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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: $x^o(1)$

0

$begingroup$

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

share|cite|improve this question

edited Mar 10 at 20:36

Dr. Mathva

2,341526

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

0

$begingroup$

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

share|cite|improve this question

edited Mar 10 at 20:36

Dr. Mathva

2,341526

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

share|cite|improve this question

edited Mar 10 at 20:36

Dr. Mathva

2,341526

asked Mar 10 at 20:31

user649882user649882

133

$endgroup$

share|cite|improve this question

edited Mar 10 at 20:36

Dr. Mathva

2,341526

asked Mar 10 at 20:31

user649882user649882

133

share|cite|improve this question

edited Mar 10 at 20:36

Dr. Mathva

2,341526

asked Mar 10 at 20:31

user649882user649882

133

edited Mar 10 at 20:36

Dr. Mathva

2,341526

edited Mar 10 at 20:36

Dr. Mathva

2,341526

asked Mar 10 at 20:31

user649882user649882

133

asked Mar 10 at 20:31

user649882user649882

133

3 Answers
3

active

oldest

votes

1

$begingroup$

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 $;nlog(log n)=obigl(log^2nbigr)$.

share|cite|improve this answer

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 | 

0

$begingroup$

Hint:
begineqnarray*
fracln nln ln n le fracnln n.
endeqnarray*

share|cite|improve this answer

answered Mar 10 at 20:38

Donald SplutterwitDonald Splutterwit

22.9k21446

$endgroup$

add a comment | 

0

$begingroup$

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.)

share|cite|improve this answer

answered Mar 10 at 20:59

eyeballfrogeyeballfrog

6,446629

$endgroup$

add a comment | 

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1

$begingroup$

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 $;nlog(log n)=obigl(log^2nbigr)$.

share|cite|improve this answer

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 | 

1

$begingroup$

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 $;nlog(log n)=obigl(log^2nbigr)$.

share|cite|improve this answer

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 $;nlog(log n)=obigl(log^2nbigr)$.

share|cite|improve this answer

edited Mar 10 at 20:57

Robert Howard

2,2112933

answered Mar 10 at 20:50

BernardBernard

122k741116

$endgroup$

share|cite|improve this answer

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

answered Mar 10 at 20:50

BernardBernard

122k741116

answered Mar 10 at 20:50

BernardBernard

122k741116

0

$begingroup$

Hint:
begineqnarray*
fracln nln ln n le fracnln n.
endeqnarray*

share|cite|improve this answer

answered Mar 10 at 20:38

Donald SplutterwitDonald Splutterwit

22.9k21446

$endgroup$

add a comment | 

0

$begingroup$

Hint:
begineqnarray*
fracln nln ln n le fracnln n.
endeqnarray*

share|cite|improve this answer

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*

share|cite|improve this answer

answered Mar 10 at 20:38

Donald SplutterwitDonald Splutterwit

22.9k21446

$endgroup$

share|cite|improve this answer

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

0

$begingroup$

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.)

share|cite|improve this answer

answered Mar 10 at 20:59

eyeballfrogeyeballfrog

6,446629

$endgroup$

add a comment | 

0

$begingroup$

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.)

share|cite|improve this answer

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.)

share|cite|improve this answer

answered Mar 10 at 20:59

eyeballfrogeyeballfrog

6,446629

$endgroup$

share|cite|improve this answer

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