Riemann Hypothesis Abstract

Copyright © A. A. Frempong
Riemann Hypothesis
An Original Approach: A Ratio Method
Abstract
The simplest solution is usually the best solution---Albert Einstein
A Breath of Fresh Air
Assuming the sum of the original Riemann series is L, a ratio
method was used to split-up the series equation into subequations and each sub-equation was solved in terms of L, and
ratio terms. It is to be noted that unquestionably, each term of
the series equation contributes to the sum, L, of the series. There
are infinitely many sub-equations and solutions corresponding to
the infinitely many terms of the series equation. After the sum,
L, and the ratio terms have been determined and substituted in
the corresponding equations, the Riemann hypothesis would
surely be either proved or disproved, since the original equation
is being solved. Using the original series equation eliminates
possible hidden flaws in derived equations and consequent
solutions.
1
Riemann Hypothesis
An Original Approach: A Ratio Method
Preliminaries
Here, one covers examples to illustrate the mathematical validity of how one splits up the series
equation to form sub-equations.
Let one think like a child - Albert Einstein. Actually, one can think like an eighth
or a ninth grader. Suppose one performs the following operations:
Example 1: 10 + 20 + 25 = 55
2
10 = 55 × 10
55 = 55 × 11
4
20 = 55 × 20
55 = 55 × 11
(1)
(2)
(3)
25 = 55 × = 55 ×
(4)
Equations (2), (3), and (4) can be written as follows:
10 = 55a
(5)
20 = 55b
(6)
25 = 55c
(7)
One will call a, b and c ratio terms.
One is dividing 55 between 10, 20, and 25 in the ratio a : b : c .
2 , b = 4 , c = 5 . Observe also that a + b + c = 1 ( 2 + 4 + 5 = 11 = 1)
Above, a = 11
11
11 11 11 11
11
One can conclude that the sum of the ratio terms is always 1.
Example 2: Solve the quadratic equation; 6 x 2 + 11x − 10 = 0 .
Method 1 (a common and straightforward method)
By factoring, 6 x 2 + 11x − 10 = 0
(3 x − 2)(2 x + 5) = 0 and solving,
(3 x − 2) = 0 or (2 x + 5) = 0
x = 2,x = − 5.
Solution set: {− 5 , 2}
3
2
2 3
2
Method 2: One applies a ratio method.
a
−
Step 2: 300
1205a + 300 = 0
2
Step 1: From 6 x 2 + 11x − 10 = 0
(1)
60 a − 241a + 60 = 0
6 x 2 + 11x = 10
241 ± 2412 − 4(60)(60)
a=
6 x 2 = 10 a ; (Here, a is a ratio term)
120
3 x 2 = 5a
(2)
a = 241 ± 43681
120
11x = 10b (Here, b is a ratio term)
241
±
209
11x = 10(1 − a)
( a + b) = 1
a=
120
11x = 10 − 10 a
a = 241 ± 209 = 241 + 209 or 241 − 209
120
120
120
x = 10 − 10 a
11
450
32
or
=
120 120
3(10 − 10 a )2 = 5a (Substituting for x in (2)
11
= 15 or 4
2)
4
15
100
200
100
−
a
+
a
3(
= 5a
121
25
55
5
11
2
Step 4: When b = − 11 , 11x = 10( − 11)
4
4
5
x=−
2
11
When b = , 11x = 10( 11 )
15
15
10
x = ( 11 ); x = 2
11 15
3
5
2
Again, one obtains the same solution set {− , } as by the factoring method.
2 3
In the above problem, one could say that if 6 x 2 and 11x are in the ratio a : b and the sum of 6 x 2
and 11x is 10, then 6 x 2 = 10 a and 11x = 10b . (Divide 10 between 6 x 2 and 11x in the ratio a : b )
Also, finding the zeros of f ( x ) = 6 x 2 + 11x − 10 is equivalent to solving 6 x 2 + 11x = 10 for x .
Step 3: Since a + b = 1, when a = 15 or 3 43
4
3
3
11
b = 1 − 3 4 = −2 4 or −
4
4
when a = , b = 1 − 4 = 11
15 15
15
Example 3: A grandmother left $45,000 in her will to be divided between eight grandchildren,
Betsy, Comfort, Elaine, Ingrid, Elizabeth, Maureen, Ramona, Marilyn, in
the ratio 1 : 1 : 1 : 1 : 5 : 1 : 7 : 2 . (Note: 1 + 1 + 1 + 1 + 5 + 1 + 7 + 2 = 1)
36 18 12 9 36 6 36 9
36 18 12 9 36 6 36 9
How much does each receive?
Solution:
Betsy's share of $45,000 = 1 × $45,000 = $1, 250
36
Comfort's share of $45,000 = 1 × $45,000 = $2, 500
18
Elaine's share of $45,000 = 1 × $45,000 = $3, 750
12
1
Ingrid's share of $45,000 = × $45,000 = $5, 000
9
Elizabeth's share of $45,000 = 5 × $45,000 = $6, 250
36
1
Maureen's share of $45,000 = × $45,000 = $7, 500
6
Ramona's share of $45,000 = 7 × $45,000 = $8,750
36
2
Marilyn's share of $45,000 = × $45,000 = $10, 000
9
Check: Sum of shares = $45,000
Sum of the fractions = 1
Example 4: The returns on investments A, B, C, D are in the ratio a : b : c : d . If the total return
on these four investments is P dollars, what is the return on each of these investments?
( a + b + c + d = 1)
Solution Return on investment A = aP dollars
Return on investment B = bP dollars
Return on investment C = cP dollars
Return on investment D = dP dollars
Check: aP + bP + cP + dP = P
P( a + b + c + d ) = P
a + b + c + d = 1 (dividing both sides by P)
3
Example 5: The returns on investments 1z , 1z , 1z , 1z , are in the ratio λ2 : λ3 : λ4 : λ5 .
2 3 4 5
If the total return on these four investments is L, what is the return on each of these
investments? ( λ2 + λ3 + λ4 + λ5 = 1 ).
Solution Return on investment 1z = λ2 L
2
Return on investment 1z = λ3 L
3
Return on investment 1z = λ4 L
4
Return on investment 1z = λ5 L
5
Check: λ2 L + λ3 L + λ4 L + λ5 L = L
L ( λ 2 + λ3 + λ 4 + λ 5 ) = L
( λ2 + λ3 + λ4 + λ5 = 1 ) (dividing both sides by L)
Example 6: Sir Isaac Newton left ρgx units in his will to be divided between − µ
∂ 2 Vx
∂ 2 Vx
,
,
−
µ
∂x 2
∂y 2
∂ 2 Vx ∂p
∂V
∂V
∂V
∂V
, , ρ x , ρVx x , ρVy x , ρVz x in the ratio a : b : c : d : f : h : m : n .
∂x
∂z
∂t
∂y
∂z 2 ∂x
where a + b + c + d + f + h + m + n = 1. How much does each receive?
∂ 2V
Solution − µ 2x 's share of ρgx units = aρgx units
∂x
∂ 2 Vx
− µ 2 's share of ρgx units = bρgx units
∂y
∂ 2 Vx
− µ 2 's share of ρgx units = cρgx units
∂z
∂p
's share of ρgx units = dρgx units
∂x
∂V
ρ x 's share of ρgx units = fρgx units
∂t
∂Vx
ρVx
's share of ρgx units = hρgx units
∂x
∂V
ρVy x 's share of ρgx units = mρgx units
∂y
∂V
ρVz x 's share of ρgx units = nρgx units
∂z
Sum of shares = ρgx units Note: a + b + c + d + f + h + m + n = 1
−µ
4
Example 7: Professor Bernhard Riemann left the sum, L, of the comvergent series,
∞
∑ n1z
in his will to be divided between the terms, 1z , 1z , 1z , 1z , ... of the series in
2 3 4 5
n=1
the ratio λ2 : λ3 : λ4 : λ5 ,... where the sum of the λ' s equals 1.
How much does each term receive?
1
Solution z ' s share of L = λ2 L
2
1 ' s share of L = λ L
3
3z
1 ' s share of L = λ L
4
4z
1 ' s share of L = λ L
5
5z
..................................
.
.... .............................
Sum of the terms = L Note: The sum of the λ' s .equals 1.
The objective of presenting examples 1, 2, 3, 4, 5, and 6 was to convince the reader that the
principles to be used in splitting the terms of the convergent Riemann series, are valid.
In Example 2, one could have used the quadratic formula directly to solve for x , without finding a
and b first. The objective was to convince the reader that the introduction of a and b did not
change the solution set of the original equation.
Nota bene
1. Finding the zeros of f ( x ) = 6 x 2 + 11x − 10 is equivalent to solving 6 x 2 + 11x = 10 for x .
(Note: f ( x ) = 6 x 2 + 11x − 10 = 0 )
2. Similarly, finding the zeros of f ( z ) =
n =1
∞
equivalent to solving
∞
∑ n1z = 11z + 21z + 31z + ... is
∑ n1z = 11z + 21z + 31z + ... = L for z , where L is the sum of the series.
n =1
(Note:
∞
∑ n1z = 11z + 21z + 31z + ... − L = 0 )
n =1
5
Riemann Hypothesis
An Original Approach: A Ratio Method
∞
Given: f ( z ) = ∑ 1z = 1z + 1z + 1z + ... Re( z ) = x > 1
n =1
n
1
2
3
z = x + iy
Required: To find the zeros of f ( z ) .
Plan: One will assume the sum of the above series is L, and equivalently
∞
solve
∑ n1z = 11z + 21z + 31z + ... = L for z , where L is the sum of the series.
n =1
One will split-up the series equation to form sub-equations using ratio terms and
then solve for z .
Let the sum, L, of the convergent series be divided between the terms, 1z , 1z , 1z , 1z , ... of the
2 3 4 5
series in the ratio λ2 : λ3 : λ4 : λ5 ,... where the sum of the λ' s equals 1.
Two types of ratio terms will be used, namely, primary ratio terms and secondary ratio terms.
The primary ratio terms will be symbolized by λ' s , and the secondary ratio terms will be
symbolized by β' s.
(Note that unquestionably, each term of the series contributes to the sum, L, of the series.)
Each sub-equation from the series yields a relation, and the relations for which Re( z ) = x > 1
will be accepted as solutions. From the terms of the series and the ratio terms, one obtains the
following sub-equations which are the for the first four terms, skipping n = 1. There are
infinitely many sub-equations and solutions corresponding to the infinitely many terms of the
series.
1. 1z = λ2 L ; 2. 1z = λ3 L ; 3. 1z = λ4 L ; 4. 1z = λ5 L . ( λ2 , λ3 , λ4 and λ5 are ratio terms)
4
2
3
5
Solutions of the Sub-equations
Case 1: For n = 2 , 1z = λ2 L
2
Step 2:
Step 1:
∞
log λ2 L
z = x + iy = −
Let ∑ 1z converge to the limit L.
log 2
n
n=1
log λ2 L
Then 1z = λ2 L (λ2 is a ratio term) x = − β2 log 2 ( β2 is a secondary ratio term)
2
log λ2 L
iy = − β3
( β3 is a secondary ratio term)
2 − z = λ2 L
log 2
z
−
log 2 = log λ2 L
( β2 + β3 = 1)
− z log 2 = log λ2 L
log λ2 L
y = iβ3
z log 2 = − log λ2 L
log 2
log λ2 L
z=−
log 2
Step 3: Determine the sum L of the series; determine λ2 , β2 , β3 and substitute.
To be continued.
6
Case 2: For n = 3, 1z = λ3 L
3
Step 1:
Step 2:
∞
Let
∑ n1z
n=1
converge to the limit L.
log λ3 L
log 3
log λ3 L
( β 4 is a secondary ratio term)
x = − β4
log 3
log λ3 L
( β5 is a secondary ratio term)
iy = − β5
log 3
( β 4 + β5 = 1)
log λ3 L
y = iβ5
log 3
z = x + iy = −
Then 1z = λ3 L (λ3 is a ratio term)
3
−
z
3 = λ3 L
log 3− z = log λ3 L
− z log 3 = log λ3 L
z log 3 = − log λ3 L
log λ3 L
z=−
log 3
Step 3
Determine λ3 , β 4 , β5 and substitute. (Also substitute for L from Case 1)
To be continued.
Case 3: For n = 4 1z = λ4 L
4
Step 2:
Step 1:
∞
log λ4 L
z = x + iy = −
Let ∑ 1z converge to the limit L.
log 4
n
n=1
log λ4 L
x = − β6
( β6 is a secondary ratio term)
Then 1z = λ4 L (λ4 is a ratio term)
log 4
4
log λ4 L
iy = − β 7
( β 7 is a secondary ratio term)
4 − z = λ4 L
log 4
z
−
log 4 = log λ4 L
( β6 + β 7 = 1)
− z log 4 = log λ4 L
log λ4 L
y = iβ 7
z log 4 = − log λ4 L
log 4
log λ4 L
z=−
log 4
Step 3
Determine λ4 , β6 , β 7 and substitute. (Also substitute for L from Case 1)
To be continued.
7
Case 4: For n = 5, 1z = λ5 L
5
Step 1:
Step 2:
∞
Let
∑ n1z
converge to the limit L.
n=1
log λ5 L
log 5
log λ5 L
( β8 is a secondary ratio term)
x = − β8
log 5
log λ5 L
( β9 is a secondary ratio term)
iy = − β9
log 5
( β8 + β9 = 1)
log λ5 L
y = iβ9
log 5
z = x + iy = −
Then 1z = λ5 L (λ5 is a ratio term)
5
z
−
5 = λ5 L
log 5− z = log λ5 L
− z log 5 = log λ5 L
z log 5 = − log λ5 L
log λ5 L
z=−
log 5
Step 3
Determine λ5 , β8 , β9 and substitute. (Also substitute for L from Case 1)
To be continued.
Other solutions
Similarly, as in cases 1-4, one can continue solving the sub-equations from the series for n = 6 ,
and so on. There will be infinitely many solutions.
General Case : 1z = λn L
n
Step 1:
Step 2:
∞
Let
∑ n1z
n=1
converge to the limit L.
log λn L
log n
log λn L
x = − βn
( βn is a secondary ratio term)
log n
log λn L
iy = − βn +1
( βn +1 is a secondary ratio term)
log n
( βn + βn +1 = 1)
log λn L
y = iβn +1
log n
z = x + iy = −
Then 1z = λn L (λn is a ratio term)
n
−
z
n = λn L
log n − z = log λn L
− z log n = log λn L
z log n = − log λn L
log λn L
z=−
log n
Step 3
Determine λn , βn , βn +1 and substitute. (Also substitute for L from Case 1)
To be continued.
8
Case Extra: For n = 1012
Step 1:
Step 2:
log λ1012 L
12 log10
log λ1012 L
x = − β1012
( β1012 is a ratio term)
12 log10
log λ1012 L
iy = − β1012 +1
( β1012 +1 is a ratio term)
12 log10
( β1012 + β1012 +1 = 1)
log λ1012 L
y = iβ1012 +1
12 log10
∞
Let
∑ n1z
1 = λ 12 L
10
1012 z
z = x + iy = −
converge to the limit L.
n=1
1 = λ 12 L (λ 12 is a ratio term)
10
10
1012 z
10 −12 z = λ1012 L
log10 −12 z = log λ1012 L
−12 z log10 = log λ1012 L)
log λ1012 L
z=−
12 log10
Step 3
Determine λ1012 , β1012 , β1012 +1 and substitute. (Also substitute for L from Case 1)
To be continued.
Then
Summary for the First Four Solutions ( n = 2, 3, 4, 5 )
log λ2 L


( β2 is a secondary ratio term)
x
=
−
β
2

log 2


log λ L

 y = iβ3 log 22


log λ3 L


 x = − β 4 log 3 ( β 4 is a secondary ratio term)


log λ L
 y = iβ5 log 33



log λ4 L


 x = − β6 log 4 ( β6 is a secondary ratio term)


log λ L

 y = iβ 7 log 44


log λ5 L


 x = − β8 log 5 ( β8 is a secondary ratio term)


log λ5 L

 y = iβ9
log 5


Observation of the above solutions shows that for the real parts to be equal,
β2 (log 3)(log λ2 L) = β 4 (log 2)(log λ3 L) (for n = 2, 3 )
To be continued.
9