1. transcendental numbers

TRANSCENDENTAL NUMBERS AND PROOF THAT THE ZEROS OF
RIEMANN ZETA FUNCTION ζ(s) ARE ONLY AND ONLY THOSE
WITH THE REAL PART Re=½
Ing. Pier Francesco Roggero, Dott. Michele Nardelli, P.A. Francesco Di Noto
Abstract:
In this paper we focus our attention on the behavior of transcendental
number that is a (possibly complex) number that is not algebraic— it is not a
root of a non-zero polynomial equation with rational coefficients.
Furthermore, we prove in paragraph 2 that the zeros of the Riemann zeta
function are only and only those with real part equal to Re(½).
Versione 1.0
30/01/2014
Pagina 2 di 34
Index:
1. TRANSCENDENTAL NUMBERS .................................................................................................... 3
1.1 TRANSCENDENT NUMBER e FOR EXCELLENCE ....................................................... 4
1.2 NUMBER π TRANSCENDENT........................................................................................... 6
1.3 TRANSCENDENT NUMBERS eα AND ALL NUMBERS THAT RESULTS FROM f(ex)
ARE TRANSCENDENT ............................................................................................................. 9
1.4 TRANSCENDENT NUMBERS ab ..................................................................................... 10
1.5 TRASCENDENT sen(a), cos(a), tan(a), cot(a), sec(a) e csc(a) NUMBERS ...................... 11
1.6 PROOF THAT THE NUMBERS logb (a) LOGARITHMS AT ANY BASIS b ARE
TRANSCENDENT .................................................................................................................... 13
1.7 ALL THE VALUES OF GAMMA FUNCTION Γ(z) ARE TRASCENDENT.................. 14
1.8 PROOF THAT THE CONSTANT EULER-MASCHERONI γ IS TRANSCENDENT .... 17
1.9 PROOF THAT THE VALUES OF RIEMANN ZETA FUNCTION Ζ (S) ARE
TRANSCENDENT EXCEPT CASES TRIVIAL s = -1, 0, AND ALL THE NEGATIVE
INTEGERS ODD s = -3, -5, -7. -9 ............................................................................................ 18
1.9.1 PROOF THAT THE VALUES OF RIEMANN ZETA FUNCTION ζ(s) FOR INTEGERS
ODD s = 3, 5, 7. 9, … ARE TRANSCENDENT ...................................................................... 23
2. PROOF THAT THE ZEROS OF RIEMANN ZETA FUNCTION ς(s) ARE ONLY AND ONLY
THOSE WITH THE REAL PART RE(½) ............................................................................................ 25
3. GEOMETRIC PROOF THAT TRANSCENDENTAL NUMBERS CANNOT BE DRAWN AS
SEGMENTS BUT ONLY AS ARCS OF CURVES............................................................................. 29
4. ANALYTICAL PROOF THAT TRANSCENDENTAL NUMBERS ARE ARCS OF CURVES .. 31
5. REFERENCES .................................................................................................................................. 34
Versione 1.0
30/01/2014
Pagina 3 di 34
1. TRANSCENDENTAL NUMBERS
A transcendental number is an irrational and unlimited number that is not the solution
of any polynomial equation of the form:
an x n + an−1 x n−1 + ... + a1 x + a0 = 0
where n ≥ 1 and the coefficients ai are rational numbers (ie, integers or "fractional") not
all equal to zero.
The solutions of the polynomial equation are all real or complex numbers ε C but they
are never transcendental numbers.
The first issue that proved to be transcendent without being purpose built was e, by
Charles Hermite in 1873. In 1882, Ferdinand von Lindemann published a demonstration
based on the previous work of Hermite, the transcendence of π. In 1874, Georg Cantor
had proved the existence and non-countability of transcendental numbers.
The discovery of transcendental numbers allowed the proof of the impossibility of
several ancient geometric problems relating to a construction with ruler and compass,
the most famous, the squaring of the circle, it is impossible because π is transcendental.
Versione 1.0
30/01/2014
Pagina 4 di 34
1.1 TRANSCENDENT NUMBER e FOR EXCELLENCE
e is a constant, very important because of its applications in various fields of
mathematics.
Its expression with 55 decimal places is: 2,71828 18284 59045 23536 02874 71352
66249 77572 47093 69995 95749
The Euler number is connected with the exponential function, which maps a real
number to the number given by the power e x , and with the natural logarithm function
(the inverse of the exponential function). Formally it can be defined as the value that the
exponential function assumes in x = 1 .
We will see that this number is the transcendent "for excellence", as derived from it all
the other transcendental numbers.
The number e can be defined in one of the following ways:
•
how the value of the limit
n
 1
e := lim1 +  ;
n→∞
 n
•
as the sum of the series
∞
1 1 1 1 1 1
= + + + + + ...
0! 1! 2! 3! 4!
n =0 n!
e := ∑
where n! is the factorial of n.
The derivative and the integral of the exponential function is the function itself:
Versione 1.0
30/01/2014
Pagina 5 di 34
de x
= ex
dx
∫e
x = ex
The function ex is "self-generating" or restart always.
This also happens when we have a function f(ex) or when in a function appears ex
this factor never disappears.
It is for this reason that in this paper e has been called the transcendental number
"for excellence."
Versione 1.0
30/01/2014
Pagina 6 di 34
1.2 NUMBER π TRANSCENDENT
In plane geometry, π is defined as the ratio between the measure of the length of the
circumference and the length measurement of the diameter of a circle, or even as the
area of a circle of radius 1.
The first 99 decimal digits of π:
3,14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209
74944 59230 78164 06286 20899 86280 34825 34211 7067
To draw in a Cartesian plane π we can. for example, build a quarter of a circle of radius
equal to 2.
Radius = 2
¼ of circumference = π
We note that we can draw a transcendental number as the arc of a curve.
The number π can also be expressed as number e
In fact, it has the following formula that connects the arctan (z) with the natural
complex logarithm, where z is a complex number of the form z = a ± ib:
arctan( z ) =
i
 1 − iz 
log

2  1 + iz 
Versione 1.0
30/01/2014
Pagina 7 di 34
Substituting z =1:
4 arctan(1) = 4
1− i 
i

log
2  1 + i 
 1− i 



1+ i 
4 arctan(1) = 2i log
4 arctan(1) = 2i log(− i ) = 2i
(− iπ ) = π
2
and then we find that:
4arctan(1) = π
We note that to find this remarkable result, we have used the complex logarithm which
has as its basis e.
We have thus shown that π is also a transcendental number because it comes from
e, that is the transcendental number for excellence.
We can also define π with this beautiful limit that also has a geometric meaning
remarkable considering regular polygons inscribed in a circle where n is the number of
sides:
lim n tan
n→∞
180°
=π ,
n
it is nothing more than the area of the circumscribed circle, thus confirming the intuition
that as the number of sides of the polygon grows, it go to "fill in" the circumscribed
circle with radius = 1, for a regular polygon of n sides endless.
Versione 1.0
30/01/2014
Pagina 8 di 34
Or considering the perimeter of a regular polygon of n infinite sides it can also define
the following remarkable limit:
 180° 
 =π

 n 
lim n → ∞ = nsen
Where to tend to infinity n the regular polygon inscribed in the circle, with infinite sides
"infinitesimal" is getting closer to the circle with radius = ½
Observe that both transcendental numbers e and π can be described with the limits
to infinity.
Versione 1.0
30/01/2014
Pagina 9 di 34
1.3 TRANSCENDENT NUMBERS eα AND ALL NUMBERS THAT
RESULTS FROM f(ex) ARE TRANSCENDENT
eα is a transcendental number if α is any complex number ε C (set of complex numbers)
and α ≠ 0 because it is derived from e.
So, it follows also that the constant Gelfond:
eπ is a transcendental number.
If we consider all the functions f(ex) where is eα or an integral function with ex and
all possible combinations we always have a transcendental number.
This follows from the considerations in paragraph 1.1 because the function ex is
"self-generating", which always recurs.
This is also true if we consider the natural logarithm function loge(x) function or
the more general complex logarithm function loge(x) with z = a ± ib because it is
the inverse function of ex
We will see in Section 1.6 that even a logarithm logb(x) in any base b always
generates transcendental numbers because we are using the proper base change
with the number e
So the sums, products, powers, etc ... associated with the number e transcendent for
excellence are all transcendental numbers.
For example:
π + e, π − e, πe, π/e, ππ, ee, πe, π√2, eπ2 , 3e, 2+e
are all transcendent numbers.
Versione 1.0
30/01/2014
Pagina 10 di 34
1.4 TRANSCENDENT NUMBERS ab
ab is a transcendental number if a and b are complex numbers ε C (set of complex
numbers) but where b cannot be integer or rational and also α ≠ 0 and 1.
This statement addresses the seventh problem of Hilbert.
To prove this we consider:
ab = eblog(a)
Even in this case ab always derives from the number e transcendental for excellence,
because it is simply hidden by the formula described above.
If b is a fractional number we would have an irrational number of radical type, for
example with a = 2 and b = ½
2½ = √2
Instead if b is an irrational number or transcendental we have that, for example, with a =
2 and b = √ 2:
2√2 is a transcendental number.
Versione 1.0
30/01/2014
Pagina 11 di 34
1.5 TRASCENDENT sen(a), cos(a), tan(a), cot(a), sec(a) e csc(a)
NUMBERS
The values of the trigonometric functions sen(a), cos(a), tan(a), cot(a), sec(a) and csc(a)
are transcendental numbers if a is a complex number ε C (set of complex numbers),
obviously not legitimate in cases where α = 0
To prove this we consider:
We know that the sine and cosine functions are respectively the real and the imaginary
part of the complex exponential function when its argument is an imaginary number:
eiθ = cos θ + isenθ .
This relationship was noticed for the first time by Euler and, for this reason, the identity
is known as Euler's formula and can also be expressed as:
eiπ + 1 = 0
This allows to extend the definition of the trigonometric functions to a complex
argument z:
(− 1)n
∞
senz = ∑
(2n + 1)!
n =0
z
2 n +1
eiz − e −iz
=
= −isenh(iz )
2i
(− 1)n z 2n = eiz + e −iz
2
n =0 (2 n )!
∞
cos z = ∑
= cosh (iz )
where i2 = −1 and senh(iz) and cosh(iz) are the hyperbolic sine and cosine functions.
Furthermore, for real x we have:
Versione 1.0
30/01/2014
Pagina 12 di 34
( )
( )
cos x = Re eix
senx = Im eix
Also in this case we can see that sen(z) and cos(z) depend on the complex exponential
function.
Then replacing z with any complex number a we obtained that sen(a) and cos(a) are
transcendental numbers.
Obviously this also applies to the tan(a), for the cot(a) =
csc(a) =
1
which are the opposite functions.
sen ( a )
1
1
, sec(a) =
and
tan( a )
cos(a )
Versione 1.0
30/01/2014
Pagina 13 di 34
1.6 PROOF THAT THE NUMBERS logb (a) LOGARITHMS AT ANY
BASIS b ARE TRANSCENDENT
logb(a) at any base is a transcendental number if a is a positive rational number (positive
integer or fraction) and a ≠ 1.
This arises simply because the natural log is the inverse function of the exponential
function.
Proof that logb(a) at any base is transcendental:
Just use the following formula to obtain:
log b x =
log k x
log k b
log b x =
log e x
log e b
which becomes:
And so we are brought back to the natural logarithm log, and we can say that a
logarithm in any base b with x equal to a positive integer or fraction is a transcendental
number.
Versione 1.0
30/01/2014
Pagina 14 di 34
1.7 ALL THE VALUES OF GAMMA FUNCTION Γ(z) ARE
TRASCENDENT
The Gamma function, also known as the Euler gamma function is a function
continuous on the positive real numbers, which extends the concept of factorial to
complex numbers, in the sense that for every non-negative integer n we have
Γ(n + 1) = n!
where n! denotes the factorial, which is the product of the integers from 1 to n: n! = 1 ×
2 × 3 × ... × n.
We also know that:
Γ( z + 1) = zΓ( z )
for which it has Γ(z ) = Γ(z + 1) / z . In this way, the definition of Γ can be extended from
the half-plane Rz > 0 to the strip − 1 < Rz < 0 , and subsequently to the entire plan Rz < 0 ,
with the exception of the straight lines Rz = 0,−1,−2,... or the gamma function Γ(z) does
not exist for negative integers and in 0
Since Γ(1) = 1, the relation above implies, for all natural numbers n,
Γ(n + 1) = n!
Another example is the following alternative formula COMPLETE:
+∞
Γ( z ) = ∑
n =0
(− 1)n
+∞
1
+ ∫ t z −1e −t dt .
n! z + n 1
In this formula are explicit negative integers and 0 that are not allowed.
Versione 1.0
30/01/2014
Pagina 15 di 34
Also in this case the Gamma function depends on e, transcendental number for
excellence.
Probably the most well-known Gamma function assumes that the value of non-integers
is
1
Γ  = π ,
2
Transcendental number because π in turn depends from e
In addition to this and the aforementioned value assumed on natural numbers are
interesting, the following properties, which affect the odd multiples of ½:
n 
 − 1 n − 1
 n  (n − 2 )!!
Γ  = (n−1)/ 2 π =  2 
! π
2 2
 n − 1  2
 2 
π
 n
Γ −  =
 2  − 1 

 n +1
 2 
!
 n + 1  2
 2 
where n!! denotes the semi-factorial and the parenthesis at two levels the binomial
coefficient.
But also all other values, such as
Γ(1/3), Γ(1/4) and Γ(1/6)
Versione 1.0
30/01/2014
Pagina 16 di 34
They are transcendental numbers just by virtue of the formula COMPLETE where we
have the number e transcendental number for excellence, in the complete resolution of
the gamma function.
Versione 1.0
30/01/2014
Pagina 17 di 34
1.8 PROOF THAT THE CONSTANT EULER-MASCHERONI γ IS
TRANSCENDENT
It is defined as the limit of the difference between the truncated harmonic series and the
natural logarithm:

n

γ = lim ∑ − ln n  = lim(H n − ln n )
1
n→∞
 k =1 k

n→∞
where H n is the nth harmonic number. Its approximate value is:
γ ≈ 0,57721 56649 01532 86060 65120 90082 40243 10421 59335 93992 35988
05767 23488 48677 26777 66467 09369 47063 29174 67495... (first 100 decimal
digits)
We try to prove that γ is a transcendental number.
The constant can also be defined through the integrals:
∞
∞
 1 1
− dx = − ∫ e − x ln xdx
0
 x  x 
γ = ∫ 
1
It follows immediately from this that γ is an integral transcendental number because it
always comes from the number and transcendental number for excellence.
Versione 1.0
30/01/2014
Pagina 18 di 34
1.9 PROOF THAT THE VALUES OF RIEMANN ZETA FUNCTION ζ(s)
ARE TRANSCENDENT EXCEPT CASES TRIVIAL s = -1, 0, AND ALL
THE NEGATIVE INTEGERS ODD s = -3, -5, -7. -9 ...
Currently it is known that all the values of ζ (n ) numbers are transcendental for n even,
as they are rational multiples of π
Let's see some significant values.
For all even s:
1 1
π2
+
+
...
=
≈ 1,645 ; the proof of this fact is the solution of the Basel problem.
2 2 32
6
The reciprocal ≈ 0.6, or about 60% is the probability of choosing 2 random numbers that
they are relatively prime, or coprime (eg 14 and 15 are relatively prime).
ζ (2) = 1 +
1 1
π4
+
+
...
=
≈ 1,0823
2 4 34
90
1 1
π6
≈ 1,0173
ζ (6) = 1 + 6 + 6 + ... =
2 3
945
ζ (4) = 1 +
More generally, we know that:
ζ (2n ) =
2 2 n−1π 2 n B2 n
(2n )!
where Bn is the 'nth Bernoulli number.
Instead, the exact calculation of the odd values of s still creates enormous difficulties.
For negative integers, we have:
Versione 1.0
30/01/2014
Pagina 19 di 34
ζ (− n ) = −
Bn+1
n +1
Valid for n ≥ 1.
In particular ζ (− n ) = 0 for all negative even integers because Bm = 0 for all odd m except
m=1, and then for n = 3, 5, 7, 9, ….
We do not know an exact formula that makes the values of the zeta function at s = 3 or
for some other odd value of s. They know only approximate values:
ζ (3) = 1 +
1 1
+ + ... ≈ 1,202
23 33
1 1
ζ (5) = 1 + 5 + 5 + ... ≈ 1,0369
2 3
1 1
ζ (7 ) = 1 + 7 + 7 + ... ≈ 1,0083
2
3
Only in 1978 Apéry Roger was able to prove that ζ(3) is an irrational number. For this
ζ(3) is called Apéry’s constant.
ζ (− 1) = −
1
12
which gives a way to assign a finite result to the divergent series 1+2 +3 +4 + • •
•, which is used in string theory.
Indeed, remember that the number 24 (12 = 24 / 2) is related to the physical vibrations
of the bosonic strings by the following Ramanujan function:
∞ cos πtxw'


− πx 2 w '
e
dx 
∫0 cosh πx

142
4 anti log
⋅ 2
πt 2
−
w'

 t w'
4
(
)
e
φ
itw
'
w
'

24 = 
,
  10 + 11 2 
 10 + 7 2  
+ 

log  



4
4
 




Versione 1.0
30/01/2014
Pagina 20 di 34
We have the following mathematical connection:
∞ cos πtxw'


−πx 2 w '
e
dx 
∫0 cosh πx

142
2 anti log
⋅ 2
πt 2
−
w'

 t w'
4
(
)
e
φ
itw
'
w'
1

ζ (− 1) = − ⇒ − 1 / 
.
12
  10 + 11 2 

 10 + 7 2 
+ 

log  



4
4
 




ζ (0) = − ;
1
2
ζ (1 / 2) ≈ −1.4603545 ;
1 1
ζ (1) = 1 + + + ... = ∞ ;
2 3
For the harmonic series, its main value is:
lim
ε →0
ζ (1 + ε ) + ζ (1 − ε )
2
Exists and is the Euler-Mascheroni constant γ ≈ 0.57721 ...
Furthermore, we have that:
ζ (3 / 2) ≈ 2,612
ζ (5 / 2) ≈ 1,341
ζ (7 / 2) ≈ 1,127
Versione 1.0
30/01/2014
Pagina 21 di 34
We can also define the zeta function for all s ∈ C \ {1}, s=a+ib except the value s = 1, with
the following integral formula:
ζ (s ) =
∞ sin (s arctan t )
2 s −1
− 2s ∫
dt ,
s
0
s −1
2 2 πt
1+ t
e +1
(
)(
)
It follows immediately from this formula that all the numbers ζ (s ) are transcendental
numbers because they are all derived from the number e transcendental number for
excellence.
The trivial zeros at s =-a with a even negative integer, or s = -2, -4, -6, -8, ... are very
easily to find, replaced with b = 0 and ζ (s ) = 0 :
∞ − sin( a arctan t )
1
= (− a − 1)∫
dt
a
0
2
2 −2
πt
(1 + t ) (e + 1)
For example with a = 2, (s = -2), we have:
∞ − sin( 2 arctan t )
1
= −3∫
dt
0
2
(1 + t 2 )−1 (eπt + 1)
1
 1
= −3 −  .
2
 6
Thence, we have:
∞ − sin( 2 arctan t )
1
 1
= −3∫
dt = −3 −  .
−1 πt
2
0
2
 6
(1 + t ) (e + 1)
Multiplying both the sides for
1
, we obtain the following equivalent expression:
4
1
3 ∞ − sin(2 arctan t )
 1  3 1
=− ∫
dt = −3 −  =
= .
−1 πt
2
0
8
4
 24  24 8
1+ t
e +1
(
)(
)
Versione 1.0
30/01/2014
Pagina 22 di 34
This expression can be related with the number 8, i.e. the “modes” that correspond to
the physical vibrations of a superstring by the following Ramanujan function:
∞ cos πtxw'


− πx 2 w '
∫0 cosh πx e dx  142

4 anti log
⋅ 2
πt 2
t w'
−
w'

e 4 φw' (itw') 
1 
8=
.
3
  10 + 11 2 
 10 + 7 2  
+ 

log  



4
4
 




In conclusion, we have the following mathematical connection:
∞ cos πtxw'


−πx 2 w '
∫0 cosh πx e dx  142

4 anti log
⋅ 2
πt 2
−
w'

 t w'
4
(
)
e
itw
φ
'
∞
w'
1
3 − sin(2 arctan t )
1 
 1  3

=− ∫
dt = −3 −  =
.
⇒ 1/
−1 πt
2
0
8
4
3
  10 + 11 2 
 24  24
1+ t
e +1
 10 + 7 2  
+ 

log  



4
4
 




(
)(
)
More in general with a even negative integer we have:
∞ − sin( a arctan t )
(− a − 1)
1
= (− a − 1)∫
dt =
a
0
2
(1 + t 2 )− 2 (eπt + 1) 2(−a − 1)
1 1
=
2 2
Versione 1.0
30/01/2014
Pagina 23 di 34
1.9.1 PROOF THAT THE VALUES OF RIEMANN ZETA FUNCTION
ζ(s) FOR INTEGERS ODD s = 3, 5, 7. 9, … ARE TRANSCENDENT
We know that ζ(3) is called Apéry’s constant.
It is defined as a particular value assumed by the Riemann zeta function, ζ(3),
ζ (3) = 1 +
1 1 1
+ + + ...
23 33 43
For its value in decimal form we have:
ζ (3) = 1,20205690315959428539...
But from the integral formula of paragraph 1.9, we can write with s = 3
ζ (s ) =
∞ sin (s arctan t )
2 s −1
− 2s ∫
dt
s
0
s −1
2 2 πt
1+ t
e +1
(
)(
)
Substituting, we have:
ζ (3) = 2 − 8∫
∞
0
sin(3 arctan t )
(1 + t ) (e
3
2 2
πt
dt
+ 1) )
The formula is actually a little complicated but perfectly solved by numerical methods.
What is important to note that ζ (3) is a transcendental number because always derives
from e.
Obviously, the same formulas can be written for ζ (5), ζ (7), ζ (9), ...all transcendent
numbers.
Versione 1.0
30/01/2014
Pagina 24 di 34
We take the above formula:
ζ (3) = 2 − 8∫
∞
0
sin(3 arctan t )
(1 + t ) (e
3
2 2
πt
dt
+ 1) )
We remember that the number 8, is connected with the “modes” that correspond to the
physical vibrations of a superstring by the following Ramanujan function:
∞ cos πtxw'


− πx 2 w '
∫0 cosh πx e dx  142

4 anti log
⋅ 2
πt 2
t w'
−
w'

e 4 φw' (itw') 
1 
8=
,
3
  10 + 11 2 
 10 + 7 2  
+ 

log  



4
4
 




We have the following mathematical connection:
∞ cos πtxw'


−πx 2 w '
∫0 cosh πx e dx  142

4 anti log
⋅ 2
πt 2
t w'
w'
−

e 4 φw' (itw') 
1 
ζ (3) = 2 −
3
  10 + 11 2 
 10 + 7 2  
+ 

log  



4
4
 




∫
∞
0
sin(3 arctan t )
(1 + t ) (e
3
2 2
πt
+ 1)
dt
)
Versione 1.0
30/01/2014
Pagina 25 di 34
2. PROOF THAT THE ZEROS OF RIEMANN ZETA FUNCTION ζ(s)
ARE ONLY AND ONLY THOSE WITH THE REAL PART RE(½)
Always using the integral formula for each s ∈ C \ {1} of paragraph 1.9, we can write
s=½+jb assuming that they are all with part equal to Real part equal Re(½):
∞ sin (s arctan t )
2 s −1
− 2s ∫
dt ,
s
0
s −1
2 2 πt
1+ t
e +1
ζ (s ) =
(
)(
)
Having to find the zeros of the function ζ (s ) we must put ζ (s ) = 0
Substituting and setting ζ (s ) = 0 , we have:
ζ (1 / 2 + jb ) =
2
1
− + ib
2
1
− + ib
2
2
1
− +ib
2
1
− + ib
2
=2
1
+ ib
2
−2
∫
1
+ib
2
1

sin( + ib  arctan t )
∞
2

dt = 0
1 b
∫0
+
i
2 4 2
πt
(1 + t ) (e + 1)
1

sin( + ib  arctan t )
∞
2

dt
0
1  1

=  − + ib  ∫
2  2
0
∞
(1 + t ) (e
1 b
2 4 +i 2
πt
+ 1)
1

sin( + ib  arctan t )
2

dt
(1 + t ) (e
1 b
2 4 +i 2
πt
+ 1)
The formula is extremely complicated but still perfectly solved by numerical methods.
It is to solve an equation where b is the unknown and using numerical methods we find
all zeros that we know of the function ζ (s ) .
For instance, even with the extremes of the integral (0, 999) we find replacing to
b=14.134725142 an approximate value ½ to 0.4998 ...
Versione 1.0
30/01/2014
Pagina 26 di 34
But let us ask whether there are other zeros in the strip (0.1), we try to apply the
formula, for example with s=1/3+ib
ζ (1 / 3 + jb ) =
2
2
− + ib
3
2
− + ib
3
2
2
− +ib
3
2
− + ib
3
−2
=2
1
+ ib
3
1
+ib
3
1

sin( + ib  arctan t )
∞
3

dt = 0
1 b
∫0
+
i
2 6 2
πt
(1 + t ) (e + 1)
∫
1

sin( + ib  arctan t )
∞
3

dt
0
1  2

=  − + ib  ∫
2  3
0
∞
(1 + t ) (e
1 b
2 6 +i 2
πt
+ 1)
1

sin( + ib  arctan t )
3

dt
(1 + t ) (e
1 b
2 6 +i 3
πt
+ 1)
But we cannot have two identical equations that would lead to ½.
If we take the last expression and multiply both sides for
1
, we obtain the following
16
equivalent expression:
1  1 ib 
= −
+ 
32  24 16  ∫0
∞
1

sin( + ib  arctan t )
3

dt .
(1 + t ) (e
1 b
2 6 +i 3
πt
)
+1
We remember that the number 24 represent the “modes” that correspond to the physical
vibrations of the bosonic strings by the following Ramanujan function:
Versione 1.0
30/01/2014
Pagina 27 di 34
∞ cos πtxw'


− πx 2 w '
∫0 cosh πx e dx  142

4 anti log
⋅ 2
πt 2
t w'
−
w'

e 4 φw' (itw') 

24 =
,
  10 + 11 2 
 10 + 7 2  
+ 

log  



4
4
 




Thence, we obtain the following mathematical connection:
∞ cos πtxw'


−πx 2 w '
∫0 cosh πx e dx  142

4 anti log
⋅ 2
πt 2
1

− w'
sin( + ib  arctan t )

 t w'
4
(
)
e
φ
itw
'
∞
w'
1  1 ib 
3


= −
+ ∫
dt ⇒ 
.
1 b
0
32  24 16 
  10 + 11 2 

2 6 +i 3 πt


10
+
7
2
e +1
1+ t
+ 

log  



4
4
 




(
) (
)
This observation also applies in case we choose any value s included in the strip (0.1)
and then with s=a+ib and with the limitation that a ε (0,1)
We have:
∞ sin( (a + ib ) arctan t )
2
ζ (a + jb ) =
− 2 a +ib ∫
dt = 0
a b
0
a − 1 + ib
2 2 +i 2
πt
(1 + t ) (e + 1)
a −1+ ib
∞ sin( (a + ib ) arctan t )
2
= 2 a +ib ∫
dt
a b
0
a − 1 + ib
2 2 +i 2
πt
(1 + t ) (e + 1)
a −1+ ib
∞ sin( (a + ib ) arctan t )
1
= (a − 1 + ib )∫
dt
a b
0
2
2 2 +i 2
πt
(1 + t ) (e + 1)
Versione 1.0
30/01/2014
Pagina 28 di 34
We can also remove the restriction that a ε (0,1) and consider all values of a ε R (set of
real number) then all s=a+ib except of course the only value that is not valid because
ζ (1 + ) = +∞ and ζ (1 − ) = −∞
So we have endless equations identical to the result ½ which is absurd.
The proof is then carried to absurd, since the only values that cancel the zeta
function ζ (s ) = 0 are only and only those with s=½±jb, necessary and sufficient
condition.
Versione 1.0
30/01/2014
Pagina 29 di 34
3. GEOMETRIC PROOF THAT TRANSCENDENTAL NUMBERS
CANNOT BE DRAWN AS SEGMENTS BUT ONLY AS ARCS OF
CURVES
Consider the transcendental numbers e, π and the irrational golden ratio Φ
Φ =
1+ 5
= 1,6180339887498948482045868343656381 ….( first 34 decimal places)
2
The golden number corresponds to one of the two possible solutions of the quadratic
equation x 2 − x − 1 = 0 , whose roots are:
1 ± 5 1,6180...
=
2
− 0,6180...
Now consider the right triangle thus formed:
The 2 catheti as values Φ and e, while the hypotenuse is worth approximately π.
To be precise, we have:
hypotenuse =
e2 + φ 2 =
10,007... = 3,1634….
A value that differs very little from the true value of π with an absolute error of only
0.69%
Versione 1.0
30/01/2014
Pagina 30 di 34
From these considerations and also from paragraph 1.2 relative to the fourth part
of a circle which defines the length of the curve equal to π, we obtain that the
transcendental numbers can be drawn only as arcs of curves.
If we had designed a spherical triangle on a sphere, then we would have a triangle
with two arcs of curves exactly equal to π and e (because they are transcendental)
while the third side of the triangle can also be a segment (because it is irrational)
but by drawing however on a sphere becomes an arc of a curve.
From this it is clear that the transcendental numbers are arcs of curves in the plan
and irrational numbers, rational or entire are segments in a two-dimensional plane
as the Cartesian plane.
Mind you that the exponential function ex at the point x = 1 is obviously equal to e, and
then we might object that is equal to the vertical segment but this is FALSE, because the
function ex that we draw is only an approximation using a value that is just e number
transcendental for excellence.
Versione 1.0
30/01/2014
Pagina 31 di 34
4. ANALYTICAL PROOF THAT TRANSCENDENTAL NUMBERS
ARE ARCS OF CURVES
The length of an arc is a positive real number that intuitively measure the extension of
an arc or a curve.
To calculate the length of a curve it is necessary that the curve is differentiable in the
interval that we consider.
So if we know the function f(t) we have that the arc length in the interval (a, b) with b>a
in which it is derivable is the real value named L(φ):
L(ϕ ) = ∫
1 + f ' (t ) dt
b
2
a
Let, for example, to apply it to calculate the length of a quarter of the circumference
introduced in paragraph 1.2 about π.
The equation of the circle with center at the origin and radius r = 2 is the following:
x2 + y2 = 4
y2 = 4 - x2
When considering the positive part of the 1st quadrant, we have:
y = + 4 − x2
Applying the formula:
L=
∫
2
0

x2 
1 +
dx =
4 − x 2 

∫
2
0
 4 
dx = 2arcsen(1) = π

2 
4− x 
We remember the famous fractal Ramanujan identity (Hardy 1927):
Versione 1.0
30/01/2014
Pagina 32 di 34
0,618033 = 1 / φ =
5 −1
= R(q) +
2
5
 1 q f 5 (−t ) dt 
3+ 5

1+
exp
1/ 5
4/5 
∫
2
 5 0 f ( −t ) t 
,




3 
5
,
R(q) +
π = 2Φ −
20 
 1 q f 5 (−t ) dt  
3+ 5

1+
exp

1/ 5
4/5  
∫
2
 5 0 f (−t ) t  

and
Φ=
where
5 +1
.
2
Furthermore, we remember that π arises also from the following identities
(Ramanujan’s paper: “Modular equations and approximations to π” Quarterly Journal of
Mathematics, 45 (1914), 350-372.):
π=
(
)(
)

 2 + 5 3 + 13 
12
24

,
and
log 
π
=
log

130
2
142




 10 + 11 2 

+


4


 10 + 7 2  

 .


4


Thence, we have the following mathematical connections:
L=
∫
2
0

x2 
1 +
dx =
2 
4
−
x


∫
2
0
 4 
dx = 2arcsen(1) = π ⇒

2 
4− x 




3 
5
⇒
⇒ π = 2Φ −
R(q) +
5
q
20 


1
f (−t ) dt 
3+ 5

1+
exp

∫
0 f ( −t 1 / 5 ) t 4 / 5 
2
5


 
  10 + 11 2 


24
 +  10 + 7 2   .
⇒π =
log  



4
4
142
 




Versione 1.0
30/01/2014
Pagina 33 di 34
Thence, mathematical connections with Φ =
5 +1
, i.e. the aurea ratio and with the
2
number 24, that is connected with the “modes” that correspond to the physical
vibrations of the bosonic strings.
This proves for the case of the transcendental number π that in a two-dimensional
Cartesian plane it can draw only as a arc of a circumference and not as a segment
(because it is not real number).
In fact, these curves are not polynomial equations of the type:
an x n + an−1 x n−1 + ... + a1 x + a0 = 0
Versione 1.0
30/01/2014
Pagina 34 di 34
5. REFERENCES
1) Wikipedia
2) From the weak Goldbach’s Conjecture to the strong Conjecture (hints to the RH1)
Gruppo “B. Riemann”* Francesco Di Noto, Michele Nardelli
http://www.scribd.com/doc/186505846/Francesco-Di-Noto-Michele-Nardelli-From-theweak-Goldbach%E2%80%99s-Conjecture-to-the-strong-Conjecture-hints-to-the-RH1#

Download Report