Exploring the Riemann Zeta Function and the Riemann Hypothesis

BC Calculus Project: Ryan Smolowyk and Samuel Stringer

In this video, we explore the Zeta Function. This function has many unique properties and is largely connected to the prime numbers.

[vaclavpeleska]

Gauss discovered that 1+2+3+...+n=(n+1)*n/2 when he was 8! = 40320 years old

[ephemera]

Zeta (-1) does not "add up to" -1/12. It has an analytical relationship to -1/12. It adds up to ♾️. For the Riemann Zeta function, when dealing with n<=0 the formula is

[lawless]

Euler enter the room
Solve the basel problem
Leaves.
CHADEST MATHEMATICIAN

[RSLT]

Would you have any advice on my zeta function? Transcendental Zeta Function

[Anteater]

10:00 that’s not what zeta(-1) is. Zeta is defined only by analytic continuation for Re(s)<1

[boomzmm]

Sir why did you use cos sine tretra hybide in this process we could have simply used zeta lite time kind equation towards solving the graph. And secondly can you please research and make a video on brother half kind equation the one introduce in 1917 needed to find more about it

[madvoice]

Love you BRO, your vedio helped me to complete my research

[juliehorinkovamiseon]

May I ask why Euler started with sin x?

[gricchastain]

Listen at 6:58..
I see the 60.7927..
No I see 6079 is prime.
And what. Do I do.
I devide by 2 3039 and +2 is a prime subtract 2 to get 3039.Devide by 3 it's a prime. .And devide by 2 again 506.5.... multiply by 36... Add 1. To see 18, 207.prime.

[dirac]

This video is good and you should feel good

[Stephen-Harding]

Right off the bat, at approx. 0:36, I was confused. You state that the function was originally only defined for when (s) had a real part greater than 1. It seems to me that the Zeta function is perfectly well defined at (s)=1, where the function becomes 1 plus the reciprocal of the counting numbers. At (s)=0, which is certainly a real part not greater than 1, then the function becomes the infinite addition of 1+1. Hmmm, where am I going wrong?

[qbhmqnf]

to get an easyer expression of the Riemann-Zetafunction try Zeta (x) = 0.5 + i(e^(x+pi*pi-6)). If you put the n (element of natural numbers) as start of its Indefinite integral and n+1 as end you will get the square of the n-th prime number. This is a proof of the Riemann hypothesis. If you want another proof draw a circle with any radius >1 from the origin (0/0) then draw a complex conjunction of this [p2(0/1)] the intersections are the ony viable points and the real part is always 0.5 -> qed

[zzeuqdhd]

Tous les explication ce ni que superficielle la question consiste de la formule général de la série ' une idée F(s)=la somme 1/n^s=la somme 1/n^a ×e^-ibln(n)=la somme 1/n^a(cosbln(n) - sinbln(n)) =

[imethanOW]

Ryan why did your Calc BC final project end up getting thousands of views?

[meenadeshpande]

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

[prajaktadange]

Riemann zeta function is convergent or divergent

[峰雄-vk]

🍎Hyper charge.
The electron charge -1 is equal to -1 hypercharge.
So -1=12ζ(-1). It's 12 dimensions.

Quark and lepton have a total of 12.

The hypercharge -1/2=-6/12=6ζ(-1) is 6th dimension.
The hypercharge -1/4=3ζ(-1) is 3D.
Color charge is 3D.
The hypercharge -1/3=4ζ(-1) is 4th dimension.
The hypercharge -5/12=5ζ(-1) is 5th dimension.
Or -5/6 = 10 dimensions of 10ζ(-1).

For hypercharge -70/12,
-70/12=
-35/6=
-5×7/6=
70ζ(-1).
It's 70 dimensions.

The hypercharge -2/3=8ζ(-1) is 8th dimension.
It's 8 gluons.
The hypercharge -1/2=6ζ(-1) is 6th dimension.
It's a quark lepton.
The hypercharge -3/4=9ζ(-1) is 9th dimension.
The hypercharge -5/6=10ζ(-1) is 10th dimension.
The hypercharge -11/12=11ζ(-1) is 11 dimensions. Einstein thinks
the 4th dimensional time space is 12 dimensional.
The hypercharge -2=24ζ(-1) is 24th dimension. Graviton.
Einstein's 4th dimensional time space is
12 dimensional.
And Leech Λ24 Lattice.
It's in 24th dimension.
It's on the 70th dimension.
The prince must kiss
196, 560 times to revive
the captive Snow White of Leech Λ24 Lattice.
The black hole will settle to the Euler constant γ.

The Arc of the Blue Sky was painted with π,
and
the rainbow color was painted with 3 colors of RGB,
and
8 Gluons were Divided into
The Heavens and Earth of
The Universe.
Everything is Due to The Grace of The Graviton.

[brendawilliams]

It’s a problem that you have to read up on a lot about.

[artificialresearching]

P.S. A piece of advice: make video 1.5 faster, I speak very slowly)

[alexanderealley]

You might find this useful.
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.