Gamma function study notes tute0062
video list , how to download video , update 2015-02-14
Zeta function: 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 11, 12, 13, 14, 15, 16, 17
Gamma function: 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 11, 12
Why gamma function this way? Γ(s) = ∫[t=0,t=∞]{ts-1*e-t*dt} eq.CG01
Integral Test, Euler's formula, Integration by parts
How to prove eq.CG16? Taylor series expansion equation




<a name="Gamma_Indx"> 
2015-01-01-15-18 start 
On 
2014-03-25-20-36 Liu,Hsinhan access
http://www.youtube.com/playlist?list=PL3E4136E122545FBE
find and download next 12 video files. 

Gamma Function 
•by MrYouMath
•12 videos
•5,940 views
•1 hour, 48 minutes
<a name="GammaIndex"> 
Gamma Function - Part 1 - Functional Equation
http://www.youtube.com/watch?v=2iBNo4j3vRo
Gamma Function - Part 2 - Gauss Representation
http://www.youtube.com/watch?v=a_Dlx7TTjkI
Gamma Function - Part 3 - Weierstrass Representation
http://www.youtube.com/watch?v=d9Oz62Ioue0
Gamma Function - Part 4 - Relationship to Sine
http://www.youtube.com/watch?v=W3ulxj3s90U
Gamma Function - Part 5 - Gamma of 0.5 ( one half)
http://www.youtube.com/watch?v=_vwqsJNKY-c
Gamma Function - Part 6 - Stirling's Approximation
http://www.youtube.com/watch?v=MuAb2dnPD0Q
An intuitive derivation of Stirling’s formula
http://www.rowan.edu/open/depts/math/osler/Stirlings_formula.pdf

Gamma Function - Part 7 - Euler Integral I
http://www.youtube.com/watch?v=VF7ud3Al6d8
Gamma Function - Part 8 - Euler Integral II The Sinc-Function
http://www.youtube.com/watch?v=h4wQ_F1s1UI
Gamma Function - Part 9 - Euler Integral III Fresnel Integral
http://www.youtube.com/watch?v=H0kT-EKbUzM
Gamma Function - Part 10 - Beta Function
http://www.youtube.com/watch?v=Korx_G7eySQ
Gamma Function - Part 11 - Legendre Duplication Formula
http://www.youtube.com/watch?v=yu9NeDXalpA
Gamma Function - Part 12 - Relation to Zeta Function
http://www.youtube.com/watch?v=jQAPlsNY_P0
Above red link has complete lecture. Blue not.
Gauss multiplication formula half way, not done!
https://proofwiki.org/wiki/Gauss_Multiplication_Formula
Γ(z) has a simple pole with residue (-1)n/n! 
at z=-n for n = 0, 1, 2, ...
http://www.math.leidenuniv.nl/~evertse/ant13-8.pdf
2015-01-01-15-38 stop

<a name="ZetaIndex"> 
2014-10-29-07-56
https://www.youtube.com/playlist?list=PL32446FDD4DA932C9
2014-10-29-08-24
MrYouMath_17_files_Riemann.htm
"Zeta Function - Part 1 - Convergence" 
http://www.youtube.com/watch?v=ZlYfEqdlhk0
"Zeta Function - Part 2 - Euler Product Representation" 
http://www.youtube.com/watch?v=I3qSCWNXZKg
"Zeta Function - Part 3 - Euler Product (easy)" 
http://www.youtube.com/watch?v=TDdGisWD5OU
<a name="docA007">
"Zeta Function - Part 4 - Infinitude of Prime Numbers" 
http://www.youtube.com/watch?v=SKa7b-3C32A
"Zeta Function - Part 5 - Prime Zeta Function" 
http://www.youtube.com/watch?v=3eN9tQX3JJ4
"Zeta Function - Part 6 - The Prime Counting Function" 
http://www.youtube.com/watch?v=U16_KTTKtb0
"Zeta Function - Part 7 - Zeta of 2 aka The Basel Problem" 
http://www.youtube.com/watch?v=GeKDmoYHiAk
"Zeta Function - Part 8 - Zeta of 2n - Part 1" 
http://www.youtube.com/watch?v=axQqExF7NsU
<a name="docA008">
"Zeta Function - Part 8 - Zeta of 2n - Part 2" 
http://www.youtube.com/watch?v=XHQ0OzqTjd0 9/17
"Zeta Function - Part 8 - Zeta of 2n - Part 3" 
http://www.youtube.com/watch?v=1f24RZfP6m8 10/17
"Zeta Function - Part 9 - Relation to Gamma Function" 
http://www.youtube.com/watch?v=UEZ4ClCdog8 11/17
"Zeta Function - Part 10 - Jacobi Theta Function" 
http://www.youtube.com/watch?v=-GQFljOVZ7I 12/17
"Zeta Function - Part 11 - Riemann Functional Equation I" 
http://www.youtube.com/watch?v=K6L4Ez4ZVZc 13/17
<a name="docA009">
"Zeta Function - Part 12 - Riemann Functional Equation II" 
http://www.youtube.com/watch?v=TnRnlJBecRg 14/17
"Zeta Function - Part 13 - Trivial Zeros of the Zeta Function" 
http://www.youtube.com/watch?v=G-fqe3BkBnE 15/17
"Zeta Function - Part 14 - Riemann Xi Function" 
http://www.youtube.com/watch?v=QfDbF_qlp58 16/17
"Sine Function Product Formula (Hadamard Factorization Theorem)"
https://www.youtube.com/watch?v=h3Hcioh2F9I 17/17
"What is a function? Why 1+2+3+4+5+.... not equals -1/12 = Zeta(-1)" 
http://www.youtube.com/watch?v=wt6ngy6pDws This video is easy. skipped a312190901

<a name="docAe01"> 
2015-01-08-18-12 start 
Next is study notes of 
Gamma Function - Part 1 - Functional Equation
http://www.youtube.com/watch?v=2iBNo4j3vRo

Lecturer is MrYouMath. 
Notes recorder is Liu,Hsinhan.
Recorded notes is about 60% match, not exact.
Liu,Hsinhan inserted text is blue Italic.
LiuHH notes may contain error!

<a name="docAe02"> Begin video 1 of 12
Hello and welcome you guys to my new territorial 
about the Euler Gamma function. This video I will 
teach you how to derive the functional equation 
of this crazy function.
<a name="docAe03">
Gamma function Γ(s) definition. s is complex number. s real part must be >0
variable s show up at right side once. Right side variable t be dummy variable.
Integrate from t=0 to t=∞, t is "all things considered", t character disappear.
Key point is that Γ(s) is not a function of t any more.
 
Γ(s)
=
t=∞
t=0
ts-1*e-t*dt
  ▐  
Real(s)>0
--- eq.CG01
eq.CG01 mean equation, Complex Gamma, sequence 01.
width of above equation a401081831
<a name="docAe04">
2015-01-08-18-40 here 
Let us just have a look at eq.CG01 Left side is 
Γ(s). Right side has integral from t=0 to t=∞ .
Integrand is ts-1*e-t*dt . Very important this 
integral only converging if complex s real 
part is greater than zero. Let us start how do 
we derive function equation for Γ(s). Let us 
just try eq.CG02 below. 
<a name="docAe05">
 
Γ(s+1)
=
t=∞
t=0
t(s+1)-1*e▬t*dt
=
t=∞
t=0
ts*d( e▬t)
--- eq.CG02
 
Γ(s+1)
∫p
=
-e-t*ts
t=∞
|
t=0
t=∞
t=0
( e-t)*
d[ts]
--- eq.CG03
 
Γ(s+1)
=
t=∞
t=0
e-t[s*ts-1*dt]
= s*
t=∞
t=0
ts-1*e-t*dt
--- eq.CG04
eq.CG02 come from integrating e▬t*dt ;
eq.CG03 left come from eq.LA15 ;
eq.CG03 right come from eq.CG02
eq.CG03 and cancel to '+' . //a401101458
width of above equation a401081909
<a name="docAe06"> 
2015-01-08-19-14 here
Instead of plug in s, we plug in s+1. In 
eq.CG02 t power become (s+1)-1 which simplify 
to ts as eq.CG02 right side shown. Next 
just do a partial integration on eq.CG02 
right side integral. Very important the ts 
in eq.CG02 is down integrated and e-t part 
is up integrated. In eq.CG03 we get boundary 
term -e-t*ts|[t=0,t=∞] In eq.CG03 we get new 
<a name="docAe07">
integral. Differentiate ts with respect to t. 
e-t in eq.CG02 integrated to -e-t this 
integrated -e-t stay in eq.CG03 not change 
any thing on that. Now eq.CG03 boundary term 
-e-t*ts|[t=0,t=∞] become zero. If you plug 
t=0 to -e-t*ts|[t=0,t=∞] , 0s is zero. If 
plug t=∞ to -e-t*ts|[t=0,t=∞] it is a little 
bit harder. You will get e-t tend to zero and 
ts tend to infinity. If use l'Hosptial's rule
you will see -e-t*ts|[t=∞] converging to zero.
<a name="docAe08">
So, in eq.CG03 -e-t*ts|[t=0,t=∞] is out of 
equation. eq.CG03 integral term has two '-' 
which cancel. We get eq.CG04 In eq.CG04 left 
integral has 's'. Now I just take this 's' 
out of integral, become eq.CG04 right integral.
Because 's' is not integration variable 't'. 
Now let us look what we have in eq.CG04 right 
integral. This is actually the Gamma function 
we had in above eq.CG01 . We have 

<a name="docAe09"> Gamma Functional Equation Γ(s+1) = s*Γ(s) --- eq.CG05

That is all the magic behind that. Now there is a little bit about this eq.CG05. You might know the motivation was find an elliptical function that is equal to factorial function Γ(n+1)=n*Γ(n)=n*(n-1)*Γ(n-1) --- eq.CG06 for integer s. Actually if you do that you just plug in s for integer number, you apply eq.CG05 again and again and again, <a name="docAe10"> you will end up having Γ(n+1)=n! --- eq.CG07 At the end, you get Γ(1) and you will see Γ(1) (see eq.CG01) is equal to one. This is the end, where we stop. This conclude this lecture. 2015-01-08-19-59 stop <a name="a40109a"> 2015-01-09-08-37 start Why gamma function this way? This topic is Liu,Hsinhan own words, LiuHH notes may contain error! Please verify, proofread. Thank you. Define a function eq.CG01 why this function value pass all factorial 3!, 8!, n! values? Next find out why. Gamma function definition is eq.CG01
<a name="a40109b">
Why gamma function this way?
 
Γ(s)
=
t=∞
t=0
ts-1*e-t*dt
  ▐  
Real(s)>0
--- eq.CG01
width of above equation a401090843
2015-01-09-08-45 here
This explanation use Integration by parts
<a name="a40109c">
Goto Integration by parts derivation begin ; priority table 1,2,3
Integration by parts change independent variable from x to t.
 
t=b
t=a
U(t)v(t)dt =
U(t)V(t)
t=b
t=a
t=b
t=a
V(t)u(t)dt
--- eq.LAa0
width of above equation a401090849
<a name="a40109d">
2015-01-09-08-54 here
See Gamma Functional Equation eq.CG05
Two key equation are 
  Γ(n+1)=n*Γ(n)=n*(n-1)*Γ(n-1) --- eq.CG06 
  Γ(n+1)=n! --- eq.CG07
eq.CG01 apply Integration by parts eq.LA15 
get eq.CG06 and eq.CG07. 
eq.CG01 ts-1 is U(x) will be differentiated.
eq.CG01 e-t is v(x) will be integrated.
eq.CG01 boundary ∫[t=0,t=∞] assign t from 0 to ∞.

<a name="a40109e"> 
Three key points must be resolved. 
■■ Key point one.
Compare eq.LA15 with eq.CG06 and eq.CG07, the 
boundary term must vanish. If not vanish then 
boundary term show up, see eq.LA39 .../3 and .../9 
terms and see eq.LA42 log(x)*x5/3/(5/3) term 
To wipe out boundary term, rely upon eq.CG01 
boundary ∫[t=0,t=∞] assign t from 0 to ∞.
Because apply eq.LA15 to eq.CG01 the boundary 
is U(t)V(t)|[t=0,t=∞] , that is 
<a name="a40109f"> 
ts-1*(-e-t)|[t=0,t=∞]
when t=0, ts-1*(-e-t)=0s-1*(-e-0)=0*1=0 --- eq.CG08 
when t→∞, ts-1*(-e-t)=ts-1/(-e+t)|[t→∞]=∞/(-∞) --- eq.CG09 
Apply l'Hosptial's rule get 
ts-1/(-e+t)|[t→∞]=[d(ts-1)/dt]/[d(-e+t)/dt]=... --- eq.CG10 
=(s-1)*(s-2)*...*1/(-e+t) this term t→∞ value 
is finite/∞ = 0. Now boundary term be zero at 
both t=0 and t=∞. Boundary term never show up. 
Because eq.CG01 boundary ∫[t=0,t=∞] assign t 
from 0 to ∞. We see only Γ(n+1)=n! , and never 
see Γ(n+1)= boundary term + n! 

<a name="a40109g"> 
■■ Key point two.
Next see eq.LA15 has ∫[x=a,b]V(t)u(t)dt 
why negative sign  never show up in eq.CG06 ? 
In eq.CG01 the negative sign in purple e-t do 
the job. purple e-t will be integrated (take 
antiderivative) The integrated value is 
∫[x=a,b](e-t)*u(t)dt //see eq.CG03 integral 
<a name="a40109h">
During integration, purple  create red  and
red  cancel blue , then 
∫[x=a,b](e-t)*u(t)dt --- eq.CG11
become  //from eq.CG03 to eq.CG04 
∫[x=a,b](e-t)*u(t)dt --- eq.CG12
Integration become positive again. 

<a name="a40109i"> 
■■ Key point three.
eq.CG06 has n*(n-1)*(n-2)*(n-3)*... 
in eq.CG01 who generate these number 
n,(n-1),(n-2),(n-3),...?
To answer this question, see eq.CG01 ts-1
ts-1 will be differentiated, 
First differentiation 
  d(ts-1)/dt=(s-1)*(ts-2) --- eq.CG13
Second differentiation 
  d[(s-1)*ts-2]/dt=(s-1)*(s-2)*(ts-3) --- eq.CG14
If s is integer, continue this way finally 
get (n-1)*(n-2)*...*1 there is no t in it, 
next differentiation is meaningless. This 
continuous ts-1 differentiation generate 
n*(n-1)*(n-2)*(n-3)*... 

<a name="a40109j">
Combine above considerations, January 8, 1730 
Euler created Gamma function eq.CG15 (eq.CG01)
Date "January 8, 1730" see next pdf page 1. 
2014-12-31-00-41 Liu,Hsinhan access 
http://www.csie.ntu.edu.tw/~b89089/link/gammaFunction.pdf
Above pdf say Euler proposed 
Γ(x) = ∫[t=0,t=1]{[−log(t)]x-1dt} --- eq.CG15
and eq.CG15 is equivalent to eq.CG01. 
2015-01-09-10-30 stop 
LiuHH notes may contain error!
2015-01-10-14-50 corrected careless error at eq.CG03

<a name="a40119a">
2015-01-19-20-30 start 
Above has ■■ Key point one/two/three. Another 
point is also in LiuHH mind, but did not find 
a satisfactory answer. Question is 
Why define Gamma function as eq.CG01 ? 
Why let eq.CG01 Γ(s) correspond to ts-1 ? 
Why not eq.CG01 Γ(s) correspond to ts ?

<a name="a40119b">
2014-08-30-04-04 LiuHH access 
https://uqu.edu.sa/files2/tiny_mce/plugins/filemanager/files/4282164/Gamma%20Function%20(1).pdf
uqu.edu.sa_GammaFunction(1).pdf
page 6/12 has 
[[
Pi function
An alternative notation which was originally
 introduced by Gauss and which is sometimes
 used is the Pi function, which in terms of 
 the Gamma function is
  Π(z)=Γ(z+1)=zΓ(z) 
so that
  Π(z)=n!
]]
Here record Pi function. If future find answer, 
come back again.
2015-01-19-20-42 stop

<a name="a40119c">
2015-01-20-09-05 start 
about 
why-is-the-gamma-function-shifted-from-the-factorial-by-1
please visit next four pages 

2015-01-20-08-28 LiuHH access 
https://www.google.com/search?as_q=Gamma+function+&as_epq=Pi+function+&as_oq=&as_eq=&as_nlo=&as_nhi=&lr=&cr=&as_qdr=all&as_sitesearch=&as_occt=any&safe=images&tbs=&as_filetype=&as_rights=&gws_rd=ssl

2015-01-20-08-30 
http://math.stackexchange.com/questions/267989/the-gamma-function-and-the-pi-function

2015-01-20-08-34
http://mathoverflow.net/questions/20960/why-is-the-gamma-function-shifted-from-the-factorial-by-1

2015-01-20-08-49
http://stackoverflow.com/questions/8276435/how-should-the-standard-factorial-function-behave
2015-01-20-09-08 stop 


<a name="docAe31"> 2015-01-09-18-17 start Next is study notes of Gamma Function - Part 2 - Gauss Representation http://www.youtube.com/watch?v=a_Dlx7TTjkI Lecturer is MrYouMath. Notes recorder is Liu,Hsinhan. Recorded notes is about 60% match, not exact. Liu,Hsinhan inserted text is blue Italic. LiuHH notes may contain error! <a name="docAe32"> Begin video 2 of 12 Hello and welcome to the new video about Gauss Representation of Gamma Function. In order to derive this representation we will start of the definition of Gamma Function.
<a name="docAe33">
 
Γ(s)
=
t=∞
t=0
ts-1*e-t*dt
  ▐  
Real(s)>0
--- eq.CG01
width of above equation a401091822
<a name="docAe34">
2015-01-09-18-22 here 
The Gamma function is defined as eq.CG01 shown. 
Very important is that for complex s, its real 
part must be greater than zero Real(s)>0 for 
absolute convergence. Now what Gauss get to 
his representation of Gamma function , he just 
replace e-t with next expression 
<a name="docAe35">
Show eq.CG16 is true at a40111e
 
e-t
=
limit
n→∞
[
1 +
t

n
]
n

--- eq.CG16
width of above equation a401091832
<a name="docAe36"> 
2015-01-09-18-35 here
We will plug eq.CG16 into eq.CG01 . 
 
Γ(s)
=
t=∞
t=0
ts-1
limit
n→∞
[
1
t

n
]
n

*dt
--- eq.CG17
width of above equation a401091841
<a name="docAe37">
2015-01-09-18-42 here 
We will get eq.CG17 . Now remember that 
integral is absolute converging for Re(s)>0 
and the limit expression eq.CG16 is also 
absolute converging . What we can do is 
that we can take the limit expression out 
of integral as shown below 
<a name="docAe38">
 
Γ(s)
=
limit
n→∞
t=n
t=0
ts-1
[
1 -
t

n
]
n

*dt
--- eq.CG18
width of above equation a401091848
<a name="docAe39">
2015-01-09-18-53 here
Now pay some attention, because I change 
something very important. So the integral 
start from zero and not to infinity, but 
to n. We can do that because n go to 
infinity anyway. This right hand part did 
not change anything. Next step what I am 
doing is actually do partial integration 
to the integral of eq.CG18 
ts-1 will be integrated and 
[1-t/n]n will be differentiated.
Now look what happen.
<a name="docAe40">
 
t=n
t=0
ts-1
[
1 -
t

n
]
n

dt
=
t=n
t=0
[
1 -
t

n
]
n

d
ts

s
--- eq.CG19
 
∫p
=
[
1 -
t

n
]
n

*
ts

s
t=n
t=0
t=n
t=0
ts

s
d
[
1 -
t

n
]
n

--- eq.CG20
 
=
0*
ts

s
0
t=n
t=0
ts

s
n
[
1 -
t

n
]
n-1

d
t

n
--- eq.CG21
 
t=n
t=0
ts-1
[
1 -
t

n
]
n

dt
= +
t=n
t=0
ts

s
[
1 -
t

n
]
n-1

dt
--- eq.CG22
width of above equation a401091938
<a name="docAe41"> 
2015-01-09-19-40 here
Liu,Hsinhan inserted text is blue Italic.
LiuHH notes may contain error!
eq.CG19 left side come from eq.CG18 integral 
term. 
eq.CG19 right integrate ts-1dt to d[ts/s]
From eq.CG19 to eq.CG20 apply Integration by 
parts formula eq.LA15. 
eq.CG20 red  is a result of applying eq.LA15.
<a name="docAe42">
Evaluate eq.CG20 get eq.CG21. In eq.CG21 
boundary terms be zero, and boundary will 
be zero in all following calculation. 
eq.CG21 left  come from eq.CG20 red 
eq.CG21 right  come from differentiation 
chain rule 
eq.CG21  and  cancel. 
eq.CG22 left term is eq.CG19 left term. 
eq.CG22 right term is result term after 
first applying Integration by parts formula. 

<a name="docAe43">
2015-01-09-19-56 here video 3:20/9:46 
In eq.CG20 the boundary vanish and does not 
matter any more. Let us look at second 
part. In eq.CG21  and  cancel. We get 
eq.CG22 right side expression. .....
Let us re-write eq.CG22
2015-01-09-20-01 here 
<a name="docAe44">
eq.CG23 left side is eq.CG19 left side. eq.CG23 right side is eq.CG22 right side.
eq.CG23 is first term = last term in eq.CG19 to eq.CG22 equation block.
eq.CG24 "∫p over =" applied Integration by parts eq.LA15
eq.CG24 right side boundary vanish and does not matter any more.
 
t=n
t=0
ts-1
[
1 -
t

n
]
n

dt
=
n

s*n
t=n
t=0
ts
[
1 -
t

n
]
n-1

dt
--- eq.CG23
 
=
n

s*n
t=n
t=0
[
1 -
t

n
]
n-1

d
ts+1

s+1
∫p
=
n

s*n
t=n
t=0
ts+1

s+1
d
[
1 -
t

n
]
n-1

--- eq.CG24
 
= boundary
be zero.
n

s*n
t=n
t=0
ts+1*(n-1)

s+1
[
1 -
t

n
]
n-2

d
t

n
--- eq.CG25
 
t=n
t=0
ts-1
[
1 -
t

n
]
n

dt
=
n

s*n
n-1

(s+1)*n
t=n
t=0
ts+1
[
1 -
t

n
]
n-2

dt
--- eq.CG26
width of above equation a401092033
<a name="docAe45">
2015-01-09-20-34 here 
Compare eq.CG23 right side and eq.CG26 right 
side. We raise ts power to ts+1 and reduce 
[1-(t/n)]n-1 power to [1-(t/n)]n-2. We will 
just repeat this. Next iteration is 
 
t=n
t=0
ts-1
[
1 -
t

n
]
n

dt
=
n

s*n
n-1

(s+1)n
n-2

(s+2)n
t=n
t=0
ts+2
[
1 -
t

n
]
n-3

dt
--- eq.CG27
width of above equation a401092042
<a name="docAe46"> 
2015-01-09-20-43 here 
You can repeat this procedure until [1-(t/n)]n
vanishes [1-(t/n)]0 , you can do it n times,
you will end up having this 
 
t=n
t=0
ts-1
[
1 -
t

n
]
n

dt
=
n

s*n
n-1

(s+1)n
n-2

(s+2)n
...
1

(s+n-1)n
t=n
t=0
ts+n-1
dt
--- eq.CG28
width of above equation a401092050
<a name="docAe47">
2015-01-09-20-51 here 
Attention: from eq.CG28 to eq.CG29, eq.CG01 
integration sign is removed !! //a401220808
eq.CG28 right side integration term become 
simple integral. After integration, get 
 
t=n
t=0
ts-1
[
1 -
t

n
]
n

dt
=
n

s*n
n-1

(s+1)n
n-2

(s+2)n
...
1

(s+n-1)n
ts+n

s+n
t=n
t=0
--- eq.CG29
width of above equation a401092056
<a name="docAe48">
2015-01-09-20-58 here 
In eq.CG29 plug in ts+n|[t=0,t=n] 
t=0 gives zero, t=n gives ns+n. 
eq.CG29 numerator has factorial of n, n!.
eq.CG29 denominator has nn 
eq.CG29 denominator has s*(s+1)*...*(s+n-1)*(s+n)
The result equation is 
<a name="docAe49">
 
t=n
t=0
ts-1
[
1 -
t

n
]
n

dt
=
n!

nn
ns+n
i=n
i=0
(s+i)-1
--- eq.CG30
 
t=n
t=0
ts-1
[
1 -
t

n
]
n

dt
=


n!*ns

i=n
i=0
(s+i)
--- eq.CG31
width of above equation a401092123
<a name="docAe50">
2015-01-09-21-24 here 
eq.CG30 left side is eq.CG18 limit integral. 
This integral value is eq.CG30 right side .
In eq.CG30 numerator and denominator nn 
cancel. Rewrite eq.CG30 as eq.CG31 
Now I am doing a little trick. 
In eq.CG31 ∏[i=0,n](s+i) isolate i=0, write
∏[i=0,n](s+i) as s*∏[i=1,n](s+i)
In eq.CG31 write n! as ∏[i=1,n](i) , get 
<a name="docAe51">
 
t=n
t=0
ts-1
[
1 -
t

n
]
n

dt
=
ns

s
i=n
i=1
i

i=n
i=1
(s+i)
--- eq.CG32
<a name="docAe52">
Gamma Function, Gauss Representation
 
Γ(s)
=
limit
n→∞
[
ns

s
i=n
i=1
i

s+i
]
--- eq.CG33
width of above equation a401092145
<a name="docAe53">
2015-01-09-21-46 here 
In eq.CG31, n! is n factorial, it is 
n!= 1*2*3*4*...*(n-1)*n that is n!=∏[i=1,n]i
In eq.CG32 numerator n!=∏[i=1,n]i combine 
with denominator ∏[i=1,n](s+i) get eq.CG33 
∏[i=1,n] term.
We will use eq.CG33 in the later video. 
eq.CG33 conclude this lecture. 
2015-01-09-21-53 
2015-01-10-10-42 done first proofread 
2015-01-10-17-27 done second proofread 


<a name="a40111a"> 2015-01-11-14-18 how to prove eq.CG16 ?
 
e-t
=
limit
n→∞
[
1 +
t

n
]
n

--- eq.CG16
 
ex
= 1+x +
x2

2!
+
x3

3!
+
x4

4!
+
x5

5!
+
x6

6!
+
x7

7!
●●●
--- eq.LAa0
 
n=1 ⇒
[
1 +
x

1
]
1

= 1+x
--- eq.LAa1
 
n=2 ⇒
[
1 +
x

2
]
2

=
1 +
2*
x1

21
+
x2

22
--- eq.LAa2
a401111450
 
n=3 ⇒
[
1 +
x

3
]
3

=
1 +
3*
x1

21
+ 3*
x2

32
+
x3

33
a40111b
--- eq.LAa3
 
n=4 ⇒
[
1 +
x

4
]
4

=
1 +
4*
x1

41
+ 6*
x2

42
+ 4*
x3

43
+
x4

44
--- eq.LAa4
 
n=5 ⇒
[
1 +
x

5
]
5

=
1 +
5x1

51
+
10x2

52
+
10x3

53
+
5x4

54
+
x5

55
--- eq.LAa5
 
n=6 ⇒
1 +
6x1

61
+
15x2

62
+
20x3

63
+
15x4

64
+
6x5

65
+
x6

66
--- eq.LAa6
 
ex
= 1+x +
x2

2!
+
x3

3!
+
x4

4!
+
x5

5!
+
x6

6!
+
x7

7!
●●●
--- eq.LAa0
width of above equation a401111512
<a name="a40111c">
n=50, x^3 coef is 19600
19600*x^3/50/50/50 approach 3!=6 ??
6*19600/50/50/50
0.9408
2015-01-11-15-17

<a name="a40111d"> 
2015-01-11-18-46
How to prove eq.CG16? 
2015-01-11-17-24 LiuHH access 
The Number e and the Exponential Function
http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/Exponential_Function.htm
<a name="a40111e">
 
e-t
=
limit
n→∞
[
1 +
t

n
]
n

--- eq.CG16
width of above equation a401111852

To prove eq.CG16 consider eq.LAa7 below.
<a name="a40111f">
From eq.LAa7 to eq.LAa8 , change y to x/n.
Binomial coeffcients appear in Pascal's triangle. 2015-02-02-23-02
 
[
1 +
y
]
n

=
1 +
n*y
+
n(n-1)

2!
y2
+
n(n-1)(n-2)

3!
y3
+ ●●●
--- eq.LAa7
 
[
1 +
x

n
]
n

=
1 +
n
x

n
+
n(n-1)

2!
x2

n2
+
n(n-1)(n-2)

3!
x3

n3
+ ●●●
--- eq.LAa8
 
[
1 +
x

n
]
n

=
1 +
x

1!
+
n2-n

n2
x2

2!
+
n3-3n2+2n

n3
x3

3!
+ ●●●
--- eq.LAa9
width of above equation a401111911
<a name="a40111g">
2015-01-11-19-11 here 
When n→∞, in eq.LAa9 
n2-n → n2, then (n2-n)/n2 → 1 
n3-3n2+2n → n3, then (n3-3n2+2n)/n3 → 1 
similarly, other term n coefficient approach 
to one. eq.LAa9 approach to 
<a name="a40111h">
 
lim
n→∞
[
1 +
x

n
]
n

=
1 +
x

1!
+
1*
x2

2!
+
1*
x3

3!
+ ●●●
--- eq.LAb0
width of above equation a401111924

Start point 2015-01-25-16-54 add eq.LEb0
 
lim
n→∞
[
1 +
x

n
]
n

=
ex
--- eq.LEb0
width of above equation a401251651

2015-01-11-19-25 here
eq.LAb0 is same as eq.LAa0 and eq.LAa0 is ex
therefore eq.LAb0 is ex. 
Replace x by -t get eq.CG16 .

<a name="a40111i">
About "eq.LAa0 is ex", please refer to calculus 
textbook. For example, George B. Thomas, Jr. 
and Ross L. Finney Calculus and Analytic 
Geometry, sixth edition, ISBN 0-201-16290-3
page 665, example 1 and page 666 equation (10).
2015-01-11-19-27 stop

<a name="a40111j">
2015-01-11-22-17 start //How to prove eq.CG16? 
Today 2015-01-11 LiuHH receive brother email 
ask LiuHH write a javascript program for him.
Liu,Hsinhan math study notes tute0062.htm will 
stop few days.
2015-01-11-22-20 stop

<a name="a40122a">
2015-01-22-08-25 start 
Brother requested javascript program is done 
2015-01-18 upload to next URL 
http://freeman2.com/utility4.htm
Assume input 
<tr> <td> data1 </td> <td> data2 </td> <td> data3 </td> </tr> 
<tr> <td> data4 </td> <td> data5 </td> <td> data6 </td> </tr> 
Program output to 
data1 data2 data3 newline 
data4 data5 data6 
Program extract data from <table>
2015-01-22-08-30 stop 


<a name="docAe61"> 2015-01-19-13-40 start Next is study notes of Gamma Function - Part 3 - Weierstrass Representation http://www.youtube.com/watch?v=d9Oz62Ioue0 Lecturer is MrYouMath. Notes recorder is Liu,Hsinhan. Recorded notes is about 60% match, not exact. Liu,Hsinhan inserted text is blue Italic. LiuHH notes may contain error! <a name="docAe62"> Begin video 3 of 12 Hello and welcome to my video on Gamma Function. In this particular video we will talk about the deriving of Weierstrass Representation of Gamma Function. Weierstrass Representation look like following.
<a name="docAe63">
Gamma Function, Weierstrass Representation
 
Γ(s)
=
1

s
e-γs
i=∞
i=1
es/i
[
1+
s

i
]
-1


--- eq.CG34
e=Euler's number=2.718281828459045...
γ=Euler-Mascheroni constant 0.57721566490153286061...
s=complex number, real(s)>0
width of above equation a401191400
<a name="docAe64">
2015-01-19-15-08 start
Small γ in e-γs is Euler-Mascheroni constant 
0.57721566490153286061
Euler-Mashroni? go online find out 

2015-01-19-15-23
http://www.google.com/advanced_search
constant 0.577215
https://www.google.com/search?as_q=constant+0.577215&as_epq=&as_oq=&as_eq=&as_nlo=&as_nhi=&lr=&cr=&as_qdr=all&as_sitesearch=&as_occt=any&safe=images&tbs=&as_filetype=&as_rights=&gws_rd=ssl

2015-01-19-15-25
http://www.ebyte.it/library/educards/constants/MathConstants.html
Euler-Mascheroni_constant_www.ebyte.it_MathConstants.html

2015-01-19-15-31
http://mathworld.wolfram.com/Euler-MascheroniConstant.html
Euler-Mascheroni_constant_mathworld.wolfram.com_Euler-MascheroniConstant.html

<a name="docAe65">
2015-01-19-16-02 start
Weierstrass Representation eq.CG34 look 
pretty similar to Gauss Representation 
eq.CG33. Actually this is a point where 
we have to start.  
<a name="docAe66">
Gamma Function eq.CG01
Gauss Representation eq.CG33
Weierstrass Representation eq.CG34
 
Γ(s)
=
t=∞
t=0
ts-1*e-t*dt
  ▐  
Real(s)>0
--- eq.CG01
 
Γ(s)
=
limit
n→∞
[
ns

s
i=n
i=1
i

s+i
]
--- eq.CG33
--- proof
 
Γ(s)
=
1

s
e-γs
i=∞
i=1
es/i
[
1+
s

i
]
-1


--- eq.CG34
from eq.CG01 to eq.CG33 integral sign is removed
e=Euler's number=2.718281828459045...
γ=Euler-Mascheroni constant 0.57721566490153286061...
s=complex number, real(s)>0
width of above equation a401191611
<a name="docAe67">
2015-01-19-16-14 here
Start with Gauss Representation of Gamma 
Function eq.CG33 In eq.CG33 , we divide i 
at i/(s+i) both numerator and denominator.
<a name="docAe68">
 
Γ(s)
=
limit
n→∞
[
ns

s
i=n
i=1
i

s+i
]
--- eq.CG33
 
Γ(s)
=
limit
n→∞
[
ns

s
i=n
i=1
i/i

(s+i)/i
]
--- eq.CG35
 
Γ(s)
=
limit
n→∞
[
ns

s
i=n
i=1
1

(s/i)+1
]
docAe69
--- eq.CG36
 
Γ(s)
=
limit
n→∞
[
ns

s
i=n
i=1
(
1+
s

i
)
-1


]
--- eq.CG37
width of above equation a401191624
<a name="docAe70">
2015-01-19-16-27 here
You see the similarity between eq.CG37 and 
eq.CG34 [1 + s/i]-1 Now we have to manipulate 
ns in eq.CG37 in order to get e-γs
Let us just do it, We rewrite ns as es*log(n) 
as shown in eq.CG39 below. 
log() and exp() are inverse operation, that is 
M=exp(log(M)), let M be ns then ns=exp(log(ns))
Apply log law log(ns)=s*log(n) get 
ns=exp(s*log(n)) --- eq.CG38 
<a name="docAe71">
 
Γ(s)
=
limit
n→∞
[
es*log(n)

s
i=n
i=1
(
1+
s

i
)
-1


]
--- eq.CG39
 
Γ(s)
=
limit
n→∞
[
e0es*log(n)

s
i=n
i=1
(
1+
s

i
)
-1


]
--- eq.CG40
width of above equation a401191648

<a name="docAe72">
 
Γ(s)
=
lim
n→∞
[
e∑[i=1,n]s/i-∑[i=1,n]s/ies*log(n)

s
i=n
i=1
(
1+
s

i
)
-1


]
--- eq.CG41
 
Γ(s)
=
lim
n→∞
[
e∑[i=1,n]s/i e-∑[i=1,n]s/i+s*log(n)

s
i=n
i=1
(
1+
s

i
)
-1


]
--- eq.CG42
width of above equation a401191701
<a name="docAe73">
2015-01-19-18-15 here 
After eq.CG39 the next step is very very trivial. 
We just write e0 , it does not matter if multiply 
something with one. (see eq.CG40) Next step we 
want to introduce partial sum. Because zero can 
be written as something minus something, which 
give you zero again.
(see eq.CG41 ∑[i=1,n]s/i - ∑[i=1,n]s/i is zero)
<a name="docAe74">
In eq.CG42, if you look up the red term 
-∑[i=1,n]s/i+s*log(n) with lim[n→∞] We get 
lowercase gamma Euler-Mascheroni constant 
γ=lim[n→∞]{∑[i=1,n](1/i)-log(n)} --- eq.CG43 
γ=0.577 215 664 901 532 860 606 512 ••• 
γ=0.577215664901532860606512 ••• 
2015-01-19-15-31 LiuHH access 
http://mathworld.wolfram.com/Euler-MascheroniConstant.html
Substitute eq.CG43 into eq.CG42 get eq.CG44 below.
<a name="docAe75">
 
Γ(s)
=
lim
n→∞
[
e∑[i=1,n]s/i -γs

s
i=n
i=1
(
1+
s

i
)
-1


]
--- eq.CG44
 
e∑[i=1,n]s/i
=
i=n
i=1
es/i
    ;    
ea+b+c=ea*eb*ec
--- eq.CG45
 
Γ(s)
=
lim
n→∞
[
e-γs

s
i=n
i=1
es/i
(
1+
s

i
)
-1


]
--- eq.CG46
width of above equation a401191736
<a name="docAe76"> 
2015-01-19-18-35 here 
In eq.CG44 we use the last thing we need to know. 
Red term in eq.CG44 can be written as a product. 
See eq.CG45 left side. eq.CG45 right side is a 
simplified expression. In eq.CG46 e-γs/s term do 
not have n in it. We can move e-γs/s term to out 
side of lim[n→∞], see eq.CG34 below. On the 
other hand, in eq.CG44 e∑[i=1,n]s/i term has 
∑[i=1,n] , we can take this product e∑[i=1,n]s/i 
into the other product ∑[i=1,n](1+s/i)-1. See 
eq.CG46 
<a name="docAe77">
Actually eq.CG46 is almost the final step, we 
only need write n to infinity. In eq.CG46, only 
∑[i=1,n] contain 'n'. Write eq.CG46 as eq.CG34 
below. 
<a name="docAe78">
Gamma Function, Weierstrass Representation
 
Γ(s)
=
1

s
e-γs
i=∞
i=1
es/i
[
1+
s

i
]
-1


--- eq.CG34
e=Euler's number=2.718281828459045...
γ=Euler-Mascheroni constant 0.57721566490153286061...
s=complex number, real(s)>0
width of above equation a401191740

2015-01-19-18-53 here 
You are ready to see one of the awesome 
formula in calculus eq.CG34 . This is the 
end of this lecture. 
2015-01-19-18-58 stop


<a name="docAf01"> 2015-01-20-17-18 start Next is study notes of Gamma Function - Part 4 - Relationship to Sine http://www.youtube.com/watch?v=W3ulxj3s90U Lecturer is MrYouMath. Notes recorder is Liu,Hsinhan. Recorded notes is about 60% match, not exact. Liu,Hsinhan inserted text is blue Italic. LiuHH notes may contain error! <a name="docAf02"> Begin video 4 of 12 Hello and welcome you guys to the new video about Gamma Function and its relationship to sinus function. We will need Gauss representation of the Gamma function. Because it is easiest way I know to derive this.
<a name="docAf03">
Gamma Function, Gauss Representation
 
Γ(s)
=
limit
n→∞
[
ns

s
i=n
i=1
i

s+i
]
--- eq.CG33
 
Γ(s)
=
limit
n→∞
[
ns

s
i=n
i=1
i

i(1+s/i)
]
--- eq.CG47
 
Γ(-s)
=
limit
n→∞
[
n-s

-s
i=n
i=1
i

i(1-s/i)
]
--- eq.CG48
width of above equation a401201745
<a name="docAf04">
2015-01-20-17-46 here
From eq.CG33  to eq.CG47, we take i out of (s+i). 
We are not only use this Γ(s), we will use Γ(-s).
What can we do with that? You are looking at ns
and n-s, multiply them together, then they will 
cancel. Multiply eq.CG47 with eq.CG48 get eq.CG49
In eq.CG49, already cancel i/i to 1 (i/i exist in 
eq.CG47 and eq.CG48)
<a name="docAf05">
 
Γ(s)Γ(-s)
=
limit
n→∞
[
ns

s
n-s

-s
i=n
i=1
1

1+s/i
1

1-s/i
]
--- eq.CG49
 
Γ(s)Γ(-s)*(-s)
=
limit
n→∞
[
1

s
i=n
i=1
1

1-s2/i2
]
--- eq.CG50
<a name="docAf06">
Gamma Functional Equation       Γ(s+1) = s*Γ(s) --- eq.CG05
Gamma Functional Equation       Γ(-s+1) = -s*Γ(-s) --- eq.CG51
 
Γ(s)Γ(1-s)
=
1

s
i=∞
i=1
1

1-s2/i2
               
--- eq.CG52
width of above equation a401201804
<a name="docAf07">
2015-01-21-12-14 start 
In eq.CG47 and eq.CG48, both limit exist. After 
eq.CG47 multiply eq.CG48, we can take limit sign 
together, see eq.CG49. 
In eq.CG49 ns*n-s cancel to one, see eq.CG50.
In eq.CG49 (1+s/i)*(1-s/i) get eq.CG50 (1-s2/i2)
I will do another trick. In eq.CG49 take -s out 
of limit sign and move -s to left hand side, see 
eq.CG50. I can do that because s is not part of 
the limit. (Limit sign control n, not control s)
After we did all of these, get eq.CG50. In 
eq.CG50 left hand side term Γ(-s)*(-s) is last 
trick I want to use before get final formula. 
<a name="docAf08">
If you look at Gamma Functional Equation      
  Γ(s+1) = s*Γ(s) --- eq.CG05 
and change +s to -s 
  Γ(-s+1) = -s*Γ(-s) --- eq.CG51 
Substitute eq.CG51 into eq.CG50 left hand side 
get eq.CG52  
From eq.CG50 to eq.CG52, there is another 
change. In eq.CG50 limit[n→∞]∏[i=1,i=n] 
merge to one symbol ∏[i=1,i=∞] in eq.CG52.

<a name="docAf09">
Now let us exam eq.CG52 right hand side.
when s=0, eq.CG52 right hand side 1/s is a pole.
s=1, eq.CG52 right hand side 1/(1-12/12) is a pole.
s=2, eq.CG52 right hand side 1/(1-22/22) is a pole.
s=3, eq.CG52 right hand side 1/(1-32/32) is a pole.
s=-4,eq.CG52 right side 1/[1-(-4)2/(-4)2] is a pole.
eq.CG52 right side has pole at all integer s. 
Infinite many zeros at all integer number s. 
There is a famous function the sin(sπ) has zeros 
at all integer number s. I will prove this in a 
separate video. (See eq.cz93 proof at 
http://freeman2.com/tute0060.htm#docAd02 )
<a name="docAf10">
Sine Function Product Formula
 
sin(πs)

πs
=
(
1 -
s2

12
)
(
1 -
s2

22
)
(
1 -
s2

32
)
●●●
--- eq.cz93
--- repeat
 
sin(πz)

πz
    =    
n=∞
n=1
[
1 -
z2

n2
]
     
--- eq.czP5
---Euler use
 
π

sin(πs)
=
1

s
i=∞
i=1
1

1-s2/i2
               
--- eq.CG53
--- here use
2014-12-17-09-53 https://www.youtube.com/watch?v=h3Hcioh2F9I
width of above equation a401201832
<a name="docAf11"> 
2015-01-21-13-11 here
Above eq.cz93 is proved at 
http://freeman2.com/tute0060.htm#docAd02
Euler use eq.czP5 above. 
Here  use eq.CG53 above. 
In eq.CG53 if you enter 1,2,3 for s. Both 
side have same zeros at denominator. 
Substitute eq.CG53 to eq.CG52 get eq.CG54
<a name="docAf12">
Gamma Function Relationship to Sine
 
Γ(s)Γ(1-s)
=
π

sin(πs)
--- eq.CG54

width of above equation a401201840
<a name="docAf13">
eq.CG54 conclude our lecture. The only part 
I did not prove is eq.CG53 (proved at next URL
http://freeman2.com/tute0060.htm#docAd02 )
I will prove in a separate video. 
That conclude the lecture. See you guys.
2015-01-21-13-25 stop

<a name="a40121a"> 
2015-01-21-13-28 start
Review earlier work, 
http://freeman2.com/tute0060.htm#docAa53
has 
[[
<a name="docAa53">
eq.czL4 is Euler reflection formula. a312152030 
Please watch //2014-12-15-22-55 add link 
Gamma Function - Part 4 - Relationship to Sine 
http://www.youtube.com/watch?v=W3ulxj3s90U 
eq.czL4 modify to eq.czL8 , 
eq.czL8 is used at docAa58 from eq.czL9 to eq.czM0 
RIGHT hand side.    
Γ(s)*Γ(1-s)=π/sin(πs) --- eq.czL4  
]]
2015-01-21-13-33 stop

<a name="a40121b">
2015-01-21-18-28 start
Above prove Gamma Function Relationship to Sine 
eq.CG54 start from Gauss Representation eq.CG33.
Below prove Gamma Function Relationship to Sine 
start from Weierstrass Representation eq.CG34.
Γ(s) is eq.CG34, Γ(-s) is eq.LAb1.
<a name="a40121c">
Gamma Function, Weierstrass Representation
 
Γ(s)
=
1

s
e-γs
i=∞
i=1
es/i
[
1+
s

i
]
-1


--- eq.CG34
 
Γ(-s)
=
1

-s
e-γ(-s)
i=∞
i=1
e-s/i
[
1+
-s

i
]
-1


--- eq.LAb1
e=Euler's number=2.718281828459045...
γ=Euler-Mascheroni constant 0.57721566490153286061...
s=complex number, real(s)>0
width of above equation a401211835

<a name="a40121d">
 
Γ(s)Γ(-s)
=
e-γs

s
e+γs

-s
i=∞
i=1
es/i - s/i
[
1

1+s/i
1

1-s/i
]
--- eq.LAb2
 
Γ(s)Γ(-s)*(-s)
=
limit
n→∞
[
1

s
i=n
i=1
1

1-s2/i2
]
--- eq.CG50
width of above equation a401211848
<a name="a40121e">
2015-01-21-19-05 here 
Multiply Γ(s) eq.CG34 with Γ(-s) eq.LAb1 get 
eq.LAb2. In eq.LAb2 eγs*e-γs cancel to one, 
and es/i - s/i cancel to one.
In eq.LAb2 move right side denominator (-s) 
to left side numerator (-s), same as eq.CG50.
In eq.LAb2 right side denominator multiply 
(1+s/i) with (1-s/i), same as eq.CG50 right 
side denominator term. 
<a name="a40121f">
In eq.LAb2 ∏[i=1,∞] is same as eq.CG50 
lim[n→∞]∏[i=1,n]. After above comparison, 
eq.LAb2 is same as eq.CG50 and the following 
Weierstrass Representation derivation is same 
as Gauss Representation derivation and get 
same conclusion.
2015-01-21-19-17 stop
2015-01-22-10-58 done first proofread
2015-01-22-13-15 done second proofread


<a name="docAf31"> 2015-01-22-13-41 start Next is study notes of Gamma Function - Part 5 - Gamma of 0.5 ( one half) http://www.youtube.com/watch?v=_vwqsJNKY-c Lecturer is MrYouMath. Notes recorder is Liu,Hsinhan. Recorded notes is about 60% match, not exact. Liu,Hsinhan inserted text is blue Italic. LiuHH notes may contain error! <a name="docAf32"> Begin video 5 of 12 Hello and welcome you guys to the new video about Gamma Function. In this video we will talk about the value of one half Γ(1/2) This integral is very important in physics. You will see why. Just let us start. Our starting point is eq.CG54 below.
<a name="docAf33">
Gamma Function Relationship to Sine
 
Γ(s)Γ(1-s)
=
π

sin(πs)
--- eq.CG54
In eq.CG54 let s=1/2 get eq.CG55.
 
Γ(1/2)Γ(1-1/2)
=
π

sin(π/2)
--- eq.CG55

width of above equation a401221401
<a name="docAf34">
2015-01-22-14-05 here
In eq.CG55 left hand side is Γ(1/2)*Γ(1/2). 
Right hand side is π/sin(π/2). But sin(π/2)=1.
eq.CG55 is 
  Γ2(1/2)=π --- eq.CG56 
We take square root get
  Γ(1/2)=√π --- eq.CG57 
I hope you can not imagine what it means. That 
is the reason why I will show you what does 
eq.CG57 mean. We will use the definition of 
Gamma function. 
<a name="docAf35">
 
Γ(s)
=
t=∞
t=0
ts-1*e-t*dt
  ▐  
Real(s)>0
--- eq.CG01
 
Γ(1/2)
=
t=∞
t=0
t0.5-1*e-t*dt
=
t=∞
t=0
e-t*dt

√t
--- eq.CG58
width of above equation a401221435
<a name="docAf36"> 
2015-01-22-14-36 here
eq.CG58 right side integral will give you 
a value of square root of PI. That is not 
actually important for us. 
 
√π
=
t=∞
t=0
e-t*dt

√t
--- eq.CG59
width of above equation a401221449

<a name="docAf37">
We will do a little substitution here. See eq.CG60.
  t=pu2 --- eq.CG60
  dt=2pudu --- eq.CG61
p>0 p is a positive constant number.
Substitute t and dt to eq.CG59 get next

<a name="docAf37a">
From 2015-01-22-14-50 to 2015-01-22-15-00
realtor agent and two buyer02 enter house. See
http://freeman2.com/bioge010.htm#a40122d

 
u=∞
u=0
(e-pu2)*2pudu

√(pu2)
=
2√p
u=∞
u=0
(e-pu2)du

=
√π
--- eq.CG62
width of above equation a401221555
<a name="docAf38">
2015-01-22-16-01 here
In eq.CG62 left side, u is integration variable. 
2 is constant and p is constant. We can take 2,p 
out of integral sign. Cancel √p then get eq.CG62 
middle term 2√p. eq.CG62 left has denominator 
√(u2) which cancel numerator u. 
eq.CG59 variable t boundary is from t=0 to t=∞
eq.CG62 variable u boundary is from u=0 to u=∞
You may ask why change variable eq.CG60, the 
boundary not change? Because t=0, u still 0 and 
t=∞ u still ∞ . u range from 0 to ∞. u is never 
zero. 

<a name="docAf39">
Now I am doing another trick. eq.CG62 integral is 
an even integral (2∫u2du) We can cancel '2' and 
extend integral from u=0 to u=∞ to u=-∞ to u=∞ 
Finally we get 
 
π

p
=
u=∞
u=-∞
e-pu2*du

--- eq.CG63
width of above equation a401221635
<a name="docAf40">
2015-01-22-16-38 here
eq.CG63 is very important integral in physics. 
eq.CG63 is Gauss integral even if this formula 
was known by Euler hundred years ago. 
eq.CG63 is part of normal distribution. 
That is it. See you guys. 
2015-01-22-16-42 stop
2015-01-22-19-26 done proofread


<a name="docAf61"> 2015-01-23-13-23 start Next is study notes of Gamma Function - Part 6 - Stirling's Approximation http://www.youtube.com/watch?v=MuAb2dnPD0Q Lecturer is MrYouMath. Notes recorder is Liu,Hsinhan. Recorded notes is about 60% match, not exact. Liu,Hsinhan inserted text is blue Italic. LiuHH notes may contain error! <a name="docAf62"> Begin video 6 of 12 Hello and welcome you guys to my new video about Gamma Function. In this video we will derive the famous Stirling's Approximation which you can see here beneath.
<a name="docAf63"> improved formula eq.LAk2
Stirling's Approximation formula
 
n!
nn

en
√(2nπ)
  ▐  
n → ∞
--- eq.CG64
width of above equation a401231349
<a name="docAf64">
2015-01-23-20-27 start 
If you remember previous video which we derived 
the formula for Γ(1/2) video 5/12, eq.CG57 
  Γ(1/2)=√π --- eq.CG57
You saw something like this √π. You can imagine 
eq.CG64 has something to do with Gamma function. 
Let us look how can we derive that. Start point 
is eq.CG65 below.
<a name="docAf65">
 
Γ(n+1)
= n! =
t=∞
t=0
tn*e-t*dt
  ▐  
n=integer
--- eq.CG65
width of above equation a401232046
<a name="docAf66"> 
2015-01-23-20-48 here
Factorial of n can be expressed as eq.CG65 
integration term. This is the most important 
thing we can notice here. First step I want 
do is I do not want to look at the integral 
right now. I just want to exam eq.CG66 below 
  log(tn*e-t)=n*log(t)-t --- eq.CG66 
LiuHH add detail steps below 
  log(tn*e-t)=log(tn)+log(e-t) --- eq.CG67
  log(tn*e-t)=n*log(t)+(-t)*log(e) --- eq.CG68 
  log(tn*e-t)=n*log(t)-t*1 --- eq.CG69 
  log(tn*e-t)=n*log(t)-t --- eq.CG66 
<a name="docAf67">
If you do not know above steps, please look up 
calculus logarithm laws. Now we will do a 
little bit substitution 
  n*log(t)-t=n*log(n+ε)-(n+ε) --- eq.CG70 
We replace t in eq.CG66 with a new variable 
(n+ε). 
  t=n+ε --- eq.CG70A  //a401241536 use eq.CG70A 
We can do that, there is no problem. 
Now I want concentrate on log(n+ε) part. 
  log(n+ε)=log[n*(1+ε/n)] --- eq.CG71 
Take n out of (n+ε), [ n*(1+ε/n)=(n+ε) ]
<a name="docAf68">
Use logarithm law, write eq.CG71 as eq.CG72 
  log(n+ε)=log(n)+log(1+ε/n) --- eq.CG72 
Now this looks nice. What I am going to do 
right now is that I want to introduce some 
thing.
For very large n, we can guarantee that 
  ε/n < 1 --- eq.CG73 
We can do special sum of logarithm 
<a name="docAf69">
 
log
(
1 +
ε

n
)
=
k=∞
k=1
(-1)k+1

k
εk

nk
--- eq.CG74
width of above equation a401232122
<a name="docAf70">
2015-01-23-21-24 here 
I proved eq.CG74 in my special video about 
Taylor series about logarithm function. You 
can look at my channel and find out. 
2015-01-23-21-28 stop 
2015-01-24-02-38 start
Normally we have xk, but in eq.CG74 we have 
to plug in εk/nk. (-1)k+1 is alternating sum.
Sign change all the time. 
Very important, we start from 1 ∑[k=1,k=∞]
We will plug eq.CG74 into eq.CG72 log(1+ε/n) 
Ultimately will plug eq.CG74 into eq.CG70 
n*log(n+ε) term. Let us see what happen.
<a name="docAf71">
Substitute eq.CG74 into eq.CG72 , then into eq.CG70 ,
eq.CG70 become eq.CG75 below.
 
n*log(n+ε)-(n+ε)
=
n*
[
log(n)
+
k=∞
k=1
(-1)k+1

k
εk

nk
]
-n
--- eq.CG75
 
n*log(n+ε)-(n+ε)
=
n*log(n) -n
+
k=∞
k=1
(-1)k+1

k
nk

nk
--- eq.CG76
 
n*log(n+ε)-(n+ε)
=
n*log(n) -n
+
k=∞
k=1
(-1)k+1

k
εk

nk-1
--- docAf72
--- eq.CG77
 
n*log(n+ε)-(n+ε)
=
n*log(n) -n
+
k=∞
k=2
(-1)k+1

k
εk

nk-1
--- eq.CG78
width of above equation a401240315
<a name="docAf73">
2015-01-24-03-17 here 
When n is very large than ε, it is good to write 
as power sum. Now I do another step. I look the 
summation sign ∑[k=1,∞] in eq.CG77 If I plug in 
k=1 what will happen? (-1)k+1=(-1)1+1 gives 
just one, and when k=1 eq.CG77 denominator 1/k 
is one. The term εk/nk-1 is ε1/n1-1=ε/1=ε 
This ε1/n1-1 cancel eq.CG77 right most -ε . 
Then eq.CG77 change ∑[k=1,∞] to ∑[k=2,∞] and drop 
right most -ε. We have eq.CG78. Now eq.CG78 is 
awesome. In eq.CG78 Stirling's Approximation only 
use the first part of the strange sum. First write 
down eq.CG78 in expanded form like eq.CG79 below. 
2015-01-24-03-39 here
<a name="docAf74">
 
n*log(n+ε)-(n+ε)
=
n*log(n)-n
ε2

2n
ε3

3n2
ε4

4n3
± ●●●
--- eq.CG79
width of above equation a401240344

<a name="docAf75">
Stirling's Approximation drop all purple terms in eq.CG79
Remember eq.CG66 and eq.CG70

log(tn*e-t)=n*log(t)-t --- eq.CG66
n*log(t)-t=n*log(n+ε)-(n+ε) --- eq.CG70

Now truncated eq.CG79 is eq.CG80 next
eq.CG80 take exp() operation get eq.CG81.


<a name="docAf76">
 
log(tn*e-t)
n*log(n)-n
ε2

2n
--- eq.CG80
 
tn*e-t
nn

en
e2/(2n)
       
--- eq.CG81
exp(eq.CG80) get eq.CG81
width of above equation a401240416
<a name="docAf77">
2015-01-24-09-41 start 
In eq.CG79 you can see the pattern here. It is 
alternating sign sum. The powers are increasing. 
Now I just take this expanded sum and say we 
stop at -ε2/(2n) and drop all other terms. 
Now we have eq.CG80 (after drop terms, use )
Now what I will do I will exponentiate both side 
of eq.CG80 get eq.CG81 . What we will do ultimately 
we will plug eq.CG81 into eq.CG65. See what happen? 
You plug eq.CG81 into eq.CG65 get eq.CG82 . 
<a name="docAf78">
In eq.CG82 our integration variable become ε .
 
Γ(n+1)
= n! =
t=∞
t=0
tn*e-t*dt
  ▐  
n=integer
--- eq.CG65
 
n!
ε=∞
ε=-n
nn

en
e2/(2n)
▐   let n→∞ then
▐  ε=-n to ε→-∞
--- eq.CG82
 
n!
nn

en
ε=∞
ε=-∞
e2/(2n)


▐   n is NOT ε
▐   move n out
--- docAf79
--- eq.CG83
 
π

p
=
u=∞
u=-∞
e-pu2*du

  ▐   see video for Γ(1/2)
  ▐   p=1/(2n) in eq.CG83
--- eq.CG63
eq.CG82 boundary not compatible with eq.CG63
eq.CG83 boundary IS compatible with eq.CG63
width of above equation a401240448
<a name="docAf80">
2015-01-24-09-55 here
2015-01-24-09-55 Sister told LiuHH that 
realtor agent just called sister, one 
buyer03 will come see house soon. Only buyer03 
come see house, agent not follow. LiuHH 
and sister clean up house immediately. 
2015-01-24-10-21 buyer03 and agent come 
2015-01-24-10-57 buyer03 and agent left
Agent name Linda Garcia at Pinnacle 
Buyer03 name Bob. Bob has more possibility 
(than buyer02) to buy this house for 
some one else move in. Agent Linda say 
she will give us information about rent 
another house. See 
http://freeman2.com/bioge010.htm#a40124b
2015-01-24-11-02 record stop

<a name="docAf81"> 
2015-01-24-11-13 start 
Now in eq.CG82 integration variable ε have to 
start from ε=-n. Because remember in eq.CG65 
variable t was equal n+ε 
 t=n+ε --- eq.CG70A 
In order to have first value t=0, 
n+ε=0 get ε=-n as integration start point. Now 
let us say, let us make n infinity large. 
(Stirling's Approximation formula is designed 
 for n → ∞ , see eq.CG64. Therefore 
 "let us make n infinity large" is consistent.
 Why let n → ∞ ? See eq.CG63, integral boundary 
 is ∫[u=-∞,u=+∞] . 
 eq.CG82 boundary not compatible with eq.CG63
 eq.CG83 boundary IS  compatible with eq.CG63)
<a name="docAf82">
After n → ∞ we have eq.CG83 
From eq.CG82 to eq.CG83 there is another change.
In eq.CG82 nn/en is not depend on ε. We can move 
nn/en out of integral. Result is eq.CG83
We know from previous video we have eq.CG63 
Compare eq.CG83 with eq.CG63 We find p in eq.CG63 
is 1/(2n) in eq.CG83. 
  p = 1/(2n) --- eq.CG84 
After p replacement, eq.CG83 become eq.CG64 below.
<a name="docAf83"> improved formula eq.LAk2
Stirling's Approximation formula
 
n!
nn

en
√(2nπ)
  ▐  
n → ∞
--- eq.CG64
width of above equation a401240452
2015-01-24-04-52 build equation stop.
<a name="docAf84">
2015-01-24-11-36 here
eq.CG64 is famous Stirling's Approximation formula 
I hope you have fun on this video. Please subscribe 
if you want to see my new upcoming videos. 
See you guys.
2015-01-24-11-38 stop
2015-01-24-16-04 done first proofread 
2015-01-24-17-05 done second proofread 


<a name="a40124a"> 2015-01-24-21-35 start //OSLER 1,2,3 2015-01-24-17-33 upload MrYouMath video lecture to http://freeman2.com/tute0062.htm 2015-01-24-18-14 Liu,Hsinhan access http://www.rowan.edu/open/depts/math/osler/Stirlings_formula.pdf The following is study notes following [[ Another intuitive approach to Stirling’s formula THOMAS J. OSLER Mathematics Department, Rowan University, Glassboro, NJ 08028, USA E-mail: Osler@rowan.edu ]] From page 1/8 to 3/8. 2. An intuitive derivation of Stirling’s formula <a name="a40124b"> Following study notes is LiuHH own words. LiuHH notes may contain error! If given a problem, ask find the answer of 1+2+3+4+5+6+7+8+9+10 we can do the following adding the given formula and reverse order formula COLUMNWISE 1+ 2+ 3+ 4+ 5+ 6+ 7+ 8+ 9+10 10+ 9+ 8+ 7+ 6+ 5+ 4+ 3+ 2+ 1 ▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬ 11+11+11+11+11+11+11+11+11+11 <a name="a40124c"> each column sum to 10+1, there are total 10 such number. The total number is 10*(10+1) which is 110. But we add same problem twice. Divide the answer with 2 get 10*(10+1)/2 which is 55. This result 55 is correct answer of original problem. This formula 10*(10+1)/2 is a general formula. Which can be written as n*(n+1)/2 and n is the maximum number 10 of given problem 1+ 2+ 3+ 4+ 5+ 6+ 7+ 8+ 9+10 <a name="a40124d"> Now if ask sum 1+ 2+ 3+ ... 999+1000 In the new problem, maximum number is 1000 Use the formula n*(n+1)/2 find 1000*(1000+1)/2 = 1001*500 = 500500 This is correct answer. The formula n*(n+1)/2 is easier than 1+ 2+ 3+ ... 999+1000 . Because we need to know only the maximum number n to get correct answer. We do not need adding one thousand times. Above is summation 1+2+3+4+5+6+7+8+9+10 Below is multiplication 1*2*3*4*5*6*7*8*9*10 <a name="a40124e"> What is the answer of 1*2*3*4*5*6*7*8*9*10 ? Direct calculation get 3628800 What is the answer of 1*2*3*...*999*1000 ? Direct calculation get 4.023872600770938E2677 Is there an easier formula to find answer for multiplication start from one, base on only the largest number 1000 to get answer? No! at this moment. But there is an estimation formula. It is called Stirling's Approximation formula //proof 1, 2 n!≈[√(2πn)]*nn*e-n --- eq.LAb3 //eq.CG64 π=PI=3.141592653589793... //eq.LAk2 e= 2.718281828459045... //log(e)=1 n=factorial multiplication largest number <a name="a40124f"> Why use Stirling’s formula eq.LAb3 ? Because eq.LAb3 right side is simpler. It has only n,π,e. However eq.LAb3 left side is harder. To calculate n!, for example 10!, need ten numbers to get 1*2*3*4*5*6*7*8*9*10=3628800 But eq.LAb3 need only n,π,e three numbers to get approximate value 3598695.61874 compare with exact 3628800. 3598695.61874/3628800 = 0.991704039555776 <a name="a40124g"> Above is 10!. If ask for value of 10000! then Stirling's Approximation is much easier. Still three numbers 10000,π,e. But exact calculation need 10000 numbers ! 10000!=2.84625 96809 17055 E 35659 2015-01-24-22-00 stop <a name="a40125a"> 2015-01-25-11-32 start //'T'=THOMAS ; improved formula An intuitive derivation of Stirling’s formula The following is a non-rigorous derivations of Stirling’s formula follow THOMAS J. OSLER paper.
 
Γ(x)
=
t=∞
t=0
tx-1*e-t*dt
  ▐  
Real(x)>0
--- eq.CG01
width of above equation a401251138
<a name="a40125b">
2015-01-25-11-42 here 
Gamma function at half integer Γ(n+1/2) is 
 
Γ(n+1/2)
=
1*3*5*...*(2n-1)

2n
√π
=
(2n)!

22n*n!
√π
  ▐  n=0,1,2,...
--- eq.LAb4
width of above equation a401251155
2015-01-25-11-55 stop 
<a name="a40125c">
2015-01-25-12-26 start 
Why eq.LAb4 is correct? See next two equations
  Γ(s+1) = s*Γ(s) --- eq.CG05 
  Γ(1/2)=√π --- eq.CG57 
In eq.LAb4 'n' is abstract. Let n=6 then 
  Γ(n+1/2)=Γ(6.5)=Γ(5.5+1)=5.5*Γ(5.5)=5.5*4.5*Γ(4.5) 
         =...=5.5*4.5*3.5*2.5*1.5*0.5*Γ(0.5) 
  Γ(6.5) =(11/2)*(9/2)*(7/2)*(5/2)*(3/2)*(1/2)*Γ(0.5) 
  Γ(6.5) =[(2*6-1)*9*7*5*3*1]*Γ(0.5)/26 --- eq.LAb5 
eq.LAb5 is same as eq.LAb4 with n=6
eq.LAb4 has √π, eq.LAb5 has Γ(0.5), they are one 
thing, see eq.CG57. Above verified eq.LAb4 left 
equality. How about right equality?

<a name="a40125d">
Second step, eq.LAb4 middle term equal right term.
That is easy. In eq.LAb4 middle term numerator and 
denominator insert 2*4*6*...*2n/(2*4*6*...*2n) 
This inserted value is one. Equation multiply by 
one not change its value. 
eq.LAb4 middle term numerator become 
 [1*3*5*...*(2n-1)]*[2*4*6*...*2n] which is (2n)!
eq.LAb4 middle term demoninator become 
2n*[2*4*6*...*2n] which is 2n*2n*[1*2*3*...*n] 
which is 22n*n! this is exactly eq.LAb4 right term 
demoninator. So this eq.LAb4 is reasonable. 
2015-01-25-12-48 stop

<a name="a40125e">
2015-01-25-13-48 start 
eq.LAb4 right hand side has √π and 22n, it is easy 
to calculate √π and 22n. But the factorial (2n)!/n! 
is a trouble. If ask factorial(million), we do not 
want multiply a million numbers. Next step, try to 
dissolve factorial. Because multiplication is hard 
to deal, for example 5!=1*2*3*4*5. But addition is 
easier to work with. How to change multiplication 
to addition? Log() function do this job. 
<a name="a40125f"> 
We know 
  log(a*b)=log(a)+log(b) --- eq.LAb6 
  log(c/d)=log(c)-log(d) --- eq.LAb7 
Let us take log of (2n)!/n! get 
  log[(2n)!/n!]=log[(2n)!]-log[n!]
     =log(n+1)+log(n+2)+log(n+3)+...+log(2n) --- eq.LAb8
Although summation is better than multiplication. 
But summation is not as good as integration.
Because integration has closed formula for many 
math terms. We want change summation to integration. 
<a name="a40125g">
See next log(n+k) summation and log(x) integration

http://freeman2.com/tut062a1.gif
<a name="a40125h">
Above Figure 1 is curve of  
  y=log(x) --- eq.LAb9 
To find the integral value of log(x), we use 
vertical stripe summation method. Each stripe 
has height log(n+integer) and width=1. 

<a name="a40125i">
Let us write next equation 
 
log
(2n)!

n!
=
log(1)+log(2)+log(3)+●●●+log(2n-1)+log(2n)
-log(1)-log(2)-log(3)-●●●-log(n-1)-log(n)
--- eq.LAc0
 
log
(2n)!

n!
=
k=2n
k=n+1
log(k)
x=2n+1/2
x=n+1/2
log(x)
dx
         
--- eq.LAc1
width of above equation a401251431
<a name="a40125j">
2015-01-25-14-32 here 
eq.LAc1 right side no multiplication, no summation 
only integration which is easiest. BUT the price 
paid is that the step from summation to integration 
formula is not exact. Because y=log(x) is not a 
straight line. Vertical stripe rectangle area is 
NOT the area under curve. We have to use the  
symbol. Next do the integration. Apply 
Integration by parts formula tute0061.htm#a40103a
  ∫udv=uv-∫vdu --- eq.LAc2 
Compare eq.LAc2 left side with eq.LAc1 right side.
<a name="a40125k"> 
Let udv be log(x)dx then 
  ∫log(x)dx=[log(x)]*x-∫x*dlog(x)
  ∫log(x)dx=[log(x)]*x-∫x*(1/x)*dx
  ∫log(x)dx=[log(x)]*x-x |[x=n+1/2,x=2n+1/2]
           =[log(x)]*x-x |[x=2n+1/2]
           -[log(x)]*x+x |[x=n+1/2]
           =[log(2n+1/2)]*(2n+1/2)-(2n+1/2)
           -[log( n+1/2)]*( n+1/2)+( n+1/2)
  ∫log(x)dx=[log(2n+1/2)]*(2n+1/2)-[log(n+1/2)]*(n+1/2)-n --- eq.LAc3
Combine eq.LAc1 and eq.LAc3 get next equation
<a name="a40125l">
Pay attention to eq.LAc4 to eq.LAc7 technique.
How to apply eq.LAe2 at later step. a401291031
 
log
(2n)!

n!
x=2n+1/2
x=n+1/2
log(x)
dx
[log(2n+1/2)]*(2n+1/2)
-[log(n+1/2)]*(n+1/2)-n
--- eq.LAc4
 
(2n+1/2)
log
[
2n
(
1+
1

2*2n
)
]
-(n+1/2)
log
[
n
(
1+
1

2n
)
]
-n
--- eq.LAc5
 
From eq.LAc4 to eq.LAc5, write log(2n+1/2) as log{2n[1+1/(4n)]} //a40125m
this is an important step. Allow (1+x/n)n ≈ ex when n→∞ to work
eq.LAc5 and eq.LAc6 are identical. delete eq.LAc6. 2015-01-26-11-31
--- eq.LAc6
 
(2n+1/2)
[
log
(2n)
+ log
(
1+
1

4n
)
]
-(n+1/2)
[
log
(n)
+ log
(
1+
1

2n
)
]
-n
--- eq.LAc7
width of above equation a401251520
<a name="a40125n">
2015-01-25-15-33 here 
Next step, we will take exponential to eq.LAc7 
It is convenient to write eq.LAc7 as eq.LAc8 
below. 
 
log
(2n)!

n!
+ (2n+1/2)*log(2n) + (2n+1/2)*log[1+1/(4n)]
-(n+1/2)*log(n) - (n+1/2)*log[1+1/(2n)] -n
--- eq.LAc8
width of above equation a401251531
<a name="a40125o">
2015-01-25-15-37 here 
Originally we have eq.LAb4 which is an equality 
equation. We take log() operation to eq.LAb4 
factorial part [(2n)!/n!] see eq.LAc0 
Convert from hardiest multiplication to easier 
addition of log(). Next change easier addition 
to easiest integration see eq.LAc1 . This step 
introduced small error. Equality equation become 
approximation equation . Next step apply 
exponentiation to eq.LAc8, back to before-log 
condition. Reader must know log() and exp() are 
inverse operation, that is 
  M=exp(log(M)) --- eq.LAc9 
<a name="a40125p"> 
Now apply exponentiation to eq.LAc8 one term by 
one term. 
  exp{log[(2n)!/n!]}=(2n)!/n! --- eq.LAd0 
eq.LAd0 is true base on eq.LAc9. Next term 
  exp{(2n+1/2)*log(2n)}=exp{log[(2n)(2n+1/2)]} --- eq.LAd1
  exp{(2n+1/2)*log(2n)}=(2n)(2n+1/2) --- eq.LAd2
eq.LAd1 left side we can NOT apply eq.LAc9 to it. 
Because eq.LAc9 ask for simple M. Not allow 
multiplication like (2n+1/2)*log(2n) But 
logarithm law allow us write (2n+1/2)*log(2n) 
as log[(2n)(2n+1/2)] and (2n)(2n+1/2) is simple M.
eq.LAd2 is the correct operation.
<a name="a40125q">
Similarily 
  exp{(2n+1/2)*log[1+1/(4n)]}=[1+1/(4n)](2n+1/2) --- eq.LAd3
  exp{-(n+1/2)*log(n)}=n-(n+1/2) --- eq.LAd4
  exp{-(n+1/2)*log[1+1/(2n)]}=[1+1/(2n)]-(n+1/2) --- eq.LAd5
  exp{-n}=e-n --- eq.LAd6
Remember 
  exp(A+B-C)=exp(A)*exp(B)/exp(C) --- eq.LAd7 
  exp[log(D*E/F)]=D*E/F --- eq.LAd8 
Next put eq.LAd0 eq.LAd2 to eq.LAd6 together 
<a name="a40125r">
 
(2n)!

n!
(2n)(2n+1/2)*[1+1/(4n)](2n+1/2)*e-n

n+(n+1/2)*[1+1/(2n)]+(n+1/2)
--- eq.LAd9
 
(2n)!

n!
2(2n+1/2)*n(2n+1/2)*[1+1/(2*2n)](2n)*√[1+1/(4n)]*e-n

n(n+1/2)*[1+1/(2n)]n*√[1+1/(2n)]
--- eq.LAe0
 
(2n)!

n!
√2*22n*nn*[1+(1/2)/(2n)]2n*√[1+1/(4n)]*e-n

[1+(1/2)/n]n*√[1+1/(2n)]
--- eq.LAe1
width of above equation a401251643
<a name="a40125s">
2015-01-25-16-58 here 
Refer to eq.LEb0 
  (1+x/n)n ≈ ex when n→∞ --- eq.LAe2
eq.LAe0 red terms indicate match eq.LAe2 pattern. 
Our goal is find Stirling's Approximation formula 
for very large n. 
In eq.LAe1 let n approach to infinity.
Apply eq.LAe2 to eq.LAe1 red terms. Both red are 
e1/2, denominator e1/2 cancel numerator e1/2.

<a name="a40125t">
It is easy to understand eq.LAe3, because a/∞=0 
  √(1+a/n) ≈ 1 when n→∞ --- eq.LAe3
Next apply eq.LAe3 to eq.LAe1 blue terms. Both 
blue are one. eq.LAe1 red,blue are all gone. 
eq.LAe1 become 
   [(2n)!]/[n!]≈√2*22n*nn*e-n --- eq.LAe4 
Substitute eq.LAe4 to Γ(n+1/2) eq.LAb4 
  Γ(n+1/2)=[(2n)!]*√π/[22n*n!] --- eq.LAb4
get //factorial in eq.LAb4 is gone!
  Γ(n+1/2)≈[√2*22n*nn*e-n]*√π/[22n]
<a name="a40125u"> 
cancel 22n and merge √2 with √π get 
  Γ(n+1/2)≈√(2π)*nn*e-n --- eq.LAe5
Compare eq.LAe5 with eq.CG64
eq.LAe5 is Γ(n+1/2) ≈ √(2π)*nn*e-n
eq.CG64 is Γ(n+1)   ≈ √(2nπ)*nn*e-n
There is difference √n 
2015-01-25-17-28 stop 

<a name="a40125v">
2015-01-25-18-06 start
From the continuity of the gamma function makes 
it reasonable to replace n with k+1/2. 
eq.LAe5 become 
  Γ(k+1/2+1/2)≈√(2π)*(k+1/2)(k+1/2)*e-(k+1/2) --- eq.LAe6
Rewrite (k+1/2) as k*[1+1/(2k)]
  Γ(k+1)≈√(2π)*k(k+1/2)*[1+1/(2k)](k+1/2)*e-(k+1/2) --- eq.LAe7
  Γ(k+1)≈√(2π)*k1/2*kk*[1+(1/2)/k]k*[1+1/(2k)]1/2*e-k*e-1/2 --- eq.LAe8
<a name="a40125w">
Apply eq.LAe2 get [1+(1/2)/k]ke1/2 when →∞
Apply eq.LAe3 get [1+1/(2k)]1/21 when →∞
  Γ(k+1)≈√(2kπ)*kk*e1/2*1*e-k*e-1/2 --- eq.LAe9
Cancel e+1/2 and e-1/2 get //improved formula 
Stirling's Approximation formula
  Γ(k+1)≈√(2kπ)*kk*e-k --- eq.LAf0
k is dummy variable, change k to n if you like. 
Compare eq.LAf0 with eq.CG64. //Γ(k+1)=k! 

Above is page 1/8 to 3/8 of next pdf file 
http://www.rowan.edu/open/depts/math/osler/Stirlings_formula.pdf
2015-01-25-18-30 stop


<a name="a40126a"> 2015-01-26-16-09 start LiuHH notes may contain error! Below is page 3/8 to 5/8 of next pdf file by THOMAS J. OSLER http://www.rowan.edu/open/depts/math/osler/Stirlings_formula.pdf An improvement in Stirling’s formula Below discuss how to improve Stirling’s formula. First review the integration graph. <a name="a40126b"> http://freeman2.com/tut062a1.gif <a name="a40126c"> THOMAS J. OSLER paper start from eq.LAb4 Γ(n+1/2)=[(2n)!]*√π/[22n*n!] --- eq.LAb4 √π is one calculation and 22n is one calculation. But red factorial invlove many multiplication. OSLER paper convert red factorial to log(factorial) See log[(2n)!/n!]=log[(2n)!]-log[n!] =log(n+1)+log(n+2)+log(n+3)+...+log(2n) --- eq.LAb8 <a name="a40126d"> This step convert continuous multiplication to continuous summation. Summation is close to integration and integration is easy to calculate. This improvement in Stirling’s formula made change at summation and integration step. No improvement summation to integration step is eq.LAc1 Improvement is made at adding additional correction term. <a name="a40126e"> Numerical integration midpoint rule states that
 
x=a+nh
x=a
f(x)*dx
=
h*
k=n
k=1
f(a+kh-h/2)
+
h3

24
k=n
k=1
f ''(ck)
         
--- eq.LAf1
width of above equation a401261643
<a name="a40126f">
2015-01-26-16-50 here 
eq.LAf1 is a general formula. Neighbor sample 
points distance is h. OSLER paper choose h be 
one. Integration start from x=a to x=a+nh. 
eq.LAf1 left side is integral value. Right 
side is summation of trapezoidal area. 
Trapezoid base length is h. Midpoint height 
is f(a+kh-h/2) Trapezoidal area=h*f(a+kh-h/2)
Where k is variable start at k=1, stop at k=n. 
k=1: Trapezoidal area=h*f(a+1h-h/2)
k=2: Trapezoidal area=h*f(a+2h-h/2) etc. 
<a name="a40126g">
Compare with figure 1. 
Variable is k. Start from k=n to k=2n. 
k=n:   Rectangle area=((n+1)-n)*log(n+1)
k=n+1: Rectangle area=((n+2)-(n+1))*log(n+2)
etc.
2015-01-26-17-05 stop 
2015-01-26-17-56 start 
Taylor series expansion for f(x) at x=a is  
<a name="a40126h"> Next is eq.LAf2 (too long, "eq.LAf2" is hidden)
Taylor series expansion for f(x) at x=a ; 2015-01-26-18-12
 
f(x)
=
f(a)
+ (x-a)*f '(a)
+
(x-a)2

2!
d2f(x)

(dx)2
x=a
+
(x-a)3

3!
d3f(x)

(dx)3
x=a
...
+
(x-a)n

n!
dnf(x)

(dx)n
x=a
...
--- eq.LAf2
Taylor series expansion for function f(x) at point x=0 is named
Maclaurin series expansion. It is next equation.
f(x)=f(0) +f'(0)*x/1! +f''(0)*x2/2! +f'''(0)*x3/3! + ... +f(n-1)(0)*xn-1/(n-1)! +Rn --- eq.dzA0
where Rn=f(n)(θx)*xn/(n)! ; 0<θ<1
width of above equation a401261836
<a name="a40126i">
2015-01-26-18-38 here 
Figure 1 is adding rectangular area to estimate 
smooth curve area. Rectangular area top side is 
horizontal, correspond to f(a) in eq.LAf2 . 
eq.LAf1 middle term is adding trapezoidal area 
to estimate smooth curve area. Trapezoidal area 
top side is a straight line (not horizontal), 
correspond to f'(a) in eq.LAf2 . Next higher 
term is second differentiation f''(x) evaluated 
at x=a : [(x-a)2/2!]*f''(x) evaluated at x=a 
This second differentiation term is used as a 
correction term, eq.LAf1 right most term. 

<a name="a40126j">
Next work is from eq.LAf2 d2f(x)/(dx)2 term to 
eq.LAf1 right most term. 
First discuss ck in eq.LAf1. 
eq.LAf2 has higher order terms d'''(x), d''''f(x) 
etc up to infinity differentiation. 
OSLER paper say //see eq.LAf1 
"Up to this point no approximations have been made"
In eq.LAf1 second differentiation term 
d2f(x)/(dx)2 evaluated at x=ck 
How can second differentiation term declare 
"exact" (no approximations have been made
OSLER paper page 4/8) 
while Taylor series expansion take infinity 
many differentiation? 
<a name="a40126k">
The difference (the answer) is that 
OSLER paper second differentiation term 
evaluated at x=ck in k-th interval. 
Taylor series expansion evaluated at x=a. 
Point 'a' is given, 
point ck is unknown. 
ck come from Mean value theorem. 
Above is first puzzle, LiuHH made clear. 
LiuHH notes may contain error!
2015-01-26-19-29 stop 

<a name="a40126l">
2015-01-26-21-44 start 
OSLER paper said in eq.LAf1 f(x) is a 
"twice differentiable function, and 
 a+(k-1)d ≦ ck ≦ a+kd. We now have, using 
 this midpoint rule with a=n+1/2, f(x)=log(x) 
 and d=1" (LiuHH change all 'd' to 'h', 
 because 'd' is used for differentiation)
In OSLER paper the following equation is 
<a name="a40126m">
 
log
(2n)!

n!
=
k=2n
k=n+1
log(k)
x=2n+1/2
x=n+1/2
log(x)
dx
+
k=n
k=1
1

24ck2
--- eq.LAf3
OSLER paper say eq.LAf3 is equality, not Approximation.
width of above equation a401262201
<a name="a40126n">
2015-01-26-22-08 here 
Compare eq.LAf3 with eq.LAc1
eq.LAf3 added ∑[k=1,n]{1/[24ck2]}
eq.LAf3 use , eq.LAc1 use 
LiuHH second puzzle is how to get eq.LAf3 right 
end term ∑[k=1,n]{1/[24ck2]} ? This term is 
"improvement in Stirling’s formula" heart part. 
2015-01-26-22-15 stop 

<a name="a40127a"> 
2015-01-27-15-10 start
2015-01-25-21-23 access 
http://www.siam.org/books/ot103/OT103%20Dahlquist%20Chapter%205.pdf
page 8/87 say 
The midpoint rule approximation can be interpreted 
as the area of the trapezium defined by the tangent 
of f at the midpoint xi+1/2
The remainder term in Taylor’s formula gives
 f(x)-[fi+1/2+(x-xi+1/2)*f'i+1/2]=(1/2)*(x-xi+1/2)2f''(ζx) --- eq.LAf4 
where ζx in [xi,xi+1/2]
In eq.LAf4 , x is variable, xi and xi+1/2 are 
selected domain points. 
<a name="a40127b">
f(x) is variable function value. 
fi+1/2 is function value evaluated at x=xi+1/2
f'i+1/2 is function first derivative value 
evaluated at x=xi+1/2
f''(ζx) is function second derivative value 
evaluated at x=ζx. ζx is an unspecified constant 
value in [xi,xi+1/2]
eq.LAf4 say exact function value f(x) and numerical 
value [fi+1/2+(x-xi+1/2)*f'i+1/2] difference (error) 
is (1/2)*(x-xi+1/2)2f''(ζx) This red error term 
is eq.LAf3 right end improvement term. 

<a name="a40127c">
Now define 
  h=xi+1-xi --- eq.LAf5 
  t=(x-xi+1/2)/h --- eq.LAf6 
h is constant (interval) and x,t both be variable.
xi+1 be interval upper end, xi+1/2 be midpoint and 
xi be interval lower end. 

When x=xi+1/2, then t become t0 
  t0=(xi+1/2-xi+1/2)/(xi+1-xi)=0 --- eq.LAf7 

<a name="a40127d">
Integration start at x=xi then t start at 
  t-=(xi-xi+1/2)/(xi+1-xi)=-0.5*h/h=-1/2 --- eq.LAf8 

Integration end at x=xi+1 then t end at 
  t+=(xi+1-xi+1/2)/(xi+1-xi)=+0.5*h/h=+1/2 --- eq.LAf9 

Integrate this improvement term over one interval 
from x=xi to x=xi+1 get 
<a name="a40127e"> Next is eq.LAg0+eq.LAg1
 
Ri
=
x=xi+1
x=xi
f ''(ζx)[x-xi+1/2]2*dx

2
=
f ''(ζi)

2
x=xi+1
x=xi
h2*[x-xi+1/2]2

h2
h*d(x-xi+1/2)

h
--- eq.LAg0
 
Ri
=
f ''(ζi)

2
t=+1/2
t=-1/2
h2*t2
h*dt
=
h3*f ''(ζi)

2
t=+1/2
t=-1/2
t2*dt
=
h3*f ''(ζi)

24
--- eq.LAg1
http://www.siam.org/books/ot103/ paper say Ri is remainder term.
http://www.rowan.edu/open/depts/math/osler/ use Ri as an improvement.
width of above equation a401271620
<a name="a40127f"> 
2015-01-27-16-21 here
In eq.LAg0 , red h are inserted one, twice. 
In eq.LAg0 , ζx is an unspecified constant value 
and move f''(ζx) out of integral. 
In eq.LAg1 , apply eq.LAf6 change variable x to t. 
In eq.LAg1 , move constant h out of integral. 
In eq.LAg1 , ∫[t=-1/2,t=+1/2]{t*t*dt} is 1/12 
eq.LAg1 right end term f ''(ζi) is 
log''(ζi) or d2[log(x)]/dx2 evaluate at x=ζi
d1[log(x)]/dx1=1/x
d2[log(x)]/dx2=-1/x2
d2[log(x)]/dx2 evaluate at x=ζi get -1/(ζi)2
<a name="a40127g">
This improvement term is h3*[-1/(ζi)2]/24
Use interval h=1 (OSLER paper page 4/8 "and d=1")
eq.LAf3 right end is 1/[24*ck2] 
Liu,Hsinhan get -1/[24*(ζi)2] 
Why negative sign difference? 
LiuHH calculation contain error!
Please help find out what is wrong.
Thank you. 
Despite LiuHH negative sign difference calculation, 
the following continue with OSLER paper improvement 
term, use positive sign.  
2015-01-27-16-46 stop 

<a name="a40127h">
2015-01-27-19-15 start 
Now found the improvement term eq.LAf1 right end term. 
Add this term to eq.LAc7 get next equation eq.LAg2 
 
log
(2n)!

n!
=
(2n+1/2)
[
log(2n)
+ log
(
1+
1

4n
)
]
(n+1/2)
[
log(n)
+ log
(
1+
1

2n
)
]
-n
+
k=n
k=1
1

24ck2






--- eq.LAg2
was eq.LAc7
eq.LAg2 added equation end red term, let change to =.
width of above equation a401271935
<a name="a40127i">
2015-01-27-19-41 here
It is convenient to write eq.LAg2 as eq.LAg3
below. //parallel to eq.LAc8 
log[(2n)!/n!] =
+ (2n+1/2)*log(2n) + (2n+1/2)*log[1+1/(4n)] 
- (n+1/2)*log(n)  - (n+1/2)*log[1+1/(2n)] -n 
+ ∑[k=1,n]{1/(24ck2)} --- eq.LAg3
<a name="a40127j"> Next is eq.LAg4 
Next apply exponentiation to eq.LAg3 and recover 
(2n)!/n! as next  //parallel to eq.LAd9 
 
(2n)!

n!
=
(2n)(2n+1/2)*[1+1/(4n)](2n+1/2)*e-n

n+(n+1/2)*[1+1/(2n)]+(n+1/2)
*
exp
(
k=n
k=1
1

24ck2
)
--- eq.LAg4
eq.LAg4 added equation end red term, let change to =.
width of above equation a401272006
<a name="a40127k"> 
Blue terms in eq.LAg4 was red and blue in eq.LAe1. 
In eq.LAe1, red and blue cancel to 1. See "eq.LAe1 
red,blue are all gone" Here is improvement section. 
Following we will squeeze a little bit information 
out of blue terms. eq.LAg4 blue terms are treated 
slight differently, following equations are a little 
bit complicated than no-improvement discussion. 
2015-01-27-20-18 stop

<a name="a40127l">
2015-01-27-22-34 start 
Rewrite eq.LAg4 such that final improvement can be 
clearer. Change [1+1/(4n)](2n+1/2), 1/[1+1/(2n)](n+1/2)
to [√(term1)]*[term2]n 
<a name="a40127m"> Next is eq.LAg5 to eq.LAg8
 
(2n)!

n!
=
(√2)*22n*n(2n+1/2)-(n+1/2)*e-n
*
exp
(
k=n
k=1
1

24ck2
)
--- eq.LAg5
 
continue
from above
*
[1+1/(4n)]1/2*[1+1/(4n)]2n

[1+1/(2n)]1/2*[1+1/(2n)]n
       
--- eq.LAg6
 
(2n)!

n!
=
(√2)*22n*nn*e-n
*
exp
(
k=n
k=1
1

24ck2
)
               
--- a40127n
--- eq.LAg7
 
continue
from above
*
[(4n+1)/(4n)]1/2*[(4n+1)/(4n)]2n

[(2n+1)/(2n)]1/2*[(2n+1)/(2n)]n
               
--- eq.LAg8
Red term is improvement eq.LAf3 right end term taking exp().
Blue terms squeeze a little bit information out of blue for improvement.
width of above equation a401272311

2015-01-27-23-12 stop 

<a name="a40128a"> 
2015-01-28-15-06 start 
eq.LAg4 has terms to power of (2n+1/2) and to power 
of (n+1/2). From eq.LAg4 blue to eq.LAg6 blue split 
(2n+1/2) power to 2n and 1/2 and split (n+1/2) power 
to n and 1/2 . Isolate square root term from whole 
number power 2n & n.
From eq.LAg5 to eq.LAg7 simplify to nn 
From eq.LAg6 to eq.LAg8 reduction of fractions to 
a common denominator. (通分)

<a name="a40128b">
eq.LAg8 blue term is analyzed below.
 
blue term =
[
4n+1

4n
2n

2n+1
]
*
[
(4n+1)2

(4n)2
2n

2n+1
]
n


--- eq.LAg9
 
blue term =
[
4n+2-1

4n+2
]
*
[
16n2+8n+1

8n*(2n+1)
]
n


--- eq.LAh0
 
blue term =
[
1 -
1

4n+2
]
*
[
1 +
1

8n*(2n+1)
]
n


--- eq.LAh1
width of above equation a401272335
<a name="a40128c">
2015-01-28-15-23 here 
eq.LAg8 to eq.LAg9 collect square root term together.
Collect power n term together. Change those power 2n 
term to [(term)2]n to keep power n form. 
eq.LAg9 to eq.LAh0 is common denominator operation.
eq.LAh0 to eq.LAh1 write fraction1 to 1-fraction2
and write fraction3 to 1+fraction4

<a name="a40128d">
eq.LAh1 is eq.LAg8 blue term. Rewrite eq.LAg7+eq.LAg8
as eq.LAh2+eq.LAh3 below.
 
(2n)!

n!
=
(√2)*22n*nn*e-n
*
exp
(
k=n
k=1
1

24ck2
)
               
--- eq.LAh2
 
continue
from above
*
[
1 -
1

4n+2
]
*
[
1 +
1

8n*(2n+1)
]
n


               
--- eq.LAh3
width of above equation a401272341
<a name="a40128e">
2015-01-28-15-31 here 
eq.LAh2+eq.LAh3 is [(2n)!]/n! It is partial of 
Γ(n+1/2) see eq.LAb4
  Γ(n+1/2)=[(2n)!]*√π/[22n*n!] --- eq.LAb4
Next substitute eq.LAh2+eq.LAh3 to eq.LAb4 get 
Γ(n+1/2) below. 
<a name="a40128f">
 
Γ(n+1/2)=
(2n)!

n!
√π

22n
=
(√2)*22n*nn*e-n
exp
(
k=n
k=1
1

24ck2
)
√π

22n
--- eq.LAh4
 
continue
from above
*
[
1 -
1

4n+2
]
*
[
1 +
1

8n*(2n+1)
]
n


       
--- eq.LAh5
width of above equation a401280231
<a name="a40128g">
2015-01-28-16-10 here 
eq.LAh4+eq.LAh5 is improved Γ(n+1/2) 
Red term is improvement adding. 
Blue term is Γ(n+1/2) and (2n)!/n! difference. 
In eq.LAh4 cancel 22n/22n
and merge √2*√π to √(2π) get eq.LAh6 below. 
<a name="a40128h"> Next is eq.LAh6
 
Γ(n+1/2)=
√(2π)*nn*e-n
[
1 -
1

4n+2
]
*
[
1 +
1

8n*(2n+1)
]
n


exp
(
k=n
k=1
1

24ck2
)
--- eq.LAh6
http://www.rowan.edu/.../Stirlings_formula.pdf page 4/8 equation (10)
width of above equation a401280238
<a name="a40128i">
2015-01-28-16-17 here
Our goal is Stirling's Approximation formula eq.CG64, 
which is inequality or it is an approximation formula 
However eq.LAh6 is an equality, not an approximation.
The reason is that eq.LAh6 involve ck. Mean value 
theorem promiss equality. But ck is a not specified 
value. Cannot calculate ck value. Following estimate 
ck value and begin approximation formula. ck first 
show up at eq.LAf1 ck is discussed at a40126j 
"d2f(x)/(dx)2 evaluated at x=ck" We know ck play the 
role of x. The estimation for ∑[k=1,n]{1/[24*ck2]}
<a name="a40128j">
See OSLER paper page 4/8 line -3
http://www.rowan.edu/open/depts/math/osler/Stirlings_formula.pdf
OSLER paper use integration of {1/[24*x2]} //a401281641
Next is eq.LAh7+eq.LAh8
 
k=n
k=1
1

24ck2
x=2n+1/2
x=n+1/2
1

24x2
dx
=
x=2n+1/2
x=n+1/2
d
-1

24x
=
-1

24(2n+1/2)
-1

24(n+1/2)
--- eq.LAh7
 
k=n
k=1
1

24ck2
1

12(2n+1)
1

12(4n+1)
=
(4n+1)-(2n+1)

12(2n+1)(4n+1)
=
n

6(2n+1)(4n+1)
--- eq.LAh8
Approximation enter equation.
width of above equation a401280859
<a name="a40128k"> 
2015-01-28-16-49 here 
eq.LAh8 is improvement term calculation. Take 
exponentiation to improvement term and put 
back to eq.LAh6, get eq.LAh9 below 
 
Γ(n+1/2)
√(2π)*nn

en
[
1 -
1

4n+2
]
*
[
1 +
1

8n*(2n+1)
]
n


exp
(
n

6(2n+1)(4n+1)
)
--- eq.LAh9
Approximation enter equation.
http://www.rowan.edu/.../Stirlings_formula.pdf page 4/8 equation (11)
width of above equation a401280903
<a name="a40128l">
2015-01-28-16-54 here 
eq.LAh9 is Γ(n+1/2) , but Stirling's formula is 
whole number Γ(n+1). The following change variable 
  n=z+1/2   --- eq.LAi0 
  n+1/2=z+1 --- eq.LAi1 
We use 'n' for integer number. But Gamma function 
allow decimal number, even allow complex number. 
OSLER paper say in eq.LAi0, let z be non-integer.
See OSLER paper page 5/8 line 1. 
After change variable from 'n' to 'z', Γ(n+1/2) 
become Γ(z+1/2+1/2) that is Γ(z+1). Following is 
equation for Γ(z+1).
<a name="a40128m"> next is eq.LAi2+eq.LAi3
 
Γ(z+1)
√(2π)*(z+1/2)z+1/2*e-z-1/2
*
[
1 -
1

4(z+1/2)+2
]
--- eq.LAi2
 
continue
from above
*
[
1 +
1

8(z+1/2)*(2(z+1/2)+1)
]
z+1/2


exp
(
(z+1/2)

6(2(z+1/2)+1)(4(z+1/2)+1)
)
--- eq.LAi3
width of above equation a401281055
<a name="a40128n">
2015-01-28-17-01 here 
eq.LAi2+eq.LAi3 is equation for Γ(z+1) still with 
Γ(n+1/2) shadow. Following change Γ(n+1/2) shadow
equation to standard Γ(z+1) equation. 

Stirling's Approximation formula use nn that is 
use zz, not use (z+1/2)z+1/2.
First job is straighten out (z+1/2)z+1/2 separately. 
Below zz/zz is inserted number one.
<a name="a40128o"> //next is eq.LAi4
 
(z+1/2)z+1/2
=
(z+1/2)z*√(z+1/2)
zz

zz
=
zz
√(z+1/2)
(z+1/2)z

zz
=
zz
[
1+
1

2z
]
z


√(
z+
1

2
)
--- eq.LAi4
width of above equation a401281145
<a name="a40128p"> 
2015-01-28-17-09 here 
Substitute eq.LAi4 to eq.LAi2+eq.LAi3 get 
eq.LAi5+eq.LAi6 below. eq.LAi5 inserted z/z 
which is one. z/z is used at eq.LAi7. 
Follow blue z.
<a name="a40128q"> Next is eq.LAi5+eq.LAi6
 
Γ(z+1)
√(2π)
z

z
e-z-1/2
zz
[
1+
1

2z
]
z


√(
z+
1

2
)
[
1 -
1

4z+2+2
]
--- eq.LAi5
 
continue
from above
*
[
1 +
1

(8z+4)*(2z+1+1)
]
z+1/2


exp
(
(2z+1)/2

6(2z+1+1)*(4z+2+1)
)
--- eq.LAi6
width of above equation a401281150
<a name="a40128r">
2015-01-28-17-12 here
From eq.LAi5 to eq.LAi7 , fabricated √(2πz)*zz*e-z. 
√(2πz)*zz*e-z is Stirling's Approximation formula 
eq.CG64 In eq.LAi7+eq.LAi8 other terms are all 
modification terms. Red term come from major 
improvement eq.LAf3 right end term. 
In eq.LAi7+eq.LAi8 {...} and √...√... and [...]z+1/2 
all come from eq.LAg4 blue terms.
See "squeeze a little bit information" 
<a name="a40128s"> Next is eq.LAi7+eq.LAi8
 
Γ(z+1)
√(2πz)
zz

ez
{
e-1/2
[
1+
1

2z
]
z


}
√(
z

z
+
1

2z
)
[
1 -
1

4(z+1)
]
--- eq.LAi7
 
continue
from above
*
[
1 +
1

16(z+1/2)*(z+1)
]
z+1/2


exp
(
(2z+1)/2

6(2z+2)(4z+3)
)
--- eq.LAi8
width of above equation a401281220
<a name="a40128t">
2015-01-28-17-27 here
eq.LAi7+eq.LAi8 is too long. OSLER paper define 
f(z),g(z),h(z),k(z) for easier discussion. 
Next is eq.LAi9 and eq.LAj0
 
f(z)
define
===
e-1/2
[
1+
1

2z
]
z


g(z)
define
===
√{[
1+
1

2z
]*[
1-
1

4(z+1)
]}
--- eq.LAi9
 
h(z)
define
===
[
1 +
1

16(z+1)*(z+1/2)
]
z+1/2


k(z)
define
===
exp
(
2z+1

24(z+1)(4z+3)
)
--- eq.LAj0
width of above equation a401281256
<a name="a40128u"> 
2015-01-28-17-29 here 
After f(z),g(z),h(z),k(z) definition, 
eq.LAi7+eq.LAi8 can be written as eq.LAj1 below.
Γ(z+1) ≈ √(2πz)*zz*e-z*f(z)*g(z)*h(z)*k(z) --- eq.LAj1 
2015-01-28-17-31 stop

<a name="a40128v">
2015-01-28-19-19 start 
Next exam f(z),g(z),h(z),k(z) four equations.
See their behavior when z approach to infinity.
We need eq.LAe2 and eq.LAe3 two equations. 
Both are n approach to infinity limit equation.
Refer to eq.LEb0 
  (1+x/n)n ≈ ex when n→∞ --- eq.LAe2
It is easy to understand eq.LAe3, because a/∞=0 
  √(1+a/n) ≈ 1 when n→∞ --- eq.LAe3

f(z) use eq.LAe2. When z approach to infinity 
f(z) = e-1/2[1+(1/2)/z]z  = e-1/2*e1/2 = 1 --- eq.LAj2

g(z) use eq.LAe3. When z approach to infinity 
g(z) = √{[1+0]*[1-0]} = 1  --- eq.LAj3

h(z) use eq.LAe2. When z approach to infinity 
h(z) = {1+[1/16(z+1)]/(z+1/2)}(z+1/2)  = e-16(z+1) = 1 --- eq.LAj4

<a name="a40128w">
k(z) use eq.LAe3. When z approach to infinity 
k(z) ≈ exp[1/(48z)] = 1 --- eq.LAj5
How k(z) become 1 ? See next four lines
k(z) = exp{(2z+1)/[24(z+1)(4z+3)]} --- eq.LAj6
k(z) = exp{(2z+1)/[24(4z2+7z+3)]} --- eq.LAj7
When z approach to infinity, drop lower order term 
k(z) ≈ exp{(2z)/[24(4z2)]}=exp[1/(48z)]  --- eq.LAj8
When z approach to infinity 
exp[1/(48z)]→exp[1/∞]=exp[0]=1 --- eq.LAj5
2015-01-28-19-55 here 

<a name="a40128x">
OSLER paper let f(z)≈1 and g(z)≈1. 
OSLER paper consider h(z) and k(z) 
h(z) = e-16(z+1) ≈ e-16(z) = exp[1/(16z)] --- eq.LAj9
k(z) ≈ exp[1/(48z)]  --- eq.LAk0

Now substitute f(z),g(z),h(z),k(z) back to eq.LAj1 
Γ(z+1) ≈ √(2πz)*zz*e-z*1*1*exp[1/(16z)]*exp[1/(48z)] --- eq.LAk1 
Γ(z+1) ≈ √(2πz)*zz*e-z*1*1*exp[1/(16z) + 1/(48z)] 
Γ(z+1) ≈ √(2πz)*zz*e-z*1*1*exp[3/(48z) + 1/(48z)] 
Γ(z+1) ≈ √(2πz)*zz*e-z*1*1*exp[4/(48z)] 
Γ(z+1) ≈ √(2πz)*zz*e-z*1*1*exp[1/(12z)] 

<a name="StirlingOsler"> 
We get another very simple but very improved 
approximation eq.LAk2 . //'O'=OSLER 
Stirling's Approximation formula very improved
 
Γ(z+1) =
z!
zz

ez
√(2zπ)
* exp
(
1

12z
)
  ▐  
z → ∞
--- eq.LAk2
width of above equation a401282016
<a name="a40128y">
2015-01-28-20-20 here 
eq.LAk2 is improved Stirling's Approximation formula 
New added term is red color term in eq.LAk2. 
Compare input number and Stirling answer, 
exact answer, improved answer are listed in 
http://www.rowan.edu/open/depts/math/osler/Stirlings_formula.pdf
page 6/8. Copy this comparison table below.
<a name="a40128z">
Compare Stirling, exact, improved
 
number
method
result
1
Stirling
0. 92213 70088
1
exact
1.0
1
improved
1. 00227 44449
10!
Stirling
3 598 695. 61874
10!
exact
3 628 800. 00000
10!
improved
3 628 810. 06142
100!
Stirling
9. 32484 76252 00000 E 157
100!
exact
9. 33262 15443 94415 E 157
100!
improved
9. 33262 15703 00000 E 157
1000!
Stirling
4. 02353 72920 00000 E 2567
1000!
exact
4. 02387 26007 70938 E 2677
1000!
improved
4. 02387 26007 82115 E 2567
10000!
Stirling
2. 84623 59621 00000 E 35659
10000!
exact
2. 84625 96809 17055 E 35659
10000!
improved
2. 84625 96809 17062 E 35659
100000!
Stirling
2. 82422 70544 00000 00000 E 456573
100000!
exact
2. 82422 94079 60347 87429 E 456573
100000!
improved
2. 82422 94079 60347 88214 E 456573
"00000" are space filling. Other bytes get un-even column.
width of above equation a401282052
<a name="a40128_">
Above is page 3/8 to 5/8 of next pdf file 
http://www.rowan.edu/open/depts/math/osler/Stirlings_formula.pdf

Here is the end of the study notes 
http://freeman2.com/tute0062.htm
LiuHH notes may contain error!
Thank you for visiting Liu,Hsinhan's page.
2015-01-28-21-00 stop
2015-01-29-12-11 done first proofread 
2015-01-29-17-11 done second proofread 

<a name="a40129a">
2015-01-29-18-50 start 
page 5/8 of next pdf file 
http://www.rowan.edu/open/depts/math/osler/Stirlings_formula.pdf
say 
It is known that the asymptotic expansion for 
the gamma function is eq.LAk3 below. 
<a name="a40129b">
more precise version of the Stirling formula
 
Γ(z+1) =
z!
=
zz

ez
√(2zπ)
(
1 +
1

12z
+
1

288z2
139

51840z3
571

2488320z4
+ ...
)
--- eq.LAk3
width of above equation a401291907

2015-01-29-19-09 here 
In eq.LAk3, OSLER paper use equality, LiuHH 
follows.
2015-01-29-19-10 stop 

<a name="a40214a">
2015-02-14-13-01 start 
update 2015-02-14 add links to all Gamma Function
Index topics.
Newer sreies is Taylor Series study notes 
http://freeman2.com/tute0064.htm
2015-02-14-13-02 stop 







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