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Approximating $log(n!)$

Stirling's formula: proof?Intuition behind logarithm inequality: $1 - frac1x leq log x leq x-1$How do I bound $n log n choose log nfrac1n ^ log n$ tightly?Unlike Jensen's Inequality, can we upper bound $log sum_iu_i exp(x_i)$?a log inequalityHow to show $fraccn leq log(1+fraccn-c)$Prove: if $n = frack^k-1k-1$ then $k=Omegaleft(fraclognloglognright)$Inequalities from Taylor expansions of $log$ functionsProve that $sum_i=1^i=n leftlfloor fracni rightrfloor = Theta(n log n)$$a_n$ is a sequence of nonnegative real numbers satisfying $a_n+1 leq a_n + frac(-1)^nn$. Prove convergence.Lower bound on $log(1-x)$ for $x$ near $0$

1

$begingroup$

Prove that $nlog(n)-n+1 leq log(n!) leq (n+1)log(n+1)-(n+1)-(2log(2)-2)$.

Approximating $int_1^nlog(x)dx$ using a riemman sum at right endpoints is $log(2) + log(3)+dots +log(n) = log(n!)$ which overestimates so the claim for the lower bound follows. For the upper bound we integrate $int_2^n+1log(x)dx$.
I can prove the upper bound if I can show that $log(1+frac1n) leq frac1n$. I know that the Taylor expansion of $log(1+frac1n)$ is an alternating series but I am not sure how to show this.
I hope someone can provide some guidance and also clarify why is
$int_1^nlog(x)dx leq int_2^n+1log(x)dx$. Is this because $log(x)$ is an increasing function?

calculus combinatorics inequality logarithms

share|cite|improve this question

edited Mar 29 at 9:40

Kenta S

1,4821518

asked Mar 31 '18 at 6:07

rachelhowardrachelhoward

748

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Related: math.stackexchange.com/a/1409131/44121
  $endgroup$
  – Jack D'Aurizio
  Mar 31 '18 at 19:22
  

  
  
  
  

add a comment | 

1

$begingroup$

Prove that $nlog(n)-n+1 leq log(n!) leq (n+1)log(n+1)-(n+1)-(2log(2)-2)$.

Approximating $int_1^nlog(x)dx$ using a riemman sum at right endpoints is $log(2) + log(3)+dots +log(n) = log(n!)$ which overestimates so the claim for the lower bound follows. For the upper bound we integrate $int_2^n+1log(x)dx$.
I can prove the upper bound if I can show that $log(1+frac1n) leq frac1n$. I know that the Taylor expansion of $log(1+frac1n)$ is an alternating series but I am not sure how to show this.
I hope someone can provide some guidance and also clarify why is
$int_1^nlog(x)dx leq int_2^n+1log(x)dx$. Is this because $log(x)$ is an increasing function?

calculus combinatorics inequality logarithms

share|cite|improve this question

edited Mar 29 at 9:40

Kenta S

1,4821518

asked Mar 31 '18 at 6:07

rachelhowardrachelhoward

748

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Related: math.stackexchange.com/a/1409131/44121
  $endgroup$
  – Jack D'Aurizio
  Mar 31 '18 at 19:22
  

  
  
  
  

add a comment | 

$begingroup$

Prove that $nlog(n)-n+1 leq log(n!) leq (n+1)log(n+1)-(n+1)-(2log(2)-2)$.

Approximating $int_1^nlog(x)dx$ using a riemman sum at right endpoints is $log(2) + log(3)+dots +log(n) = log(n!)$ which overestimates so the claim for the lower bound follows. For the upper bound we integrate $int_2^n+1log(x)dx$.
I can prove the upper bound if I can show that $log(1+frac1n) leq frac1n$. I know that the Taylor expansion of $log(1+frac1n)$ is an alternating series but I am not sure how to show this.
I hope someone can provide some guidance and also clarify why is
$int_1^nlog(x)dx leq int_2^n+1log(x)dx$. Is this because $log(x)$ is an increasing function?

calculus combinatorics inequality logarithms

share|cite|improve this question

edited Mar 29 at 9:40

Kenta S

1,4821518

asked Mar 31 '18 at 6:07

rachelhowardrachelhoward

748

$endgroup$

share|cite|improve this question

edited Mar 29 at 9:40

Kenta S

1,4821518

asked Mar 31 '18 at 6:07

rachelhowardrachelhoward

748

share|cite|improve this question

edited Mar 29 at 9:40

Kenta S

1,4821518

asked Mar 31 '18 at 6:07

rachelhowardrachelhoward

748

edited Mar 29 at 9:40

Kenta S

1,4821518

edited Mar 29 at 9:40

Kenta S

1,4821518

asked Mar 31 '18 at 6:07

rachelhowardrachelhoward

748

asked Mar 31 '18 at 6:07

rachelhowardrachelhoward

748

1 Answer
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active

oldest

votes

1

$begingroup$

The fact that $log(1+x) leq x$ for $x>-1$ is a consequence of just concavity, no higher than second derivatives are required. This is a general principle: concave functions lie below their tangent lines.

In general for an increasing function $f$ and a fixed $b>0$, $F(a)=int_a^a+b f(x) dx$ is an increasing function of $a$ (when it makes sense). Thus in your case it is important that the lengths of the two intervals of integration are the same.

share|cite|improve this answer

edited Mar 31 '18 at 19:02

answered Mar 31 '18 at 17:18

IanIan

68.9k25392

$endgroup$

add a comment | 

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1

$begingroup$

The fact that $log(1+x) leq x$ for $x>-1$ is a consequence of just concavity, no higher than second derivatives are required. This is a general principle: concave functions lie below their tangent lines.

In general for an increasing function $f$ and a fixed $b>0$, $F(a)=int_a^a+b f(x) dx$ is an increasing function of $a$ (when it makes sense). Thus in your case it is important that the lengths of the two intervals of integration are the same.

share|cite|improve this answer

edited Mar 31 '18 at 19:02

answered Mar 31 '18 at 17:18

IanIan

68.9k25392

$endgroup$

add a comment | 

1

$begingroup$

The fact that $log(1+x) leq x$ for $x>-1$ is a consequence of just concavity, no higher than second derivatives are required. This is a general principle: concave functions lie below their tangent lines.

In general for an increasing function $f$ and a fixed $b>0$, $F(a)=int_a^a+b f(x) dx$ is an increasing function of $a$ (when it makes sense). Thus in your case it is important that the lengths of the two intervals of integration are the same.

share|cite|improve this answer

edited Mar 31 '18 at 19:02

answered Mar 31 '18 at 17:18

IanIan

68.9k25392

$endgroup$

add a comment | 

$begingroup$

The fact that $log(1+x) leq x$ for $x>-1$ is a consequence of just concavity, no higher than second derivatives are required. This is a general principle: concave functions lie below their tangent lines.

In general for an increasing function $f$ and a fixed $b>0$, $F(a)=int_a^a+b f(x) dx$ is an increasing function of $a$ (when it makes sense). Thus in your case it is important that the lengths of the two intervals of integration are the same.

share|cite|improve this answer

edited Mar 31 '18 at 19:02

answered Mar 31 '18 at 17:18

IanIan

68.9k25392

$endgroup$

share|cite|improve this answer

edited Mar 31 '18 at 19:02

answered Mar 31 '18 at 17:18

IanIan

68.9k25392

answered Mar 31 '18 at 17:18

IanIan

68.9k25392

answered Mar 31 '18 at 17:18

IanIan

68.9k25392