Riemann hypothesis
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- Опубліковано 24 сер 2019
- The Riemann hypothesis is widely regarded as the greatest unsolved problem in mathematics.
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![[Visual] The Riemann Zeta Function Visualised](http://i.ytimg.com/vi/GcbML8bGSu8/mqdefault.jpg)
A Brilliant and wonderful explanation. This makes me love math and R.H. even more. There is so much beauty and mystery in the Riemann hypothesis and prime numbers patterns.
A wonderful summary of this intriguing hypothesis. I like the apparent connection to real world physical systems.
Hello Professor,
Would Quantum Computers and Quantum A.I. will solve these types of problems like Riemann Hypothesis or Navier Stroke Equations...?
@@MOHNAKHAN How do you know he Is a professor lmao. Just because he has a good grasp on English.
@@strangerdanger7616 that's what you say and think...
He is connected with me on Social Media that's the reason I know 🥴🥴🥴🥴🥴
I solved and proved the Reimann zeta function, check it out. I posted it yesterday!
@@reggyreptinall9598 send link
I was really impressed by the detailed explanation.
Thank you very much.
literally couldn't understand what Riemann hypothesis was from all the other videos on youtube, thank you for this video !!!!!
I watched many talks regarding the Riemann Hypothesis, but this one is one of the two best ones I ever watched. Thank-you very much. I will subscribe your channel.
Thank you.
I've watched many videos on this subject, but i really feel like yours gives me the best picture of what this all means.
Your videos on mathematicians are amazing. Please continue this work.
That's very fascinating.
The Riemann hypothesis is named after the German mathematician Bernhard Riemann, who proposed it in 1859, and it has to do with the distribution of prime numbers. A prime number is a number that can be divided only by itself and one without leaving a remainder. Prime numbers are fascinating, not only because they are of fundamental importance in math, but also because although we can't predict where the next one will turn up on the number line, they don't occur completely at random. There are rules governing their overall distribution. The most important of these rules has been proved today, and it is called "the prime number theorem." It states that for any number
the best lay introduction to the Riemann-Hypothesis i have ever seen , great job sir , you deserve many more views
Thank you for your kind comment.
A Solution for the RIEMANN ZETA FUNCTION is extremely valuable because It also point to Solutions for enhancing the HAMILTON GEOMETRZATION Poincare conjecture, Hodge Invariance conjecture as it relates to PRIME NUMBERS, TAYLOR INFINITE SERIES ,MACULIN SERIES and Doing Arithmetic past ZERO or Singularity as it is called in Analytic Geometry , and Algebraic Geometry, and it Directly points to the Prime factorization Algorithm , the Division algorithm, and the QUADRIATIC FORMULA This Solves many DIMENSIONS and RANK IN THE COMPLEX FUNCTION PLANE for MANIFOLD like The Kahler MANIFOLD ,CALIBU YAU MANIFOLD simeoustanesly and Points to Soulutions to the entire Millennium Prize Problems proposed by The Early 20th Century Philospher and Mathematician David HILBERT , Including the YANG-MILL Mass GAP , and the NP COMPUTATION time space COMPLEXITY problem also know as the Traveling Salesman problem
Absolutey great content, thank you for this Sir
Thank you, Steve.
I love your video, full of wisdom as the topic itself.
You’re very eloquent. I would really like to hear from you what a mathematical proof is and maybe give a couple examples. Thank you for the video.
Fantastic explanation!
A Solution for the RIEMANN ZETA FUNCTION is extremely valuable because It also point to Solutions for enhancing the HAMILTON GEOMETRZATION Poincare conjecture, Hodge Invariance conjecture as it relates to PRIME NUMBERS, TAYLOR INFINITE SERIES ,MACULIN SERIES and Doing Arithmetic past ZERO or Singularity as it is called in Analytic Geometry , and Algebraic Geometry, and it Directly points to the Prime factorization Algorithm , the Division algorithm, and the QUADRIATIC FORMULA This Solves many DIMENSIONS and RANK IN THE COMPLEX FUNCTION PLANE for MANIFOLD like The Kahler MANIFOLD ,CALIBU YAU MANIFOLD simeoustanesly and Points to Soulutions to the entire Millennium Prize Problems proposed by The Early 20th Century Philospher and Mathematician David HILBERT , Including the YANG-MILL Mass GAP , and the NP COMPUTATION time space COMPLEXITY problem also know as the Traveling Salesman problem
There is no such thing as the MACULIN SERIES!
@@azzteke Funny Can't you Say Something More Constructive !Than pointing out a Simple Spelling Error??
Good conclusion of the Riemann hypothesis, I'd like to ask you about to make a video about, how can the man calculate the first non trivial zero 1/2+i14... by hand, Thank you!
Thank you sir ,for such an outstanding lucid lecture ....about the intriguing hypothesis
Thanks for your kind words, Sivaditya.
@@discovermaths All twin primes lie at multiples of 15.
Odd numbers will be where 1,3 and 7,9
Even numbers will be twins 9,1.
Euler Euclid PNT Q= P1,P2,P3.... P24
First three primes 2,3,5= 30. So you will only find primes 9,1.
I have the full proof written down and I will upload the video in the next 23 hours.
Fastest way to confirm this is find know twins. And divide the number ending in 5 after the twins 1,3. And dividing numbers ending in 5 before twins 7,9
And the number in between twins 9,1 which will end in 0.
This is partial proof of the riemann hypothesis. I have the full proof
@@MarkusDarkess Least insane riemann hypothesis "prover"
Anyone after solving it? 😅
Fantastic.
I’ve always wanted a video on how the zeta function actually revealed the secrets of prime numbers - and your video delivered! Thank you. I’ve subscribed :)
hi
i hope i can help answer your question by directing you to the zetamath channel
their last video about integrals in the complex world shows why this is the case at a certain point in the video
Very good!
Thanks!
Primes arranged in a particular way do show a pattern and indeed display this order amongst all the randomness. Trouble is the pattern is beautiful to see but not repetitive. Maybe we haven't put the numbers into a coordinate system where a repeated pattern is possible, yet.
C'mon man, how will you not show the proof to the solution, you're really keeping me on the edge of my seat.
Best explaination of the Riemann hypothesis. Simple and straight forward.
Thank you!
@@discovermaths I solved the riemann hypothesis. There are 7 lines of proof. And I know the true fibonacci ratio. It's on my channel
Good presentation, and fascinating topic. This $1Million prize though has fairly declined in value since first instituted - could they perhaps inflation adjust it ? What with the sky-rocketing prices of fuel and energy especially it is not going to go too far !
Dear Professor. Is there perhaps a website or book listing the failed evidence of the Riemann-hypothesis ?
I'll get on it as soon as I'm done proving the solvability of the Generalized Navier-Stokes Equation.
I have developed a new theory, I have called Partitions Trigonometric and I have discovered something amazing. I can do X Rays with these equations applied to Z Riemann Equation.
If I said I understood more than the first minute of this, I would be lying.
Thanks for the video. I notice you have a diagram at 5:07 that shows a pole at 1; surprisingly you also have a pole at 0. Unless I'm mistaken, 0 has a value of -1/2.
No, you're completely correct. Most representations of the function that use the right half to reflect onto the left have a hole there, so in this case while the function isn't continuous there the limit of the function for x-->0 tends towards -1/2. Most definitions of the zeta function patch up this hole though, so I'm guessing it was left in for the symmetry of the diagram and mislabelled.
@@glumbortango7182 That is not correct.The ζ-function is continuous at s = 0, where its value is ζ(0) = -½.
@@tommyrjensen
Frequently the functional equation is used for continuation.
ζ(s) = 2ˢ * πˢ⁻¹ * sin(πs/2) * Γ(1- s) * ζ(1- s)
Plugging in zero for s gives 2ˢ = 1, πˢ⁻¹ = 1/π, sin(πs/2) = 0, Γ(1- s) = 1, ζ(1- s) = ∞
Now sin(πs/2) ≈ πs/2 for small values of s, and ζ(1- s) ≈ 1/-s due to the residue 1 of the zeta-Funktion at s=1. So we have
ζ(0) = 1 * 1/π * π/2 * lim[s->0] (s * 1/-s) = -1/2.
So the pole of ζ(s) at s=1 and the zero of sin(πs/2) cancel each other perfectly. The pole at s=1 might tempt someone to believe that there is a discontinuity at s=0 but there is none.
Surely a disproof would be more interesting as so much work "assumes" the Riemann Hypothesis
Its a way to get a me measurement to calculated a place in space
"'Sblood, there is something in this more than natural, if philosophy could find it out."
Where can one look at the evidence of the statements of 1915, 1989 and 1995? Could you please write an answer-Thank you.
So underrated
Great video. Thanks!
What is the smallest imaginary number on the 1/2 real number strip ever found?
14.134725: z=1/2 + 14.134725 i is the "smallest" non-trivial zero of the zeta function
Here are the first 6 zeros (with Re = 1/2, numerically found many years ago)
1 1/2+14.134725 i
(* "smallest")
2 1/2+21.022040 i
3 1/2+25.010858 i
4 1/2+30.424876 i
5 1/2+32.935062 i
6 1/2+37.586178 i
One can easily show for all non-trivial zeros (in { 00 (real)
The functional equation of the zeta function shows a symmetry of the zeta function at the axis Re(z) = 1/2
=> If z is a zero, then 1-z is also a zero. together with the above mentioned (conjugated complex) we have:
Let be z=1/2+delta + it with zeta(z)= 0. Then there are always 4 different "related" zeros forming a rectangle (z1,z2,z3,z4) with
z1= 1/2+ delta + it
z2= 1/2+ delta - it
z3= 1/2 - delta - it
z4= 1/2 - delta +it
zeta(z1)=zeta(z2)=zeta(z3)=zeta(z4) = 0
If the RH is true, then this delta is always equal to 0 (delta=0, no pure rectangle)
@@scp3178 wow. Thank you.
It is moving that's a stranger dedicate his time for me.... Thanks again 😍
How to find next primes?
2x=3,x=1.5
2x*3x =5,x=5/6x
2x*3x*5x =7,x=7/30x
2x*3x*5x7x=11,x=11/210x
If these x value have a rule, we can find next primes?
For example when 2 3 5 7 11 13 17 19 primes what is next prime?
What if I have found a method to exactly calculate all Pseudoprimes chronologicaly and can therefore tell you were all prime numbers are standing up...Would that be the long awaited solution?
Multiplicative primes for mass and addition primes for time. Spin quantum is ratio. And when ratios match quantum entangled. 7 2×7-1 3×7-2. Numbers of Doppler effects.
That word salad is delicious. Have any more?
Great video! Only at 1:20 Riemann picture, he was born in 1826, not 1822. Died when he was 39 in 1866.
You can find it zeta of function (1/2 + 14.13 ... i) = 0
One thing I don’t get as an amateur mathematician, viz., the concept of the first so-many zeros of the function (about 8 minutes in). I presume the implied ordering relates to progression in one direction along the critical line starting at the real axis (say going up), BUT ... if there was an accumulation point of such zeros when approached from above then it wouldn’t make sense to refer to the first n zeros for a sufficiently large n (like it doesn’t make sense to refer to the first n rational numbers bigger than 0). I presume there isn’t and can’t be any such accumulation point of zeros of the Riemann zeta function but is that obvious, or is it a deep result on its own?
Good observation!
By a well known result any analytic function can have only isolated zeros (unless it's the constant function 0).
The Zeta function is analytic (i.e. it can be described locally by a power series). Thus this is not a concern.
Proof sketch:
Suppose f is analytic and has an accumulation of zeros at z=0 (without loss of generality, otherwise we just apply the argument to a translation of f). By analyticity we can expand
f(z) = c0 + c1z + c2z² + ...
in an open neighborhood of 0.
By taking limits along a sequence of zeros we immediately see that c0 = f(0) = 0. It then follows that f(z)/z is also analytic.
f(z)/z = c1 + c2z + c3z² + ...
We can apply the exact same argument again to show that c1 = 0. And continuing inductively we in fact get all coefficients equal to 0. You can convince yourself that this means f is identically 0.
there is a very simple solution to this issue and a way to predict all prime numbers, prime numbers in sequence and all twin prime numbers also, it also accounts for the gaps between primes to, would like to chat more about this if anyone wishes to see some proof of this concept but so far it can count the primes in only the prime sequence e.g: 30000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000223 is a direct result of 223 and also there are 97 0's in the above and 97 is the 25th prime the opposite to that wave function is 223 or more so 3 being the opposite to 7 were 5 is the mid point or decimal or 0 so 22 = 0 0 .5 variation in that wave function so also this long number tells me that 31223 is also prime but this can be done in sequence also so from 31223 (3364th prime) the next prime is 31231 (3365th prime) the next would be 31237 (3366th prime) the next is 31247 (3367th prime) next = 31249 (3368th prime) etc counting in sequence with the right base functions, there are wave functions within the whole number as a single wave function, anyways if anyone is open to talking about how this is achieved then let me know because i'm eager to work with someone that wants to figure this math problem out as much as I do.
ζ/12 is Riemann hypothesis zeta twelve.thanks.
Do you have a podcast?
Can you review to Beimar Lopez, he's a Bolivian and he found a new formula to understand and solve more fast.
I've seen several claims from people that supposedly solved it. There are many here in UA-cam, what's your take on that? Is there a possibility that the actual level of knowledge in mathematics prevents us from either confirming or rejecting certain solutions?
As you'll know, most claims by people who aren't professional mathematicians of having solved long-standing problems don't stand up to close scrutiny. Some problems - and the Riemann hypothesis may be one of them - probably require further advances in maths to make them tractable. On the other hand, as we saw with Fermat's Last Theorem, unexpected solutions can suddenly come to light.
@@discovermaths Thank you!
The are just attention seekers who don't understand the math and don't provide any proof.
Please one और video lacture on riamann hypothisis
The birth year given in the graphic is incorrect. Riemann was born in 1826, not 1822.
Good Catch!
Sir, please analyze PRMO question paper and please make a series to help us prepare for it. (PRMO - Pre regional Mathematic Olympiad)
7:07 - the 1915 & Hardy looks a little out of context here :)
I solved Riemann's hypothesis that every zero does end up on the line and I have proof to show you in a way for you to check and see if every zero ends up on a critical line
TRUE
Saddened to hear that Mr Dyson of Dyson Sphere fame (and Im sure other ideas) has passed not too long ago. We need more great minds like him..
Indeed. He was one of the great intellects of the 20th century.
good
Nice ;D
does that mean each non trivial zero corresponds to a prime number?
-_- No words
i can proof riemann hypothesis. where do i began?
Welcome
The explanation is beautiful and clear, but I have a lot of questions about the topic. Can I send you a private message to answer them....
You are my only hope
The solution is quadrature of the circle
Sir can you explain that function theory and Function we want to understand that Function and want to reached on it
I know the answer and i can make any number less than one a critical line
Prove by induction.
Does anyone think Leonhard Euler could have solved the Riemann Hypothesis?
Probably not, because Euler had not enough complex analysis skills.
He worked on zeta(x in R) = Sum (1/n^x) (n=1.. inf) with x R (real numbers), specially gave very smart proofs of zeta(2k) = constant(k) * pi^(2k) (constant(k) is rational, k natural number, with real analysis tools) (solution of Basel Problem: zeta(2) = pi^2/6)
We therefore know: zeta(2k) is irrational, and even more: zeta(2k) is transcendental (bc. pi=> pi^m, is!)
The analytic continuation of the zeta function to the whole complex plain (without the simple pole at s=1) and the residue theorem (complex analysis, Euler didn't know) gives an expression related to the "weighted prime number counting" function, where the non-trival zeroes of the zeta function appear. That's the key point of the relation of the "Riemann Hypothesis" and the "prime number theory" (best possible approximation of the "number of primes
I’ve figured it out, wrote a paper on it and have no idea where to submit my paper.
Oh, I get it. So basically the Riemann hypothesis is a mathematical hypothesis created by a mathematician named Bernhard Riemann. According to what your saying. He created it the 1800s. Roughly 200 years later it's still unsolved. There's a million dollar prize for the one who solves it. The equation dealt with prime numbers and complex analysis. Along with complex functions. Thank you for the video. I hope the Riemann hypothesis gets solved soon.
You misspelled Riemann's date of birth.
الترجمه
It was solved
KRISHNA.
I solved the RH. But there's not enough space in the comment section for me to write it.
Why not to tell everyone
I solved it and it's on my channel.
Nice Fermat reference
Interesting stories.
By a bit of a random path around math-physics and a lot of Feynman type guesses, the congruence of approaches, "top down" math-experiment discrimination of proof/disproof "wave-package" superimposed on a "bottom up" reciprocal physics measurement analysis, is not so surprising as a comparison of reciprocals in the naturally occurring e-Pi-i interference, temporal positioning resonance Principle, derived format.
But this is Amateur Observation, about as relevant as reading about someone else's hard work.
Lo siento pero parece que un boliviano Beimar López encontró una fórmula para dicho problema del milenio.
Saludos 🖖 desde Bolivia 🇧🇴.
the greatest experts in this area must evaluate the work rigorously, nothing is said yet, we must be responsible
all is prime numbers.
P^all
Huh?
The 1000000001st zero sits on 0.4
1000000021 this would be the next prime for 1000000001, not sure if that helps but that's a result of the prime 127 remove the 0's for example the next prime is 127 and the opposite to that 7 is 3 were 5 is the mid point of that wave function, again not sure if that helps with anything
Amazing , I solved in 2010 and mathematicians are using it to come up with alternative explanations and their new conjectures < Peter Sarnak , Barry Mazur , Michael Atiyah . Anyway all zeros have real part Zero and roots are in form of S= i (pi+- 2kpi )/lnp^2^n . Pere Sarnak had it because I send it to Annals of Mathematics and they had my manuscript for 9 months !!!
Send it in and get millions dollars, bruh
Indian person has solved it, what's your opinion on his solution??? Do you agree with him ,if not then why???
P.S.P. in my humble opinion (Professional Science Popularization),
however the Riemann-zeta function has only one pole at 1,
so there is no pole at 0 where its value is -1/2.
Tibor Banjai: (-1/2) is equal to the square root of negative 1 which is i. The entire critical line is equal to i which is equal to zero.
@@williejohnson5172, good joke!
So, i is equal to zero, and i is square root of negativ 1, so the negative 1 must be equal to zero, because only zero is the number of which square root is zero... Good to know that (I'm joking as well). :)
@@vocnus : Take a deep breath, and think a bit.
1. I hope we both agree that Zeta(0)=(-1/2).
2. Now in the unit circle starting at zero degrees it takes two 90 degrees rotations of i to generate -1. Therefore i=-1/2.
3. In the unit circle the diameter of the unit circle along the entire y axis is 2i. The midpoint of that diameter is i which is exactly the origin of the circle which is zero. Therefore i=0.
4. Therefore i is equivalent to negative one half which is generated by Zeta(0).
5. You cannot refute these first 4 steps mathematically. Period.
6. Try logic instead of sarcasm. It works so much better when solving math problems.
@@williejohnson5172, I'm too low to this to catch up with these squared circles... though maybe I'm reaching 1/2 i high already, so I totally agree with myself that I need just once more this amount to grow to reach the fullness of that one crazy big I which You talking about my friend. Bytheway, if I will reach fully full I, will I forget the real numbers at all... a bit confused about that, because I fear no one would take some newly crypted i dollars at shop when I want to buy something... They are so small, aren't they, don't want to grow to higher math and to hyper truths of nature... LoOoL
@@vocnus : Yep. Just as I thought. You have a good day.
F
I watched the entire video and I only have one question:
what?
This problem was solved recently by indian mathematician Dr Kumar Eswaran.
Not fack, it is open to challenge world top mathamacian to debate with him.
Last words, only Bharat know very well "Ghanit"
He _claimed_ that he proved it, and so did some nincompoop Indian journalists. Nevertheless, his "proof" was never accepted by the international experts, and the prize from the Clay Mathematics Institute has not been awarded to him.
I = √-1=-1^1/2×4×1/4=-1^2×1/4=-(-1^1/2)^1/4=1^1/4="+1" or "-1". Hence there is no imaginary number
Is the hypothesis maybe unprovable? Has its decidability been established? Has anyone proved that it is not unprovable?
Nobody knows. Some years ago, many people conjectured that the Fermat conjecture might be _unprovable and unrefutable._ That was before Andrew Wiles found his sensational proof in 1993, unfortunately removing all what had been mystical about it.
There's an interesting idea that this may fall into the Godel cracks in provable theorems. (That is, there are true statements that cannot be proven.) If THAT can be proven then we would know that Riemann is true, but we can't prove it directly.
Problem solved by hyderabad India this is so easy😎 problem
😂
@@shivabollam4081 🤣
SADNESS OF
MATH YET TO
PROVE IT'S ALL
IN RETURNING OF
JESUS
Hi This is Abhijeet Deshpande and....
This is how to understand the theorem....
Points:
1.) From 1 to 100, calculate the number of odds and evens
2.) Now for every single of the odd and even numbers, measure and write down the number of steps, for both to go to the number 4.
3.) For both the odd and even numbers, calculate individually as below,
a.) Add the number of steps to get a total of both odd and even
b.) Get a total of odd / even numbers by addition
i.e.
a.) How many numbers are odd and even
b.) And what the the sum total of odd and even by addition
c.) What is the sum total of odd and even by division
d.) What is the sum total of odd and even numbers by substraction
4.) Divide the number of steps with the number of odd / even numbers wiithin 1 to 100
5.) Now upon fiding the value of the division of both odd and even numbers,
Use the above results of calculations to calculate with the results to determine the base structure or the point of average c
divisions or calculations, where both the calculations of odd and even align,
And Voila, you have a symmetry of calculative set of equations that would determine the results of any similar supposedly unsolvable equations.
These equative calculations of mine can also solve the problems of Rieman hypothesis of prime numbers as well.
As such I am eligible to win the seed of Clay Institute for of and towards the same.
3x+1, Rieman Hypothesis
© Abhijeet Deshpande, 2021
30001500000000000000590009500000000000093 just noticed also this is a sequence of Pi π (also a prime number)
It is solved by telugu man ..andhra in india ..
The solution to the Riemann Hypothesis will almost certainly be found by a lone genius, like Andrew Wiles (Fermat's Last Theorem), working in secret, bringing together several seemingly unrelated branches of math. That's the beauty of it.
I agree. and should I say Good luck ?
This problem solved by
Doctor kumar ishwaran
From
India
I believe I have solved it. And yes the proof is as beautiful and mystifying as the problem.
Send your proposed solution to the Clay Mathematics Institute. There's a million-dollar prize on offer! www.claymath.org/millennium-problems/riemann-hypothesis
@@discovermaths the CLAY said not to send proposed solutions to them, but rather publish it in a renowned mathematical paper. I know I it has to be formated a certain way, So I have to get familiar with that process first.
@@2piee I haven't sent yet, the proof itself is just a by-product of another research I'm working on. Hopefully I'll get it finished soon.
I found a really wonderful proof already 30 years ago, when I was sitting in an inn. I wanted to write it down immediately. Unfortunately this beer mat was too small.
It was solved by a telugu state professor, who live is Hyderabad, india and he has been awarded 7crores by Cambridge
It is solved by an indian 4 days ago
I might be wrong, but this could work: ua-cam.com/video/PvUrbpsXZLU/v-deo.html
P.S. A piece of advice: make video 1.5 faster, I speak very slowly)
I like the show but there were many too commercials so I turned it off and unsubscribed.
This is proved
I solved it already. -2x+iy+ai+c=0 Quadratic function of Decimal will never give you the imaginary integral Value of ZERO(0). A summation of something can mathematically NEVER Equal (=) equal... Trust me I failed Alegebra 1 &2...25 zeros will lower the average but the 25th primordial number in the first 100 of the Number Matrix is 97... So I technically got an A+ not a 37% ( another prime number) but we live in a Matriarchy full of bitchness and emotions. So fuck em
Nonsense with a side of misogyny. Lovely.
The answer to the Riemann Hypothesis is Infinity.
Infinity times infinity equals infinity to the power of infinity.
Infinity squared equals infinity to the power of infinity.
If 2 is a prime then so is infinity.
You are all welcome.