
Dr. Mathva
2,316526
asked Mar 10 at 20:31
user649882user649882
133
edited Mar 10 at 20:36
Dr. Mathva
2,316526
asked Mar 10 at 20:31
user649882user649882
133
edited Mar 10 at 20:36
Dr. Mathva
2,316526
edited Mar 10 at 20:36
Dr. Mathva
2,316526
edited Mar 10 at 20:36
Dr. Mathva
2,316526
2,316526
asked Mar 10 at 20:31
user649882user649882
133
asked Mar 10 at 20:31
user649882user649882
133
asked Mar 10 at 20:31
user649882user649882
133
133
2
$begingroup$
Hint: write both as $e^textsomething$.
$endgroup$
– Wojowu
Mar 10 at 20:32
$begingroup$
@Wojowu in this case $n^logn$ is faster but I checked their graphics and $n^logn$ seems to be slower
$endgroup$
– user649882
Mar 10 at 20:33
$begingroup$
@user649882 Really? $n^log n = 2^log^2 n$, how would it be "faster" than $(log n)^n = 2^nlog n$?
$endgroup$
– Clement C.
Mar 10 at 20:36
$begingroup$
$overbrace fraclog(n)^2n ^frac1nlogleft(color#C00n^log(n)right)to0$ and $overbracelog(log(n))vphantomfrac(n)^2n^frac1nlogleft(color#C00log(n)^nright)toinfty$, so it would seem that $log(n)^n$ grows faster.
$endgroup$
– robjohn♦
Mar 10 at 20:38
$begingroup$
@ClementC. to see which function is the fastest between two on a graphic like on desmos.com/calculator you have to see which function is closer to the y axis as it grows, am I wrong?
$endgroup$
– user649882
Mar 10 at 20:42
|
show 3 more comments
2
$begingroup$
Hint: write both as $e^textsomething$.
$endgroup$
– Wojowu
Mar 10 at 20:32
$begingroup$
@Wojowu in this case $n^logn$ is faster but I checked their graphics and $n^logn$ seems to be slower
$endgroup$
– user649882
Mar 10 at 20:33
$begingroup$
@user649882 Really? $n^log n = 2^log^2 n$, how would it be "faster" than $(log n)^n = 2^nlog n$?
$endgroup$
– Clement C.
Mar 10 at 20:36
$begingroup$
$overbrace fraclog(n)^2n ^frac1nlogleft(color#C00n^log(n)right)to0$ and $overbracelog(log(n))vphantomfrac(n)^2n^frac1nlogleft(color#C00log(n)^nright)toinfty$, so it would seem that $log(n)^n$ grows faster.
$endgroup$
– robjohn♦
Mar 10 at 20:38
$begingroup$
@ClementC. to see which function is the fastest between two on a graphic like on desmos.com/calculator you have to see which function is closer to the y axis as it grows, am I wrong?
$endgroup$
– user649882
Mar 10 at 20:42
2
2
$begingroup$
Hint: write both as $e^textsomething$.
$endgroup$
– Wojowu
Mar 10 at 20:32
$begingroup$
Hint: write both as $e^textsomething$.
$endgroup$
– Wojowu
Mar 10 at 20:32
$begingroup$
@Wojowu in this case $n^logn$ is faster but I checked their graphics and $n^logn$ seems to be slower
$endgroup$
– user649882
Mar 10 at 20:33
$begingroup$
@Wojowu in this case $n^logn$ is faster but I checked their graphics and $n^logn$ seems to be slower
$endgroup$
– user649882
Mar 10 at 20:33
$begingroup$
@user649882 Really? $n^log n = 2^log^2 n$, how would it be "faster" than $(log n)^n = 2^nlog n$?
$endgroup$
– Clement C.
Mar 10 at 20:36
$begingroup$
@user649882 Really? $n^log n = 2^log^2 n$, how would it be "faster" than $(log n)^n = 2^nlog n$?
$endgroup$
– Clement C.
Mar 10 at 20:36
$begingroup$
$overbrace fraclog(n)^2n ^frac1nlogleft(color#C00n^log(n)right)to0$ and $overbracelog(log(n))vphantomfrac(n)^2n^frac1nlogleft(color#C00log(n)^nright)toinfty$, so it would seem that $log(n)^n$ grows faster.
$endgroup$
– robjohn♦
Mar 10 at 20:38
$begingroup$
$overbrace fraclog(n)^2n ^frac1nlogleft(color#C00n^log(n)right)to0$ and $overbracelog(log(n))vphantomfrac(n)^2n^frac1nlogleft(color#C00log(n)^nright)toinfty$, so it would seem that $log(n)^n$ grows faster.
$endgroup$
– robjohn♦
Mar 10 at 20:38
$begingroup$
@ClementC. to see which function is the fastest between two on a graphic like on desmos.com/calculator you have to see which function is closer to the y axis as it grows, am I wrong?
$endgroup$
– user649882
Mar 10 at 20:42
$begingroup$
@ClementC. to see which function is the fastest between two on a graphic like on desmos.com/calculator you have to see which function is closer to the y axis as it grows, am I wrong?
$endgroup$
– user649882
Mar 10 at 20:42
|
show 3 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Convert to a standard exponential:
$$n^log n=mathrm e^log^2n, quad (log n)^n=mathrm e^nlog(log n)$$
then check whether $;log^2n=obigl(nlog(log n)bigr):$ or $;nlog(log n)=obigl(log^2nbigr)$.
edited Mar 10 at 20:57
Robert Howard
2,2112933
answered Mar 10 at 20:50
BernardBernard
122k741116
$endgroup$
$begingroup$
is $n^n$ faster than $n^logn$?
$endgroup$
– user649882
Mar 10 at 20:53
$begingroup$
@user649882, Yes, because the exponent of $n^n$ is much larger than that of $n^log n$.
$endgroup$
– Sangchul Lee
Mar 10 at 21:08
add a comment |
$begingroup$
Hint:
begineqnarray*
fracln nln ln n le fracnln n.
endeqnarray*
answered Mar 10 at 20:38
Donald SplutterwitDonald Splutterwit
22.9k21446
$endgroup$
add a comment |
$begingroup$
If $2le a < b$, then $b^a < a^b$. So $n^log n < (log n)^n$ for all $n>7$.
(This bound isn't tight--it's actually true for all $n > 5$. But it should still be good enough.)
answered Mar 10 at 20:59
eyeballfrogeyeballfrog
6,446629
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Convert to a standard exponential:
$$n^log n=mathrm e^log^2n, quad (log n)^n=mathrm e^nlog(log n)$$
then check whether $;log^2n=obigl(nlog(log n)bigr):$ or $;nlog(log n)=obigl(log^2nbigr)$.
edited Mar 10 at 20:57
Robert Howard
2,2112933
answered Mar 10 at 20:50
BernardBernard
122k741116
$endgroup$
$begingroup$
is $n^n$ faster than $n^logn$?
$endgroup$
– user649882
Mar 10 at 20:53
$begingroup$
@user649882, Yes, because the exponent of $n^n$ is much larger than that of $n^log n$.
$endgroup$
– Sangchul Lee
Mar 10 at 21:08
add a comment |
$begingroup$
Convert to a standard exponential:
$$n^log n=mathrm e^log^2n, quad (log n)^n=mathrm e^nlog(log n)$$
then check whether $;log^2n=obigl(nlog(log n)bigr):$ or $;nlog(log n)=obigl(log^2nbigr)$.
edited Mar 10 at 20:57
Robert Howard
2,2112933
answered Mar 10 at 20:50
BernardBernard
122k741116
$endgroup$
$begingroup$
is $n^n$ faster than $n^logn$?
$endgroup$
– user649882
Mar 10 at 20:53
$begingroup$
@user649882, Yes, because the exponent of $n^n$ is much larger than that of $n^log n$.
$endgroup$
– Sangchul Lee
Mar 10 at 21:08
add a comment |
$begingroup$
Convert to a standard exponential:
$$n^log n=mathrm e^log^2n, quad (log n)^n=mathrm e^nlog(log n)$$
then check whether $;log^2n=obigl(nlog(log n)bigr):$ or $;nlog(log n)=obigl(log^2nbigr)$.
edited Mar 10 at 20:57
Robert Howard
2,2112933
answered Mar 10 at 20:50
BernardBernard
122k741116
$endgroup$
Convert to a standard exponential:
$$n^log n=mathrm e^log^2n, quad (log n)^n=mathrm e^nlog(log n)$$
then check whether $;log^2n=obigl(nlog(log n)bigr):$ or $;nlog(log n)=obigl(log^2nbigr)$.
edited Mar 10 at 20:57
Robert Howard
2,2112933
answered Mar 10 at 20:50
BernardBernard
122k741116
edited Mar 10 at 20:57
Robert Howard
2,2112933
edited Mar 10 at 20:57
Robert Howard
2,2112933
edited Mar 10 at 20:57
Robert Howard
2,2112933
2,2112933
answered Mar 10 at 20:50
BernardBernard
122k741116
answered Mar 10 at 20:50
BernardBernard
122k741116
answered Mar 10 at 20:50
BernardBernard
122k741116
122k741116
$begingroup$
is $n^n$ faster than $n^logn$?
$endgroup$
– user649882
Mar 10 at 20:53
$begingroup$
@user649882, Yes, because the exponent of $n^n$ is much larger than that of $n^log n$.
$endgroup$
– Sangchul Lee
Mar 10 at 21:08
add a comment |
$begingroup$
is $n^n$ faster than $n^logn$?
$endgroup$
– user649882
Mar 10 at 20:53
$begingroup$
@user649882, Yes, because the exponent of $n^n$ is much larger than that of $n^log n$.
$endgroup$
– Sangchul Lee
Mar 10 at 21:08
$begingroup$
is $n^n$ faster than $n^logn$?
$endgroup$
– user649882
Mar 10 at 20:53
$begingroup$
is $n^n$ faster than $n^logn$?
$endgroup$
– user649882
Mar 10 at 20:53
$begingroup$
@user649882, Yes, because the exponent of $n^n$ is much larger than that of $n^log n$.
$endgroup$
– Sangchul Lee
Mar 10 at 21:08
$begingroup$
@user649882, Yes, because the exponent of $n^n$ is much larger than that of $n^log n$.
$endgroup$
– Sangchul Lee
Mar 10 at 21:08
add a comment |
$begingroup$
Hint:
begineqnarray*
fracln nln ln n le fracnln n.
endeqnarray*
answered Mar 10 at 20:38
Donald SplutterwitDonald Splutterwit
22.9k21446
$endgroup$
add a comment |
$begingroup$
Hint:
begineqnarray*
fracln nln ln n le fracnln n.
endeqnarray*
answered Mar 10 at 20:38
Donald SplutterwitDonald Splutterwit
22.9k21446
$endgroup$
add a comment |
$begingroup$
Hint:
begineqnarray*
fracln nln ln n le fracnln n.
endeqnarray*
answered Mar 10 at 20:38
Donald SplutterwitDonald Splutterwit
22.9k21446
$endgroup$
Hint:
begineqnarray*
fracln nln ln n le fracnln n.
endeqnarray*
answered Mar 10 at 20:38
Donald SplutterwitDonald Splutterwit
22.9k21446
answered Mar 10 at 20:38
Donald SplutterwitDonald Splutterwit
22.9k21446
answered Mar 10 at 20:38
Donald SplutterwitDonald Splutterwit
22.9k21446
answered Mar 10 at 20:38
Donald SplutterwitDonald Splutterwit
22.9k21446
22.9k21446
add a comment |
add a comment |
$begingroup$
If $2le a < b$, then $b^a < a^b$. So $n^log n < (log n)^n$ for all $n>7$.
(This bound isn't tight--it's actually true for all $n > 5$. But it should still be good enough.)
answered Mar 10 at 20:59
eyeballfrogeyeballfrog
6,446629
$endgroup$
add a comment |
$begingroup$
If $2le a < b$, then $b^a < a^b$. So $n^log n < (log n)^n$ for all $n>7$.
(This bound isn't tight--it's actually true for all $n > 5$. But it should still be good enough.)
answered Mar 10 at 20:59
eyeballfrogeyeballfrog
6,446629
$endgroup$
add a comment |
$begingroup$
If $2le a < b$, then $b^a < a^b$. So $n^log n < (log n)^n$ for all $n>7$.
(This bound isn't tight--it's actually true for all $n > 5$. But it should still be good enough.)
answered Mar 10 at 20:59
eyeballfrogeyeballfrog
6,446629
$endgroup$
If $2le a < b$, then $b^a < a^b$. So $n^log n < (log n)^n$ for all $n>7$.
(This bound isn't tight--it's actually true for all $n > 5$. But it should still be good enough.)
answered Mar 10 at 20:59
eyeballfrogeyeballfrog
6,446629
answered Mar 10 at 20:59
eyeballfrogeyeballfrog
6,446629
answered Mar 10 at 20:59
eyeballfrogeyeballfrog
6,446629
answered Mar 10 at 20:59
eyeballfrogeyeballfrog
6,446629
6,446629
add a comment |
add a comment |
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2
$begingroup$
Hint: write both as $e^textsomething$.
$endgroup$
– Wojowu
Mar 10 at 20:32
$begingroup$
@Wojowu in this case $n^logn$ is faster but I checked their graphics and $n^logn$ seems to be slower
$endgroup$
– user649882
Mar 10 at 20:33
$begingroup$
@user649882 Really? $n^log n = 2^log^2 n$, how would it be "faster" than $(log n)^n = 2^nlog n$?
$endgroup$
– Clement C.
Mar 10 at 20:36
$begingroup$
$overbrace fraclog(n)^2n ^frac1nlogleft(color#C00n^log(n)right)to0$ and $overbracelog(log(n))vphantomfrac(n)^2n^frac1nlogleft(color#C00log(n)^nright)toinfty$, so it would seem that $log(n)^n$ grows faster.
$endgroup$
– robjohn♦
Mar 10 at 20:38
$begingroup$
@ClementC. to see which function is the fastest between two on a graphic like on desmos.com/calculator you have to see which function is closer to the y axis as it grows, am I wrong?
$endgroup$
– user649882
Mar 10 at 20:42