An approximation by Stirling's formulaLet 0 < α < 1 be a real number and...

Joel French

Joel French

Answered

2022-07-08

An approximation by Stirling's formula
Let 0 < α < 1 be a real number and αn be α n integer. I want to prove the following fact
( n α n ) = 1 + o ( 1 ) 2 π α ( 1 α ) n 2 n H ( α ) ,
where H ( α ) = α log 2 α ( 1 α ) log 2 ( 1 α ). To do this, I've used the Stirling's formula for the factorial:
n ! = ( n e ) n 2 π n e r ( n ) ,,
where 1 / ( 12 n + 1 ) < r ( n ) < 1 / 12 n.
What I've tried:
( n α n ) = n ! ( α n ) ! ( ( 1 α ) n ) ! = ( n e ) n 2 π n e r ( n ) ( α n e ) α n 2 π α n e r ( α n ) ( ( 1 α ) n e ) ( 1 α ) n 2 π ( 1 α ) n e r ( ( 1 α ) n ) e r ( n ) r ( α n ) r ( ( 1 α ) n ) ( α α ( 1 α ) ( 1 α ) ) n 2 π α ( 1 α ) n = e r ( n ) r ( α n ) r ( ( 1 α ) n ) 2 π α ( 1 α ) n 2 n H ( α ) .
Now how can I show that e r ( n ) r ( α n ) r ( ( 1 α ) n ) = 1 + o ( 1 )?

Answer & Explanation

poquetahr

poquetahr

Expert

2022-07-09Added 18 answers

Step 1
You are only having difficulty because the version of Stirling you're using is too precise. Instead, use n ! = ( 1 + o ( 1 ) ) 2 π n ( n e ) n - replacing all of your e r ( n ) factors by 1 + o ( 1 ).
Step 2
The step you're stuck on, where you're trying to show that e r ( n ) r ( α n ) r ( ( 1 α ) n ) = 1 + o ( 1 ), becomes 1 + o ( 1 ) ( 1 + o ( 1 ) ) ( 1 + o ( 1 ) ) = 1 + o ( 1 ), which is straightforward.
(Technically, the 1 + o ( 1 ) factor we get from Stirling's approximation is 1 + o n ( 1 ), so we also have to observe that 1 + o α n ( 1 ) and 1 + o ( 1 α ) n ( 1 ) are the same asymptotically. This is fine as long as α is a constant, or even more generally than that.)
Waldronjw

Waldronjw

Expert

2022-07-10Added 2 answers

Step 1
Since f ( n ) = o ( 1 ) just means f ( n ) 1 0 as n (i.e. f ( n ) = 1 + o ( 1 ) means f ( n ) 1), our goal is to show that e r ( n ) r ( α n ) r ( ( 1 α ) n ) 1, which can be done by showing r ( n ) r ( α n ) r ( ( 1 α ) n ) 0. We can use the bounds on r(n) to show that r ( n ) r ( α n ) r ( ( 1 α ) n ) < 1 12 n 1 12 α n + 1 1 12 ( 1 α ) n + 1 n 0 and r ( n ) r ( α n ) r ( ( 1 α ) n ) > 1 12 n + 1 1 12 α n 1 12 ( 1 α ) n n 0
Step 2
So since r ( n ) r ( α n ) r ( ( 1 α ) n ) is squeezed between 0 on both sides asymptotically, indeed we have r ( n ) r ( α n ) r ( ( 1 α ) n ) 0, and we are done.

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