Stirling’s Formula: an Approximation of the Factorial · • Stirling’s formula. ... (a,b) if...
Transcript of Stirling’s Formula: an Approximation of the Factorial · • Stirling’s formula. ... (a,b) if...
Outline
• Introduction of formula• Convex and log convex functions• The gamma function• Stirling’s formula
Outline
• Introduction of formula• Convex and log convex functions• The gamma function• Stirling’s formula
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
• Introduction of formula• Convex and log convex functions• The gamma function• Stirling’s formula
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 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 ),(
Log Convex Functions -Properties
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
• Introduction of formula• Convex and log convex functions• The gamma function• Stirling’s formula
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.
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)
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)(2) The domain of f contains all x>0 and f(x)
is log convex
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)(2) The domain of f contains all x>0 and f(x)
is log convex(3) f(1)=1
The Gamma Function -Uniqueness
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)!
The Gamma Function -Uniqueness
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.
The Gamma Function -Uniqueness
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
The Gamma Function -Uniqueness
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
nx
nnxfn ln)!1ln()(ln)1ln( ≤−−+
≤−
The Gamma Function -Uniqueness
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
nx
nnxfn ln)!1ln()(ln)1ln( ≤−−+
≤−
])!1(ln[)(ln])!1()1ln[( −≤+≤−− nnnxfnn xx
The Gamma Function -Uniqueness
By property (2) f(x) is log convex, so by definition of convexity
The logarithm is monotonic so
nnnfnf
nnxnfnxf
nnnfnf
−+−+
≤−+−+
≤−+−−+−
)1()(ln)1(ln
)()(ln)(ln
)1()(ln)1(ln
nx
nnxfn ln)!1ln()(ln)1ln( ≤−−+
≤−
])!1(ln[)(ln])!1()1ln[( −≤+≤−− nnnxfnn xx
)!1()()!1()1( −≤+≤−− nnnxfnn xx
The Gamma Function -Uniqueness
Since f(x+n)=x(x+1)…(x+n-1)f(x) then
)!1()()1)...(1()!1()1( −≤−++≤−− nnxfnxxxnn xx
The Gamma Function -Uniqueness
Since f(x+n)=x(x+1)…(x+n-1)f(x) then
)!1()()1)...(1()!1()1( −≤−++≤−− nnxfnxxxnn xx
nnxxnxnn
nxxxnnxf
xx
nxxxnn x
)...()(!
)1)...(1()!1()()1)...(1(
)!1()1(
++
=−++
−≤≤−++
−−
The Gamma Function -Uniqueness
Since f(x+n)=x(x+1)…(x+n-1)f(x) then
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(!
+++
≤≤++
The Gamma Function -Uniqueness
This simplifies to
As n goes to infinity,)(
))...(1(!)( xf
nxxxnn
nxnxf
x
≤++
≤+
))...(1(!)( lim nxxx
nnxfx
n ++=
∞→
The Gamma Function -Uniqueness
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
• Introduction of formula• Convex and log convex functions• The gamma function• Stirling’s formula
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
Stirling’s Formulas
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
Stirling’s Formula
Simplifying,
Guess a function to approximate Γ(x)
nnnn eenneen −+− ≤+Γ≤⇒ 1)1(
)(2/1)( xxx eexxf μ−−=
nnnn eenneen −+− ≤≤ 1!
Stirling’s Formula
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
Stirling’s Formulas
Simplifying f(x+1)/f(x) gives
)(1)/11ln()2/1()()1( xgxxxx ≡−++=−+⇒ μμ
xeex xxx =+ −+−+ )()1(12/1)/11( μμ
Stirling’s Formulas
Simplifying f(x+1)/f(x) gives
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
Stirling’s Formula
Consider the Taylor expansionwith...
531]
11ln[2/1
53
+++=−+ yyy
yy
121+
=x
y
...)12(3
112
1)/11ln(2/1 3 +++
+=+⇒
xxx
Stirling’s Formula
Consider the Taylor expansionwith...
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
Stirling’s Formula
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
Stirling’s Formula
g(x) is a geometric series, thus
)1(121
121
)12(11
1)12(3
1)(2
2 +−=
+−+
≤xx
xx
xg
Stirling’s Formula
g(x) is a geometric series, thus
))1(12
1)(12
1()()(00 ++
−+
≤+= ∑∑∞
=
∞
= nxnxnxgx
nn
μ
)1(121
121
)12(11
1)12(3
1)(2
2 +−=
+−+
≤xx
xx
xg
Stirling’s Formula
g(x) is a geometric series, thus
))1(12
1)(12
1()()(00 ++
−+
≤+= ∑∑∞
=
∞
= nxnxnxgx
nn
μ
)1(121
121
)12(11
1)12(3
1)(2
2 +−=
+−+
≤xx
xx
xg
x121
=
Stirling’s Formula
Since g(x) is positive then
for someFor property (2) we need to show f(x) is log convex
xx
121)(0 << μ
⇒x
x12
)( θμ = 10 << θ
Stirling’s Formula
Since g(x) is positive then
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+−−=
Stirling’s Formula
is convex because g(x) is the sum of convex functions
is convex since its second
derivative is positive for positive x.
)(xμ
xxx −− ln)2/1(
2211xx
+
Stirling’s Formula
is convex because g(x) is the sum of convex functions
is convex since its second
derivative is positive for positive x.
Thus f(x) is log convex.
)(xμ
xxx −− ln)2/1(
2211xx
+
Stirling’s Formulas
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
Stirling’s Formulas
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
Stirling’s Formula
)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
Stirling’s Formula
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
Stirling’s Formula
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
−+
−+
∞→
=
Stirling’s Formula
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 −
∞→=
Stirling’s Formula
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=⇒
Stirling’s Formula
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