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  4. What makes Riemann's Hypothesis Hard to Prove?
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What makes Riemann's Hypothesis Hard to Prove?

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Offline hamdani yusuf (OP)

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What makes Riemann's Hypothesis Hard to Prove?
« on: 27/06/2022 15:40:12 »
What makes Riemann's Hypothesis Hard to Prove?
First, an introduction.
Quote
https://en.wikipedia.org/wiki/Riemann_hypothesis

In mathematics, the Riemann hypothesis is a conjecture that the Riemann zeta function has its zeros only at the negative even integers and complex numbers with real part
1/2
. Many consider it to be the most important unsolved problem in pure mathematics.[1] It is of great interest in number theory because it implies results about the distribution of prime numbers. It was proposed by Bernhard Riemann (1859), after whom it is named.

The Riemann hypothesis and some of its generalizations, along with Goldbach's conjecture and the twin prime conjecture, make up Hilbert's eighth problem in David Hilbert's list of twenty-three unsolved problems; it is also one of the Clay Mathematics Institute's Millennium Prize Problems, which offers a million dollars to anyone who solves any of them. The name is also used for some closely related analogues, such as the Riemann hypothesis for curves over finite fields.

The Riemann zeta function ζ(s) is a function whose argument s may be any complex number other than 1, and whose values are also complex. It has zeros at the negative even integers; that is, ζ(s) = 0 when s is one of −2, −4, −6, .... These are called its trivial zeros. The zeta function is also zero for other values of s, which are called nontrivial zeros. The Riemann hypothesis is concerned with the locations of these nontrivial zeros, and states that:

The real part of every nontrivial zero of the Riemann zeta function is
1/2
.

Thus, if the hypothesis is correct, all the nontrivial zeros lie on the critical line consisting of the complex numbers
1/2 + i t, where t is a real number and i is the imaginary unit.

You may find the visualization in this video by 3blue1brown to be helpful.
Quote
Visualizing the Riemann zeta function and analytic continuation
« Last Edit: 04/05/2024 15:09:06 by hamdani yusuf »
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Offline hamdani yusuf (OP)

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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #1 on: 27/06/2022 15:43:52 »
A Youtube channel seems to be dedicated to explain this problem, and I find it as one of the best explanation online.
https://www.youtube.com/c/zetamath/videos
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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #2 on: 27/06/2022 15:54:10 »
Someone thinks that he might have solved the problem
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Offline hamdani yusuf (OP)

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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #3 on: 28/06/2022 11:36:53 »
Quote from: hamdani yusuf on 27/06/2022 15:43:52
A Youtube channel seems to be dedicated to explain this problem, and I find it as one of the best explanation online.
https://www.youtube.com/c/zetamath/videos
If you are serious to understand this problem, this video is a good place to start.

Quote
Factorials, prime numbers, and the Riemann Hypothesis

Today we introduce some of the ideas of analytic number theory, and employ them to help us understand the size of n!.  We use that understanding to discover a surprisingly accurate picture of the distribution of the prime numbers, and explore how this fits into the broader context of one of the most important unsolved problems in mathematics, the Riemann Hypothesis.
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Offline hamdani yusuf (OP)

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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #4 on: 28/06/2022 11:37:15 »
Quote
The Basel Problem Part 1: Euler-Maclaurin Approximation
This is the first video in a two part series explaining how Euler discovered that the sum of the reciprocals of the square numbers is π^2/6, leading him to define the zeta function, and how Riemann discovered the surprising connection between the zeroes of the zeta function and the distribution of the primes, leading ultimately to his statement of the Riemann Hypothesis.  This video focuses on how Euler developed a method to approximate this sum to 17 decimal places, as well as how the Bernoulli numbers naturally appear as part of this problem.



Quote
The Basel Problem Part 2: Euler's Proof and the Riemann Hypothesis

In this video, I present Euler's proof that the solution to the Basel problem is pi^2/6. I discuss a surprising connection Euler discovered between a generalization of the Basel problem and the Bernoulli numbers, as well as his invention of the zeta function. I explain Euler's discovery of the connection between the zeta function and the prime numbers, and I discuss how Riemann's continuation of Euler's work led him to state the Riemann hypothesis, one of the most important conjectures in the entire history of mathematics.
Sections of this video:
00:00 Intro
01:24 Euler's Basel proof
23:20 The zeta function and the Bernoulli numbers
32:01 Zeta and the primes
48:15 The Riemann hypothesis

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Offline hamdani yusuf (OP)

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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #5 on: 28/06/2022 11:38:32 »
Quote
Analytic Continuation and the Zeta Function

Where do complex functions come from? In this video we explore the idea of analytic continuation, a powerful technique which allows us to extend functions such as sin(x) from the real numbers into the complex plane. Using analytic continuation we can finally define the zeta function for complex inputs and make sense of what it is the Riemann Hypothesis is claiming.

Chapters:
00:00 zetamath does puzzles
00:23 Recap
02:40 Bombelli and the cubic formula
08:45 Evaluating real functions at complex numbers
12:33 Maclaurin series
21:22 Taylor series
27:19 Analytic continuation
35:57 What goes wrong
48:19 Next time
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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #6 on: 28/06/2022 11:40:43 »

Quote
Complex Integration and Finding Zeros of the Zeta Function

In this video we examine the other half of complex calculus: integration. We explain how the idea of a complex line integral arises naturally from real definite integrals via Riemann sums, and we examine some of the properties of this new sort of integral. In particular, we consider some complications that arise when trying to apply the fundamental theorem of calculus to complex functions.

We then bring these ideas to the central focus of this series: the zeta function and the Riemann hypothesis. By the end of the video, we will be able to use complex integrals to approximate the location of the zeroes of the zeta function (or those of any other complex function for that matter)!


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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #7 on: 29/06/2022 17:17:50 »
Here are some interesting results which might be useful in solving the problem.
Quote
https://en.wikipedia.org/wiki/Riemann_zeta_function#Riemann's_functional_equation

This zeta function satisfies the functional equation

where Γ(s) is the gamma function. This is an equality of meromorphic functions valid on the whole complex plane. The equation relates values of the Riemann zeta function at the points s and 1 − s, in particular relating even positive integers with odd negative integers. Owing to the zeros of the sine function, the functional equation implies that ζ(s) has a simple zero at each even negative integer s = −2n, known as the trivial zeros of ζ(s). When s is an even positive integer, the product sin(πs/2)Γ(1 − s) on the right is non-zero because Γ(1 − s) has a simple pole, which cancels the simple zero of the sine factor.

Quote
https://en.wikipedia.org/wiki/Riemann_zeta_function#Other_results
The fact that

for all complex s ≠ 1 implies that the zeros of the Riemann zeta function are symmetric about the real axis. Combining this symmetry with the functional equation, furthermore, one sees that the non-trivial zeros are symmetric about the critical line Re(s) = ½

It is also known that no zeros lie on a line with real part 1.


Quote
https://en.wikipedia.org/wiki/Riemann_hypothesis#Zeros_on_the_critical_line
Hardy (1914) and Hardy & Littlewood (1921) showed there are infinitely many zeros on the critical line, by considering moments of certain functions related to the zeta function. Selberg (1942) proved that at least a (small) positive proportion of zeros lie on the line. Levinson (1974) improved this to one-third of the zeros by relating the zeros of the zeta function to those of its derivative, and Conrey (1989) improved this further to two-fifths. In 2020, this estimate was extended to five-twelfths by Pratt, Robles, Zaharescu and Zeindler[22] by considering extended mollifiers that can accommodate higher order derivatives of the zeta function and their associated Kloosterman sums.

Quote
https://mathworld.wolfram.com/RiemannHypothesis.html
 It is known that the zeros are symmetrically placed about the line I(s)=0. This follows from the fact that, for all complex numbers s,

1. s and the complex conjugate s* are symmetrically placed about this line.

2. From the definition (1), the Riemann zeta function satisfies zeta(s*)=zeta(s)*, so that if s is a zero, so is s*, since then zeta(s*)=zeta(s)*=0*=0.

It is also known that the nontrivial zeros are symmetrically placed about the critical line R(s)=1/2, a result which follows from the functional equation and the symmetry about the line I(s)=0. For if s is a nontrivial zero, then 1-s is also a zero (by the functional equation), and then 1-s* is another zero. But s and 1-s* are symmetrically placed about the line R(s)=1/2, since 1-(x+iy)*=(1-x)+iy, and if x=1/2+x', then 1-x=1/2-x'.

From above results, it can be inferred that for any point in the critical strip,
ζ(s*)=ζ(1-s) if and only if s is in critical line,
where s*=complex conjugate of s.
« Last Edit: 13/07/2022 03:55:41 by hamdani yusuf »
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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #8 on: 01/07/2022 05:59:51 »
Basically, proofing Riemann's Zeta function must demonstrate that assuming the existence of non-trivial zero of Riemann's Zeta function with real component other than 0.5 leads to contradiction. Thousands of top mathematicians have tried to solve it unsuccessfully. Millions of math enthusiasts might have tried their luck attacking the problem from various directions, but it still withstands.

At this point, it should be obvious that direct attack on the problem is impossible. There must be some missing key ingredients not yet thought to be related to the problem.
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Offline Bored chemist

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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #9 on: 01/07/2022 13:32:43 »
Quote from: hamdani yusuf on 01/07/2022 05:59:51
There must be some missing key ingredients not yet thought to be related to the problem.
No
It may not be provable.
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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #10 on: 02/07/2022 06:07:27 »
Quote from: Bored chemist on 01/07/2022 13:32:43
Quote from: hamdani yusuf on 01/07/2022 05:59:51
There must be some missing key ingredients not yet thought to be related to the problem.
No
It may not be provable.
Why? Do you have any reason to think that way? Or is it just your intuition?
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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #11 on: 02/07/2022 12:17:55 »
Quote from: hamdani yusuf on 02/07/2022 06:07:27
Quote from: Bored chemist on 01/07/2022 13:32:43
Quote from: hamdani yusuf on 01/07/2022 05:59:51
There must be some missing key ingredients not yet thought to be related to the problem.
No
It may not be provable.
Why? Do you have any reason to think that way? Or is it just your intuition?
It is known that some problems are impossible to solve- That's the incompleteness theorem.
I'm simply stating that this problem might be one of them.

And if that's true then this statement

Quote from: hamdani yusuf on 01/07/2022 05:59:51
There must be some missing key ingredients not yet thought to be related to the problem.
is false.
There may not be any missing ingredient that would allow you to solve it.
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Offline evan_au

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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #12 on: 03/07/2022 10:45:45 »
Quote from: Bored Chemist
some problems are impossible to solve- That's the incompleteness theorem
The Incompleteness Theorem applies within a specific domain of mathematics - it may be impossible to prove some true statements within the axioms of that system.

However, some mathematical breakthroughs occur when applying results from a quite different domain of mathematics.
- Effectively, this extends the original set of axioms with an additional set of axioms over a different domain
- There may be additional true statements within the new, extended set of axioms that are unproveable within that extended set of axioms.
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Offline hamdani yusuf (OP)

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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #13 on: 13/07/2022 03:51:56 »
Quote from: hamdani yusuf on 01/07/2022 05:59:51
At this point, it should be obvious that direct attack on the problem is impossible.

But it doesn't stop us from trying anyway. We can start from the functional equation.

Quote from: hamdani yusuf on 29/06/2022 17:17:50
Here are some interesting results which might be useful in solving the problem.
Quote
https://en.wikipedia.org/wiki/Riemann_zeta_function#Riemann's_functional_equation

This zeta function satisfies the functional equation

where Γ(s) is the gamma function. This is an equality of meromorphic functions valid on the whole complex plane. The equation relates values of the Riemann zeta function at the points s and 1 − s, in particular relating even positive integers with odd negative integers. Owing to the zeros of the sine function, the functional equation implies that ζ(s) has a simple zero at each even negative integer s = −2n, known as the trivial zeros of ζ(s). When s is an even positive integer, the product sin(πs/2)Γ(1 − s) on the right is non-zero because Γ(1 − s) has a simple pole, which cancels the simple zero of the sine factor.

When Riemann zeta function produces 0 result, ζ(s) = 0, at least one of these terms is 0
1) 2s  = 0 →  s = -∞
2) πs-1  = 0 →  s = -∞
3) sin(πs/2).Γ(1-s) = 0  →  s ∈ {negative even numbers}
4) ζ(1-s) = 0 = ζ(s)

Point #3 gives trivial zeros, while point #4 gives non-trivial zeros.
when 1-s=s  → 1=2s  →  s=1/2 
But  ζ(1/2) <> 0
« Last Edit: 11/05/2024 14:10:34 by hamdani yusuf »
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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #14 on: 13/07/2022 08:52:30 »
Quote from: evan_au on 03/07/2022 10:45:45
There may be additional true statements within the new, extended set of axioms that are unproveable within that extended set of axioms.
And the RH might be one of them.
Is there proof that there is a proof, or is it possible that the OP is wasting his time?
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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #15 on: 13/07/2022 09:17:04 »
Quote from: hamdani yusuf on 29/06/2022 17:17:50
From above results, it can be inferred that for any point in the critical strip,
ζ(s*)=ζ(1-s) if and only if s is in critical line,
where s*=complex conjugate of s.

Quote from: hamdani yusuf on 13/07/2022 03:51:56
4) ζ(1-s) = 0 = ζ(s)

Quote from: hamdani yusuf on 29/06/2022 17:17:50
The fact that
ζ(s) = (ζ(s*))*

But we only care about the case where ζ(s) = 0.
0* = 0
 →  ζ(s) = ζ(s*) = ζ(1-s) = 0

when s* = 1-s  →  s+s* = 1 
→  Re(s)+Im(s).i + Re(s)-Im(s).i = 1  →  2.Re(s) = 1 
→  Re(s) = ½
« Last Edit: 14/07/2022 03:42:58 by hamdani yusuf »
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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #16 on: 14/07/2022 06:16:23 »
For the sake of the argument, let's assume that Riemann's hypothesis is false. It implies that there are ζ(s)=0 where Re(s)≠½.
Consequently, s*≠1-s, and there will be 2 zeros with the same imaginary parts.
« Last Edit: 15/07/2022 05:42:24 by hamdani yusuf »
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Offline hamdani yusuf (OP)

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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #17 on: 19/07/2022 18:58:30 »
Quote from: Bored chemist on 13/07/2022 08:52:30
Quote from: evan_au on 03/07/2022 10:45:45
There may be additional true statements within the new, extended set of axioms that are unproveable within that extended set of axioms.
And the RH might be one of them.
Is there proof that there is a proof, or is it possible that the OP is wasting his time?
Let's learn what history can teach us.
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Offline hamdani yusuf (OP)

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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #18 on: 21/07/2022 04:09:15 »
From the functional equation


Quote from: hamdani yusuf on 13/07/2022 03:51:56
When Riemann zeta function produces 0 result, ζ(s) = 0, at least one of these terms is 0
1) 2s  = 0 →  s = -∞
2) πs-1  = 0 →  s = -∞
3) sin(πs/2).Γ(1-s) = 0  →  s ∈ {negative even numbers}
4) ζ(1-s) = 0 = ζ(s)

Point #3 gives trivial zeros, while point #4 gives non-trivial zeros.
when 1-s=s  → 1=2s  →  s=½
But  ζ(½) <> 0
Riemann's hypothesis only cares about point #4, which gives non-trivial zeros.

Quote
https://en.wikipedia.org/wiki/Riemann_zeta_function#Other_results
The fact that

Or if written in alternative notation, ζ(s)=(ζ(s*))*
for all complex s ≠ 1 implies that the zeros of the Riemann zeta function are symmetric about the real axis. Combining this symmetry with the functional equation, furthermore, one sees that the non-trivial zeros are symmetric about the critical line Re(s) = ½

It is also known that no zeros lie on a line with real part 1.
Using above results, proving Riemann's hypothesis can be done by following these reasonings
  • For every s non-trivial zeros of Riemann's Zeta function, it applies that ζ(s)=0=ζ(1-s*), because 0*=0
  • In case ζ(s)=0 where Re(s)≠½, there would be at least two different zeros with the same imaginary part, i.e.
    s and 1-s*
    or,
    Re(s)+Im(s) and Re(1-s)+Im(s)
  • Riemann's hypothesis can be proven by showing that ζ(s) - ζ(1-s*) = 0 if and only if s=1-s*, which implies that Re(s)=½




« Last Edit: 22/07/2022 17:42:28 by hamdani yusuf »
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Offline hamdani yusuf (OP)

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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #19 on: 21/07/2022 12:48:17 »
Let's start by observing how the equation ζ(s) looks like for several different parameters.
First, let's see the plot of Zeta function in critical line.
In WolframAlpha, type
plot zeta (0.5+si) from 0 to 30
https://www.wolframalpha.com/input?i=plot+zeta+%280.5%2Bsi%29++from+0+to+30

Then compare to it's absolute value
plot |zeta (0.5+si)|  from 0 to 30
https://www.wolframalpha.com/input?i=plot+%7Czeta+%280.5%2Bsi%29%7C++from+0+to+30

Then compare to absolute value of points outside the critical line
plot |zeta (0.6+si)|  from 0 to 30
https://www.wolframalpha.com/input?i=plot+%7Czeta+%280.6%2Bsi%29%7C++from+0+to+30

plot |zeta (0.3+si)|  from 0 to 30
https://www.wolframalpha.com/input?i=plot+%7Czeta+%280.3%2Bsi%29%7C++from+0+to+30

plot |zeta (0.7+si)|  from 0 to 30
https://www.wolframalpha.com/input?i=plot+%7Czeta+%280.7%2Bsi%29%7C++from+0+to+30
plot |zeta (0.9+si)|  from 0 to 30
https://www.wolframalpha.com/input?i=plot+%7Czeta+%280.9%2Bsi%29%7C++from+0+to+30

The pattern we can observe is that for the same imaginary part, the plot's minima of zeta function gets closer to 0 when the real part of z is closer to 0.5
It means if there exist zero of zeta function where the real part of the variable is not equal to 0.5, the function will also produce zero when the real part of the variable is changed to get closer to 0.5
By reflection, if there exist zero of zeta function where the real part of the variable is not equal to 0.5, the function will also produce zero when the real part of the variable is changed to get further away from 0.5
But somehow it's no longer zero when Re(s)=1
These result in contradiction that there would be infinitely many zeros with the same imaginary parts, which brings us to conclude that Non-trivial zeros of Riemann's hypothesis must be located at the critical line, Re(s)=0.5, which means that Riemann's hypothesis must be true.
« Last Edit: 22/07/2022 05:03:19 by hamdani yusuf »
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