Stirling’s Formula: an Approximation of the Factorial · • Stirling’s formula. ... (a,b) if the function Is a monotonically increasing function of on the interval 1 2 1 2

Stirling’s Formula: an Approximation of the Factorial

Eric Gilbertson

Outline

• Introduction of formula• Convex and log convex functions• The gamma function• Stirling’s formula

Outline


Introduction of Formula

In the early 18th century James Stirling proved the following formula:

For some

nnn enn 12/2/12! θπ +−+=10 << θ

Introduction of Formula

In the early 18th century James Stirling proved the following formula:

For some

This means that as

nnn enn 12/2/12! θπ +−+=10 << θ

nn enn −+ 2/12~! π

∞→n

Outline


Convex Functions

A function f(x) is called convex on the interval (a,b) if the function

Is a monotonically increasing function of on the interval

21

2121

)()(),(xx

xfxfxx−−

=φ

1x

2x

A function f(x) is called convex on the interval (a,b) if the function

Is a monotonically increasing function of on the interval

Ex: is convexis not convex on the interval (-1,0)

Convex Functions

21

2121

)()(),(xx

xfxfxx−−

=φ

1x

3x

Convex Functions - Properties

If f(x) and g(x) are convex then f(x)+g(x) is also convex

If f(x) is twice differentiable and the second derivative of f is positive on an interval, then f(x) is convex on the interval

Log Convex Functions

A positive-valued function f(x) is called log convex on the interval (a,b) if the function is convex on the interval.

)(ln xf

Log Convex Functions

A positive-valued function f(x) is called log convex on the interval (a,b) if the function is convex on the interval.

Ex: is log convex

)(ln xf

2xe

Log Convex Functions -Properties

The product of log convex functions is log convex


The product of log convex functions is log convexIf f(t,x) is a log convex function twice differentiable in

x, for t in the interval [a,b] and x in any interval then

is a log convex function of x∫b

adtxtf ),(


The product of log convex functions is log convexIf f(t,x) is a log convex function twice differentiable in

x, for t in the interval [a,b] and x in any interval then

is a log convex function of x

Ex: is log convexdtteb

a

xt∫ −− 1

∫b

adtxtf ),(

Outline


The Gamma Function

An extension of the factorial to all positive real numbers is the gamma function where

∫∞ −−=Γ

0

1)( dttex xt

The Gamma Function

An extension of the factorial to all positive real numbers is the gamma function where

Using integration by parts, for integer n

∫∞ −−=Γ

0

1)( dttex xt

)!1()( −=Γ nn

The Gamma Function -Properties

• The Gamma function is log convexthe integrand is twice differentiable on ),0[ ∞

The Gamma Function -Properties

• The Gamma function is log convexthe integrand is twice differentiable on

• And

),0[ ∞

1)1(0

==Γ ∫∞

− dte t

The Gamma Function -Uniqueness

Theorem: If a function f(x) satisfies the following three conditions then it is identical to the gamma function.



(1) f(x+1) = f(x)



(1) f(x+1) = f(x)(2) The domain of f contains all x>0 and f(x)

is log convex



(1) f(x+1) = f(x)(2) The domain of f contains all x>0 and f(x)

is log convex(3) f(1)=1


Proof: Suppose f(x) satisfies the three properties. Then since f(1)=1 and f(x+1)=xf(x), for integer n≥2,

f(x+n)=(x+n-1)(x+n-2)…(x+1)xf(x)f(n) = (n-1)!


Proof: suppose f(x) satisfies the three properties. Then since f(1)=1 and f(x+1)=xf(x), for integer n ≥ 2,

f(x+n)=(x+n-1)(x+n-2)…(x+1)xf(x).f(n) = (n-1)!Now if we can show Γ(x) and f(x) agree on

[0,1], then by these properties they agree everywhere.


By property (2) f(x) is log convex, so by definition of convexity

nnnfnf

nnxnfnxf

nnnfnf

−+−+

≤−+−+

≤−+−−+−

)1()(ln)1(ln

)()(ln)(ln

)1()(ln)1(ln



nnnfnf

nnxnfnxf

nnnfnf

−+−+

≤−+−+

≤−+−−+−

)1()(ln)1(ln

)()(ln)(ln

)1()(ln)1(ln

nx

nnxfn ln)!1ln()(ln)1ln( ≤−−+

≤−



nnnfnf

nnxnfnxf

nnnfnf

−+−+

≤−+−+

≤−+−−+−

)1()(ln)1(ln

)()(ln)(ln

)1()(ln)1(ln

nx


≤−

])!1(ln[)(ln])!1()1ln[( −≤+≤−− nnnxfnn xx



The logarithm is monotonic so

nnnfnf

nnxnfnxf

nnnfnf

−+−+

≤−+−+

≤−+−−+−

)1()(ln)1(ln

)()(ln)(ln

)1()(ln)1(ln

nx


≤−

])!1(ln[)(ln])!1()1ln[( −≤+≤−− nnnxfnn xx

)!1()()!1()1( −≤+≤−− nnnxfnn xx


Since f(x+n)=x(x+1)…(x+n-1)f(x) then

)!1()()1)...(1()!1()1( −≤−++≤−− nnxfnxxxnn xx



)!1()()1)...(1()!1()1( −≤−++≤−− nnxfnxxxnn xx

nnxxnxnn

nxxxnnxf

xx

nxxxnn x

)...()(!

)1)...(1()!1()()1)...(1(

)!1()1(

++

=−++

−≤≤−++

−−



Since this holds for n ≥ 2, we can replace n by n+1 on the right. Then

)!1()()1)...(1()!1()1( −≤−++≤−− nnxfnxxxnn xx

nnxxnxnn

nxxxnnxf

xx

nxxxnn x

)...()(!

)1)...(1()!1()()1)...(1(

)!1()1(

++

=−++

−≤≤−++

−−

nnxxxnxnnxf

nxxxnn xx

))...(1()(!)(

))...(1(!

+++

≤≤++


This simplifies to )(

))...(1(!)( xf

nxxxnn

nxnxf

x

≤++

≤+


This simplifies to

As n goes to infinity,)(

))...(1(!)( xf

nxxxnn

nxnxf

x

≤++

≤+

))...(1(!)( lim nxxx

nnxfx

n ++=

∞→


This simplifies to

As n goes to infinity,

Since Gamma(x) satisfies the 3 properties of the theorem, then it satisfies this limit also.

Thus Γ(x) = f(x)

))...(1(!)(

)())...(1(

!)(

lim nxxxnnxf

xfnxxx

nnnx

nxf

x

n

x

++=

≤++

≤+

∞→

)())...(1(

!)( xfnxxx

nnnx

nxfx

≤++

≤+

))...(1(!)( lim nxxx

nnxfx

n ++=

∞→

Outline


Stirling’s Formula

Goal: Find upper and lower bounds for Γ(x)

Stirling’s Formulas

Goal: Find upper and lower bounds for Γ(x)From the definition of e, for k=1,2,…,(n-1)

1)/11()/11( ++≤≤+ kk kek


Goal: Find upper and lower bounds for Gamma(x)From the definition of e, for k=1,2,…,(n-1)

Multiply all of these together to get

1)/11()/11( ++≤≤+ kk kek

...)21()

1(...)

21()

1( 1121 −−−−

−−

−≤≤

−−

−nnnnn

nn

nne

nn

nn


Simplifying,

nnnn eenneen −+− ≤≤ 1!


Simplifying,

nnnn eenneen −+− ≤+Γ≤⇒ 1)1(



Simplifying,

Guess a function to approximate Γ(x)

nnnn eenneen −+− ≤+Γ≤⇒ 1)1(

)(2/1)( xxx eexxf μ−−=



μ(x) must be chosen so that(1) f(x+1)=xf(x)(2) f(x) is log convex(3) f(1)=1


μ(x) must be chosen so that(1) f(x+1)=xf(x)(2) f(x) is log convex(3) f(1)=1

(1) Is equivalent to f(x+1)/f(x)=x


Simplifying f(x+1)/f(x) gives

xeex xxx =+ −+−+ )()1(12/1)/11( μμ



)(1)/11ln()2/1()()1( xgxxxx ≡−++=−+⇒ μμ

xeex xxx =+ −+−+ )()1(12/1)/11( μμ



Then satisfies the equation

)(1)/11ln()2/1()()1( xgxxxx ≡−++=−+⇒ μμ

∑∞

=

+=0

)()(n

nxgxμ

)()()1(0 0

xgnxgnxgn n

=+−++∑ ∑∞

=

∞

=

xeex xxx =+ −+−+ )()1(12/1)/11( μμ


Consider the Taylor expansionwith...

531]

11ln[2/1

53

+++=−+ yyy

yy

121+

=x

y



531]

11ln[2/1

53

+++=−+ yyy

yy

121+

=x

y

...)12(3

112

1)/11ln(2/1 3 +++

+=+⇒

xxx



531]

11ln[2/1

53

+++=−+ yyy

yy

121+

=x

y

...)12(3

112

1)/11ln(2/1 3 +++

+=+⇒

xxx

...)12(5

1)12(3

1)(1)/11ln()2/1( 42 ++

++

==−++⇒xx

xgxx


Consider the Taylor expansionwith

Replacing the denominators with all 3’s gives

...531

]11ln[2/1

53

+++=−+ yyy

yy

121+

=x

y

...)12(3

112

1)/11ln(2/1 3 +++

+=+⇒

xxx

...)12(5

1)12(3

1)(1)/11ln()2/1( 42 ++

++

==−++⇒xx

xgxx

...)12(3

1)12(3

1)( 42 ++

++

≤xx

xg


g(x) is a geometric series, thus

)1(121

121

)12(11

1)12(3

1)(2

2 +−=

+−+

≤xx

xx

xg



))1(12

1)(12

1()()(00 ++

−+

≤+= ∑∑∞

=

∞

= nxnxnxgx

nn

μ

)1(121

121

)12(11

1)12(3

1)(2

2 +−=

+−+

≤xx

xx

xg



))1(12

1)(12

1()()(00 ++

−+

≤+= ∑∑∞

=

∞

= nxnxnxgx

nn

μ

)1(121

121

)12(11

1)12(3

1)(2

2 +−=

+−+

≤xx

xx

xg

x121

=


Since g(x) is positive then x

x12

1)(0 << μ


Since g(x) is positive then

for some

xx

121)(0 << μ

⇒x

x12

)( θμ = 10 << θ



for someFor property (2) we need to show f(x) is log convex

xx

121)(0 << μ

⇒x

x12

)( θμ = 10 << θ



for someFor property (2) we need to show f(x) is log convexThis meansmust be convex.

xx

121)(0 << μ

⇒x

x12

)( θμ = 10 << θ

)(ln)21()(ln xxxxxf μ+−−=


is convex because g(x) is the sum of convex functions

)(xμ



is convex since its second

derivative is positive for positive x.

)(xμ

xxx −− ln)2/1(

2211xx

+



is convex since its second

derivative is positive for positive x.

Thus f(x) is log convex.

)(xμ

xxx −− ln)2/1(

2211xx

+


For property (3) we need f(1)=1 so thatxxx eaxx 12/2/1)( θ+−−=Γ


For property (3) we need f(1)=1 so that

Consider the function

)2

1()2/(2)( +ΓΓ=

xxxh x

xxx eaxx 12/2/1)( θ+−−=Γ


For property (3) we need f(1)=1 so that

Consider the function

h(x) is log convex since the second derivative of is nonnegative and the gamma function is log convex.

)2

1()2/(2)( +ΓΓ=

xxxh x

xxx eaxx 12/2/1)( θ+−−=Γ

x2ln


)12/()2

1(2)1( 1 +Γ+

Γ=+ + xxxh x


)12/()2

1(2)1( 1 +Γ+

Γ=+ + xxxh x

)2

1()2/(22

2 +ΓΓ=

xxx x


)12/()2

1(2)1( 1 +Γ+

Γ=+ + xxxh x

)2

1()2/(22

2 +ΓΓ=

xxx x

)()1( xxhxh =+⇒


)12/()2

1(2)1( 1 +Γ+

Γ=+ + xxxh x

)2

1()2/(22

2 +ΓΓ=

xxx x

)()1( xxhxh =+⇒This means h(x) satisfies properties (1) and (2) of the uniqueness theorem, thus

)()( 2 xaxh Γ= for some constant 2a


Setting x=1 gives

)1()1()2/1(2)1( 2Γ=ΓΓ= ah


Setting x=1 gives

)1()1()2/1(2)1( 2Γ=ΓΓ= ah)1()2/1(22 ΓΓ=⇒ a


Setting x=1 gives

Using the limit equation for the gamma function

)1()1()2/1(2)1( 2Γ=ΓΓ= ah)1()2/1(22 ΓΓ=⇒ a

)!22(2!

lim2122/3

)1()2/1(2+

+=ΓΓ

∞→ nnn n

n


2/1

22

2 )!2(2)!(lim2nn

nan

n ∞→=


2/1

22

2 )!2(2)!(lim2nn

nan

n ∞→=

2/124/22/12

212/22122

])2([2][

2

1

lim2 neenaeena

nnn

nnnn

nθ

θ

−+

−+

∞→

=


2/1

22

2 )!2(2)!(lim2nn

nan

n ∞→=

2/124/22/12

212/22122

])2([2][

2

1

lim2 neenaeena

nnn

nnnn

nθ

θ

−+

−+

∞→

=

nn

nea 24/12/2 21lim2 θθ −

∞→=


2/1

22

2 )!2(2)!(lim2nn

nan

n ∞→=

2/124/22/12

212/22122

])2([2][

2

1

lim2 neenaeena

nnn

nnnn

nθ

θ

−+

−+

∞→

=

nn

nea 24/12/2 21lim2 θθ −

∞→=

2)2/1(22 aa =Γ=⇒


2/1

22

2 )!2(2)!(lim2nn

nan

n ∞→=

2/124/22/12

212/22122

])2([2][

2

1

lim2 neenaeena

nnn

nnnn

nθ

θ

−+

−+

∞→

=

nn

nea 24/12/2 21lim2 θθ −

∞→=

2)2/1(22 aa =Γ=⇒ π2)2/1(2 =Γ=⇒ a


Substituting a in the formula for f gives Stirling’sfinal result:

for some

nnn enn 12/2/12! θπ +−+=

10 << θ

Documents

Stirling’s Formula: an Approximation of the Factorial · • Stirling’s formula. ... (a,b) if the function Is a monotonically increasing function of on the interval 1 2 1 2