<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
t^{s-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 t^{s-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.
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 t^{s} as eq.CG02 right side shown. Next
just do a partial integration on eq.CG02
right side integral. Very important the t^{s}
in eq.CG02 is down integrated and e^{-t} part
is up integrated. In eq.CG03 we get boundary
term －e^{-t}*t^{s}|[t=0,t=∞] In eq.CG03 we get new
<a name="docAe07">
integral. Differentiate t^{s} 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}*t^{s}|[t=0,t=∞] become zero. If you plug
t=0 to －e^{-t}*t^{s}|[t=0,t=∞] , 0^{s} is zero. If
plug t=∞ to －e^{-t}*t^{s}|[t=0,t=∞] it is a little
bit harder. You will get e^{-t} tend to zero and
t^{s} tend to infinity. If use l'Hosptial's rule
you will see －e^{-t}*t^{s}|[t=∞] converging to zero.
<a name="docAe08">
So, in eq.CG03 －e^{-t}*t^{s}|[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
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
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.CG07eq.CG01 apply Integration by parts eq.LA15
get eq.CG06 and eq.CG07.
eq.CG01 t^{s-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)*x^{5/3}/(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">t^{s-1}*(-e^{-t})|[t=0,t=∞]
when t=0, t^{s-1}*(-e^{-t})=0^{s-1}*(-e^{-0})=0*1=0 --- eq.CG08
when t→∞, t^{s-1}*(-e^{-t})=t^{s-1}/(-e^{+t})|[t→∞]=∞/(-∞) --- eq.CG09
Apply l'Hosptial's rule get
t^{s-1}/(-e^{+t})|[t→∞]=[d(t^{s-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 t^{s-1}t^{s-1} will be differentiated,
First differentiation
d(t^{s-1})/dt=(s-1)*(t^{s-2}) --- eq.CG13
Second differentiation
d[(s-1)*t^{s-2}]/dt=(s-1)*(s-2)*(t^{s-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 t^{s-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-1}dt} --- 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 t^{s-1} ?
Why not eq.CG01 Γ(s) correspond to t^{s} ?
<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="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="docAe36">
2015-01-09-18-35 here
We will plug eq.CG16 into eq.CG01 .
Γ(s)
=
t=∞
∫
t=0
t^{s-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="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
t^{s-1} will be integrated and
[1-t/n]^{n} will be differentiated.
Now look what happen.
<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 t^{s-1}dt to d[t^{s}/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.
<a name="docAe45">
2015-01-09-20-34 here
Compare eq.CG23 right side and eq.CG26 right
side. We raise t^{s} power to t^{s+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
t^{s-1}
[
1 -
t
n
]
n
dt
=
n
s*n
n-1
(s+1)n
n-2
(s+2)n
t=n
∫
t=0
t^{s+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
t^{s-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
t^{s+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
t^{s-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^{s+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 t^{s+n}|[t=0,t=n]
t=0 gives zero, t=n gives n^{s+n}.
eq.CG29 numerator has factorial of n, n!.
eq.CG29 denominator has n^{n}
eq.CG29 denominator has s*(s+1)*...*(s+n-1)*(s+n)
The result equation is
<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 n^{n}
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="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 ?
<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
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!
y^{2}
+
n(n-1)(n-2)
3!
y^{3}
+ ●●●
--- eq.LAa7
[
1 +
x
n
]
n
=
1 +
n
x
n
+
n(n-1)
2!
x^{2}
n^{2}
+
n(n-1)(n-2)
3!
x^{3}
n^{3}
+ ●●●
--- eq.LAa8
[
1 +
x
n
]
n
=
1 +
x
1!
+
n^{2}-n
n^{2}
x^{2}
2!
+
n^{3}-3n^{2}+2n
n^{3}
x^{3}
3!
+ ●●●
--- eq.LAa9
width of above equation
a401111911
<a name="a40111g">
2015-01-11-19-11 here
When n→∞, in eq.LAa9
n^{2}-n → n^{2}, then (n^{2}-n)/n^{2} → 1
n^{3}-3n^{2}+2n → n^{3}, then (n^{3}-3n^{2}+2n)/n^{3} → 1
similarly, other term n coefficient approach
to one. eq.LAa9 approach to
2015-01-11-19-25 here
eq.LAb0 is same as eq.LAa0 and eq.LAa0 is e^{x}
therefore eq.LAb0 is e^{x}.
Replace x by -t get eq.CG16 .
<a name="a40111i">
About "eq.LAa0 is e^{x}", 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="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.
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="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
n^{s} in eq.CG37 in order to get e^{－γs}
Let us just do it, We rewrite n^{s} as e^{s*log(n)}
as shown in eq.CG39 below.
log() and exp() are inverse operation, that is
M=exp(log(M)), let M be n^{s} then n^{s}=exp(log(n^{s}))
Apply log law log(n^{s})=s*log(n) get
n^{s}=exp(s*log(n)) --- eq.CG38
<a name="docAe73">
2015-01-19-18-15 here
After eq.CG39 the next step is very very trivial.
We just write e^{0} , 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="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.
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="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 n^{s}
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="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 n^{s}*n^{-s} cancel to one, see eq.CG50.
In eq.CG49 (1+s/i)*(1-s/i) get eq.CG50 (1-s^{2}/i^{2})
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/(1-1^{2}/1^{2}) is a pole.
s=2, eq.CG52 right hand side 1/(1-2^{2}/2^{2}) is a pole.
s=3, eq.CG52 right hand side 1/(1-3^{2}/3^{2}) 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="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="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="a40121e">
2015-01-21-19-05 here
Multiply Γ(s) eq.CG34 with Γ(-s) eq.LAb1 get
eq.LAb2. In eq.LAb2 e^{γs}*e^{-γs} cancel to one,
and e^{s/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="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
Γ^{2}(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="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=pu^{2} --- 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
√(pu^{2})
=
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
√(u^{2}) 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∫u^{2}du) 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="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="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(t^{n}*e^{-t})=n*log(t)-t --- eq.CG66
LiuHH add detail steps below
log(t^{n}*e^{-t})=log(t^{n})+log(e^{-t}) --- eq.CG67
log(t^{n}*e^{-t})=n*log(t)+(-t)*log(e) --- eq.CG68
log(t^{n}*e^{-t})=n*log(t)-t*1 --- eq.CG69
log(t^{n}*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="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 x^{k}, but in eq.CG74 we have
to plug in ε^{k}/n^{k}. (-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.
<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 ε^{k}/n^{k-1} is ε^{1}/n^{1-1}=ε/1=ε
This ε^{1}/n^{1-1} 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="docAf75">
Stirling's Approximation drop all purple terms in eq.CG79
Remember eq.CG66 and eq.CG70
log(t^{n}*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.
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 -ε^{2}/(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 .
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 n^{n}/e^{n} is not depend on ε. We can move
n^{n}/e^{n} 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.
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)]*n^{n}*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 formulaAn 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
t^{x-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)
2^{n}
√π
=
(2n)!
2^{2n}*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)/2^{6} --- 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
2^{n}*[2*4*6*...*2n] which is 2^{n}*2^{n}*[1*2*3*...*n]
which is 2^{2n}*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 2^{2n}, it is easy
to calculate √π and 2^{2n}. 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} ≈ e^{x} 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.
<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="a40125s">
2015-01-25-16-58 here
Refer to eq.LEb0
(1+x/n)^{n} ≈ e^{x} 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
e^{1/2}, denominator e^{1/2} cancel numerator e^{1/2}.
<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*2^{2n}*n^{n}*e^{-n} --- eq.LAe4
Substitute eq.LAe4 to Γ(n+1/2) eq.LAb4
Γ(n+1/2)=[(2n)!]*√π/[2^{2n}*n!] --- eq.LAb4
get //factorial in eq.LAb4 is gone!
Γ(n+1/2)≈[√2*2^{2n}*n^{n}*e^{-n}]*√π/[2^{2n}]
<a name="a40125u">
cancel 2^{2n} and merge √2 with √π get
Γ(n+1/2)≈√(2π)*n^{n}*e^{-n} --- eq.LAe5
Compare eq.LAe5 with eq.CG64
eq.LAe5 is Γ(n+1/2) ≈ √(2π)*n^{n}*e^{-n}
eq.CG64 is Γ(n+1) ≈ √(2nπ)*n^{n}*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)*k^{1/2}*k^{k}*[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}→e^{1/2} when →∞
Apply eq.LAe3 get [1+1/(2k)]^{1/2}→1 when →∞
Γ(k+1)≈√(2kπ)*k^{k}*e^{1/2}*1*e^{-k}*e^{-1/2} --- eq.LAe9
Cancel e^{+1/2} and e^{-1/2} get //improved formulaStirling's Approximation formula
Γ(k+1)≈√(2kπ)*k^{k}*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/[2^{2n}*n!] --- eq.LAb4
√π is one calculation and 2^{2n} 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)
+
h^{3}
24
k=n
∑
k=1
f ''(c_{k})
--- 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!
d^{2}f(x)
(dx)^{2}
│
x=a
+
(x-a)^{3}
3!
d^{3}f(x)
(dx)^{3}
│
x=a
...
+
(x-a)^{n}
n!
d^{n}f(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)*x^{2}/2! +f'''(0)*x^{3}/3! + ... +f^{(n-1)}(0)*x^{n-1}/(n-1)! +R_{n}--- eq.dzA0
where R_{n}=f^{(n)}(θx)*x^{n}/(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 d^{2}f(x)/(dx)^{2} term to
eq.LAf1 right most term.
First discuss c_{k} 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
d^{2}f(x)/(dx)^{2} evaluated at x=c_{k}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=c_{k} in k-th interval.
Taylor series expansion evaluated at x=a.
Point 'a' is given,
point c_{k} is unknown.
c_{k} 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 ≦ c_{k} ≦ 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
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/[24c_{k}^{2}]}
eq.LAf3 use ＝, eq.LAc1 use ≈LiuHH second puzzle is how to get eq.LAf3 right
end term ∑[k=1,n]{1/[24c_{k}^{2}]} ? 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 x_{i+1/2}
The remainder term in Taylor’s formula gives
f(x)-[f_{i+1/2}+(x-x_{i+1/2})*f'_{i+1/2}]=(1/2)*(x-x_{i+1/2})^{2}f''(ζ_{x}) --- eq.LAf4
where ζ_{x} in [x_{i},x_{i+1/2}]
In eq.LAf4 , x is variable, x_{i} and x_{i+1/2} are
selected domain points.
<a name="a40127b">
f(x) is variable function value.
f_{i+1/2} is function value evaluated at x=x_{i+1/2}
f'_{i+1/2} is function first derivative value
evaluated at x=x_{i+1/2}
f''(ζ_{x}) is function second derivative value
evaluated at x=ζ_{x}. ζ_{x} is an unspecified constant
value in [x_{i},x_{i+1/2}]
eq.LAf4 say exact function value f(x) and numerical
value [f_{i+1/2}+(x-x_{i+1/2})*f'_{i+1/2}] difference (error)
is (1/2)*(x-x_{i+1/2})^{2}f''(ζ_{x}) This red error term
is eq.LAf3 right end improvement term.
<a name="a40127c">
Now define
h=x_{i+1}-x_{i} --- eq.LAf5
t=(x-x_{i+1/2})/h --- eq.LAf6
h is constant (interval) and x,t both be variable.
x_{i+1} be interval upper end, x_{i+1/2} be midpoint and
x_{i} be interval lower end.
When x=x_{i+1/2}, then t become t_{0}
t_{0}=(x_{i+1/2}-x_{i+1/2})/(x_{i+1}-x_{i})=0 --- eq.LAf7
<a name="a40127d">
Integration start at x=x_{i} then t start at
t_{-}=(x_{i}-x_{i+1/2})/(x_{i+1}-x_{i})=-0.5*h/h=-1/2 --- eq.LAf8
Integration end at x=x_{i+1} then t end at
t_{+}=(x_{i+1}-x_{i+1/2})/(x_{i+1}-x_{i})=+0.5*h/h=+1/2 --- eq.LAf9
Integrate this improvement term over one interval
from x=x_{i} to x=x_{i+1} get
http://www.siam.org/books/ot103/ paper say R_{i} is remainder term.
http://www.rowan.edu/open/depts/math/osler/ use R_{i} 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 , ζ_{x} is an unspecified constant value
and move f''(ζ_{x}) 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 ''(ζ_{i}) is
log''(ζ_{i}) or d^{2}[log(x)]/dx^{2} evaluate at x=ζ_{i}
d^{1}[log(x)]/dx^{1}=1/x
d^{2}[log(x)]/dx^{2}=-1/x^{2}
d^{2}[log(x)]/dx^{2} evaluate at x=ζ_{i} get -1/(ζ_{i})^{2}<a name="a40127g">
This improvement term is h^{3}*[-1/(ζ_{i})^{2}]/24
Use interval h=1 (OSLER paper page 4/8 "and d=1")
eq.LAf3 right end is 1/[24*c_{k}^{2}]
Liu,Hsinhan get -1/[24*(ζ_{i})^{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
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/(24c_{k}^{2})} --- 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
24c_{k}^{2}
)
--- 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}
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 n^{n}
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
]
*
[
16n^{2}+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)*2^{2n}*n^{n}*e^{-n}
*
exp
(
k=n
∑
k=1
1
24c_{k}^{2}
)
--- 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/[2^{2n}*n!] --- eq.LAb4
Next substitute eq.LAh2+eq.LAh3 to eq.LAb4 get
Γ(n+1/2) below.
<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 2^{2n}/2^{2n}
and merge √2*√π to √(2π) get eq.LAh6 below.
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 c_{k}. Mean value
theorem promiss equality. But c_{k} is a not specified
value. Cannot calculate c_{k} value. Following estimate
c_{k} value and begin approximation formula. c_{k} first
show up at eq.LAf1 c_{k} is discussed at a40126j
"d^{2}f(x)/(dx)^{2} evaluated at x=c_{k}" We know c_{k} play the
role of x. The estimation for ∑[k=1,n]{1/[24*c_{k}^{2}]}
<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*x^{2}]} //a401281641
Next is eq.LAh7+eq.LAh8
k=n
∑
k=1
1
24c_{k}^{2}
≈
x=2n+1/2
∫
x=n+1/2
1
24x^{2}
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
24c_{k}^{2}
≈
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)*n^{n}
e^{n}
√
[
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)
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<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="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 n^{n} that is
use z^{z}, not use (z+1/2)^{z+1/2}.
First job is straighten out (z+1/2)^{z+1/2} separately.
Below z^{z}/z^{z} is inserted number one.
<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="a40128r">
2015-01-28-17-12 here
From eq.LAi5 to eq.LAi7 , fabricated √(2πz)*z^{z}*e^{-z}.
√(2πz)*z^{z}*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="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
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<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)*z^{z}*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} ≈ e^{x} 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}*e^{1/2} = 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(4z^{2}+7z+3)]} --- eq.LAj7
When z approach to infinity, drop lower order term
k(z) ≈ exp{(2z)/[24(4z^{2})]}=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)*z^{z}*e^{-z}*1*1*exp[1/(16z)]*exp[1/(48z)] --- eq.LAk1
Γ(z+1) ≈ √(2πz)*z^{z}*e^{-z}*1*1*exp[1/(16z) + 1/(48z)]
Γ(z+1) ≈ √(2πz)*z^{z}*e^{-z}*1*1*exp[3/(48z) + 1/(48z)]
Γ(z+1) ≈ √(2πz)*z^{z}*e^{-z}*1*1*exp[4/(48z)]
Γ(z+1) ≈ √(2πz)*z^{z}*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!
≈
z^{z}
e^{z}
√(2zπ)
* exp
(
1
12z
)
▐
z → ∞
--- eq.LAk2
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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.
"00000" are space filling. Other bytes get un-even column.
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<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.
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