gamma function
DESCRIPTION
Gamma Function mathematics and history. Please send comments and suggestions for improvements to [email protected]. Thanks. More presentations on different subjects can be found on my website at http://www.solohermelin.com.TRANSCRIPT
1
Gamma Function
SOLO HERMELIN
Updated 28.10.12
2
SOLO
TABLE OF CONTENT Gamma Function
Gamma Function HistoryGamma Function: Euler’s Second IntegralProperties of Gamma Function
Other Gamma Function Definitions: Gauss’ Formula
Some Special Values of Gamma Function:Bohr-Mollerup-Artin Theorem
Other Gamma Function Definitions: Weierstrass’ Formula
Differentiation of Gamma Function
Beta Function: Euler’s First Integral
Euler Reflection Formula
Duplication and Multiplication Formula
Stirling Approximation Formula
References
3
SOLO
Gamma Function History
The Gamma Function was first introduced by the Swiss mathematician Leonhard Euler (1707 – 1783). His goal was to generalize the factorial to non-integer values. Later, it was studied by Adrien-Marie Legendre (1752-1833), Carl Friedrich Gauss (1777-1855), Christoph Gudermann (1798-1852), Joseph Liouville (1809 – 1882), Karl Weierstrass (1815- 1897), Charles Hermite (1822-1901),…and others
Leonhard Euler( 1707– 1783)
0ln1
0
1
xtdtzt
t
x
Adrien-Marie Legendre (1752 – 1833 )
The problem of extending the factorial to non-integer arguments was apparently first considered by Daniel Bernoulli and Christian Goldbach in the 1720s, and was solved at the end of the same decade by Leonhard Euler. Euler gave two different definitions: the first was not his integral but an infinite product,
1 1
11
!k
n
knk
n
of which he informed Goldbach in a letter dated October 13, 1729. He wrote to Goldbach again on January 8, 1730, to announce his discovery of the integral representation
Gamma Function
4
SOLO
Gamma Function History
Leonhard Euler( 1707– 1783)
0ln1
0
1
xtdtzt
t
x
During the years 1729 and 1730, Euler introduced the following analytic function,
By changing of variables we can obtain more known forms
0ln0
101
1
0
1
xtd
e
ttdetuduz
t
tt
xt
t
txeu
dtedu
u
u
xt
t
022ln0
120
12
2
1
0
1 22
2
2
xtdettdettuduz
t
t
txt
t
txeu
dtetdu
u
u
xt
t
The notation Γ (x) is due to Legendre in 1809, while Gauss used Π (x) = Γ (x+1)
Carl Friedrich Gauss
)1777 – 1855(
Adrien-Marie Legendre (1752 – 1833 )
Gamma Function
5
SOLO
t
tt
z
tde
tz
0
1
Proof:
Gamma Function
0& xyixz
t
tt
zt
tt
zt
tt
z
tde
ttd
e
ttd
e
t
1
11
0
1
0
1
For the first part:
xt
xxtx
tdttde
ttd
e
t x
t
t
t
xt
t
xet
tt
yixt
tt
z t 1lim
1110
1
0
1
0
111
0
11
0
1
The first integral converges for any x ≥ δ > 0.
For the second integral, using integration by parts:
t
tt
x
e
t
t
txedv
tu
t
tt
x
e
t
t
txedv
tu
t
tt
xt
tt
yixt
tt
z
tde
txxetx
e
tde
txettd
e
ttd
e
ttd
e
t
t
x
t
x
1
3
/1
1
2
1
2
/1
1
1
1
1
1
1
1
1
2111
1
2
1
Euler’s Second IntegralGamma integral is defined, and converges uniformly for x > 0.
Gamma Function
6
SOLO
t
tt
z
tde
tz
0
1
Proof (continue):
Gamma Function
0& xyixz
For the second integral, using integration by parts:
t
tt
x
e
t
t
txedv
tu
t
tt
x
e
t
t
txedv
tu
t
tt
xt
tt
yixt
tt
z
tde
txxetx
e
tde
txettd
e
ttd
e
ttd
e
t
t
x
t
x
1
3
/1
1
2
1
2
/1
1
1
1
1
1
1
1
1
2111
1
2
1
After [x] (the integer defined such that x-[x] < 1) such integration the power of t in the integrand becomes x-[x]-1 < 0. and we have:
t
tt
t
ttxx
tde
xxxxtdet
xxxx11
1
121
121
Therefore the Gamma integral is defined, and converges uniformly for x > 0.
Gamma integral is defined, and converges uniformly for x > 0.
q.e.d.
Gamma Function
Return to Table of Content
7
SOLO
t
tt
z
tde
tz
0
1
Proof :
Gamma Function
0& xyixz
zzz 1
zztdetztdtzeettdetzt
t
tzt
t ud
z
v
t
v
t
u
zdtedvtu
partsby
t
t
tztz
0
1
0
1
0
,
nintegratio0
01
Properties of Gamma Function : 1
Note that for the evaluation of Gamma Function for a Positive Real Number we need to know only the value of Γ (x) for 0 < x < 1
xxxnxnxnx 121
121
nxnxxx
nxx
For x < 0 with –n < x < -n+1 or 0 < x+n < 1, we define
We can see that for x = 0 or a negative integer the denominator of the right side is zero, and so Γ (x) is undefined (goes to infinity)
Gamma Function
,2,1,0!1 nnn
8
SOLO
t
tt
z
tde
tz
0
1
Proof :
Gamma Function
!1
1Residue
1
1
nz
n
nzResidues of Gamma Function at x = 0,-1, -2,---,-n,..:
121
nxnxxx
nxx
q.e.d.
!1
1
121
1
1211limResidue
11
11
nnn
nxnxxx
nxnxx
n
nxnx
Gamma Function
9
SOLO
Gamma Function Γ (x) and its Inverse 1/Γ (x) Gamma Function
10
SOLO
t
tt
z
tde
tz
0
1
Gamma Function
Absolute value |Γ )z(|
Real value ReΓ )z(
Imaginary value ImΓ )z(
Gamma Function
11
SOLO
t
tt
z
tde
tz
0
1
Gamma Function
Absolute value |Γ )z(|
Gamma Function
12
SOLO
t
tt
z
tde
tz
0
1
Gamma Function
zzz 1
Let compute
110
0
tt
t
t etde
Therefore for any n positive integer:
!1122112111 nnnnnnnnn
Properties of Gamma Function : 1
2
q.e.d.
Gamma Function
13
SOLO Primes
Second definition identical to First
bayxallyfxfyxf ,,1,011
xa by yx 1
yxf 1
yfxf 1
Convex Function :
A Function f (x) is called Convex in an interval (a,b) if for every x,y ϵ (a,b) we have
A Function f (x), defined for x > 0, is called Convex, if the corresponding function
y
xfyxfy
defined for all y > -x, y ≠ 0, is monotonic Increasing throughout the range of definition.
x yx y
yxf
xf
If 0 < x1 < x < x2, are given by choosing y1 = x1 – x < 0, y2 = x2 – x > 0, we express the condition of convexity as
xx
xfxfy
xx
xfxfy
2
22
1
11
xxxfxfxxxfxf 1221
1
12
12
12
21 xx
xxxf
xx
xxxfxf
One other equivalent definition:
14
SOLO Primes
1,0ln1ln1ln yfxfyxf
Logarithmic Convex Function :
A Function f (x)>0 is called logarithmic-convex or simply log-convex if ln (f (x) ) is convex or
This is equivalent to 1ln1ln yfxfyxf
Since the logarithm is a momotonic increasing function we obtain
yxyfxfyxf ,1,01 1
15
SOLO Primes
t
tt
z
tde
tz
0
1
Proof :
Gamma Function
0& xyixz
1,0ln1ln1ln baba
Properties of Gamma Function :
3Gamma is a Log Convex Function
1
1
0
1
0
1
0
111
0
111
badtetdtet
dtetetdtetba
tbtaInequalityHolder
tbtatba
q.e.d.
Return to Table of Content
16
SOLO Primes
t
tt
z
tde
tz
0
1
Proof :
Gamma Function
Other Gamma Function Definitions:
nxxx
nnx
x
n
1
!limGauss’ Formula
Since the Gamma Function is monotonically increasing the logarithm of Gamma Function is also monotonic increasing and for 0 < x < 1 and any n > 2 we have
nnx
nnx
lnln
!1
!ln
!2
!1ln
1
!1ln!ln!1lnln
1
!1ln!2ln
n
n
n
n
nn
x
nnxnn
n
xn
nx
n ln!1
ln1ln
x1 1
yln
0
1
1
ln1ln
x
nn
nn nn
nnx
1
ln1ln1
Carl Friedrich Gauss)1777 – 1855(
17
SOLO Primes
t
tt
z
tde
tz
0
1
Proof (continue - 1) :
Gamma Function
Other Gamma Function Definitions:
Since the Gamma Function is monotonically increasing the logarithm of Gamma Function is also monotonic increasing and for 0 < x < 1 and any n > 2 we have
n
xn
nx
n ln!1
ln1ln
xx nn
nxn ln
!1ln1ln
10 x
!1!11 nnnxnn xx
Use xxxnxnxnx
0
121
xxnxnx
nnx
xxnxnx
nn xx
121
!1
121
!11
nxxx
nnx
x
n
1
!limGauss’ Formula
Euler 1729Gauss 1811
18
SOLO Primes
t
tt
z
tde
tz
0
1
Proof (continue - 2) :
Gamma Function
Other Gamma Function Definitions:
xxnxnx
nnx
xxnxnx
nn xx
121
!1
121
!11
xxnxnx
nnx
xxnxnx
nn xx
11
!1
11
!
Take the limit n → ∞
xxnxnx
nn
nx
xxnxnx
nn x
n
x
n
x
n 11
!lim
11lim
11
!lim
1
1,011
!lim
x
xxnxnx
nnx
x
n
Substitute n+1 for n
nxxx
nnx
x
n
1
!limGauss’ Formula
19
SOLO Primes
t
tt
z
tde
tz
0
1
Let substitute x + 1 for x
Gamma Function
Other Gamma Function Definitions:
1,011
!lim
x
xxnxnx
nnx
x
x
n
n
q.e.d
nxxx
nnx
x
n
1
!limGauss’ Formula
Proof (continue - 3) :
1,011
!lim
1lim
11
!lim1
1
1
xxxxxnxnx
nn
nx
nx
xnxnx
nnx
x
x
nn
x
n
The right side is defined for 0 < x <1. The left side extend the definition for(1 , 2). Therefore the result is true for all x , but 0 and negative integers.
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20
SOLO
t
tt
z
tde
tz
0
1
Gamma Function
Other Gamma Function Definitios:
Start from Gauss Formula xx nn
lim
q.e.d
constantMascheroni-Euler57721566.0ln1
2
11lim
11
nn
kx
e
x
ex
n
k
k
xx
Weierstrass’ Formula
Proof :
nx
nxx
x
eeee
xx
nx
nx
n
xxnxnx
nnx
n
xxx
nnx
xx
n
11
11
1
11
111
11
!:
211
2
11ln
11
1
2
11ln
11
1limlim
k
k
xxn
k
k
x
nnx
nn
n
kx
e
x
e
kx
e
xexx
Karl Theodor Wilhelm Weierstrass
(1815 – 11897)
Gamma Function
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21
SOLO Primes
t
tt
z
tde
tz
0
1
Gamma Function
Some Special Values of Gamma Function:
q.e.d
2222/1
02
0
22 t
t
uut
duudt
t
t
t
udetdt
e
n
nnnnnn
2
125312/12/112/32/12/12/12/1
12531
21
2/12/32/1
2/1
2/1
2/32/1
nnnn
nn
nn
2/1
n
nn
2
125312/1
12531
212/1
n
nnn
Proof:
Return to Table of Content
22
SOLO
Harald August Bohr ) 1887 – 1951(
Proof:Choose n > 2, and 0 < x < 1 and let 11 nxnnn
By logarithmic convexity of f (x), we get
nn
nfnf
nxn
nfxnf
nn
nfnf
1
ln1lnlnln
1
ln1ln
1
!1ln!ln!1lnln
1
!1ln!2ln
nn
x
nxnfnn
By the second property !1,!1,!21 nnfnnfnnf xfxxxnxnxnf 121
xx nn
xnfn ln
!1ln1ln
Emil Artin(1898 – 1962)
Hamburg University
Johannes Mollerup(1872 – 1937)
Gamma Function Bohr-Mollerup-Artin Theorem:
The theorem characterizes the Gamma Function, defined for x > 0 by
as the only function f (x) on the interval x > 0 that simultaneously has the three properties• f (1) = 1• f (1+x) = x f (x) for x > 0• f is logarithmically convex
or Gauss Formula
t
t
tz tdetz0
1 nxxx
nnz
x
n
1
!lim
23
SOLO
Bohr-Mollerup-Artin Theorem:
Harald August Bohr ) 1887 – 1951(
The theorem characterizes the Gamma Function, defined for x > 0 by
as the only function f (x) on the interval x > 0 that simultaneously has the three properties• f (1) = 1• f (1+x) = x f (x) for x > 0• f is logarithmically convex
Proof (continue-1):
By the second property xfxxxnxnxnf 121
xx nn
xfxxxnxnn ln
!1
121ln1ln
We found
Since lan is a monotonic increasing function, we have
121
!1
121
!11
xnxnxx
nnxf
xnxnxx
nn xx
x
xxx
n
n
xnxnxx
nnxf
xnxnxx
nn 1
11
!
11
!
n
n
1
t
t
tz tdetz0
1 nxxx
nnx
x
n
1
!limor Gauss Formula
Emil Artin(1898 – 1962)
Hamburg University
Johannes Mollerup(1872 – 1937)
Gamma Function
24
SOLO
Bohr-Mollerup-Artin Theorem:
q.e.d.
Harald August Bohr ) 1887 – 1951(
The theorem characterizes the Gamma Function, defined for x > 0 by
as the only function f (x) on the interval x > 0 that simultaneously has the three properties• f (1) = 1• f (1+x) = x f (x) for x > 0• f is logarithmically convex Johannes Mollerup
(1872 – 1937)
Proof (continue - 2):
xxx
nxnxnxx
nnxf
xnxnxx
nn
1
111
!
11
!
t
t
tz tdetz0
1 nxxx
nnx
x
n
1
!limor Gauss Formula
By taking n → ∞ we obtain
1
11lim
11
!lim
11
!lim
x
n
x
x
n
x
x
n nxnxnxx
nnxf
xnxnxx
nn
But this is possible only if xxf
Emil Artin(1898 – 1962)
Hamburg University
Gamma Function
25
SOLO
t
tt
z
tde
tz
0
1
Gamma Function Gamma integral is defined, and converges uniformly for x > 0.
Differentiation of Gamma Function:
q.e.d
0,2!11'
ln
01'''
ln
constantMascheroni-Euler57721566.0111'
ln
11
1
122
2
2
2
1
xnkx
n
x
x
xd
dx
xd
d
kxx
xxxx
xd
d
kxkxx
xx
xd
d
kn
n
n
n
n
n
k
k
Proof :
Start from Weierstrass Formula
1 1k
k
xx
kx
e
x
ex
11
1lnlnlnkk k
x
k
xxxx
11 1
111
lnkk
kxk
kxx
xd
d
0111111
ln0
21
221
2
2
kkk kxkxxkxkxxd
dx
xd
d
0
1
1 !11'ln
kn
n
n
n
n
n
kx
n
x
x
xd
dx
xd
d
Gamma Function
We can see that
1
11
1 1
11lim
1
1
1
1'1ln
n
n
kn kk
xxd
d
Return to Table of Content
26
SOLO
1
0
11 1,s
s
zy sdsszyBBeta Function
Beta Function is related to Gamma Function:
u
u
uy
duudt
utt
t
ty udeutdety0
12
20
1 22
2
zy
zyzyB
,
Proof:
In the same way:
v
v
vz vdevz0
12 2
2
u
u
v
v
vuuzy vdudevuzy0 0
1212 22
4
Use polar coordinates:
drdrdrdr
rdrd
vrv
uruvdud
rv
ru
cossin
sincos
//
//
sin
cos
2/
0
1212
0
12
0
2/
0
121212
sincos22
sincos4
2
2
drder
drderzy
zy
zy
r
r
rzy
r
r
rzyzy
Euler’s First Integral
Gamma Function
27
SOLO
1
0
11 1,s
s
zy sdsszyBBeta Function Euler’s First Integral
Beta Function is related to Gamma Function: zy
zyzyB
,
Proof (continue):
2/
0
1212 sincos2
dzyzy zy
Change variables in the integral using dsds cossin2sin 2
zyBsdssds
s
yzzy ,1sincos21
0
112/
0
1212
zyBzyzy ,Therefore q.e.d.
Use z→y and y → 1 - z
u
u
zu
u
z
z
zu
us
u
udsd
s
s
zz
udu
u
u
ud
u
u
u
u
dssszzBzz
0
1
021
11
1
1
0
1
1111
1
11,11
2 q.e.d.
Gamma Function
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28
SOLO
Proof
yzBzyzyBzyyzyzBzyB
,,,,
Use y → 1 - z
u
u
zu
u
z
z
zu
us
u
udsd
s
s
zz
udu
u
u
ud
u
u
u
u
dssszzBzz
0
1
021
11
1
1
0
1
1111
1
11,11
2
t
tt
z
tde
tz
0
1
Gamma Function
Other Gamma Function Properties: zzz
sin
1 Euler Reflection Formula
Gamma Function
29
SOLO
Proof (continue - 1)
u
u
x
udu
uxx
0
1
11
t
tt
z
tde
tz
0
1
Gamma Function
Other Gamma Function Properties:
R
RC
C
1
planeu
uRe
uIm
Replace the path from 0 to ∞ by the Hankel contour Hε
in the Figure, described by four paths, traveled in counterclockwise direction: 1. going counterclockwise above the real axis, (u = |u|)2. along the circular path CR, 3. bellow the real axis, (u= |u|e -2πi )4. along the circular path Cε.
C
yR yyi
C
yR y
udu
uud
u
ueud
u
uud
u
u
R1111
2
Define y = 1 – x, and assume x,y ϵ (0,1)
zzz
sin
1 Euler Reflection Formula
Gamma Function
30
SOLO
Proof (continue - 1)
u
u
x
udu
uxx
0
1
11
t
tt
z
tde
tz
0
1
Gamma Function
Other Gamma Function Properties:
R
RC
C
1
planeu
uRe
uIm
This path encloses the pole u=-1 of that has the residue1
u
u y
yi
eu
yy
euu
ui
11
Residue
By the Residue Theorem
For z ≠ 0 we have
yzyzyzyy zeeez lnlnReln
zzz
sin
1 Euler Reflection Formula
yiy
eu
y
C
yR yiy
C
yR y
eiu
uui
u
uizd
z
zud
u
uezd
z
zud
u
u
i
R
21
1lim2
1Residue2
1111
1
2
Gamma Function
31
SOLO
Proof (continue - 2)
t
tt
z
tde
tz
0
1
Gamma Function
Other Gamma Function Properties:
R
RC
C
1
planeu
uRe
uIm
yi
C
yR yiy
C
yR y
eizdz
zud
u
uezd
z
zud
u
u
R
2
11112
For the second and forth integral we have
0lnlnReln zzeeezyzyzyzyy
z
z
z
z
z
zyyy
111
Hence for small ε we have:
and for large R we have:
01
21
01
y
C
y
zdz
z
01
21
1
Ry
C
y
R
Rzd
z
z
R
Therefore the integrals on the circular paths are zero for ε→0 and R →∞
zzz
sin
1 Euler Reflection Formula
Gamma Function
32
SOLO
Proof (continue - 3)
t
tt
z
tde
tz
0
1
Gamma Function
Other Gamma Function Properties:
R
RC
C
1
planeu
uRe
uIm
yiy
iyy
eiudu
ueud
u
u
2
11 0
2
0
We obtain
Multiply both sides by yie
iudu
uee
yiyiy 2
10
yee
iud
u
uiyiy
y
sin
2
10
Rearranging we obtain
Since both sides of this equation are meromorphic (analytic) in x ϵ (0,1) we can extend the result for all analytic parts of z ϵ C (complex plane).
1,0sin1sin11
10
1
0
1
xxx
udu
uud
u
uxx
u
u
yxyu
u
x
Substituting y = 1 – x we obtain
zzz
sin
1 Euler Reflection Formula
Gamma Function
33
SOLO
Onother Proof
t
tt
z
tde
tz
0
1
Gamma Function
Other Gamma Function Properties:
Start with Weierstrass Gamma Formula
zzz
sin
1 Euler Reflection Formula
1 1k
k
xx
kx
e
x
ex
12
22
1
2 1111
kk k
x
k
xxx
k
xx
e
kx
e
kx
eexxx
Use the fact that Γ (-x)=- Γ (1-x)/x to obtain
12
2
11
1
k k
xx
xx
Now use the well-known infinite product
12
2
1sink k
xxx
q.e.d.
Gamma Function
34
SOLO
Proof
t
tt
z
tde
tz
0
1
Gamma Function
Other Gamma Function Properties: zzz
cos2
1
2
1
Start from
Substitute ½ +z instead of z
zzz
sin
1
zz
zz
cos
21
sin2
1
2
1
q.e.d.
Gamma Function
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35
SOLO
t
tt
z
tde
tz
0
1
Gamma Function
Stirling Approximation Formula:
121 xexxx xx
u
u
uuxxx
u
u
xuxxxu
u
xxuxuxt
udxtd
t
t
xt
udeex
udueexudxuxetdtex
1
1ln1
1
1
1
11
1
111Proof:
The function f(u) = -u + ln (1 + u) equals zero for u = 0. For other values of u we havef(u) < 0. This implies that the integrand of the last integral equals 1 at u = 0 and that thisintegrand becomes very small for large values of x at other values of u. So for large values of x we only have to deal with the integrand near u = 0. Note that we have
02
1
2
11ln 2222 uforuuuuuuuuuf
This implies that
xforduedueu
u
uxu
u
uux 2/
1
1ln 2
James Stirling, a contemporary of Euler, also attempted to find a continuous expression for the factorial and came up with what is now known as Stirling's formula. Although Stirling's formula gives a good estimate of , also for non-integers, it does not provide the exact value. Extensions of his formula that correct the error were given by Stirling himself and by Jacques Philippe Marie Binetן
Gamma Function
36
SOLO
t
tt
z
tde
tz
0
1
Gamma Function
Stirling Approximation Formula:
121 xexxx xxProof (continue):
xfordueexudeexxu
u
uxxxu
u
uuxxx 2/1
1
1ln1 2
1
xforxdtexduet
t
txtu
xtdud
u
u
ux
22 2/12/1/2
/2
2/ 22
If we set we have by using the normal integralxtu /2
therefore:
xexxx xx21
q.e.d.
Gamma Function
37
SOLO
t
tt
z
tde
tz
0
1
Gamma Function
Duplication and Multiplication Formula:
0Re222
112
zzzz
z
Legendre Duplication Formula1809
Adrien-Marie Legendre (1752 – 1833 )
Proof:
2/1,2sin22sin2
2sin22sincos2,
212/
0
1221
0
1221
2/
0
12212/
0
1212
zBdd
ddzzB
zzzzz
zzzz
0Re2/1
2/122/1,2,
22121
zz
zzBzzB
z
zz zzWe have
therefore
q.e.d
0Re222
112
2
1
zzzzz
Gamma Function
38
SOLO
t
tt
z
tde
tz
0
1
Gamma Function
Duplication and Multiplication Formula:
znnn
nz
nz
nzz znn
2/12/12
121
Gauss Multiplication
Formula
Proof:
nz
1
Carl Friedrich Gauss)1777 – 1855(
nn
n
nn
n 2/12121
Euler
Multiplication Formula
Gamma Function
Define the function:
n
nx
n
x
n
xnxf x 11
:
This function has the following properties:
1
xfxn
x
n
x
n
nx
n
x
n
xnn
n
nx
n
nx
n
x
n
xnxf
x
x
121
1211 1
39
SOLO
t
tt
z
tde
tz
0
1
Gamma Function
Duplication and Multiplication Formula:
znnn
nz
nz
nzz znn
2/12/12
121 Gauss
Multiplication Formula
Proof (continue – 1):Carl Friedrich Gauss
)1777 – 1855(
Gamma Function
Since (ln nx)”=(x ln n)”=(ln n)’=0, and each Γ ((x+k)/k) is log convex.f (x) is log convex.
n
n
nnnaaf nn
2111
So using Bohr-Mollerup-Artin Theorem we can write: f (x) = an Γ(x)where an is a constant, to be found, and Γ (1)=1 (the third condition of the Theorem).
2
Therefore
Use Gauss’ Formula for Gamma Function with x=k/n
pnknkk
npp
pnk
nk
nk
pp
n
k pn
k
p
n
k
p
1!lim
1
!lim
40
SOLO
t
tt
z
tde
tz
0
1
Gamma Function
Duplication and Multiplication Formula:
znnn
nz
nz
nzz znn
2/12/12
121 Gauss
Multiplication Formula
Proof (continue – 2):Carl Friedrich Gauss
)1777 – 1855(
Gamma Function
pnknkk
npp
n
k pn
k
p
1!lim
Since k = 1,2,…,p
!1!
11211
nppnn
pnnpnnnnnpnknkkp
k
!!
lim!
!lim
21 2
11
11
pnn
pnpn
pnn
pnpn
n
n
nnna
npnn
p
n
n
npnn
pn
41
SOLO
t
tt
z
tde
tz
0
1
Gamma Function
Duplication and Multiplication Formula:
znnn
nz
nz
nzz znn
2/12/12
121 Gauss
Multiplication Formula
Proof (continue – 3):Carl Friedrich Gauss
)1777 – 1855(
Gamma Function
!!
lim2
11
pnn
pnpna
npnn
pn
Use the identity
npp pnpn
pnn
pn
n
pnpn
1
!
!lim1
21
11lim1
to an to get
2
1
2
112
11
!
!lim
!
!
!
!lim1
!
!lim
n
pnn
pn
npnn
p
npnn
pn
ppn
npn
pnpn
pnn
pnn
pnpn
pnn
pnpna
42
SOLO
t
tt
z
tde
tz
0
1
Gamma Function
Duplication and Multiplication Formula:
znnn
nz
nz
nzz znn
2/12/12
121 Gauss
Multiplication Formula
Proof (continue – 4):Carl Friedrich Gauss
)1777 – 1855(
Gamma Function
to an to get
2
1
!
!lim
n
pnn
pn
ppn
npna
pepp pp
2
1
2! pepnpn pnpn2
1
2!
2
1
2
1
2
1
2
1
2
1
2
2
2
lim n
pepn
nep
nan
npnpn
pn
n
pp
pn
Use Stirling’s Approximation formula xexxx xx21
43
SOLO
t
tt
z
tde
tz
0
1
Gamma Function
Duplication and Multiplication Formula:
znnn
nz
nz
nzz znn
2/12/12
121 Gauss
Multiplication Formula
Proof (continue – 4):Carl Friedrich Gauss
)1777 – 1855(
Gamma Function
2
1
2
1
2 nan
n
xan
nx
n
x
n
xnxf nx
11:
We have
or xnn
nx
n
x
n
x xn
2
1
2
1
211
Define x = n z to obtain
znnn
nz
nzz
znn
2
1
2
1
211 q.e.d
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44
SOLO
References
Internethttp://en.wikipedia.org/wiki/
G.B. Arfken, H.J. Weber, “Mathematical Methods for Physicists”, Academic Press, Fifth Ed., 2001
http://www.frm.utn.edu.ar/analisisdsys/material/function_gamma.pdfhttp://homepage.tudelft.nl/11r49/documents/wi4006/gammabeta.pdf
Gamma Function
M.Abramowitz & I.E. Stegun, ED., “Handbook of Mathematical Functions”, Dover Publication, 1965,
H.Vic Dannon,”Riemann Zeta Function: The Riemann Hypothesis Origin, the Factoriztion Error, and the Count of the Primes”, Gauge Institute Journal, Vol. 5, No. 4, November 2009J. Baltzersen, “Hardy’s Theorem and the prime number theorem”, Thesis University of Copenhagen, June 2007D. Miličić, “Notes on Riemann Zeta Function”, http://www.math.utah.edu/~milicic/zeta.pdf
P. Garrett, “Riemann’ Explicit/Exact formula”, (October 2, 2010), http://www.math.umn.edu/~garrett/m/mfms/notes_c/mfms_notes_02.pdfhttp://homepage.tudelft.nl/11r49/documents/wi4006/gammabeta.pdf
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April 10, 2023 45
SOLO
TechnionIsraeli Institute of Technology
1964 – 1968 BSc EE1968 – 1971 MSc EE
Israeli Air Force1970 – 1974
RAFAELIsraeli Armament Development Authority
1974 – 2013
Stanford University1983 – 1986 PhD AA