Gamma function study notes tute0062
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"> t062link()
2015-01-01-15-18 start
On
2014-03-25-20-36 Liu,Hsinhan access

Gamma Function
•by MrYouMath
•12 videos
•5,940 views
•1 hour, 48 minutes
Gamma Function - Part 1 - Functional Equation
Gamma Function - Part 2 - Gauss Representation
Gamma Function - Part 3 - Weierstrass Representation
Gamma Function - Part 4 - Relationship to Sine
Gamma Function - Part 5 - Gamma of 0.5 ( one half)
Gamma Function - Part 6 - Stirling's Approximation
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
Gamma Function - Part 8 - Euler Integral II The Sinc-Function
Gamma Function - Part 9 - Euler Integral III Fresnel Integral
Gamma Function - Part 10 - Beta Function
Gamma Function - Part 11 - Legendre Duplication Formula
Gamma Function - Part 12 - Relation to Zeta Function
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

2014-10-29-07-56
2014-10-29-08-24
MrYouMath_17_files_Riemann.htm
"Zeta Function - Part 1 - Convergence"
"Zeta Function - Part 2 - Euler Product Representation"
"Zeta Function - Part 3 - Euler Product (easy)"
<a name="docA007">
"Zeta Function - Part 4 - Infinitude of Prime Numbers"
"Zeta Function - Part 5 - Prime Zeta Function"
"Zeta Function - Part 6 - The Prime Counting Function"
"Zeta Function - Part 7 - Zeta of 2 aka The Basel Problem"
"Zeta Function - Part 8 - Zeta of 2n - Part 1"
<a name="docA008">
"Zeta Function - Part 8 - Zeta of 2n - Part 2"
"Zeta Function - Part 8 - Zeta of 2n - Part 3"
"Zeta Function - Part 9 - Relation to Gamma Function"
"Zeta Function - Part 10 - Jacobi Theta Function"
"Zeta Function - Part 11 - Riemann Functional Equation I"
<a name="docA009">
"Zeta Function - Part 12 - Riemann Functional Equation II"
"Zeta Function - Part 13 - Trivial Zeros of the Zeta Function"
"Zeta Function - Part 14 - Riemann Xi Function"
"Sine Function Product Formula (Hadamard Factorization Theorem)"
"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

2015-01-08-18-12 start
Next is study notes of
Gamma Function - Part 1 - Functional Equation

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"> t062link()
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
Gamma Functional Equation
Γ(s+1) = s*Γ(s) --- eq.CG05
That is all the magic behind that. Now there
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

2015-01-09-08-37 start
Why gamma function this way?
This topic is Liu,Hsinhan own words,
LiuHH notes may contain error!
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 ∞.

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
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!

■■　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 ?
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.

■■　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
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
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
why-is-the-gamma-function-shifted-from-the-factorial-by-1

2015-01-20-08-28 LiuHH access

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

2015-01-09-18-17 start
Next is study notes of
Gamma Function - Part 2 - Gauss Representation

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"> t062link()
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"> t062link()
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"> t062link()
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

2015-01-11-14-18 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

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

 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">
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
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

2015-01-19-13-40 start
Next is study notes of
Gamma Function - Part 3 - Weierstrass Representation

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
constant 0.577215

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
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"> t062link()
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

2015-01-20-17-18 start
Next is study notes of
Gamma Function - Part 4 - Relationship to Sine

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
```
<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
width of above equation a401201832
```<a name="docAf11"> t062link()
2015-01-21-13-11 here
Above eq.cz93 is proved at
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
I will prove in a separate video.
That conclude the lecture. See you guys.
2015-01-21-13-25 stop

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
Gamma Function - Part 4 - Relationship to Sine
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

2015-01-22-13-41 start
Next is study notes of
Gamma Function - Part 5 - Gamma of 0.5 ( one half)

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"> t062link()
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

2015-01-23-13-23 start
Next is study notes of
Gamma Function - Part 6 - Stirling's Approximation

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"> t062link()
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
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-ε
--- eq.CG75

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

 n*log(n+ε)-(n+ε) = n*log(n) -n + k=∞ ∑ k=1 (-1)k+1 k εk nk-1 -ε
--- 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 e-ε2/(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 e-ε2/(2n) dε ▐   let n→∞ then ▐  ε=-n to ε→-∞
--- eq.CG82

 n! ≈ nn en ε=∞ ∫ ε=-∞ e-ε2/(2n) dε ▐   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
come see house, agent not follow. LiuHH
and sister clean up house immediately.
Agent name Linda Garcia at Pinnacle
Buyer03 name Bob. Bob has more possibility
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

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

2015-01-24-21-35 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!

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
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
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

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

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

<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.
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.
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
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
Now apply exponentiation to eq.LAc8 one term by
one term.
eq.LAd0 is true base on eq.LAc9. Next term
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.
<a name="a40125q">
Similarily
Remember
```
<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)

 (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]
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]k→e1/2 when →∞
Apply eq.LAe3 get [1+1/(2k)]1/2→1 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

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
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">t062link()
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?
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 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

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"> t062link()
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!
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"> t062link()
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

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)
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"> t062link()
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
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"> t062link()
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.
```
<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"> t062link()
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=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)]

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,
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
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Above is page 3/8 to 5/8 of next pdf file
http://www.rowan.edu/open/depts/math/osler/Stirlings_formula.pdf

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2015-01-28-21-00 stop

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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.
```
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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

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2015-02-14-13-01 start
Index topics.
Newer sreies is Taylor Series study notes
http://freeman2.com/tute0064.htm
2015-02-14-13-02 stop

```

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