<a name="docA001">
2014-10-29-08-55 start
Liu,Hsinhan write complex number calculator
http://freeman2.com/complex4.htm
include many basic complex number functions
such as complex exponential cexpf() and
complex sine csinf() etc. Now added advanced
complex number functions
complex gumma function Γ()=gamma()
in complex4.htm box3 input Γ'1.2+3.4i'
complex zeta function ζ()=czeta()
in complex4.htm box3 input ζ'2.1-4.3i'
<a name="docA002">
Liu,Hsinhan write complex zeta function
following next web page
2014-09-30-22-34 LiuHH access
http://www.donotrememberthisaddress.com/mathematics/calculatevalue.php
http://www.robertelder.ca/calculatevalue/
for complex input z real part > 0
eq.cz01, eq.cz02.
<a name="docA003">
Below is input complex real part <= 0
eq.cz03. Reference is
2014-03-26-10-20 use YTD download
http://www.youtube.com/watch?v=rGo2hsoJSbo
and
2014-03-25-18-28 MrYouMath
http://www.youtube.com/watch?v=TnRnlJBecRg&list=PL32446FDD4DA932C9&index=15
both URL one equation, input complex real part <= 0
ζ(s)=2^s*PI^(s-1)*sin(PI*s/2)*Γ(1-s)*ζ(1-s)
here 's' is complex input, such as '1.1+2.2i'
<a name="docA004">
After done complex4.htm Γ()=gamma() and
ζ()=czeta() , Liu,Hsinhan hope to know how
to derive these equations. This web page
http://freeman2.com/tute0057.htm
is study notes for Riemann Zeta function
derivation. Main textbook is MrYouMath
lecture video. URL is next.
<a name="docA005">
2014-10-29-07-56
https://www.youtube.com/playlist?list=PL32446FDD4DA932C9
save as MrYouMath_17_files_Riemann.htm
URL are next.
[[
http://www.youtube.com/watch?v=ZlYfEqdlhk0
http://www.youtube.com/watch?v=I3qSCWNXZKg
http://www.youtube.com/watch?v=TDdGisWD5OU
http://www.youtube.com/watch?v=SKa7b-3C32A
http://www.youtube.com/watch?v=3eN9tQX3JJ4
http://www.youtube.com/watch?v=U16_KTTKtb0
http://www.youtube.com/watch?v=GeKDmoYHiAk
http://www.youtube.com/watch?v=axQqExF7NsU
http://www.youtube.com/watch?v=XHQ0OzqTjd0
http://www.youtube.com/watch?v=1f24RZfP6m8
http://www.youtube.com/watch?v=UEZ4ClCdog8
http://www.youtube.com/watch?v=-GQFljOVZ7I
http://www.youtube.com/watch?v=K6L4Ez4ZVZc
http://www.youtube.com/watch?v=TnRnlJBecRg
http://www.youtube.com/watch?v=G-fqe3BkBnE
http://www.youtube.com/watch?v=QfDbF_qlp58
http://www.youtube.com/watch?v=wt6ngy6pDws
]]
<a name="docA005a">
Alert! Norton delete YTDYTDreget //2015-02-09-09-43
Reader can download MrYouMath Riemann lectures
You can try YTD downloader (YouTube Downloader)
In YTD input playlist URL
https://www.youtube.com/playlist?list=PL32446FDD4DA932C9
YTD will download all 17 videos at one click.
YTD basic (free) has limit. YTD Pro better.
2013-10-21-19-48 Liu,Hsinhan access
http://www.youtubedownloadersite.com/installers/YTDSetup.exe
10/21/2013 07:49 PM 11,574,336 YTDSetup.exe
2014-05-27-14-25 click "Place order" //$29.90
https://secure.avangate.com/order/verify.php?CART_ID=deleted_by_LiuHH
<a name="docA006">
2014-10-29-08-24
MrYouMath_17_files_Riemann.htm
[[
"Zeta Function - Part 1 - Convergence"
data-video-id="ZlYfEqdlhk0"
"Zeta Function - Part 2 - Euler Product Representation"
data-video-id="I3qSCWNXZKg"
"Zeta Function - Part 3 - Euler Product (easy)"
data-video-id="TDdGisWD5OU"
<a name="docA007">
"Zeta Function - Part 4 - Infinitude of Prime Numbers"
data-video-id="SKa7b-3C32A"
"Zeta Function - Part 5 - Prime Zeta Function"
data-video-id="3eN9tQX3JJ4"
"Zeta Function - Part 6 - The Prime Counting Function"
data-video-id="U16_KTTKtb0"
"Zeta Function - Part 7 - Zeta of 2 aka The Basel Problem"
data-video-id="GeKDmoYHiAk"
"Zeta Function - Part 8 - Zeta of 2n - Part 1"
data-video-id="axQqExF7NsU"
<a name="docA008">
"Zeta Function - Part 8 - Zeta of 2n - Part 2"
data-video-id="XHQ0OzqTjd0" 9/17
"Zeta Function - Part 8 - Zeta of 2n - Part 3"
data-video-id="1f24RZfP6m8" 10/17
"Zeta Function - Part 9 - Relation to Gamma Function"
data-video-id="UEZ4ClCdog8" 11/17
"Zeta Function - Part 10 - Jacobi Theta Function"
data-video-id="-GQFljOVZ7I" 12/17
"Zeta Function - Part 11 - Riemann Functional Equation I"
data-video-id="K6L4Ez4ZVZc" 13/17<a name="docA009">
"Zeta Function - Part 12 - Riemann Functional Equation II"
data-video-id="TnRnlJBecRg" 14/17
"Zeta Function - Part 13 - Trivial Zeros of the Zeta Function"
data-video-id="G-fqe3BkBnE" 15/17
"Zeta Function - Part 14 - Riemann Xi Function"
data-video-id="QfDbF_qlp58" 16/17
"Sine Function Product Formula (Hadamard Factorization Theorem)"
https://www.youtube.com/watch?v=h3Hcioh2F9I 17/17"What is a function? Why 1+2+3+4+5+.... not equals -1/12 = Zeta(-1)"
data-video-id="wt6ngy6pDws" This video is easy. skipped a312190901
]]
2015-02-09-09-45 add <a name="GammaIndex">Gamma Function - Part 1 - Functional Equation
http://www.youtube.com/watch?v=2iBNo4j3vRo
Gamma Function - Part 2 - Gauss Representation
http://www.youtube.com/watch?v=a_Dlx7TTjkI
Gamma Function - Part 3 - Weierstrass Representation
http://www.youtube.com/watch?v=d9Oz62Ioue0
Gamma Function - Part 4 - Relationship to Sine
http://www.youtube.com/watch?v=W3ulxj3s90U
Gamma Function - Part 5 - Gamma of 0.5 ( one half)
http://www.youtube.com/watch?v=_vwqsJNKY-c
Gamma Function - Part 6 - Stirling's Approximation
http://www.youtube.com/watch?v=MuAb2dnPD0Q
An intuitive derivation of Stirling’s formula
http://www.rowan.edu/open/depts/math/osler/Stirlings_formula.pdf
Gamma Function - Part 7 - Euler Integral I
http://www.youtube.com/watch?v=VF7ud3Al6d8
Gamma Function - Part 8 - Euler Integral II The Sinc-Function
http://www.youtube.com/watch?v=h4wQ_F1s1UI
Gamma Function - Part 9 - Euler Integral III Fresnel Integral
http://www.youtube.com/watch?v=H0kT-EKbUzM
Gamma Function - Part 10 - Beta Function
http://www.youtube.com/watch?v=Korx_G7eySQ
Gamma Function - Part 11 - Legendre Duplication Formula
http://www.youtube.com/watch?v=yu9NeDXalpA
Gamma Function - Part 12 - Relation to Zeta Function
http://www.youtube.com/watch?v=jQAPlsNY_P0
Above red link has complete lecture. Blue not.
2015-01-01-15-38 stop
<a name="docA010">
"study notes" is watching MrYouMath video,
convert lecture speech to html text string,
and add derivation details.
Copy lecture may not have error, but LiuHH
add derivation details may contain error !
Reader MUST be alert.
2014-10-29-09-23 here
<a name="docA011">
2014-10-29-16-42 include next three equations
s be a complex number, example s='1.1+2.2i'
Riemann Zeta function ζ(s), if Real(s)>1
ζ(s)=∑[n=1,∞]{1/n^s} = ∏[p=prime]{p^s/(p^s-1)} --- eq_cz01a
Next is same equation in better form.
2014-10-08-00-19
if Real(s)>1 , ζ(s) is defined by next equation.
ζ(s)
=
n=∞
∑
n=1
1
n^{s}
=
p=∞
∏
p=prime
p^{s}
p^{s}-1
--- eq.cz01
width of above equation
<a name="docA011a">
2014-11-05-12-25 next is for entry level reader.
s be a complex number, example s='1.1+2.2i'
complex number s real part MUST BE > 1.0
n=∞
∑
n=1
1
n^{s}
=
1
1^{s}
+
1
2^{s}
+
1
3^{s}
+
1
4^{s}
+
1
5^{s}
+ ...
--- eq_cz01b
width of above equation
p=∞
∏
p=prime
p^{s}
p^{s}-1
=
2^{s}
2^{s}-1
*
3^{s}
3^{s}-1
*
5^{s}
5^{s}-1
*
7^{s}
7^{s}-1
*
11^{s}
11^{s}-1
* ...
--- eq_cz01c
width of above equation
In eq_cz01b all denominator are positive integers.
In eq_cz01c all denominator are prime numbers.
Both eq_cz01b and eq_cz01c extend to infinity.
2014-11-05-12-42 above is for entry level reader.
<a name="docA012">
s be a complex number, example s='0.5-3.4i'
complex number s real part MUST BE 0<Real(s)<=1
Riemann Zeta function ζ(s), if 0<Real(s)<=1
ζ(s)=∑[n=1,∞]{-1^{n+1}/n^s}/(1-2^{1-s}) --- eq_cz02a
or
ζ(s)={[1-2^(1-s)]^(-1)}*∑[n=1,∞]{-1^{n+1}/n^s} --- eq_cz02b
Next is same equation in better form.
2014-10-08-00-38
http://www.youtube.com/watch?v=rGo2hsoJSbo
do NOT have '=' . Next line, LiuHH added '=' ;
if 0<Real(s)<=1 , ζ(s) is defined by next equation.
ζ(s)
=
1
1-2^{1-s}
*
n=∞
∑
n=1
(-1)^{n+1}
n^{s}
--- eq.cz02
width of above equation
<a name="docA013">
s be a complex number, example s='-0.78+1.23i'
complex number s real part MUST BE <=0
Riemann Zeta function ζ(s), if Real(s)<=0
ζ(s)=2^{s}*PI^{s-1}*sin(PI*s/2)*Γ(1-s)*ζ(1-s) --- eq_cz03a
or
ζ(s)=2^s*PI^(s-1)*sin(PI*s/2)*Γ(1-s)*ζ(1-s) --- eq_cz03b
here PI=π=3.14159265358979323846264338327950288419716939937510 ...
Next is same equation in better form.
2014-10-08-00-48
http://www.youtube.com/watch?v=rGo2hsoJSbo
do NOT have '=' . Next line, LiuHH added '=' ;
if Real(s)<=0 , ζ(s) is defined by next equation.
ζ(s)
=
2^{s}
*
π^{s-1}
*
sin
(
π*s
2
)
*
Γ(1-s)
*
ζ(1-s)
--- eq.cz03
width of above equation
above ζ(1-s) is Riemann Zeta function with input 1-s
above Γ(1-s) is complex Gamma function with input 1-s
Because if Real(s)<=0 , then Real(1-s)>=1
<a name="docA014">
Liu,Hsinhan access next video and find above three equations.
2014-03-25-18-28 use YTD download (YouTube Downloader)
http://www.youtube.com/watch?v=TnRnlJBecRg
MrYouMath: "Zeta Function - Part 12 - Riemann Functional Equation II"
eq.cz01 , eq.cz02 , eq.cz03 from next URL
2014-03-26-10-20 use YTD download
http://www.youtube.com/watch?v=rGo2hsoJSbo
singingbanana "The Riemann Hypothesis"
From time 11:40 to 13:05 screen has three equations .
total minute/second is 19:35 .
2014-10-29-19-29 done include and modify.
<a name="docA015">Begin video 1 of 17
2014-10-30-07-57 start
Next is study notes of
"Zeta Function - Part 1 - Convergence"
http://www.youtube.com/watch?v=ZlYfEqdlhk0
Lecturer is MrYouMath.
Notes recorder is Liu,Hsinhan.
Recorded notes is about 90% match, not exact.Liu,Hsinhan inserted text is blue Italic.
LiuHH notes may contain error!<a name="docA016">
Hello and welcome to the first video of
serial about zeta function.
What is zeta function?
zeta function is a very important function
in numerical analysis. If you want to know
more about numbers, then you have to know a
lot about Riemann zeta function. Actually the
most important part is about the distribution
of prime numbers. It gives you analytic way
to count how many prime numbers exist. So
special large numbers. In order to understand
this question, Riemann Hypothesis is a hundred
fifty years old problem.
<a name="docA017">
I thought about doing this video, because I
did not found anything similar to that on
youtube. This is the reason why I start from
very basic stuff like convergence and go step
forward, we will derive the Euler product
representation and jump to derive relationship
between zeta function and prime number zeta
function. We will show reciprocal sum of prime
number is infinitely large.
<a name="docA018">
We will calculate values of zeta function for
even numbers ζ(2) or ζ(4) etc. Odd numbers ζ(3)
or ζ(5) present big problems until today, no
one knows how to calculate them. We will do
a lot stuff, maybe at the end we will have
the Riemann Hypothesis you will understand
where it came from, what it actually means
why raise to this strange looking function.
But now let us just look at the zeta function.
//2014-10-30-08-36 here
<a name="docA019"> Euler defined s be real number. Riemann defined
s be complex number, this is why we call it as
Euler Riemann zeta function.
ζ(s)
=
n=∞
∑
n=1
1
n^{s}
--- eq.cz04
width of above equation
<a name="docA020">
2014-10-30-08-53 here
zeta function first defined like eq.cz04
There are many zeta function representation.
This eq.cz04 is first defined by Euler.
Euler solved zeta function for even numbers
ζ(2) or ζ(4) etc. Euler also try solve odd
numbers ζ(3) or ζ(5) but it is a big problem.
Not solved until today, maybe you will be
able to solve this, but no mathematician
is able to solve that. eq.cz04 is the
function, we want to know what is the
region of convergence of eq.cz04.
<a name="docA021">
You may ask why doing such easy things in
first video? I just want to show you how
to work with zeta function, how to work
with infinite sum and then you get a feel
what we will do later on. First step is to
take absolute value of eq.cz04 .
//2014-10-30-09-12 here
<a name="docA023">
2014-10-30-09-32
Taking absolute value of a sum, its value
is always less or equal to sum of absolute
value of its factors. It is called
triangular inequality. LiuHH notes may contain error!
For example //blue italic is LiuHH notes
| 1+2+3 | = 6 <= 6 = |1| + |2| + |3|
| 1-2+3 | = 2 <= 6 = |1| +|-2| + |3|
above is REAL number example. For complex
number, need triangular inequality. <a name="docA024">
Black |ζ(s)| equal eq.cz05 red term, because
of eq.cz04.
eq.cz05 red term less or equal to eq.cz05
blue term, because of triangular inequality.
eq.cz05 blue term equal to eq.cz05 purple
term, because purple numerator is 1. |1|=1 .
Triangular inequality say two sides length
is always longer than one side length.
<a name="docA025">LiuHH notes may contain error!
//blue italic is LiuHH notes.
The term n^s , n is integer, s is complex,
for example, if n=3, if s='-1.2-3.4i' then
1/(n^s)=n^(-s)=3^'1.2+3.4i'=-3.098-2.091i
result is a complex -3.097-2.090i
if n=4, if s='-1.2-3.4i' then
1/(n^s)=n^(-s)=4^'1.2+3.4i'=0.005-5.278i
result is a complex 0.005-5.278i
<a name="docA026">
Next see sum first then take absolute value
|3^'1.2+3.4i' + 4^'1.2+3.4i'|=
|-3.098-2.091i+0.005-5.278i|=|-3.093-7.369i|
=7.992
Next see take absolute value first then sum
|3^'1.2+3.4i'| + |4^'1.2+3.4i'|=
|-3.098-2.091i| + |0.005-5.278i|
3.737630 + 5.278002 = 9.015632
<a name="docA027">
for two terms, n=3 and n=4 with s='-1.2-3.4i'
take absolute value first then sum get 9.015
sum first then take absolute value get 7.992
this two terms example illustrate
triangular inequality 7.992 <= 9.015 that is
eq.cz05 red term <= eq.cz05 blue term
<a name="docA028">A complex has direction, different 1/(n^s)
has different direction, then triangular
inequality apply. On the other hand,
real numbers have SAME direction.
Complex number is a 2-D being,
real number is a 1-D being.<a name="docA029">
In complex4.htm box3 input next four lines
aa=3^'1.2+3.4i'
ab=4^'1.2+3.4i'
|aa+|ab //greater value
|(aa+ab) //smaller value
click [eval Box3] get
caddf((cabsf(aa)),(cabsf(ab)))
9.015224461938128,0
cabsf(caddf(aa,ab))
7.991268934969664
//blue italic is LiuHH notes, may be error!
<a name="docA030">
Next step I take absolute value of number one
giving one, no change. |1|=1
2014-10-30-10-03 stop at video 3:56
2014-10-30-20-29 start
Now only denominator has absolute value.
(see eq.cz05 right end purple term)
just go ahead write down what we find out.
<a name="docA032">
2014-10-30-20-42 here
In eq.cz06 we want to know the red term 1/|n^{s}|
s is a complex number. It is hard to know 1/|n^{s}|.
Euler defined s be real number. Riemann defined
s be complex number, this is why we call it as
Euler Riemann zeta function. Every complex number
can be written as x+iy, n to x+iy and take
absolute value. You can separate them n^(x+iy)
just power law to [n^x]*[n^(iy)]. First part is
very simple, 1/[n^x] is real n to real x power.
This is exponential function which take a positive
real number. The right hand side 1/[n^(iy)] is a
little bit hard to understand, but we will do a
little step in order to understand better.
<a name="docA034">
2014-10-30-21-05
Just write this n to iy, n^{iy}, as exponential
e^{iy*log(n)}.
LiuHH notes may contain error!
//blue italic is LiuHH notes.
exponential function and log function are
inverse to each other. For example exp(log(A))=A
exp(log(2.2)) = 2.2
cexpf(clogf(-1.3)) = -1.3+0i ; therefore
n^{iy} = exp(log(n^{iy})) = exp(iy*log(n)) = e^{iy*log(n)}
2014-10-30-21-19 here
<a name="docA035">
e^{iy*log(n)} is e (=2.718281828459045...) to a power
of imaginary number i*[y*log(n)], both y and n are
real. y*log(n) is real and i*[y*log(n)] is an
imaginary number. complex number "0+i*[y*log(n)]"
real part is zero. e power to an imaginary number,
its magnitude is always one.
<a name="docA036">
See Euler's formula e^{it}=cos(it)+i*sin(it) and
|e^{it}|=sqrt[cos(it)*cos(it)+sin(it)*sin(it)]
|e^{it}|=sqrt[ 1 ]= 1 //here t≡y*log(n)
In eq.cz07 from blue term to purple term, use
above explanation. |e^{it}| is one and omitted.
Above is e=2.71828... to imaginary power.
Below is real n to real x power.
|n^{x}| is real n to power of x, s=x+i*y and x be
any real number. |n^{x}| is always positive
and drop absolute sign, get eq.cz07 purple term.
absolute sign all gone.
2014-10-30-21-34 stop at video 6:43
<a name="docA037">
2014-10-31-07-36
To proof Euler's formula
exp(i*t)=cos(t)+i*sin(t) ---(A91)
Please see Tom M. Apostol, Calculus volume 1.
Second ed. 1967 page 366. LiuHH study notes
http://freeman2.com/tute0006.htm#a20402e
Please pay attention to that in equation (A91)
't' can be both real and complex.
If 't' is complex then right hand side is
complex1 + i*complex2 --- eq.cz08
//cos(complex) is another complex.
<a name="docA038">
In general eq.cz08 is NOT what we want.
All complex function derivation MUST end at
real1 + i*real2 --- eq.cz09
therefore 't' in Euler's formula is REAL
this is the case most discussion focus.
2014-10-31-07-49
<a name="docA039">
2014-10-31-07-58 start video at 6:43
Expression e^{iy*log(n)} its power is imaginary,
complex number e^{iy*log(n)} is on unit circle.
//see |e^{it}|=sqrt[1]= 1, radius 1 is unit length.
Unit circle has absolute value one. Then
|e^{iy*log(n)}| is one. //see eq.cz07 blue term
We end up find out the zeta function |ζ(s)|
is smaller or equal to ∑[n=1,∞]{1/n^{x}}
complex s = x+i*y; x is real number,
i*y contribute to length 1. "*1" is omitted.
//next red text is MrYouMath words.
The zeta function definition is ∑[n=1,∞]{1/n^{s}}
where s is complex. Now reduce to ∑[n=1,∞]{1/n^{x}}
here x is real variable. We can apply all the
formula and theorem about real sum.
<a name="docA040">
Next eq.cz10 is our first result.
2014-10-31-08-18
Alert, 's' is a complex number. s=x+iy
x and y are (MUST BE) both real
a311040901
|ζ(s)|
<=
n=∞
∑
n=1
1
n^{x}
--- eq.cz10
width of above equation
<a name="docA041">
2014-10-31-08-23
We use a very important theorem in math. It is
called Cauchy Integral. Cauchy is a Frenchman.
We use Cauchy-Integral Theorem:
Sum is convergent/divergent
if the corresponding
integral is convergent/divergent . <a name="docA042">
If you want to find the sum ∑[n=1,∞]{1/n^{x}}
is convergent or divergent, we have to look at
its corresponding integral.
You replace sum sign ∑ by integral sign ∫ .
2014-10-31-08-33
<a name="docA044">
2014-10-31-08-49
You replace n by t. Because if you plug in
n for t, you will have all the same values.
This is just approximation of the sum.
If one is converging, the other is converging.
If one is diverging , the other is diverging.
Now integrating eq.cz11 red term is very easy.
The result is eq.cz11 blue term. It is one
over 1-x and multiply with t to 1-x power.
You just differentiate eq.cz11 blue term
with respect to t and get eq.cz11 red term.
<a name="docA045">
Now we have eq.cz11 blue term. This is a
little bit problematic. First you see that
in 1/(1-x) , x is not allow to be one.
The case x=1, in ∫[t=1,t=∞]{1/t^{x}}dt we have
∫[t=1,t=∞]{1/t}dt this is logarithms, its
value is infinity, diverging.
<a name="docA046">
Now let us look the other case (x NOT =1).
If we plug in 1 fot t in eq.cz11 blue term,
there is no problem. But
If we plug in infinity ∞ fot t in eq.cz11
blue term, we have problem. Infinity to
power of positive number it will become
infinity large. This is diverge.
Infinity to power of negative number is
one over infinity which is zero. <a name="docA047">
//In eq.cz11 blue term {t^{1-x}}|[t=1,t=∞]
We have to make t power a negative number.
You write down 1-x<0 , you will get 1<x
x should be greater than one.
Remember that x is real part of complex s.
Complex number real part must be greater
than one.
This conclude first lecture.
<a name="docA048">Above is study notes of
"Zeta Function - Part 1 - Convergence"
http://www.youtube.com/watch?v=ZlYfEqdlhk0
2014-10-31-09-41 stop
2014-11-04-09-17 done first proofread, add link.
2014-11-04-10-03 done second proofread.
2014-11-04-17-46 done third proofread.
third proofread follow video lecture.
Recorded notes is about 90% match, not exact.<a name="docA049">
2014-11-05-18-00
update 2014-11-05 made minor change at
add youtubedownloadersite.com
add next is for entry level reader.
Cauchy-Integral TheoremSum is convergent/divergent
if the corresponding
integral is convergent/divergent .
Liu,Hsinhan go online search for
"Cauchy-Integral Theorem" all result are
complex number related. Finally get the
better term "Series Integral Test".
<a name="docA051">
Next is reading time and textbook name.
2014-11-05-23-06
textbook by Tom M. Apostol vol.1
second ed. page 397 has proof of
integral test.
2014-11-05-23-14
textbook by George, B. Thomas Jr. and
Ross L. Finney Calculus sixth ed.
page 624, 625 has proof of integral test.
2014-11-05-23-22
textbook by George Arfken third ed.
page 283 has proof of integral test.
If sum convergence/divergence is known, eq.cz14 conclude
integral convergence/divergence. //a311080658
width of above equation
<a name="docA053">
2014-11-07-12-53 here
Integral Test is TWO comparison tests.
In Integral Test 1 serial red area is
greater than function black area.
In Integral Test 2 serial red area
parallel move to left one unit. Then
serial sum start from 2. (not from 1)
Express serial by blue color.
a_{1} is dash blue, move out of definition
region [1, ∞)
a_{2}, a_{3} etc to a_{∞} are solid blue.
Solid blue area is less than black smooth
function curve area.
<a name="docA054">
Infinite sum is hard to calculate.
Integration is easy to calculate.
Integrate eq.cz11 red term is very easy.
The result is eq.cz11 blue term.
Infinite sum is converge? or diverge?
Rely on integration result. We need next
relation , let ∫ squeeze ∑ .
∫[x=1,∞]{f(x)*dx} > ∑[n=1,∞]a_{n} > ∫f(x)*dx --- eq.cz15 (error)
Above relation M>N>M certainly not exist.
<a name="docA055">
We need a form similar to eq.cz15 .
In eq.cz15, middle term and right term
∑[n=1,∞]a_{n} > ∫f(x)*dx --- eq.cz16
is eq.cz12
We pay attention to eq.cz13
eq.cz13 blue term is partial sum, not
whole sum. Missing a_{1}. We add a_{1} to
eq.cz13 both side get eq.cz17 next
If integral convergence/divergence is known, eq.cz19
conclude sum convergence/divergence. //a311091744
width of above equation
<a name="docA057">
2014-11-07-13-38 here
Blue term in eq.cz17 sum start from 2
(not from 1). Blue term add a_{1} become
eq.cz18 red term.
Merge eq.cz18 and eq.cz12 get eq.cz19
Now eq.cz19 is similar to eq.cz15 (error)
eq.cz19 is a correct expression.
In eq.cz19 red sum is hard to find it is
convergent or is divergent. But
red sum is less than a_{1}+∫f(x)*dx and
red sum is greater than ∫f(x)*dx
<a name="docA058">
if ∫f(x)*dx diverge, //see eq.cz19
∑a_{n} > ∫f(x)*dx force ∑a_{n} diverge.
if ∫f(x)*dx converge, //see eq.cz19
a_{1}+∫f(x)*dx > ∑a_{n} force ∑a_{n} converge.
eq.cz19 integration value tell serial sum
is convergent or divergent.
eq.cz19 is an inequality. If convergent
integration value is NOT summation value.
<a name="docA059">
Here conclude Cauchy-Integral TheoremSeries Integral Test
Sum is convergent/divergent
if the corresponding
integral is convergent/divergent .
2014-11-07-13-52 stop
draw Integral Test 1,2,3<a name="docA101">
2014-11-07-20-47 start
Next is study notes of
"Zeta Function - Part 2 -
Euler Product Representation"
http://www.youtube.com/watch?v=I3qSCWNXZKg
Lecturer is MrYouMath.
Notes recorder is Liu,Hsinhan.
Recorded notes is about 90% match, not exact.Liu,Hsinhan inserted text is blue Italic.
LiuHH notes may contain error!<a name="docA102">Begin video 2 of 17
Welcome to the new video on the zeta function.
This video we will talk about Euler product
and derive it. You might ask what is the Euler
product? and who Euler was? and how does the
Euler product look like? In this video you will
understand why the zeta function defined in the
first video has really to do something with the
prime numbers. May be it is prime number's
distribution. Let us just start have a look
what is Euler product? and what we want to
derive? Now
<a name="docA103">
s is a complex number, for example '1.5-2.6i'
n is positive integer, p is prime numbers.
if Real(s)>1 , ζ(s) is defined by next equation.
ζ(s)
=
n=∞
∑
n=1
1
n^{s}
=
p=∞
∏
p=prime
1
1-p^{-s}
--- eq.cz20
width of above equation
<a name="docA104">
2014-11-07-21-09 here
ζ(s) is defined to be ∑[n=1,∞]{1/n^{s}}
You can re-write as ∏[p=prime]{1/[1-p^(-s)]}
Euler introduced both ∑ for summation and ∏
for product. In ∏[p=prime]{1/[1-p^(-s)]}
p is prime number. eq.cz20 say you can write
sum of positive integer n; ∑[n=1,∞]{1/n^{s}} as
<a name="docA105">
product of prime number
∏[p=prime]{1/[1-p^(-s)]}
This video proof eq.cz20 This proof is given
by Euler .
∑[n=1,∞]{1/n^{s}} = ∏[p=prime]{1/[1-p^(-s)]}
Next video proof same eq.cz20 with easier
way. Now let us do the hard way.
First, let us start from eq.cz20 right hand
side. 2014-11-07-21-24 here
<a name="docA106">
s is a complex number, for example '1.5-2.6i'
k_{i} is positive integer, p_{i} is prime number.
if Real(s)>1 , ζ(s) is defined by next equation.
p=∞
∏
p=prime
1
1-p^{-s}
=
i=∞
∏
i=1
k_{i}=∞
∑
k_{i}=0
[
1
p_{i}^{s}
]
k_{i}
=
i=∞
∏
i=1
k_{i}=∞
∑
k_{i}=0
1
p_{i}^{s*ki}
--- eq.cz21
width of above equation
2014-11-07-21-40 stop at video 03:00
<a name="docA107">
2014-11-08-07-55 start
In eq.cz21 left side, p is a prime number.
s is a complex number. But let us consider
s be a real number. First video conclude
that complex s real part must be greater
than one. Then real s must be greater than
one. Prime number p start from 2,3,5 etc.
p is greater than one. p^{-s} is a number
less than one.
Example 2^{-1.1}=0.46651649576 < 1
prime p=2>1, real s=1.1>1
eq.cz21 left side is one over one minus a
number smaller than one then we can use
geometric sum formula.
2014-11-08-08-10 here
<a name="docA108">
Textbook by George, B. Thomas Jr. and
Ross L. Finney Calculus sixth ed. page
610, 611 has proof of geometric series.
2014-11-08-08-24 continue
<a name="docA109">
Thomas and Finney Calculus page 611 equation 8c=eq.cz22 . require |r| < 1.
a
*
n=∞
∑
n=0
r^{n}
=
a
1-r
set a=1, set r to p^{-s}, set n to k_{i}
--- eq.cz22
1
1-p^{-s}
=
k_{i}=∞
∑
k_{i}=0
(p^{-s})^{ki}
=
k_{i}=∞
∑
k_{i}=0
1
p_{i}^{s*ki}
--- eq.cz23
p=∞
∏
p=prime
1
1-p^{-s}
=
i=∞
∏
i=1
p=prime
k_{i}=∞
∑
k_{i}=0
1
p_{i}^{s*ki}
--- eq.cz24
width of above equation
eq.cz23 is single prime equation. eq.cz24 is all prime multiplication.
eq.cz24 is MrYouMath lecture 2 video at 2:15 second equation on screen.
<a name="docA110">
2014-11-08-09-35 here
We move k_{i} into bracket, so we do not use
bracket all the time. Next step is a little
bit confusion. 2014-11-08-09-37 here
2014-11-08-09-52 here
<a name="docA112">
2014-11-08-10-08 start
You have such product like eq.cz25 then
you can do a little trick.
=
k_{1}=∞
∑
k_{1}=0
k_{2}=∞
∑
k_{2}=0
k_{3}=∞
∑
k_{3}=0
...
[
1
p_{1}^{k1}
*
1
p_{2}^{k2}
* ...
1
p_{N}^{kN}
* ...
]
s
--- eq.cz26
width of above equation
<a name="docA113">
2014-11-08-10-20 here
You can take all sum ∑ out of square bracket.
Then write all the factors in square bracket.
Because if I would have written this way, for
example we only have large prime numbers
<a name="docA114">LiuHH notes may contain error!
//blue italic is LiuHH notes.
s is a complex number, for example '1.5-2.6i'
k_{1} is positive integer, p_{1} is prime number.
if c5 not contain k_{1}, we can move c5 in or out of summation.
c5 *
k_{1}=∞
∑
k_{1}=0
1
p_{1}^{s*k1}
=
k_{1}=∞
∑
k_{1}=0
c5
p_{1}^{s*k1}
--- eq.cz27
width of above equation
<a name="docA115">
Now consider constant c5 is ∑[k_{2}=0,∞]{1/p_{2}^{s*k2}}
Because variable k_{1} and variable k_{2} are
independent from each other, then eq.cz28 is true.
First bring c5=∑[k_{2}=0,∞]{1/p_{2}^{s*k2}} in to ∑[k_{1}=0,∞]
sum . Done in eq.cz27 . Second bring {1/p_{1}^{s*k1}}
in to ∑[k_{2}=0,∞] sum, not shown . Get next equation.
k_{1}=∞
∑
k_{1}=0
1
p_{1}^{s*k1}
*
k_{2}=∞
∑
k_{2}=0
1
p_{2}^{s*k2}
=
k_{1}=∞
∑
k_{1}=0
k_{2}=∞
∑
k_{2}=0
[
1
p_{1}^{s*k1}
*
1
p_{2}^{s*k2}
]
--- eq.cz28
width of above equation
Repeat apply eq.cz28 for all k_{n} from n=1
to n=∞, get eq.cz26 .
Above is LiuHH explanation for change from
eq.cz25 to eq.cz26 .
2014-11-08-10-54 stop at video 05:00<a name="docA116">
2014-11-08-14-20 start
..... I hope you understood that. Next I
take a little manipulation. Move s to
outside of the bracket.
see change from eq.cz25 to eq.cz26 , it is
OK to move, because s is common power factor.<a name="docA117">
Now let us look at eq.cz26 denominator.
p_{1}^{k1}
*
p_{2}^{k2}
* ...
p_{N}^{kN}
* ...
--- eq.cz29
width of above equation
<a name="docA118">
2014-11-08-14-37 here
k_{1} go from zero to infinity,
k_{2} go from zero to infinity, ...
k_{N} go from zero to infinity, ...
we have infinite many primes. Now this is
a little bit strange thing.
Fundamental Theorem of Arithmetic
Let n be a natural number, then it is
possible to write n as a product of
primes with corresponding powers. This
representation is unique!
<a name="docA121">
2014-11-08-14-51
LiuHH notes may contain error!
//blue italic is LiuHH notes.
first few primes are 2,3,5,7,11,13,
17,19,23,29,31,37,41,43,47,53,59,61,...
first prime p_{1}=2, second prime p_{2}=3,
third prime p_{3}=5, fourth prime p_{4}=7 etc.
<a name="docA122">
To understand Fundamental Theorem of
Arithmetic see few example
example 1: 1=2^{0}*3^{0}*...*59^{0}*...
example 2: 20=2*2*5=2^{2}*3^{0}*5^{1}*7^{0}...
example 3: 64=2*2*2*2*2*2=2^{6}*3^{0}*5^{0}*7^{0}...
example 4: 120=2*2*2*3*5=2^{3}*3^{1}*5^{1}*7^{0}*11^{0}...
example 5:1281=3*7*61=2^{0}*3^{1}*5^{0}*7^{1}...*59^{0}*61^{1}*...
<a name="docA123">
Each integer has unique prime power sequence
representation. For above five examples
1 representation 0,0,0,0,... (infinity long)
20 representation 2,0,1,0,... (infinity long)
64 representation 6,0,0,0,... (infinity long)
120 representation 3,1,1,0,... (infinity long)
1281 representation 0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0...
<a name="docA124">
Each integer's prime power sequence
representation is unique.
Each integer's prime power sequence
representation is infinity long. Because
there are infinity many prime numbers.
Above red number sequence is k_{1}, k_{2}, k_{N}
in eq.cz30
2014-11-08-15-29 here<a name="docA125">
This is Fundamental Theorem of Arithmetic.
What it states is if you have a natural
number n, then it is possible write n as
product of primes with corresponding powers.
Important thing is this representation is
unique. If you write it down, it look like
this
<a name="docA127">
2014-11-08-15-45 here
Product sign ∏ in eq.cz31 is not needed.
This is an error. You can drop ∏ from eq.cz31
Set n equal to prime 1 to k_{1} power multiply by
prime 2 to k_{2} power , you can go ahead to N(n).
N(n) is largest prime in n.
<a name="docA128">
2014-11-08-15-50
In
example 1: 1 largest prime not exist.
example 2: 20=2*2*5 largest prime is N(20)=5
example 3: 64=2*2*2*2*2*2 largest prime is N(64)=2
example 4: 120=2*2*2*3*5 largest prime is N(120)=5
example 5:1281=3*7*61 largest prime is N(1281)=61
<a name="docA129">
All prime greater than largest prime has
power zero and contribute nothing.
eq.cz30 is infinity long product for
arbitrary large number.
eq.cz31 is finite long product for a
given number. N(n) is largest prime with
power greater than 0. 2014-11-08-15-59<a name="docA130">
p_{N(n)}^{kN(n)} is largest prime number in n.
For example you have 18 (80). You can write
18 (80) as 2*9 (2*2*2*2*5) then eighteen is
18=2^{1}*3^{2} this representation is unique. You
cannot write a number in other sequence.
(prime number power sequence)
<a name="docA131">
There is only one representation for
each integer. If you look eq.cz31 we
have this representation in denominator,
so we are summing over ALL integers.
eq.cz30 term right to '=' has infinite
many ∑ for infinite many prime numbers.
That is "summing over ALL integers".<a name="docA132">
Each integer has a special k_{1} k_{2} sequence.
For example, if all k_{1} k_{2} are zero, we get
number one. If only k_{1}=1 all other k_{n} are
zero, we get one over two. One over two to
s power we get second part in zeta function.
(see eq_cz01b 1/2^{s})
<a name="docA133">
You go ahead for 1/3. In eq.cz30 you make
all k_{n} to zero except let k_{2} equal to one.
(because second prime p_{2} is 3 )
and you would have one over three to the s
power. (see eq_cz01b 1/3^{s})
<a name="docA134">
You can construct for example four would be
1/4^{s} let k_{1} equal to two and all other k_{n} be
zero.
("k_{1} equal to two" say first prime p_{1}=2 take
k_{1}=2 power, 2^{2}=4 get 1/4^{s}. see eq_cz01b 1/4^{s})
You can construct all the numbers out of
eq.cz30 . If we write this down
2014-11-08-16-52 here
<a name="docA135">
s is a complex number, for example '1.5-2.6i'
n is positive integer, p is prime numbers.
if Real(s)>1 , ζ(s) is defined by next equation.
<a name="docA136">
2014-11-08-16-58
We are summing all integer numbers with
power s. This is zeta function ζ(s).
I hope you can follow these arguments.
It is very hard to prove. In next video
we learn how to prove easier.
If you like my video, please subscribe.
See you guys.
2014-11-08-17-10 stop.
2014-11-08-18-12 done first proofread.
2014-11-08-18-40 done second proofread.
2014-11-08-20-41 done third proofread.
<a name="docA201">
2014-11-09-07-28 start
Next is study notes of
"Zeta Function - Part 3 - Euler Product (easy)"
http://www.youtube.com/watch?v=TDdGisWD5OU
Lecturer is MrYouMath.
Notes recorder is Liu,Hsinhan.
Recorded notes is about 90% match, not exact.Liu,Hsinhan inserted text is blue Italic.
LiuHH notes may contain error!<a name="docA202">Begin video 3 of 17
Welcome to the new video on the zeta function.
This video we will talk about an alternative
way in reducing Euler product representation
of Euler Riemann zeta function. In the video
before we somehow derived using the
Fundamental Theorem of Arithmetic. In this
video it is a little bit easier we will use
an idea which is pretty similar to the ????.
I will maybe explain in another video. You
will understand how it actually works.
2014-11-09-07-40 here
<a name="docA203">
s be a complex number, example s='1.1+2.2i'
n is an positive integer. p is a prime number.
Riemann Zeta function ζ(s).
if Real(s)>1 , ζ(s) is defined by next equation.
<a name="docA204">
2014-11-09-07-52 here
Les us have a look the equation you want
to derive. Euler Riemann Zeta function ζ(s)
is defined as eq.cz33 shown.
eq.cz33 middle term equal right term, this
is Euler Product representation. At this
point I am not intend to say there are
infinite many primes. Let us do sum, how
does this work? In order to understand
these steps which is pretty easy. Start
from ζ(s) definition equation
<a name="docA206">
2014-11-09-08-11 here
We want to take ζ(s) definition equation
eq.cz34 and divide by 2^{s}
ζ(s)
*
1
2^{s}
=
n=∞
∑
n=1
1
n^{s}
*
1
2^{s}
=
n=∞
∑
n=1
1
(2*n)^{s}
--- eq.cz35
width of above equation
<a name="docA207">
2014-11-09-08-25 here
What happen then is we can take 1/2^{s} to
the sum.
Summation ∑[n=1,∞] use variable n.
The term 1/2^{s} not contain n . Second point
n and s are independent from each other.
Therefore 1/2^{s} is a constant to ∑[n=1,∞]
It is OK to move a constant in or out of
a summation sign. 2014-11-09-08-30<a name="docA208">
Re-write [1/n^{s}]*[1/2^{s}] as 1/[(2*n)^{s}]
Next we see 1/[(2*n)^{s}] are all the even
terms in eq.cz34 . Next we take the difference
of eq.cz34 and eq.cz35
(eq.cz34 minus eq.cz35 equal eq.cz36)
ζ(s)
－
ζ(s)
*
1
2^{s}
=
n=∞
∑
n=1
1
n^{s}
－
n=∞
∑
n=1
1
(2*n)^{s}
--- eq.cz36
width of above equation
<a name="docA209">
2014-11-09-08-47 here
eq.cz36 is the result. Right hand side is
∑[n=1,∞]{1/n^{s}} and subtract all the even
numbers ∑[n=1,∞]{1/(2*n)^{s}} One can write
eq.cz36 in next form.
ζ(s)
[
1
－
1
2^{s}
]
=
n=∞
∑
n=1
n≠2k
1
n^{s}
--- eq.cz37
width of above equation
<a name="docA210">
2014-11-09-08-59 here
In eq.cz36 left hand side, take ζ(s) out
to get eq.cz37 left hand side. If you
look at eq.cz37 left hand side compare
with eq.cz33 ζ(s)=blue term , it is pretty
similar. In eq.cz37 you can write
[1 - 1/2^{s}] as [1 - 2^{-s}]
In eq.cz37, [1 - 2^{-s}] is numerator
In eq.cz33, [1 - 2^{-s}] is denominator.
Somehow look like we are getting closer to
Euler product representation. In eq.cz37
right hand side, I take new notation,
∑[n=1,∞;n≠2k]{1/n^{s}} say n go from one to
infinity, but n≠2k require n is not an
even number. Because we subtracted even
numbers.
<a name="docA211">
Next step, start from eq.cz37 and
multiply 1/3^{s} to eq.cz37 get eq.cz38.
ζ(s)
[
1
－
1
2^{s}
]
*
1
3^{s}
=
n=∞
∑
n=1
n≠2k
1
n^{s}
*
1
3^{s}
--- eq.cz38
width of above equation
<a name="docA212">
2014-11-09-09-34 here
Let us do that and see what happen. In
eq.cz38 we can write ∑[n=1,∞;n≠2k]{(1/n^{s})*(1/3^{s})}
put n and 3 together become ∑[n=1,∞;n≠2k]{1/(3n)^{s}}
eq.cz37 subtract eq.cz38 get next
ζ(s)
[
1
－
1
2^{s}
]
[
1
－
1
3^{s}
]
＝
n=∞
∑
n=1
n≠2k
1
n^{s}
－
n=∞
∑
n=1
n≠2k
1
(3n)^{s}
--- eq.cz39
width of above equation
<a name="docA213">
2014-11-09-09-47 here
eq.cz37 left hand side subtract eq.cz38
left hand side, both red term goto
eq.cz39 left hand side red term. Their
difference is eq.cz39 left hand side black
term.
Look at eq.cz39 right hand side, sum all
odd numbers ∑[n=1,∞;n≠2k]{1/n^{s}} and subtract
all the numbers that are multiple of 3.
<a name="docA214">
We can easily write down next expression.
ζ(s)
[
1
－
1
2^{s}
]
[
1
－
1
3^{s}
]
＝
n=∞
∑
n=1
n≠2k
n≠3k
1
n^{s}
--- eq.cz40
width of above equation
<a name="docA215">
2014-11-09-10-15 here
eq.cz40 is looking interesting. eq.cz40
left side have first prime 2^{-s} and
second prime 3^{-s}. We would suspect the
next part would be [1-1/5^{s}], actually
if you do that
<a name="docA217">
2014-11-09-10-26 here
We are repeating what we are doing. eq.cz41
right hand side below ∑ sign we get a list
of all the multiple of prime numbers.
(See red n≠2k, n≠3k, n≠5k ... ) Which says
that in the end we will only left with
one number which is not multiple of any
prime number. There is only one number that
is able to do that. This is one.
(see eq.cz42 right end "=1")
<a name="docA219">
2014-11-09-10-33 here
eq.cz42 right hand side we have one. Because
we took out all the multiple of prime numbers.
Only one left. Because the number which is not
product of primes only one. We devide eq.cz42
with all square bracket terms, get
<a name="docA220">
Middle term denominator use an inner <table> . a311091559
ζ(s)
＝
_{
1
[
1
－
1
2s
]
[
1
－
1
3s
]
[
1
－
1
5s
]
●●●
}
＝
p=∞
∏
p=prime
1
1-p^{-s}
--- eq.cz43
width of above equation
Liu,Hsinhan re-write eq.cz43 as eq.cz44 below. Attention: 1/2^{s}=2^{-s} Lecture 3 time 7:12 use eq.cz43. But eq.cz44 is easier to understand.
ζ(s)
＝
1
1-2^{-s}
*
1
1-3^{-s}
*
1
1-5^{-s}
*
●●●
＝
p=∞
∏
p=prime
1
1-p^{-s}
--- eq.cz44
a311092020
width of above equation
<a name="docA221">
2014-11-09-10-53 here
The denominator in eq.cz43 middle term can be
written as [1-2^{-s}]*[1-3^{-s}]*[1-5^{-s}]*...
Then eq.cz43 middle term to eq.cz43 right term
is easy to understand. (see eq.cz44 above)
We get eq.cz43 right hand side term which is
Euler representation. It is product of all
primes in the form 1/[1-p^{-s}] We proved this
in another way.
<a name="docA222">
The main idea is we take
zeta function ζ(s) and multiply with
reciprocal power of the first prime 1/2^{s},
and subtract this result from original
ζ(s). So we have sum of odd numbers for n.
We took out all multiple of prime numbers.
At the end get only one. Devide eq.cz42
with [1-2^{-s}]*[1-3^{-s}]*[1-5^{-s}]*...
get Euler product representation eq.cz43.
<a name="docA223">
Actually that was second proof of Euler
product representation. Next video we will
go ahead look what is the relation between
zeta function and prime zeta function.
You might not understand what is prime
zeta function as yet. But I can tell you
instead of taking n to be any integer
(see eq.cz34 right term denominator) You
only allow prime numbers. Then you have
prime zeta function. We want to see what
is the relationship between zeta function
to that strange looking function.
<a name="docA224">
If you like my video, please subscribe.
also comment, or give thumb up, if you
have question, feel free to ask . I am
here to help you to understand. Because
if you do not understand these steps
which we are doing now, then you might
have problems later on. Even if you can,
You might be able to solve the upcoming
problems.
This conclude the lecture.
See you guys.
2014-11-09-11-32 stop
2014-11-09-15-56 done first proofread
2014-11-09-16-31 done second proofread
<a name="docA225">
2014-11-09-19-16 start
About Liu,Hsinhan math study time.
Liu,Hsinhan has no job and sister support
LiuHH living. Then LiuHH has many free time.
Study mathematics is LiuHH's interesting.
If no other thing to do, LiuHH study math
and write notes, just like this page
http://freeman2.com/tute0057.htm
<a name="docA226">
But if brother or sister has work to do.
LiuHH first priority is help brother and
sister. Start from 2014-March, remove
back yard grass root from soil.
http://freeman2.com/mov2013e.jpg
http://freeman2.com/hpim9403.mpg
Summer time everyday spend about four
hours. Morning two hours, evening two
hours. Winter time everyday spend about
two hours. Now 2014-11-09 done about 50%
back yard area.
<a name="docA227">
2014-10 sister get 450 pound grape from
brother yard. LiuHH wash most of 450 pound.
From 2014-10-05 to about 2014-10-17 LiuHH
time is totally grape time, no math time.
Wash grape work done about 2014-10-2?
2014-11-07 LiuHH wait for hospital phone
call. All day not do yard work.
2014-11-08 and 2014-11-09 LiuHH use three
different kind prescription eye drop.
Each use four times daily. 3*4=12. Sister
help one drop one hour for LiuHH . Total
12 hours.
<a name="docA228">
Result is that 2014-11-08 and
2014-11-09 not do yard work. Tomorrow
2014-11-10 LiuHH will have cataract surgery.
After eye surgery, LiuHH may have difficulty
reading for few days. Therefore recent few
days, LiuHH write tute0057.htm study notes
fast. Next few days need more rest for eye
surgery recovery.
2014-11-09-19-48 stop
<a name="docA229">
2014-11-17-10-29 start
2014-11-10-13-?? in UCLA eye clinic
Liu,Hsinhan left eye cataract surgery.
Please see detail
http://freeman2.com/bioge010.htm#docA001<a name="docA301">
2014-11-17-10-51 start
Next is study notes of
"Zeta Function - Part 4 -
Infinitude of Prime Numbers"
http://www.youtube.com/watch?v=SKa7b-3C32A
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="docA302">Begin video 4 of 17
Welcome to the new video on the zeta function.
In this video we will go through the steps
Euler showed there are infinite many Prime.
This proof is not the standard proof. You
should not use this (Euler) way. But this
proof is very cool and short way to show
there are infinite many primes. The standard
proof of infinite prime is given by Euclid.
Euclid proof is very elementary you do not
have high order stuff, or (Euclid proof)
not have really complicated stuff.
<a name="docA303">
Let us look how Euler derive the Infinitude
of Prime Numbers.
2014-11-17-11-09 here
s is a complex number, for example '1.5-2.6i'
n is positive integer, p is prime numbers.
if Real(s)>1 , ζ(s) is defined by next equation.
<a name="docA304">
2014-11-17-11-16 here
He start out from the representation eq.cz45
which Euler find out himself. For the zeta
function ζ(s) is defined to be ∑[n=1,∞]{1/n^{s}}
The is eq.cz45 left equal sign. Euler
representation equate to ∏[p=prime]{1/[1-p^{-s}]}
The is eq.cz45 right equal sign.
Euler representation given by Euler see docA101Euler representation easy proof see docA201
Please look at them. Now what did Euler do?
Actually Euler looked at very specific value
the value of s equal one. ζ(s)=ζ(1), s=1 .
2014-11-17-11-34 here
2014-11-17-12-25 stop
2014-11-17-15-08 start
Liu,Hsinhan inserted text is blue Italic.
LiuHH notes may contain error!<a name="docA307">
To show ζ(1) has value infinite, please see
eq.cz46 to eq.cz50 . eq.cz46 right side expand
to eq.cz47 right side. "●●●" mean extend to
infinity. How do we know eq.cz47 diverge?
(diverge = sum to infinity) Please see
eq.cz48 to eq.cz50. Rewrite eq.cz47 with
three more terms get eq.cz48. Please compare
eq.cz48 with eq.cz49. Both have 1/1 + 1/2
no change (equal value). BUT
<a name="docA308">
eq.cz48 1/3 change to eq.cz49 1/4 we know
1/3 = 0.33333... > 0.25 = 1/4. Therefore
1/3 + 1/4 > 1/4 + 1/4 = 1/2
Attention: see red terms in eq.cz48, eq.cz49
and eq.cz50 .
<a name="docA309">
Similarly,
eq.cz48 1/5 change to eq.cz49 1/8 we know
1/5 = 0.20000000... > 0.125 = 1/8.
eq.cz48 1/6 change to eq.cz49 1/8 we know
1/6 = 0.16666666... > 0.125 = 1/8.
eq.cz48 1/7 change to eq.cz49 1/8 we know
1/7 = 0.14285714... > 0.125 = 1/8.
Therefore
1/5 + 1/6 + 1/7 + 1/8 > 1/8 + 1/8 + 1/8 + 1/8 = 1/2
Attention: see purple terms in eq.cz48,
eq.cz49 and eq.cz50 .
<a name="docA310">
Next comparison is
1/9 + 1/10 + 1/11 + 1/12 + 1/13 + 1/14 + 1/15 + 1/16
greater than
1/16+ 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16
Sum eight 1/16 get 1/2 . This comparison
is not shown in above equations.
Above red terms and purple terms extend
to all remaining terms, we have eq.cz50
sum infinite many 1/2. which is infinity.
<a name="docA311">
Exam eq.cz50 . Left side is ζ(1) , right
side has greater than infinity.
We MUST conclude that ζ(1) diverge.
2014-11-17-15-57 LiuHH notes <a name="docA312">
What we can do next, see eq.cz45 again.
In eq.cz45 plug in 1 for s get eq.cz51
n is positive integer .
ζ(1)
=
n=∞
∑
n=1
1
n
=
∞
--- eq.cz51
width of above equation
<a name="docA313">
Let us assume there is only finite number of
prime numbers. (false assumption on purpose)
Let us look at eq.cz45 right hand side.
(In eq.cz45, replace s by 1, get eq.cz53 below.)
<a name="docA315">
2014-11-17-16-23 here
If eq.cz53 right hand side has only finite
terms (one prime one term, assume finite
prime, then finite terms) If you get finite
product, you get finite number. But eq.cz51
tell us the result is infinite. This conflict
(infinite = finite number) tell us the
assumption that only finite set of prime
number is not true.
<a name="docA316">
The result is written out as next.
A finite set of primes would make the
product converge, but it has to diverge!
That conclude that there must be infinite
many primes. Q.E.D.<a name="docA317">
The upcoming video will also prove the sum
of reciprocal of prime number is infinitely
large.
This conclude this video.
2014-11-17-16-43 stop
2014-11-17-18-33 done proofread
<a name="a31229a">
2014-12-29-17-08
update 2014-12-29 complete the link at
tute0057.htm#docA006<a name="docA999">
∑[n=1,∞]{1/n^{s}} = ∏[p=prime]{1/[1-p^{-s}]}
∑[n=1,∞]{1/n^{s}} = ∏[p=prime]{1/[1-p^(-s)]}