Zeta function study notes tute0058
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<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">
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
]]

<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

ns
=
p=∞
p=prime
ps

ps-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

ns
=
1

1s
+
1

2s
+
1

3s
+
1

4s
+
1

5s
+ ...
--- eq_cz01b
width of above equation
 
p=∞
p=prime
ps

ps-1
=
2s

2s-1
*
3s

3s-1
*
5s

5s-1
*
7s

7s-1
*
11s

11s-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,∞]{-1n+1/n^s}/(1-21-s) --- eq_cz02a
or
ζ(s)={[1-2^(1-s)]^(-1)}*∑[n=1,∞]{-1n+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

1-21-s
*
n=∞
n=1
(-1)n+1

ns
--- 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)=2s*PIs-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) = 2s * πs-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="docA401"> 
2014-11-18-15-22 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.
<a name="docA403">
Alert: p be PRIMES 2,3,5,7,11,13,17 ... to infinity.
 
p=∞
p=Prime
1

p
→ ∞
--- eq.cz54
eq.cz54 is our goal, not start point. eq.cz54 is not given.
width of above equation
<a name="docA404">
2014-11-18-15-39 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).
 
ζ(s)
p=∞
p=prime
1

1-p-s
--- eq.cz55
width of above equation
<a name="docA407">
2014-11-18-16-14 here 
eq.cz55 is the Euler representation of the 
zeta function.
See Euler original proof, and Easier proof
What we do is we take logarithm of eq.cz55. 
 
log ζ(s) log
[
p=∞
p=prime
1

1-p-s
]
--- eq.cz56
width of above equation
<a name="docA408">
2014-11-18-16-30 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.
2014-11-18-16-38 LiuHH notes
<a name="docA409">
 
log ζ(s)
p=∞
p=prime
log
[
1

1-p-s
]
--- eq.cz57
width of above equation
<a name="docA410">
2014-11-18-16-43 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. 
<a name="docA411">
 
log
1

1-x
log
(1-x)
[
n=∞
n=1
xn

n
]
--- eq.cz58
width of above equation
2014-11-18-16-57 stop 
<a name="docA412">
2014-11-18-19-18 start 
eq.cz58 left black term can be written as 
log[(1-x)-1]. logarithm has power rule 
log[Ab]=b*log(A) --- eq.cz59 
therefore log[(1-x)-1]=-log(1-x) 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, p0 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="docA414">
 
log ζ(s)
p=∞
p=prime
n=∞
n=1
(p-s)n

n
--- eq.cz60
width of above equation
<a name="docA415">
2014-11-18-19-47 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"> 
2014-11-18-19-56 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">
2014-11-18-20-12 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. 
2014-11-18-20-33
<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.
 
log ζ(s)
p=∞
p=prime
p-s
p=∞
p=prime
n=∞
n=2
p-sn

n
--- eq.cz62
--- repeat
 
log ζ(s)
p=∞
p=prime
p-s
n=∞
n=2
p=∞
p=prime
p-sn

n
"docA421"
--- eq.cz65
 
log ζ(s)
p=∞
p=prime
p-s
n=∞
n=2
[
1

n
p=∞
p=prime
1

psn
]
--- eq.cz66
Remember: n=1 become blue term. Right end term start from n=2
width of above equation
<a name="docA422"> 
2014-11-18-20-46 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/psn} we 
can find what does ∑[p=prime,∞]{1/psn} mean? 
In order to find that out, we want to take 
the absolute value of ∑[p=prime,∞]{1/psn} 
<a name="docA425">
 
|
p=∞
p=prime
1

psn
|
p=∞
p=prime
1

| psn |
=
p=∞
p=prime
1

| p(x+iy)*n |
=
p=∞
p=prime
1

| pxn+iyn |
--- eq.cz67
width of above equation
<a name="docA426"> 
2014-11-18-21-43 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 
psn as p(x+iy)*n //s is complex (x+iy)
2014-11-18-21-52 stop at video 09:00/22:02
<a name="docA427">
2014-11-19-14-24 start 
To understand better we can write complex 
variable s as (x+iy). The absolute value of 
p to sn, |psn| 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 |pxn+iyn|
2014-11-19-14-51 here
<a name="docA428">
 
|
p=∞
p=prime
1

psn
|
p=∞
p=prime
1

| pxn |*| piyn |
=
p=∞
p=prime
1

pxn | e iyn*log(p) |
=
p=∞
p=prime
1

pxn
--- eq.cz68
Remember: e = 2.718281828459045...
Remember: | e iyn*log(p) | always = 1, and ignored
width of above equation
<a name="docA429">
2014-11-19-14-59 here
Next step, from 
eq.cz67 right end term |p(x+iy)*n| 
to eq.cz68 second term |pxn|*|piyn|. 
Breaking up (x+iy)*n into product of 
two powers. This is rule of exponential. 
|pxn| 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 |pxn| is |positive+/-| 
For example prime 3 : 
3-2=1/32=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 , |piyn|, we do same 
trick as for convergence for the zeta 
function. (see tute0057.htm#docA034) We 
introduce here an exponential. 
( |piyn| = | eiyn*log(p) | this e is 
 "introduced exponential". a311191527)
e to the logarithm of p to the iyn. {eiyn*log(p)} 
If you use the Euler's formula (tute0057.htm#docA035 )
|eiyn*log(p)| always equal to one. 
We ignore one. 
<a name="docA432">
In eq.cz68 right end, we get ∑[p=Prime,∞]{1/pxn}
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. 
2014-11-19-15-52 here
<a name="docA433">
 
|
p=∞
p=prime
1

psn
|
p=∞
p=prime
1

pxn
p=∞
p=prime
1

pn
k=∞
k=2
1

kn
--- eq.cz69
width of above equation
<a name="docA434">
2014-11-19-15-59 here 
What we do next step is 
eq.cz69 second term 1/pxn is less than 
eq.cz69 third term  1/pn . We know x is 
greater than one. (see tute0057.htm#docA047)
If you make x > 1 this value 1/pxn become 
smaller and smaller. 1/pxn upper bound is 1/pn. 
Next step, we do another thing. 
∑[p=Prime,∞]{1/pn} is less than ∑[k=2,∞]{1/kn} 
We will do a little trick, sum over prime 
must be less than sum over all the integer 
number. 2014-11-19-16-15 here
<a name="docA435">
eq.cz70 prime number sum ☺/☺ term is non-exist 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. 2014-11-19-16-32
 
p=∞
p=2
1

pn
=
1

2n
+
1

3n
+

+
1

5n
+

+ ...
--- eq.cz70
 
k=∞
k=2
1

kn
=
1

2n
+
1

3n
+
1

4n
+
1

5n
+
1

6n
+ ...
--- eq.cz71
width of above equation
<a name="docA436"> 

This is very easy to understand. Because 
prime number is just part of integer numbers. 
2014-11-19-16-35 stop, graph show up. 

<a name="docA437">
<a name="integralTest4">

Integral Test 4 ; 2014-11-19-18-10


Box1, debug ;

QAboxd01.value=''
<a name="docA438">
2014-11-19-18-30 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/kn}
 see eq.cz19 in tute0057.htm#docA056 )
<a name="docA439">
In "Integral Test 4" graph, red curve is 
y=1/xn. Look at the black rectangular area. 
Left rectangular area is height 1 * width 1.
Second rectangular area is H 1/2n * W 1.
Third rectangular area is H 1/3n * 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.)
2014-11-19-18-57 stop
<a name="docA440">
2014-11-19-19-37 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="docA441">
 
|
p=∞
p=prime
1

psn
|
k=∞
k=2
1

kn
k=∞
k=2
t=k
t=k-1
1

tn
dt
t=∞
t=1
1

tn
dt
--- eq.cz72
width of above equation
<a name="docA442">
2014-11-19-20-03 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. 
2014-11-19-20-35

<a name="docA444">
eq.cz72 right end term is an easy function 
to integrate.
 
|
p=∞
p=prime
1

psn
|
1

1 - n
t1-n
t=∞
t=1
1

n - 1
--- eq.cz73
width of above equation
<a name="docA445">
2014-11-19-20-47 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 t1-n where 1-n must be 
negative to get a finite integration value. 
We start n at 2 (first prime number) 1-n<0 
is true. In  eq.cz73 left end term value 
is smaller than 1/(n-1)
2014-11-19-20-59 stop at video 17:50/22:02

<a name="docA447">
2014-11-20-09-05 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/ps} 
 strange looking zeta= ∑[p=Prime,∞]{1/psn} 
 n = 2,3,4,5,6 ... infinity , see eq.cz62 
 eq.cz66 blue term regular prime zeta function
 eq.cz66 purple term strange prime zeta function 
 2014-11-20-09-22 here)
of f(n) is smaller or equal to 1/(n-1).
<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

psn
]
|
n=∞
n=2
1

n
|
p=∞
p=prime
1

psn
|
n=∞
n=2
1

n
1

n - 1
1
--- eq.cz74
width of above equation
<a name="docA449">
2014-11-20-09-42 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/[n-1]) = (1/[n-1]) - (1/n) --- eq.cz75
Calculate each term one by one. 
n= 2: (1/n)*(1/[n-1]) = (1/1)-(1/2)
n= 3: (1/n)*(1/[n-1]) = (1/2)-(1/3)
n= 4: (1/n)*(1/[n-1]) = (1/3)-(1/4)
n= 5: (1/n)*(1/[n-1]) = (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/[n-1])} answer is 1. 
2014-11-20-10-10 stop at video 18:00
<a name="docA452">
2014-11-20-13-45 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

ps
n=∞
n=2
[
1

n
p=∞
p=prime
1

psn
]
--- eq.cz66
--- repeat
width of above equation
<a name="docA453">
2014-11-20-13-52 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="docA454">
 
log ζ(1)
p=∞
p=prime
1

p
n=∞
n=2
[
1

n
p=∞
p=prime
1

pn
]
--- eq.cz76
from eq.cz66
and set s=1
width of above equation
<a name="docA455">
2014-11-20-14-12 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. 
2014-11-20-14-23
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!
2014-11-20-14-29 stop
2014-11-20-15-23 first proofread
2014-11-20-17-27 second proofread



<a name="primeList"> 2014-11-21-08-52 ;
Update 2014-11-22 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 prime-in-memory 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 ; 2014-11-21-16-52
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 2014-11-21-18-52
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='' ; 2014-11-21-23-20 done Prime Counting Function Graph

<a name="primcnt1">

http://freeman2.com/tute0058.htm#docA405
http://freeman2.com/tute0058.htm#primeCountDraw
<a name="docA501"> 
update 2014-11-23
2014-11-22-15-30 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. 
2014-11-22-15-42 here
<a name="docA503">
     π(x) is Prime Counting Function. smooth.

 
log ζ(s)
x=∞
x=2
s*π(x)

x(xs-1)
dx
--- eq.cz77
width of above equation
<a name="docA504">
2014-11-22-16-00 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 n-1. 
<a name="docA506"> 
This difference, π(n)-π(n-1), 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(n-1)=1 
if n is not a prime number π(n)-π(n-1)=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

1-p-s
--- eq.cz55
--- repeat
 
log ζ(s) log
[
p=∞
p=prime
1

1-p-s
]
--- eq.cz56
--- repeat
 
log ζ(s)
p=∞
p=prime
log
[
1

1-p-s
]
--- eq.cz57
--- repeat
width of above equation
<a name="docA508">
2014-11-22-16-38 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 re-write 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.
2014-11-22-16-52 stop at video 3:42/13:46 

2014-11-22-17-33 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(n-1)=0 or 1
2014-11-22-17-40 here
<a name="docA510">
eq.cz57 sum over p=prime number.
eq.cz78 sum over n=integer number.
if n=prime, {π(n)-π(n-1)} is one .
if n NOT= prime, {π(n)-π(n-1)} is zero .
 
log ζ(s)
n=∞
n=2
{π(n)-π(n-1)} * log
[
1

1-n-s
]
--- eq.cz78
width of above equation
<a name="docA511"> 
2014-11-22-17-51 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(n-1) . 
<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(n-1) is zero. Why is that so? If we 
just try this out for 2 for example, 
π(2)-π(2-1) = π(2)-π(1) = 1-0 = 1 
Let us try three, 
π(3)-π(3-1) = π(3)-π(2) = 2-1 = 1 
Now let us try four, what will happen is 
π(4)-π(4-1) = π(4)-π(3) = 2-2 = 0 
2014-11-22-18-11 stop at video 5:34/13:46 
<a name="docA513">
2014-11-22-19-25 start
π(4) we did not get a new prime number. 
If we subtract π(3) from π(4) we will get 
zero. (2-2 = 0) Now what I am doing is 
multiply things out. 
2014-11-22-19-28 here
<a name="docA514">
 
log ζ(s)
n=∞
n=2
{π(n)-π(n-1)} * log
[
1

1-n-s
]
--- eq.cz78
 
n=∞
n=2
π(n) * log
1

1-n-s
n=∞
n=2
π(n-1) * log
1

1-n-s
--- eq.cz79
width of above equation
<a name="docA515">
2014-11-22-19-41 here
eq.cz78 red {π(n)-π(n-1)} 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(n-1) become π(2-1) which is π(1)
There is no prime number smaller or equal to 
one. For n=2, π(2-1)=π(1)=0 Actually we can 
start eq.cz79 right term sum from three. 
Now we are doing another thing. Instead of 
π(n-1) all the time, just using n 
<a name="docA516"> 
2014-11-22-19-50 
In eq.cz79 change π(n-1) to π(n) do as 
following 
eq.cz79 right term, ignore minus sign, is
∑[n=2,∞]{π(n-1) * log[1/(1-ns)]} 
Now set t=n-1 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 2014-11-22-20-01 here
<a name="docA517">
log ζ(s)
 
n=∞
n=2
π(n) * log
1

1-n-s
n=∞
n=2
π(n-1) * log
1

1-n-s
--- eq.cz79
 
n=∞
n=2
π(n) * log
1

1-n-s
n=∞
n=2
π(n) * log
1

1-(n+1)-s
--- eq.cz80
width of above equation
<a name="docA518">
2014-11-22-20-04 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 
<a name="docA519">
log ζ(s)
 
n=∞
n=2
π(n) *
{
log
1

1-n-s
log
1

1-(n+1)-s
}
--- eq.cz81
 
n=∞
n=2
π(n) *
{
log
[1-(n+1)-s]
log
[1-n-s]
}
--- eq.cz82
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">
2014-11-22-20-22 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.
 
d

dx
log(1-x-s)
1

1-x-s
[
s*x-s-1
]
s

x*(xs -1)
--- eq.cz83
---whole eq
 
d

dx
log(1-x-s) =
1

1
d

dx
log(1-x-s) =
d(1-x-s)

d(1-x-s)
d

dx
log(1-x-s)
--- eq.cz83a
--- partial 1
 
=
d[log(1-x-s)]

d(1-x-s)
d(1-x-s)

dx
=
1

1-x-s
d(1-x-s)

dx
=
1

1-x-s
[0-(-s)*x-s-1]
--- eq.cz83b
--- partial 2
 
=
s*x-s-1

1-x-s
=
1

1-x-s
s

xs+1
=
s

xs+1-x-s+s+1
=
s

x*(xs -1)
--- eq.cz83c
--- partial 3
eq.cz83a + eq.cz83b + eq.cz83c = equation eq.cz83 . //a311230923 partial
width of above equation
<a name="docA523">
2014-11-22-20-47 here
You might think eq.cz83 look like differentiating.
I want show you something. If you differentiate 
log(1-x-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="docA524">
 
log ζ(s)
n=∞
n=2
π(n) *
{
log
[1-(n+1)-s]
log
[1-n-s]
}
--- eq.cz84
width of above equation
<a name="docA525">
2014-11-22-21-02 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. 
2014-11-22-21-10 stop at video 10:37/13:46 
<a name="docA526"> 
2014-11-23-13-56 start
 
log(1-x-s)
=
s

x(xs-1)
dx
+ c
--- eq.cz85
 
log ζ(s)
=
n=∞
n=2
π(n)
x=n+1
x=n
s

x(xs-1)
dx
--- eq.cz86
width of above equation
<a name="docA527">
2014-11-23-14-15 here
I am re-writing 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).
2014-11-23-14-30 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(xs-1)
dx
--- eq.cz87
 
log ζ(s)
x=∞
x=2
s*π(x)

x(xs-1)
dx
--- eq.cz77
--- repeat
width of above equation
<a name="docA530">
2014-11-23-14-42 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(n-1)=1 
if n is not a prime number π(n)-π(n-1)=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.
2014-11-23-14-55 stop
2014-11-23-17-53 done first proofread
2014-11-23-18-30 done second proofread



<a name="docA601"> update 2014-11-24 2014-11-24-14-03 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 2014-11-24-14-06 here
<a name="docA603">
 
ζ(2) =
n=∞
n=1
1

n2
=
1

12
+
1

22
+
1

32
+
1

42
+
1

52
+ ... =
π2

6
--- eq.cz88
 ζ(2) = 1.6449340668482264...
width of above equation
<a name="docA604">
2014-11-24-14-41 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="docA605">
 
sin(s) = s
s3

3!
+
s5

5!
s7

7!
+/- ...
--- eq.cz89
 
sin(s)

s
=
s

s
s3

s*3!
+
s5

s*5!
s7

s*7!
+/- ...
--- eq.cz90
divide by s
 
sin(πs)

πs
= 1
(πs)2

3!
+
(πs)4

5!
(πs)6

7!
+/- ...
--- eq.cz91
all s to πs
width of above equation
<a name="docA606"> 
2014-11-24-15-10 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 
re-write eq.cz91 
<a name="docA607">
 
sin(πs)

πs
= 1
π2

3!
s2 +
π4

5!
s4
π6

7!
s6 +/- ...
--- eq.cz92
width of above equation
<a name="docA608">
2014-11-24-15-46 here
eq.cz92 has more nice way, so we can see 
the coefficients, 1 and -π2/3! and +π4/5!
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. 
2014-11-24-16-06 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
2015-02-02-21-18
eq.cz93 proved at tute0060.htm#docAd02
 
sin(πs)

πs
=
(
1 -
s2

12
)
(
1 -
s2

22
)
(
1 -
s2

32
)
...
--- eq.cz93
2014-12-17-09-53 https://www.youtube.com/watch?v=h3Hcioh2F9I
width of above equation
<a name="docA610">
2014-11-24-16-17 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. 
2014-11-24-16-35 here 
<a name="docA612">
 
sin(πs)

πs
= 1 -
(
1

12
1

22
1

32
●●●
)
s2
--- eq.cz94a
 
continue
(
1

12*22
1

12*32
1

22*32
●●●
)
s4
--- eq.cz94b
 
continue
(
1

12*22*32
●●●
)
s6
●●●
--- eq.cz94c
width of above equation
<a name="docA613">
2014-11-24-16-49 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 s2 with all 1 in 
other bracket, you get s2 term in eq.cz94. 
If you go to infinite long, you get the square 
numbers in eq.cz94a. 
2014-11-24-16-57 stop 

<a name="docA614">
2014-11-24-17-33 start
This should remind you something. 
eq.cz94a s2 coefficient is ζ(2). This was the 
reason why Euler did it. in eq.cz94 other 
bracket you get other term like s4 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">
s4 coefficient and s6 coefficient are not 
important for our Basel Problem. But you 
can use s4 coefficient and s6 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 s2 coefficient and eq.cz94 s2 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. 
2014-11-24-18-01 stop 
2014-11-24-19-11 done first proofread
2014-11-24-19-31 done second proofread



<a name="a31229a"> 2014-12-29-17-17 update 2014-12-29 complete the link at tute0058.htm#docA006

<a name="docA999"> Following is frequently needed strings. π(n)-π(n-1) ●●● ∑[n=1,∞]{1/ns} = ∏[p=prime]{1/[1-p-s]} ∑[n=1,∞]{1/ns} = ∏[p=prime]{1/[1-p^(-s)]} ∑[p=Prime,∞]{1/ps} 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/psn} |∑{n-s}| ≦ ∑{|n-s|} a311191334 tute0057.htm#docA034 pxn | eiyn*log(p) | Euler formula Euler's formula x should be greater than one. prime zeta function






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