<a name="docA001">
201410290855 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'2.14.3i'
<a name="docA002">
Liu,Hsinhan write complex zeta function
following next web page
201409302234 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
201403261020 use YTD download
http://www.youtube.com/watch?v=rGo2hsoJSbo
and
201403251828 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^(s1)*sin(PI*s/2)*Γ(1s)*ζ(1s)
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">
201410290756
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=SKa7b3C32A
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=Gfqe3BkBnE
http://www.youtube.com/watch?v=QfDbF_qlp58
http://www.youtube.com/watch?v=wt6ngy6pDws
]]
<a name="docA005a">
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.
201310211948 Liu,Hsinhan access
http://www.youtubedownloadersite.com/installers/YTDSetup.exe
10/21/2013 07:49 PM 11,574,336 YTDSetup.exe
201405271425 click "Place order" //$29.90
https://secure.avangate.com/order/verify.php?CART_ID=deleted_by_LiuHH
<a name="docA006">
201410290824
MrYouMath_17_files_Riemann.htm
[[
"Zeta Function  Part 1  Convergence"
datavideoid="ZlYfEqdlhk0"
"Zeta Function  Part 2  Euler Product Representation"
datavideoid="I3qSCWNXZKg"
"Zeta Function  Part 3  Euler Product (easy)"
datavideoid="TDdGisWD5OU"
<a name="docA007">
"Zeta Function  Part 4  Infinitude of Prime Numbers"
datavideoid="SKa7b3C32A"
"Zeta Function  Part 5  Prime Zeta Function"
datavideoid="3eN9tQX3JJ4"
"Zeta Function  Part 6  The Prime Counting Function"
datavideoid="U16_KTTKtb0"
"Zeta Function  Part 7  Zeta of 2 aka The Basel Problem"
datavideoid="GeKDmoYHiAk"
"Zeta Function  Part 8  Zeta of 2n  Part 1"
datavideoid="axQqExF7NsU"
<a name="docA008">
"Zeta Function  Part 8  Zeta of 2n  Part 2"
datavideoid="XHQ0OzqTjd0" 9/17
"Zeta Function  Part 8  Zeta of 2n  Part 3"
datavideoid="1f24RZfP6m8" 10/17
"Zeta Function  Part 9  Relation to Gamma Function"
datavideoid="UEZ4ClCdog8" 11/17
"Zeta Function  Part 10  Jacobi Theta Function"
datavideoid="GQFljOVZ7I" 12/17
"Zeta Function  Part 11  Riemann Functional Equation I"
datavideoid="K6L4Ez4ZVZc" 13/17<a name="docA009">
"Zeta Function  Part 12  Riemann Functional Equation II"
datavideoid="TnRnlJBecRg" 14/17
"Zeta Function  Part 13  Trivial Zeros of the Zeta Function"
datavideoid="Gfqe3BkBnE" 15/17
"Zeta Function  Part 14  Riemann Xi Function"
datavideoid="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)"
datavideoid="wt6ngy6pDws" This video is easy. skipped a312190901
]]
<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.
201410290923 here
<a name="docA011">
201410291642 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^s1)}  eq_cz01a
Next is same equation in better form.
201410080019
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">
201411051225 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.
201411051242 above is for entry level reader.
<a name="docA012">
s be a complex number, example s='0.53.4i'
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}/(12^{1s})  eq_cz02a
or
ζ(s)={[12^(1s)]^(1)}*∑[n=1,∞]{1^{n+1}/n^s}  eq_cz02b
Next is same equation in better form.
201410080038
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
12^{1s}
*
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^{s1}*sin(PI*s/2)*Γ(1s)*ζ(1s)  eq_cz03a
or
ζ(s)=2^s*PI^(s1)*sin(PI*s/2)*Γ(1s)*ζ(1s)  eq_cz03b
here PI=π=3.14159265358979323846264338327950288419716939937510 ...
Next is same equation in better form.
201410080048
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^{s1}
*
sin
(
π*s
2
)
*
Γ(1s)
*
ζ(1s)
 eq.cz03
width of above equation
above ζ(1s) is Riemann Zeta function with input 1s
above Γ(1s) is complex Gamma function with input 1s
Because if Real(s)<=0 , then Real(1s)>=1
<a name="docA014">
Liu,Hsinhan access next video and find above three equations.
201403251828 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
201403261020 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 .
201410291929 done include and modify.
<a name="docA401">
201411181522 start
Next is study notes of
"Zeta Function  Part 5  Prime Zeta Function"
http://www.youtube.com/watch?v=3eN9tQX3JJ4
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="docA402">Begin video 5 of 17
Welcome to the new video on the zeta function.
In this special video we will talk about the
Prime Zeta Function which is just very similar
to the Zeta Function itself. It only summing
over the reciprocal power of prime numbers.
eq.cz54 is our goal, not start point. eq.cz54 is not given.
width of above equation
<a name="docA404">
201411181539 here
You do not take n as the denominator, you
take p be prime numbers. At the end, we
will derive a very important result, the
sum of the reciprocal of prime numbers is
tending to infinity. This is quite
interesting. If you take all the prime
numbers add their reciprocal value you
should end up on infinity. Important to
mention here, if you take same thing, the
reciprocal value of square numbers, add
them up together, you will end up to a
very specific value which is PI squared
over six.
<a name="docA405">1/1+1/4+1/9+1/16+1/25+1/36+1/49+...∞
=PI*PI/6=1.6449340668482264 //a311181557
I will prove that in upcoming video. This
state the density of prime numbers is
higher then the density of square numbers.
In another words, there are more prime
numbers than square numbers. Sounds strange,
because we know there are infinite many
prime numbers and infinite many square
numbers. But if you compare them, there
are less square numbers than prime numbers.
See prime and square, go here click [A1]. a311221407
Let us look how can we derive the result.
Our starting point is
<a name="docA406">
s be a complex number, example s='1.1+2.2i'. p is a prime number.
Riemann Zeta function ζ(s).
<a name="docA408">
201411181630 here
The logarithm of a product is equal to
the sum over logarithms.
log(6)=log(2*3)=log(2)+log(3)
log(2)=0.6931471805599453
log(3)=1.0986122886681097
1.0986122886681097+0.6931471805599453=1.791759469228055
log(2)+log(3)=1.791759469228055
log(6) =1.791759469228055
We can write eq.cz56 as eq.cz57 below.
201411181638 LiuHH notes
<a name="docA410">
201411181643 here
Now I am using the Taylor expansion for
the logarithm. Actually I am using the
Taylor expansion for the logarithm of
one minus x. (eq.cz58 red to blue)
Please watch my video series about
Power series or Taylor expansion or
Taylor series to see how to derive that.
If you look at that, you will find next.
201411181657 stop
<a name="docA412">
201411181918 start
eq.cz58 left black term can be written as
log[(1x)^{1}]. logarithm has power rule
log[A^{b}]=b*log(A)  eq.cz59
therefore log[(1x)^{1}]=log(1x) this is
eq.cz58 red term.
From eq.cz58 red term to blue term, use
Taylor series expansion for the logarithm.
<a name="docA413">
Very important from eq.cz58 red term to
blue term, the absolute value of x must
be smaller than one. Review eq.cz57 the
term p^{s} its absolute value is always
smaller than one.
(eq.cz57 p^{s} sit in eq.cz58 x seat)
If set s be zero, p^{0} is one. In eq.cz58
two minus cancel. We will plug eq.cz58
blue sum into eq.cz57 log[...] , because
they are equal to each other. After plug
in, eq.cz57 become next eq.cz60.
<a name="docA415">
201411181947 here
Very important is that x in eq.cz58 is
p^{s} in eq.cz60 . First summing over prime
number, then summing over n equals one
to infinite. Now
log
ζ(s)
＝
p=∞
∑
p=prime
n=∞
∑
n=1
p^{sn}
n
 eq.cz61
width of above equation
<a name="docA416">
201411181956 here
This is pretty interesting. In eq.cz61
I just write s and n together, we have
more compact. What we will now do is to
split up sum into two parts. It is very
easy. In eq.cz61 , n sum from one to
infinite. We take n=1 out of summation.
See next eq.cz62.
<a name="docA417">
Alert: take out blue term, right end sum change start from n=1 to n=2
Alert: blue term n=1, then p^{s*n} become p^{s}.
log
ζ(s)
＝
p=∞
∑
p=prime
p^{s}
＋
p=∞
∑
p=prime
n=∞
∑
n=2
p^{sn}
n
 eq.cz62
Remember: n=1 become blue term. Right end term start from
n=2
width of above equation
<a name="docA418">
201411182012 here
In eq.cz62 ∑[p=prime,∞]{p^{s}} is reciprocal sum
of prime numbers.
∑[p=prime,∞]{p^{s}} is called prime zeta function.
It is not full zeta function, prime zeta function
has only prime numbers.
eq.cz62 right side is this body ∑∑{p^{sn}/n}.
In eq.cz62 we know logζ(s) has infinity value.
[if set s=1, ζ(1) has infinity value]
In eq.cz62 we expect ∑[p=prime,∞]{p^{s}} has infinity
value. Then in eq.cz62 ∑∑{p^{sn}/n} has to be smaller
than some number.
<a name="docA419">LiuHH notes: If we find
infinity = ∑[p=prime,∞]{p^{s}} + finite  eq.cz63
then ∑[p=prime,∞]{p^{s}} MUST be infinite.
to satisfy whole equation infinity = infinity .
The other case, we get trouble. Now logζ(s) has
infinity value. Assume ∑∑{p^{sn}/n} has infinity
value too (error assumption) then eq.cz62 is
infinity = ∑[p=prime,∞]{p^{s}} + infinity  eq.cz64
In this (error) case,
because infinity = infinity + infinity is true
and infinity = finite + infinity is also true
we cannot conclude ∑[p=prime,∞]{p^{s}} is infinite.
201411182033<a name="docA420">
We want to find out ∑∑{p^{sn}/n} upper bound.
In order to do that, I inter change two ∑.
//see eq.cz62 change to eq.cz65 below.
Remember: n=1 become blue term. Right end term start from
n=2
width of above equation
<a name="docA422">
201411182046 here
From eq.cz65 to eq.cz66. Summation ∑[p=prime]
control only prime number. Integer n is not
a prime, then n is a constant to ∑[p=prime]
It is possible to take n out of ∑[p=prime]
as eq.cz66 right side shown. //a311182056<a name="docA423">
Then I am doing little thing. First part
∑[p=prime,∞]{p^{s}} not change at all.
You see n is not depending on p. So we can
just drag n out of eq.cz65 right end
denominator, become eq.cz66. We have sum
from n=2 You might ask why start from n=2?
Because we took n=1 case to left hand side
∑[p=prime,∞]{p^{s}} remember that. So we have
eq.cz66 square bracket term. You see the
square bracket term and ∑[p=prime,∞]{p^{s}}
seems related.
<a name="docA424">∑[p=prime,∞]{p^{s}} is prime zeta function in s.
square bracket term has prime zeta function
in sn. Its like scaling the function. Now
written in this way ∑[p=prime,∞]{1/p^{sn}} we
can find what does ∑[p=prime,∞]{1/p^{sn}} mean?
In order to find that out, we want to take
the absolute value of ∑[p=prime,∞]{1/p^{sn}}
<a name="docA426">
201411182143 here
Take sum first then take absolute value
is always smaller or equal
take absolute value first then take adding
them. It is very logic thing.
Please see tute0057.htm#docA023
Rewrite eq.cz67 second term denominator
p^{sn} as p^{(x+iy)*n}//s is complex (x+iy)
201411182152 stop at video 09:00/22:02
<a name="docA427">
201411191424 start
To understand better we can write complex
variable s as (x+iy). The absolute value of
p to sn, p^{sn} can be written as p^{(x+iy)*n}
This is just a little notation, write s in
real part and imaginary part. Then we can
work a little more on that. Next just
multiply n in, from p^{(x+iy)*n} to p^{xn+iyn}
201411191451 here
Remember: e = 2.718281828459045...
Remember:  e ^{iyn*log(p)}  always = 1, and ignored
width of above equation
<a name="docA429">
201411191459 here
Next step, from
eq.cz67 right end term p^{(x+iy)*n}
to eq.cz68 second term p^{xn}*p^{iyn}.
Breaking up (x+iy)*n into product of
two powers. This is rule of exponential.
p^{xn} is a positive number, absolute value
sign is redundant.<a name="docA430">
Prime number p is positive 2,3,5,7,11 ...
n is positive. x in (x+iy) can be positive
or negative. Then p^{xn} is positive^{+/}
For example prime 3 :
3^{2}=1/3^{2}=1/9 > 0
3^{+2}=3*3=9 > 0
Then 1/9 or 9 is same as 1/9 or 9 .
absolute value sign is redundant. a311191515<a name="docA431">
Then we have p to iyn , p^{iyn}, we do same
trick as for convergence for the zeta
function. (see tute0057.htm#docA034) We
introduce here an exponential.
( p^{iyn} =  e^{iyn*log(p)}  this e is
"introduced exponential". a311191527)
e to the logarithm of p to the iyn. {e^{iyn*log(p)}}
If you use the Euler's formula (tute0057.htm#docA035 )
e^{iyn*log(p)} always equal to one.
We ignore one.
<a name="docA432">
In eq.cz68 right end, we get ∑[p=Prime,∞]{1/p^{xn}}
What we find out is very similar to zeta
function
(zeta function eq.cz07 tute0057.htm#docA033
prime zeta function eq.cz68 tute0058.htm#docA428)
eq.cz68 right end is real function, we can
do all the stuff we want, real power sum.
201411191552 here
<a name="docA434">
201411191559 here
What we do next step is
eq.cz69 second term 1/p^{xn} is less than
eq.cz69 third term 1/p^{n} . We know x is
greater than one. (see tute0057.htm#docA047)
If you make x > 1 this value 1/p^{xn} become
smaller and smaller. 1/p^{xn} upper bound is 1/p^{n}.
Next step, we do another thing.
∑[p=Prime,∞]{1/p^{n}} is less than ∑[k=2,∞]{1/k^{n}}
We will do a little trick, sum over prime
must be less than sum over all the integer
number. 201411191615 here
<a name="docA435">
eq.cz70 prime number sum ☺/☺ term is nonexist term.
4 is not a prime, 6 is not a prime, 4,6 do not enter eq.cz70.
eq.cz71 integer number sum has more term than eq.cz70.
therefore eq.cz70 sum < eq.cz71 sum ; eq.cz69 right end
two terms inequality are confirmed. 201411191632
p=∞
∑
p=2
1
p^{n}
=
1
2^{n}
+
1
3^{n}
+
☺
☺
+
1
5^{n}
+
☺
☺
+ ...
 eq.cz70
k=∞
∑
k=2
1
k^{n}
=
1
2^{n}
+
1
3^{n}
+
1
4^{n}
+
1
5^{n}
+
1
6^{n}
+ ...
 eq.cz71
width of above equation
<a name="docA436">
This is very easy to understand. Because
prime number is just part of integer numbers.
201411191635 stop, graph show up.
<a name="docA438">
201411191830 here
This sum is very easy sum, it does not have
complex number. Does not even have x in it.
It has only n which is a fix number. We have
k, summing over k. What we are doing now we
will use integral to estimate this body.
("this body" is eq.cz69 right most term
use integral to estimate sum ∑[k=2,∞]{1/k^{n}}
see eq.cz19 in tute0057.htm#docA056 )
<a name="docA439">
In "Integral Test 4" graph, red curve is
y=1/x^{n}. Look at the black rectangular area.
Left rectangular area is height 1 * width 1.
Second rectangular area is H 1/2^{n} * W 1.
Third rectangular area is H 1/3^{n} * W 1. and
so forth. Since eq.cz69 sum start from 2.
Since we know 2 is first prime number.
In "Integral Test 4" graph, I can parallel
move all black rectangular area to left
distance one. We will get a graph similar
to tute0057.htm#integralTest3
(Before move, rectangular area above curve.
After move, rectangular area below curve.)
201411191857 stop
<a name="docA440">
201411191937 start
We know the integral is greater than the
sum.
("integral is greater than the sum"
integral is greater than the sum which
start from n=2 in eq.cz14 ; k=2 in eq.cz69
integral is smaller than the sum which
start from n=1 in eq.cz14.
please see eq.cz14 in tute0057.htm#docA052
black integral is greater than blue sum.
blue sum start from n=2 .
black integral is smaller than red sum.
red sum start from n=1 . a311191952 )
<a name="docA442">201411192003 here
eq.cz72 left two terms (≦) are result
of eq.cz69
eq.cz72 middle two terms (＜) are result
of "Integral Test 4" graph blue dash
rectangular area sum is smaller than
red curve integration value.
blue dot rectangular area out of range
(x>=2) blue dot rectangular is ignored.
<a name="docA443">
eq.cz72 right two terms (＝) are result
of adding individual integral to get a
continuous integral.
Because first integral upper bound is
same as second integral lower bound for
all boundary points. Then it is possible
to add individual integral to get a
continuous integral.
MrYouMath similar lecture omitted.
201411192035<a name="docA444">eq.cz72 right end term is an easy function
to integrate.

p=∞
∑
p=prime
1
p^{sn}

≦
1
1  n
t^{1n}
t=∞
｜
t=1
＝
1
n － 1
 eq.cz73
width of above equation
<a name="docA445">201411192047 here
Integrate eq.cz72 right end term get eq.cz73
middle term.
eq.cz73 left two terms (≦) are eq.cz72 left
term and right term carry out integration.
eq.cz73 right two terms (＝) are integration
result. <a name="docA446">
eq.cz73 middle term t^{1n} where 1n must be
negative to get a finite integration value.
We start n at 2 (first prime number) 1n<0
is true. In eq.cz73 left end term value
is smaller than 1/(n1)
201411192059 stop at video 17:50/22:02
<a name="docA447">
201411200905 start
What we found out is this strange looking
prime zeta function
(strange looking zeta is eq.cz73 left term
prime zeta function = ∑[p=Prime,∞]{1/p^{s}}
strange looking zeta= ∑[p=Prime,∞]{1/p^{sn}}
n = 2,3,4,5,6 ... infinity , see eq.cz62eq.cz66 blue term regular prime zeta function
eq.cz66 purple term strange prime zeta function
201411200922 here)
of f(n) is smaller or equal to 1/(n1).
<a name="docA448">
Let us plug it back to where we work.
(plug in eq.cz66 purple term a311200929)

n=∞
∑
n=2
[
1
n
p=∞
∑
p=prime
1
p^{sn}
]

≦
n=∞
∑
n=2
1
n

p=∞
∑
p=prime
1
p^{sn}

≦
n=∞
∑
n=2
1
n
1
n  1
＝
1
 eq.cz74
width of above equation
<a name="docA449">201411200942 here
eq.cz74 left term is eq.cz66 purple term
in absolute sign.
Compare eq.cz63 with eq.cz66 , we want to
show eq.cz66 purple term has finite value.
Then take absolute value to purple term
get eq.cz74 left term.
<a name="docA450">
eq.cz74 left two terms, left ≦, please
see triangular inequality at
tute0057.htm#docA023
eq.cz74 middle two terms, right ≦, is due
to eq.cz73 .
eq.cz74 right two terms, ＝, is due to
telescope sum below.
部分分數(式)展開 partial fraction expansion
<a name="docA451">
Partial fractioning Basic relation
(1/n)*(1/[n1]) = (1/[n1])  (1/n)  eq.cz75
Calculate each term one by one.
n= 2: (1/n)*(1/[n1]) = (1/1)(1/2)
n= 3: (1/n)*(1/[n1]) = (1/2)(1/3)
n= 4: (1/n)*(1/[n1]) = (1/3)(1/4)
n= 5: (1/n)*(1/[n1]) = (1/4)(1/5)
.....
when sum ∑[n=2,∞] to infinity
(1/2) cancel (1/2)
(1/3) cancel (1/3)
(1/4) cancel (1/4)
.....
only (1/1) left.
∑[n=2,∞]{(1/n)*(1/[n1])} answer is 1.
201411201010 stop at video 18:00<a name="docA452">
201411201345 start at video 19:18
This sound nice and good. But why is
this so important? Now let us look at
our first result eq.cz66 repeat below.
log
ζ(s)
＝
p=∞
∑
p=prime
1
p^{s}
＋
n=∞
∑
n=2
[
1
n
p=∞
∑
p=prime
1
p^{sn}
]
 eq.cz66
 repeat
width of above equation
<a name="docA453">
201411201352 here
We found our logarithm of zeta function
ζ(s) , eq.cz66 left end term log ζ(s) .
is equal to prime zeta function
eq.cz66 blue term.
added this sum, prime zeta function with
power sn . eq.cz66 purple term.
Now let us set s be one , s → 1 . We know
eq.cz66 (eq.cz76) purple term has magnitude
always smaller than one. see eq.cz74
<a name="docA455">201411201412 here
eq.cz76 left end ζ(1) please see
tute0057.htm#docA305 eq.cz46 and eq.cz47
eq.cz50 indicate ζ(1) has value infinity.
log ζ(1) is log(infinity) still infinity.
eq.cz76 is an equality equation.
eq.cz76 left side is infinity,
eq.cz76 right side must be infinity,
eq.cz76 purple term value smaller than 1.
<a name="docA456">eq.cz76 blue term must be infinity.
Please see eq.cz54 . That is,
prime zeta function must be infinity.
Conclude our goal.
201411201423
Here conclude the lecture. For MrYouMath
detail lecture, please watch the video.
Above blue italic words are all LiuHH words.
LiuHH notes may contain error!
201411201429 stop
201411201523 first proofread
201411201727 second proofread
<a name="primeList">
201411210852 ;
Update 20141122 added Prime number generator, prime counting function
list,
draw,
Program default "var newPrime=[2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61];"
Message "Start build prime table at 63" is correct. Program based
on newPrime[] build higher prime start from 63.
Box 11
First build prime number list. Second, build prime counting function list.
Prime number generator
Output to Box 11
RUN ⇒
;
Prime per line
;
Separator
Prime Upper Bound
;
Output to Box 18.
View ⇒
Here record current primeinmemory maximum prime.
Update internal data. Check
here.
tute0058.htm not use. jsprime2.htm use
Liu,Hsinhan copied code from jsprime2.htm and use code in tute0058.htm
Above, Click [Find Prime] output prime number to Box11. Below
<a name="primeCntList">
Click [prime counting function list] output prime counting function list to Box12
Prime Counting Function Upper Bound
prime counting function π(n) RUN ⇒
Box 12
Counting function output to Box 12
Box 12, "6 , 3" = "number , prime count" = "n , π(n)" ; 3 primes <= 6 ; 2,3,5 < 6
In "number , prime count" if "number" is a prime, then "prime count" add one.
If "number" is not a prime, its "prime count" not increase. IMPORTANT point.
Box 18 debug
click [find prime] output one prime one line to Box 11. ; boxc18.value='string'
click [now prime] output all prime in one line to Box 18.
<a name="primeCountDraw">
Prime Counting Function Graph ; 201411211652
Click [Draw prime] to draw default x=0,60 ; y=0,18 . You can draw larger number. But
you need click [find prime] to get larger prime list, then click [prime counting] to get counting
list. Finally click [Draw prime]. Also adjust x/y min/max. Liu,Hsinhan 201411211852
x max: box control Prime Counting Function input n . Increase x max: box value,
for different graph. But you must increase [Prime Upper Bound] value and increase
[Prime Counting Function Upper Bound] value then click [Draw prime] and click
[prime counting] Finally click [Draw prime] to get graph. Adjust y max: box too.
Draw square counter sqrt(x) function, check
here.
Prime Upper Bound
Both run button silent, not output to box.
Prime Counting Function Upper Bound
Graph area size, W:
H:
x min:
, x max:
; y min:
, y max:
; example
Change box value for different graph.
If change x max: value, also change y max: value.
Prime Counting Graph
RUN ⇒
Box21, debug
;
QBboxd01.value='' ; 201411212320 done Prime Counting Function Graph
<a name="docA501">
update 20141123
201411221530 start
Next is study notes of
"Zeta Function  Part 6 
The Prime Counting Function"
http://www.youtube.com/watch?v=U16_KTTKtb0
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="docA502">Begin video 6 of 17
Hello. Welcome to the new video on the famous
and important zeta function. In this video I
will try to show you the relationship between
Zeta Function ζ(s) and π(x) it is the Prime Counting Function. If
you plug in arbitrary number for example 5.
It is counting how many prime numbers are
below or equal this x. For example π(5) will
give you a value of three. Because we have
three prime numbers 2,3,5 below or equal 5.
Now there is a very nice integral representation.
201411221542 here
<a name="docA504">
201411221600 here
which gives us the relationship between zeta
function ζ(s) and prime counting function π(x).
Actually, one could write log ζ(s) is equal
integral from 2 to infinity ... see eq.cz77
This eq.cz77 is a very nice thing. But before
we advance, let us look a short piece of π(x).
Here show up π(60) see graph click RUN button.
I hope you can see the grid here, very small.
Lower left corner star is one. If you plug 1
to π(x), get π(1), you should get zero. Because
there are no prime numbers smaller or equal one.
<a name="docA505">
If you plug in 2 to π(x), get π(2), you get one.
Because two itself is a prime number. And so
forth. If you plug in 3 to π(x), get π(3), you
get a jump. Very importantly, we only get jump
at prime numbers. For example, four is not a
prime number. But at five get a jump, at seven
get a jump. So this thing π(x), like a step
function. The step is always on the prime numbers.
There is a very important idea. The idea is take
PI function of n, π(n), subtract the PI function
of n1.
<a name="docA506">
This difference, π(n)π(n1), is only
equal to one if n is a prime number. If not,
it equal to zero.
if n is a prime number π(n)π(n1)=1
if n is not a prime number π(n)π(n1)=0
Let us start. Our starting point is the
Euler representation of the zeta function
eq.cz55 below.
<a name="docA507">
s be a complex number, example s='1.1+2.2i'. p is a prime number.
Riemann Zeta function ζ(s).
ζ(s)
＝
p=∞
∏
p=prime
1
1p^{s}
 eq.cz55
 repeat
log
ζ(s)
＝
log
[
p=∞
∏
p=prime
1
1p^{s}
]
 eq.cz56
 repeat
log
ζ(s)
＝
p=∞
∑
p=prime
log
[
1
1p^{s}
]
 eq.cz57
 repeat
width of above equation
<a name="docA508">
201411221638 here
We do the same as the video we did before.
We take logarithm of zeta function eq.cz56.
Using logarithm rule, for example log(A*B)
[Please see log(6)=log(2*3)=log(2)+log(3)]
you can rewrite log(A*B) as log(A)+log(B)
This is what I am doing, from eq.cz56 to
eq.cz57 .
<a name="docA509">
Now the next step may be a little bit confusing.
201411221652 stop at video 3:42/13:46
201411221733 start
Attention: before this point, lecture 5 and
lecture 6 use same equation. After this point,
lecture 5 use eq.cz62 split ∑[n=1,∞] to n=1+∑[n=2,∞]
lecture 6 use eq.cz78 insert π(n)π(n1)=0 or 1
201411221740 here
<a name="docA510">
eq.cz57 sum over p=prime number.
eq.cz78 sum over n=integer number.
if n=prime, {π(n)π(n1)} is one .
if n NOT= prime, {π(n)π(n1)} is zero .
log
ζ(s)
＝
n=∞
∑
n=2
{π(n)π(n1)}
*
log
[
1
1n^{s}
]
 eq.cz78
width of above equation
<a name="docA511">
201411221751 here
We could do a little trick. This trick is very
important. Every time you sum prime numbers,
you can do that trick. We do not want sum over
primes. primes are so random, we do not know
when a prime is. I want to sum over all the
integer values. Start from two, which is first
prime number, to infinity. But I cannot do that,
for example, (to sum all the integers) I sum
over four or six or eight and so forth. These
are not prime numbers. In order to handle that
we bring in a factor π(n)π(n1) .
<a name="docA512">
This difference only equal to one if n is a
prime number. if n is not a prime number,
π(n)π(n1) is zero. Why is that so? If we
just try this out for 2 for example,
π(2)π(21) = π(2)π(1) = 10 = 1
Let us try three,
π(3)π(31) = π(3)π(2) = 21 = 1
Now let us try four, what will happen is
π(4)π(41) = π(4)π(3) = 22 = 0
201411221811 stop at video 5:34/13:46
<a name="docA513">
201411221925 start
π(4) we did not get a new prime number.
If we subtract π(3) from π(4) we will get
zero. (22 = 0) Now what I am doing is
multiply things out.
201411221928 here
<a name="docA515">
201411221941 here
eq.cz78 red {π(n)π(n1)} break to two part
as shown in eq.cz79 . eq.cz79 left term do
not do anything to it. eq.cz79 right term
looks a little bit ugly. In order to cope
with left term better, right term sum start
from n=2. π(n1) become π(21) which is π(1)
There is no prime number smaller or equal to
one. For n=2, π(21)=π(1)=0 Actually we can
start eq.cz79 right term sum from three.
Now we are doing another thing. Instead of
π(n1) all the time, just using n
<a name="docA516">201411221950
In eq.cz79 change π(n1) to π(n) do as
following
eq.cz79 right term, ignore minus sign, is
∑[n=2,∞]{π(n1) * log[1/(1n^{s})]}
Now set t=n1 then t+1=n. get
∑[t=1,∞]{π(t) * log[1/(1(t+1)^{s})]}
Because π(t)=π(1)=0, we can start from t=2
∑[t=2,∞]{π(t) * log[1/(1(t+1)^{s})]}
Symbol t is dummy, replace t with n, get
∑[n=2,∞]{π(n) * log[1/(1(n+1)^{s})]}
Above ten line are eq.cz79 right side term
with minus sign ignored. Then eq.cz79 is
same as next eq.cz80 201411222001 here
<a name="docA518">
201411222004 here
Let us check if eq.cz80 is same as eq.cz79 .
.....
The next step is, while we have two sums.
Same ∑ indices n=2 to n=∞ and same π(n)
We can write them together
eq.cz81 blue log to eq.cz82 blue log ;
eq.cz81 black log to eq.cz82 black log .
width of above equation
<a name="docA520">
201411222022 here
and factor out π(n) as eq.cz81 shown.
Now I did other trick.
logarithm of one over something is always
equal to minus of logarithm of something.
log(1/A) = log(A)
Please see eq.cz59. Example
log(1/9)
2.1972245773362195
log(9)
2.1972245773362195<a name="docA521">
eq.cz81 blue log term is actually equal to
eq.cz82 blue log term.
eq.cz81 black log term is actually equal to
eq.cz82 black log term.
Now what I will do next ? Look at this
equation.
<a name="docA522">
eq.cz83 is in lecture screen. But,
eq.cz83a,b,c are LiuHH inserted, not in lecture screen.
<a name="docA523">
201411222047 here
You might think eq.cz83 look like differentiating.
I want show you something. If you differentiate
log(1x^{s}) You have to use chain rule
of differentiation ..... get eq.cz83 right side
term.
Now substitute eq.cz83 to eq.cz82 and eq.cz82
has equality to ζ(s), then get next eq.cz84
<a name="docA525">
201411222102 here
If I integrate eq.cz83 right side term,
I will get eq.cz83 left side term
(without d/dx) If I take special boundary.
In eq.cz82 I integrate from n to n+1 I
will have the difference. I will end up
having eq.cz82.
This is what I am doing.
201411222110 stop at video 10:37/13:46
<a name="docA526">
201411231356 start
log(1x^{s})
=
∫
s
x(x^{s}1)
dx
+
c
 eq.cz85
log ζ(s)
=
n=∞
∑
n=2
π(n)
x=n+1
∫
x=n
s
x(x^{s}1)
dx
 eq.cz86
width of above equation
<a name="docA527">
201411231415 here
I am rewriting eq.cz85 as this integral
eq.cz86. Pretty nice, isn't it? We have the
strange looking π(n) in front of integral.
What can we do with that? You can think π(n)
as constant to the integral.
integration variable is x, (dx). π(n) do not
contain x. π(n) is a constant to the integral.
We do not look at the boundary actually.
Boundary n, n+1 is not variable. dx is.<a name="docA528">There is no prime number which is decimal number.
You can take constant π(n) into the integral
and replace π(n) by π(x).
201411231430 here
<a name="docA529">
π(n) is Prime Counting Function. Step type.
π(x) is Prime Counting Function. smooth.
log
ζ(s)
＝
n=∞
∑
n=2
x=n+1
∫
x=n
s*π(x)
x(x^{s}1)
dx
 eq.cz87
log
ζ(s)
＝
x=∞
∫
x=2
s*π(x)
x(x^{s}1)
dx
 eq.cz77
 repeat
width of above equation
<a name="docA530">
201411231442 here
Now there is only one step left. We see
we have integral start from n to n+1.
∑[n=2,∞] adding these integrals. 2 to 3,
3 to 4, 4 to 5, 5 to 6 and so forth. You
see all of these integrals can reduce to
a simple integral start from x=2 to x=∞.
eq.cz77 is our last result. This was a
nice proof. The important idea is next
if n is a prime number π(n)π(n1)=1
if n is not a prime number π(n)π(n1)=0 <a name="docA531">
The result eq.cz77 is strange looking.
It would be very nice to transform eq.cz77
and get π(x) out of integral.
NOT get π(n) out of integral ! After move
π(n) into integral, change π(n) to π(x).
Inside integral no π(n) any more.
What Riemann was doing in his famous work
in 1859 he did something similar to what
we did. But Riemann use another function.
Maybe I will also do that function later
on. ..... That conclude the lecture.
See you guys.
201411231455 stop
201411231753 done first proofread
201411231830 done second proofread
<a name="docA601">
update 20141124
201411241403 start
Next is study notes of
"Zeta Function  Part 7 
Zeta of 2 aka The Basel Problem"
http://www.youtube.com/watch?v=GeKDmoYHiAk
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="docA602">Begin video 7 of 17
Hello. Welcome to the new video on the Euler
and Riemann zeta function. In this video I
will try to show you the famous proof to the
Basel Problem
201411241406 here
ζ(2) = 1.6449340668482264...
width of above equation
<a name="docA604">
201411241441 here
which was first proved by Euler. Basel Problem
is the sum of this kind, see eq.cz88. Euler
found out this value is equal to PI*PI/6 If
you remember the definition of zeta function,
then eq.cz88 is just zeta function evaluated
at two. Euler first derived this and was able
to calculate all zeta value of even integers.
Let us look how did he do this.
<a name="docA606">
201411241510 here
Euler knew the Taylor expansion for sine
function eq.cz89. ... Power of these bodies
are always odd numbers. We have one, three,
five and seven. We should mention also is
that the denominator is odd factorial. We
have one factorial, three factorial, five
factorial, seven factorial in alternating
sum. What Euler did was he just take this
equation eq.cz89, divided by s (eq.cz90).
Instead of using s, he plug in πs. And he
got following thing (eq.cz91) In eq.cz91
the powers are always even. Now let us
rewrite eq.cz91
<a name="docA608">
201411241546 here
eq.cz92 has more nice way, so we can see
the coefficients, 1 and π^{2}/3! and +π^{4}/5!
and so forth, until ... if you watch this
and say OK, you did not do anything else
but dividing and plug in πs that is not a
genius trick. If Euler would have stopped
here he would not have become the famous
at that moment. What he did then is that
Euler proved a very important statement.
201411241606 here
<a name="docA609">
The full name of sinc(x)=sin(x)/x is "sine cardinal". The word
"cardinal" is synonymous with "principal", "essential" or
"fundamental". sine cardinal=正弦基數
201502022118
eq.cz93 proved at tute0060.htm#docAd02
<a name="docA610">
201411241617 here
He did not prove it rigorously. One can
extend this proof very rigorously and show
Euler's proof was right. What he said was
sin(πs)/πs can be written this way eq.cz93
If we know all the zeros of sin(πs)/πs ,
we can use fundamental theorem of algebra
At Euler time, fundamental theorem was
not proved, Euler just imply this is true.
<a name="docA611">
What he said was, if I plug in +1 or 1
to s in eq.cz93 I get zero. If I plug in
+2 or 2 to s in eq.cz93 I get zero, and
so forth. All the zeros of sin(πs)/πs are
enclosed in eq.cz93. ..... Euler said
eq.cz92 and eq.cz93 are looking the same.
Let us just expand eq.cz93. He knew if you
have Taylor expansion then the power series
all the coefficients are equal. Because
Taylor expansion is unique.
201411241635 here
<a name="docA613">
201411241649 here
What Euler did was he just multiply everything
out. You must imagine eq.cz93 goto infinity.
If in eq.cz93 you multiply all 1 each other,
you get 1 in eq.cz94.
If you multiply eq.cz93 one s^{2} with all 1 in
other bracket, you get s^{2} term in eq.cz94.
If you go to infinite long, you get the square
numbers in eq.cz94a.
201411241657 stop
<a name="docA614">
201411241733 start
This should remind you something.
eq.cz94a s^{2} coefficient is ζ(2). This was the
reason why Euler did it. in eq.cz94 other
bracket you get other term like s^{4} coefficient
and so forth, you can imagine something going
on. These are mixed quadratic terms. These are
mixed tripple quadratic terms and so forth.
<a name="docA615">
s^{4} coefficient and s^{6} coefficient are not
important for our Basel Problem. But you
can use s^{4} coefficient and s^{6} coefficient to
find other even value of zeta function.
What Euler did was, he said since Taylor
expansions are unique. These both coefficient
have to be equal to each other.
eq.cz92 s^{2} coefficient and eq.cz94 s^{2} coef.
have to equal to each other. a311241746
What Euler found out was the strange sum
equal to PI square over three factorial.
(see eq.cz88; Here screen show up eq.cz92
show up eq.cz94 and
show up eq.cz88)
<a name="docA616">
But what is this actually?
"De Summis Serierum Reciprocarum" (1735)
eq.cz88 make Euler big star in mathematics.
But he did not stop on the square case. He
also ..... Euler showed ζ(4) equal to PI to
fourth power over 90. Euler can do this for
all even values of zeta function. Euler
later also publish better proof which is
more analytical.
Let us just stop here, wait for next video
appear.
201411241801 stop
201411241911 done first proofread
201411241931 done second proofread
<a name="a31229a">
201412291717
update 20141229 complete the link at
tute0058.htm#docA006<a name="docA999">
Following is frequently needed strings.
π(n)π(n1) ●●●
∑[n=1,∞]{1/n^{s}} = ∏[p=prime]{1/[1p^{s}]}
∑[n=1,∞]{1/n^{s}} = ∏[p=prime]{1/[1p^(s)]}
∑[p=Prime,∞]{1/p^{s}}
the reciprocal power of prime numbers
Euler representation of the zeta function
logarithm
ζ(s)
Taylor series expansion for the logarithm
∑[p=prime,∞]{p^{s}}∑[p=prime,∞]{1/p^{sn}}∑{n^{s}} ≦ ∑{n^{s}} a311191334
tute0057.htm#docA034
p^{xn}  e^{iyn*log(p)} 
Euler formula Euler's formulax should be greater than one.
prime zeta function