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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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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #220 on: 08/09/2024 07:04:03 »
I asked ChatGPT, How did Riemann calculated the first non-trivial zero of zeta function?
Quote
Riemann did not explicitly calculate the first non-trivial zero of the Riemann zeta function, but he made significant contributions to understanding its properties, including the distribution of its zeros. In his famous 1859 paper, **"On the Number of Primes Less Than a Given Magnitude,"** he introduced what is now called the **Riemann Hypothesis**, proposing that all non-trivial zeros of the zeta function \zeta(s) lie on the **critical line** \Re(s) = \frac{1}{2}, where s = \sigma + it is a complex number.

However, calculating the actual zeros was a much later development. Early calculations were performed numerically by later mathematicians, such as:

- In 1903, **J. P. Gram** and **Thomas J. Stieltjes** performed the first significant numerical work on finding zeros of the zeta function.
- **G. H. Hardy** and **John E. Littlewood** furthered this in the early 20th century.
- The first actual numerical calculation of the non-trivial zeros was made by **R. J. Backlund** in 1914.

### How it was calculated (later developments):

1. **Numerical Approximation**: The zeros are calculated using computational techniques, which often rely on approximating \zeta(s) and finding where it crosses the real axis (i.e., where \zeta(s) = 0) along the critical line \Re(s) = \frac{1}{2}.
   
2. **Series Representations and Integrals**: Using various representations of the zeta function, such as the Euler-Maclaurin formula or integrals involving the zeta function, allows for efficient numerical evaluations.

3. **Gram Points**: These are specific points used to locate zeros. Gram developed a method to approximate where the zeros are by checking specific values of t in \zeta\left(\frac{1}{2} + it\right).

The first non-trivial zero, approximately t_1 \approx 14.1347, was found to satisfy \zeta\left(\frac{1}{2} + it\right) = 0.

Riemann's main contribution was setting the framework and conjecturing the connection between the zeros of the zeta function and the distribution of prime numbers, though the actual calculation of the zeros came from later work.
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Offline hamdani yusuf (OP)

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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #221 on: 08/09/2024 07:10:37 »
I asked about Gram point.
Quote
A **Gram point** is a specific value of t in the context of the Riemann zeta function, typically used to approximate the locations of non-trivial zeros of the zeta function on the critical line \Re(s) = \frac{1}{2}. It is named after the Danish mathematician **J. P. Gram**, who developed a method for identifying intervals on the critical line that are likely to contain zeros of the zeta function.

### Definition:
A **Gram point** is a value of t_n such that the function \theta(t), defined as:
\[
\theta(t) = \arg \left( \Gamma\left( \frac{1}{4} + \frac{it}{2} \right) \right) - \frac{t \log \pi}{2},
\]
where \Gamma(s) is the Gamma function, satisfies the condition:
\[
\theta(t_n) = n\pi,
\]
for some integer n.

This condition means that at a Gram point, \theta(t_n) takes values that are integer multiples of \pi.

### Role in Zeta Function Calculations:
Gram points are useful in the study of the zeta function because they help locate intervals where the zeta function might have zeros. The Gram points are distributed roughly regularly along the critical line, and the Riemann-Siegel formula, which approximates the zeta function for large t, is particularly well-behaved near Gram points.

### Gram's Law:
For a time, it was observed (and this observation is now known as **Gram's Law**) that between two successive Gram points, there is typically one zero of the zeta function on the critical line. However, this is not always the case, and counterexamples to Gram's Law have been found, though it holds true for many zeros.

### Example of a Gram Point:
The first Gram point occurs at approximately t_1 \approx 17.8456, and the first few zeros of the zeta function can be located by looking near these Gram points.

In summary, Gram points help mathematicians narrow down where the zeros of the Riemann zeta function are likely to occur, making them useful for both numerical and theoretical investigations of the zeta function and the distribution of its zeros.

Quote
**Gram's Law** states that between two successive **Gram points**, there is typically one non-trivial zero of the Riemann zeta function on the critical line \Re(s) = \frac{1}{2}. While this law holds in many cases, it is not universally true. **Counterexamples to Gram's Law** occur when the number of zeros between two successive Gram points is either zero or greater than one.

### Counterexamples to Gram's Law:
Counterexamples arise when:

1. **No Zeros Between Two Gram Points**: Sometimes, no zeros of the zeta function occur between two successive Gram points, which directly violates Gram's Law.
   
2. **More Than One Zero Between Two Gram Points**: In other cases, more than one zero might be found between successive Gram points.

### First Counterexample:
The first known counterexample to Gram's Law occurs at the 126th zero of the zeta function. Specifically, between the **126th** and **127th Gram points**, **no zero** of the zeta function is found. This was discovered in detailed numerical computations of the zeros of the zeta function.

### Another Counterexample:
The **128th** and **129th Gram points** provide a different type of counterexample, where **two zeros** of the zeta function are found between these successive Gram points, again violating Gram's Law.

### Frequency of Counterexamples:
While Gram's Law generally holds for a large number of zeros, counterexamples become more frequent as we move further along the critical line. Although the law is a good heuristic for locating zeros, these counterexamples show that the relationship between Gram points and zeros of the zeta function is more complex than initially thought.

### Summary:
- **First counterexample**: No zero between the 126th and 127th Gram points.
- **Another counterexample**: Two zeros between the 128th and 129th Gram points.
- **General trend**: Counterexamples to Gram's Law become more common as one looks at higher Gram points (for larger values of t).

This shows that while Gram's Law is a helpful tool, it is not infallible, and other methods are needed to accurately locate all zeros of the zeta function.
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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #222 on: 08/09/2024 07:14:58 »
Quote
Yes, there are known counterexamples to Gram's Law where the number of zeros between two successive Gram points is greater than two. While Gram's Law generally predicts that there should be exactly one zero between each pair of successive Gram points, in some cases more than one zero can occur, and in rare cases, more than two zeros can appear.

### Examples:
1. **First known example of more than two zeros**:
   The first known instance where **three zeros** of the Riemann zeta function were found between two successive Gram points occurs between the **169,852,770th and 169,852,771st Gram points**. This was discovered through extensive computational efforts in the search for counterexamples to Gram's Law.

2. **More than two zeros in later calculations**:
   As higher and higher Gram points are explored, computational evidence has shown that there are intervals between successive Gram points where **three or even four zeros** have been found. However, such cases are quite rare compared to the more common counterexamples with no zeros or two zeros.

### Frequency and Distribution:
While the vast majority of Gram point intervals contain exactly one zero, and the counterexamples with zero or two zeros are much more frequent, cases with more than two zeros are extremely rare. The probability of encountering more than two zeros between successive Gram points increases as we look at higher values of t, but they remain exceptional.

### Summary:
While counterexamples to Gram's Law where **two zeros** are found between successive Gram points are more common, cases with **three or more zeros** have been discovered at very high Gram points, such as the example at the 169,852,770th Gram point. These findings further emphasize that Gram's Law, though useful as a heuristic, does not perfectly capture the behavior of the zeros of the zeta function.
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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #223 on: 08/09/2024 08:59:16 »
Quote from: hamdani yusuf on 28/08/2024 16:34:35
Meanwhile, the odd curves always intersect at 0.5.
The absolute value of the function resembles V.
https://www.wolframalpha.com/input?i=plot+abs%28%2Bzeta%28s%2B14i%29-zeta%281-s%2B14i%29+%29from+0+to+1

But if the absolute function is done for individual terms of zeta functions, the result resembles S.
https://www.wolframalpha.com/input?i=plot+abs%28zeta%28s%2B14i%29%29+-+abs%28zeta%281-s%2B14i%29+%29from+0+to+1

Unlike previous V and S function, this time they don't involve logarithm. Thus the calculation can be simplified.
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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #224 on: 08/09/2024 09:04:17 »
https://www.wolframalpha.com/input?i=plot+abs%28zeta%28s%2B2.002117+pi+i%29%29+-+abs%28zeta%281-s%2B2.002117+pi+i%29+%29from+0+to+1

https://www.wolframalpha.com/input?i=plot+abs%28zeta%28s%2B2+pi+i%29%29+-+abs%28zeta%281-s%2B2+pi+i%29+%29from+0+to+1

The inflection point seems to be the same as in logarithmic S function.

https://www.wolframalpha.com/input?i=plot+re%28log%28zeta%28s%2B2.002117+pi+i%29%29%29+-+re%28log%28zeta%281-s%2B2.002117+pi+i%29%29%29from+0+to+1

https://www.wolframalpha.com/input?i=plot+re%28log%28zeta%28s%2B2+pi+i%29%29%29+-+re%28log%28zeta%281-s%2B2+pi+i%29%29%29from+0+to+1
« Last Edit: 16/09/2024 02:59:52 by hamdani yusuf »
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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #225 on: 08/09/2024 10:05:47 »
Plotting the subtracting function vertically, can help finding non-trivial zeros.
https://www.wolframalpha.com/input?i=plot+abs%28zeta%280.55%2Bs+i%29%29+-+abs%28zeta%281-0.55%2Bs+i%29+%29from+0+to+30

If the absolute function is applied after the subtraction,
https://www.wolframalpha.com/input?i=plot+abs%28zeta%280.55%2Bs+i%29-zeta%281-0.55%2Bs+i%29+%29from+0+to+40
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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #226 on: 18/09/2024 03:27:46 »
https://www.wolframalpha.com/input?i=plot+re%28log%28zeta%28s%2B14.1347+i%29%29%29+-+re%28log%28zeta%281-s%2B14.1347+i%29%29%29+%2B0.81+i%280.5-s%29from+0+to+1
Logarithmic S function looks like a straight line even for imaginary part close to a non-trivial zero of Zeta function.

Although a slight difference is still there.
https://www.wolframalpha.com/input?i=plot+re%28log%28zeta%28s%2B14.1347+i%29%29%29+-+re%28log%28zeta%281-s%2B14.1347+i%29%29%29+-0.810755%280.5-s%29from+0+to+1
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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #227 on: 23/09/2024 09:50:23 »
Another hard math problems which was unsolved for a long time can give us some lessons.
The Man Who Solved the World?s Most Famous Math Problem
How Andrew Wiles solved the problem that stumped mathematicians for 357 years.

A comment to the video.
Quote
I've watched that original documentary and its amazing how emotional Andrew gets. You can see the passion for the problem in how he barely is able to get out the words "Nothing I ever do again will...." as he almost bursts into tears. I would say 99% of people would just say "umm..its just a math problem!". To him though, this represented the culmination of an entire lifetime of dedication that has finally been realized.

A quote given by Simon Singh from what Piet Hein said: "Problems worthy of attack prove their worth by fighting back". That is this entire journey in a nutshell.
« Last Edit: 23/09/2024 10:35:24 by hamdani yusuf »
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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #228 on: 23/09/2024 10:06:00 »
Elliptic Curves and Modular Forms | The Proof of Fermat?s Last Theorem
Quote
Elliptic curves, modular forms, and the Taniyama-Shimura Conjecture: the three ingredients to Andrew Wiles? proof of Fermat?s Last Theorem.

This is by far the hardest video I've ever had to make: both in terms of learning the content and explaining it. So there a few questions I don't have answers for. If you're up for it, feel free to answer these as a YouTube comment or on Twitter (@00aleph00)!

QUESTIONS:

1. The Taniyama-Shimura Conjecture seems really contrived. We made a weirdly specific sequence from elliptic curves. We made a weirdly specific sequence from modular forms. And behold, the sequences match! It seems manufactured to work. What?s profound about it?

2. Why do we care about elliptic curves of all things? It?s described by, again, a weirdly specific equation: why is it the darling child of number theory?

3. Does the Taniyama-Shimura conjecture also guarantee uniqueness? That is, does it say that for every elliptic curve there is a unique modular form with the same sequence as it?

4. We defined how a matrix from the group SL2Z ?acts? on a complex number. Does anyone have a geometric picture for this? Does a matrix act on a complex number just like how it would act on a vector in R^2 (i.e: by rotating it)?

5. This is a more advanced question. Most elliptic curve books encode the sequence m_n of a modular form using something called a Dirichlet L-function, a generalization of the Reimann Zeta function. More precisely, instead of associating a modular form to a *sequence*, we associate it to a modified version of the Riemann Zeta Function, where the n_th coefficient of the series is the term m_n. (This is sometimes called the Hasse-Weil L-function of a modular form). This seems unnecessary. What is the benefit of doing this?

6. Does anyone understand Andrew Wiles? paper? LOL
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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #229 on: 23/09/2024 10:34:52 »
Every Unsolved Math problem that sounds Easy
Quote
These are some of the famous and toughest math problems, which are unsolved.

Timestamps

0:00 The Kissing Number
1:16 The Goldbach Conjecture
2:25 Collatz Conjecture
3:39 The Twin Prime Conjecture
5:41 The Unknotting Problem
7:05 Pi + e
8:30 Birch and Swinnerton-Dyer Conjecture
9:14 Riemann Hypothesis
11:01 The Lonely Runner Conjecture
11:46 is γ rational?
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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #230 on: 24/09/2024 03:13:02 »
Euler Product Formula
Quote
There are many beautiful math equations, but in this video I want to make the case for the Euler Product formula. In 1737 Euler proved this crazy formula that compares a sum over natural numbers with a product over all prime numbers - weird! We'll see a proof using the Sieve of Eratosthenes, see the Basel problem fall out, extend beyond the proof of infinitely many prime numbers to the sum of reciprocal primes, and talk about the Riemann-Zeta function and how it connects to the Riemann Hypothesis, one of the million dollar Millennium Problems.

0:00 Euler Product Formula
2:56 Proof of the Formula
7:52 s=1 and Reciprocal Primes
12:30 Reimann-zeta function and Riemann Hypothesis
15:48 Brilliant.org/TreforBazett
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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #231 on: 02/10/2024 06:16:47 »
I've compiled my study of Riemann's Hypothesis in my Youtube videos.
Introducing Riemann Hypothesis as an unsolved math problem.
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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #232 on: 02/10/2024 06:17:51 »
Trying to find regular patterns in Riemann's Zeta function out of the irregularities in it.
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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #233 on: 02/10/2024 06:18:28 »
Showing how functional equation of Riemann's Zeta function determines where the zeros are.
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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #234 on: 04/10/2024 14:16:44 »
Finding the nontrivial zeros of the Riemann Zeta Function using Desmos
Quote
In this video I discuss extending the Riemann Zeta Function using the Dirichlet Eta Function, and use this along with Desmos to find the first few nontrivial zeros.

All nontrivial zeros of the Riemann Zeta Function so far calculated by humans (there are about 10^13 of them) all have real part 1/2, which falls in line with the Riemann Hypothesis which says that all of these zeros have real part 1/2.

The Riemann Hypothesis is a currently unsolved math problem, and is attached to a $1,000,000 prize awarded by the Clay Mathematics Institute.
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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #235 on: 04/10/2024 14:43:07 »
A Proof of the Functional Equation for the Riemann Zeta Function using Mellin Transforms
Quote
The Riemann Zeta Function is originally defined as a series, but this ends up only being valued when the real part of the complex input, s, is greater than 1; however, it can be shown that the domain of this function can be extended beyond these values for s, and the functional equation allows us to extend its domain to essentially all complex numbers.

In this video I explore some fundamental properties of the Mellin Transform, as well the Fourier Series for the fraction part of a number, to prove the Riemann Zeta Function's functional equation. I do my best to prove each result from the ground up, while there are two results I use from previous videos (proven in those videos).

The Riemann Hypothesis is a currently open problem in mathematics and the use of the functional equation - more specifically a crucial consequence of it regarding non-trivial zeros, shown in this video - will be undoubtedly utilized in the proof of the Riemann Hypothesis, if it can be proved.
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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #236 on: 04/10/2024 14:52:46 »
Proving the Riemann Hypothesis in the most elementary way possible
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WARNING: This video is intended as an April Fool's joke, but is a phenomenal example of one simple mistake leading to results such as proving an essentially unprovable statement. can you find the fallacy? Comment below if so.
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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #237 on: 07/10/2024 09:07:52 »
A Direct Proof of the Riemann Hypothesis - 2024

Quote
Four years ago this time of year, I and my co-authors uploaded a preprint entitled, "A Direct Proof of the Riemann Hypothesis".  We have updated the paper to eliminate Lemmas 2 and 3.  The result of Lemma 3 has caused question in the mathematics community.  After Lemma 3 in the earlier version, we stated that at this point there are several ways to prove the Riemann Hypothesis.  It turns out, that the proof is implied even by Lemma 1.  We now call this simply "The Lemma".  The proof follows quickly after.  We still use Upsilon of s, or more precisely its multiplicative factor Alpha of s to illustrate the uniqueness of the zeros on the critical line.  The proof was already implied by Lemma 1.  In short, we just needed to "turn the oscillator on" to see it.  Now we show it to you.

This video is very rough, as we are just casually discussing it unscripted.  We welcome all serious eyes on this.  Feel free to skip around.  We have animated the oscillator using GeoGebra.

https://www.researchgate.net/publication/344037263_A_Direct_Proof_of_the_Riemann_Hypothesis


As mentioned in the video, AI can help in solving hard math problems, even when humans have difficulties in solving them.

« Last Edit: 07/10/2024 09:32:07 by hamdani yusuf »
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Re: What makes Riemann's Hypothesis Hard to Prove?
« Reply #238 on: 07/10/2024 23:41:59 »
Quote from: hamdani yusuf 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.
This can be shown by combining Y function with S function which have been shifted to the left by a half.
https://www.wolframalpha.com/input?i=plot+re%28log%28%28zeta%280.5%2Be+d%2B9999i%29-zeta+%280.5-e+d%2B9999i%29%29+%2F+%28zeta%280.5%2Bd%2B9999i%29-zeta+%280.5-d%2B9999i%29%29%29+%29+from+-0.2+to+0.2

Y function is defined by the logarithm of the difference between ζ(s) and ζ(1-s*)
S function is defined by the logarithm of the  ratio between ζ(s) and ζ(1-s*).
For s with high imaginary part, S function looks like a straight line with negative slope. Higher imaginary part yields steeper and more straight line. At critical line, s=1-s*, thus the ratio=1, and log(1) =0.
It's visually more convincing to show that zeta of any points on the right of critical line are always closer to zero, compared to its reflection about the critical line, providing high enough imaginary part. But one may argue that the ratio of two zeros of zeta function can have a value other than 1.
On the other hand, functional equation guarantees that any non-trivial zero of zeta function lead to negative infinity in Y function, because log(0-0) = log(0) = -~.
But it's visually less convincing to say that Y function can't have more than 1 "leg".
Combination of Y function and S function can be more convincing in both visually and algebraically.

* Screenshot 2024-10-08 213039.png (17.14 kB, 556x470 - viewed 108 times.)
« Last Edit: 08/10/2024 15:31:07 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 #239 on: 08/10/2024 15:04:58 »
Riemann Hypothesis 3 : Extracting regular pattern
Here we will try to find a process to extract regular patterns in Riemann Zeta function, and filter out the irregularities. Hopefully we can get a bigger picture on how the Riemann Zeta function will behave at high imaginary part, which is the focus of Riemann's Hypothesis.
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