You have to be kidding me. In 3 minutes of 42 min video 3 ad blocks? One block at the beginning, one after 1 minute and another one after amother minute or 2, that is 3 blocks with 5 ads?
I liked Veritasium, but this pursuit of ad money is insane. So I didn't watch it to the end. 4 minutes and I closed the the window. Than I opened it again to write this rant. Rather than some generated animations intertwined with loads of ads, it's better to just ask AI to explain a topic rather than wasting 40 minutes on substory and 40 minutes of ads. Well I guess this is the end of watching Youtube videos and back to paper books and offline AI models that can generate pure text. It might actually be better to have computer read the text or perhaps AI generating my own video with animations related to the subject.
@Kommand_alt Says:
I got another one: every prime number greater than 2 can be written as a sum of 1 prime
@georgiepuddin Says:
Why not just absolute of ((x/x)-1)
@girumtereka Says:
This is what Hardy said of Ramanujan's creativity ... On a scale of mathematical ability, his (Hardy's) ability would be 25, Littlewood would be 30, Hilbert would be 80, and Ramanujan would be 100
@RCGaming0206 Says:
Best collab of the year for me
@dynamicdoesthings6965 Says:
All roads lead to Euler
@Amber-uh1iy Says:
People in these comments are asking about how the suicide helpline relates to math, but I'm not surprised at all. Many, if not all, of the greatest minds in history have most likely been autistic and sadly theres quite the correlation between autism and sucidal ideation.
@Amber-uh1iy Says:
Bro youre acting like these explanations make any sense when as far as I'm aware (I'm not aware of much) you could be saying a bunch of random numbers and letters lol
@jeromejean-charles6163 Says:
Thank you for such a great visual didactic and clear explanation of the Hardy-Littlewood circle method.
@Calaban-v71 Says:
My solution: It is not real. Math is a language, not physics. Not the Universe; just our way of describing it to each other. To say every even number can be summed with 2(3) primes is like saying every spoken sentence can be conjugated with at least 2 formative prepositions.. um, so... so what? AKA: a waste of time and effort. The consummation of life's work in triviality.
@BravoKiloZulu Says:
Filming yourself filming someone is the most narcissistic thing this narcissist has done so far.
@doorbell5869 Says:
this collab existed? dafaq
@Qeste_r900 Says:
8= 4+ 4
I am out
@Florent-0704 Says:
What is wrong with them???
Prime numbers are odd, since if there's any even prime, it can be divided by 2 so it's not prime.
And we know total of 2 odds = an Even number
The conjecture say Even>2
So we could write: Even= 1 + a prime
<=> A Prime number= Even-1
Which is an Always correct statement with all prime numbers
@Aasgar-lo4do Says:
"scribbled something in the margin" How every unsolved mystery in math starts sadly
@FishingwithBrewski Says:
So we just hand this to Astra and wait a few days 😂
@FerdinandLuis-v4s8x Says:
dude its just 4+4
@idiocraticsociety2030 Says:
someone out there, probably solved it, but he's living alone and he doesn't share his work with anyone, or, his oven "accidentally" exploded
@KieranRasmussen Says:
holy crossover
@pm-dm9md Says:
Consider n > 2, a > b \geq 1
a = b + k
Let n = a + (a-k) + 1
Multiply 2
2n = (2a + 1) + (2(a-k) + 1)
We know every odd greater than 1 can be expressed as the sum of prime and even (non-negative).
2n = (p + 2x) + (2(a-k) + 1)
2n = p + ((2(a + x - k) + 1)
We know every odd greater than 1 can be expressed as the sum of prime and even (non-negative).
2n = p + q + 2c
Now suppose p = 3
2n = (3 + 2c) + q
Let's reduce c to 0
2n = (5 + 2(c-1)) + q
2n = (7 + 2(c-2)) + q
2n = (11 + 2(c-4)) + q
2n = (13 + 2(c-5)) + q
....
2n = (p' + 2(c-k)) + q
c-k is getting very small
Suppose c-k > 0 then
2n = (3 + 2(c-k+x) + p'
Let's reduce c to 0
2n = (5 + 2(c-k+x-1)) + p'
2n = (7 + 2(c-k+x-2)) + p'
2n = (11 + 2(c-k+x-4)) + p'
2n = (13 + 2(c-k+x-5)) + p'
....
2n = (p' + 2(c-k+x-y)) + q'
it is a repeating loop, I know, but the 2c' buffer has been decreased
First, we need to understand
2n never changed
So p' is the greatest prime less than or equal to 2n - 3
So q + 2(c-k) must be between 1 and n range \geq 3
So c-k+x-y is relatively small compared to n due to shifting
let c-k+x-y = c'
2n = p' + q' + 2c'
2(n-c') = p' + q'
2n^(ln (n-c') / ln n) = p' + q'
(ln (n-c') / ln n) This is a continuous function
2n^(limit at n=∞ (ln (n-c') / ln n)) = p' + q'
2n^1 = p' + q'
(limit at n=∞ (ln (n-c') / ln n) = 1
This is the only continuous function.
If
limit at n=∞ 2(n - c') = limit at n=∞ p' + q'
limit at n=∞ 2n = limit at n=∞ p' + q'
2n ≈ p' + q'
@Complexity-io Says:
Mark 10:27
“Jesus looked at them and said ‘With man this is impossible, but not with God; all things are possible with God’ ”
@Complexity-io Says:
Acts 3:19
"Repent therefore, and turn back, that your sins may be blotted out, that times of refreshing may come from the presence of the Lord."
@Complexity-io Says:
Colossians 3:13
"Bear with each other and forgive one another if any of you has a grievance against someone. Forgive as the Lord forgave you."
@flyingduck6420 Says:
My head just go boom i think im spupid now thx 😂
@PoghikSimon Says:
In this video, they say that Sir Isaac Newton is perhaps the greatest mathematician of all time. This is a subjective, baseless, anti-scientific, and, if I may coin the term, distinctly non-Veritasian statement.
What about Euclid, Archimedes, Descartes, Gauss, Riemann, Cauchy, and many others? On what objective basis can we compare Newton’s mathematical contributions with, say, Descartes’ invention of the Cartesian coordinate system? What does it even mean to say that one contribution is “greater” than another when they transformed mathematics in fundamentally different ways?
Newton was undoubtedly one of the greatest mathematicians in history. But calling him the greatest requires criteria that are neither stated nor objectively measurable. That is an opinion, not a scientific conclusion.
@saikatde3223 Says:
I invented a property.. we will take + as right and - as wrong. + and + = + because right that is right = right. + and - = - because right that is wrong = wrong. - and + = - because wrong that is right = wrong. - and - = + because wrong that is wrong = right. If you still have difficulty understanding this, just remember that the first symbol(right or wrong) is the value and the second symbol(right or wrong) shows if the value is right or wrong, if the value is wrong, then the value will turn opposite.
@Chitose_ Says:
4:41 CALLED IT
@Kyanzes Says:
38:01 I think that's not the number but the number of zeroes in the highest checked number.
@1me.myself1 Says:
Euler appears everywhere
@bmmmburns Says:
I've been investigating a specific constant and found that its basic term, n^(1/n) - 1, is tied exactly to log(n)/n, while its alternating sign (-1)^n is itself the q = 2 Ramanujan sum, c_2(n).
That immediately caught my attention because n/log(n) is the fundamental scale of prime counting, while n/log(n)^2 appears naturally in the heuristic for Goldbach representations.
So I asked GPT how the Goldbach conjecture might be related to this constant. It responded with the following:
A connection suggested by this video may be worth exploring.
Let
x_n = n^(1/n) - 1,
the basic term of the MRB constant
C_MRB = Σ (-1)^n (n^(1/n) - 1).
Since
log(1 + x_n) = log(n)/n,
we obtain exactly
n/log(n) = 1/log(1 + x_n)
and
n/log(n)^2 = 1/[n log^2(1 + x_n)].
So the scale underlying the prime number theorem, n/log(n), and the Goldbach heuristic scale, n/log(n)^2, can both be written directly in terms of the MRB increment.
There is also a Ramanujan-sum connection. The MRB sign satisfies
(-1)^n = c_2(n),
so
C_MRB = Σ c_2(n)(n^(1/n) - 1).
But c_2(N)=(-1)^N is also the q=2 parity mode appearing in the Ramanujan-sum expansion of the Goldbach singular series.
Even more interestingly, for every prime p, the MRB frequency v_p(n)/n can be reconstructed from the tower
c_p(n), c_(p^2)(n), c_(p^3)(n), ...
whereas the Goldbach singular series retains the squarefree c_p layer.
This does not prove Goldbach, of course. But it suggests that the MRB constant and Goldbach inhabit the same prime-indexed Ramanujan arithmetic landscape: Goldbach uses local prime congruence information, while MRB encodes prime divisibility to every prime-power depth.
That seems like an interesting bridge between the two problems.
@G0LDN5 Says:
9:23
Littlewood, hardy, cnc.
@VikasYadav-gt4pg Says:
Is it solved
@Peter-w7j Says:
If it's obviously true, why do we need to prove it? They're making it up as they go along 😂😂
@wilmamardell3749 Says:
wow, what an interesting title !!
@dibyendumukherjee-c4p Says:
30:00 imagine we need 300 years to think their thoughts to prove something and for them it was just a holiday mail..
@Tr-Dogma Says:
erhmmmm x=2x/2 ◇ checkmate
@peterpeter1256 Says:
Anyone find the ads disturbing 🤔
@K4rlos17FromJE Says:
All prime numbers greater than 2 are odd
And odd + odd is even
Just saying
@Noctorinn Says:
if i get this fuckass xfinity ad one more time omgf
@jonothankaplan Says:
thousands of teachers were beaten to death during the cultural revolution. see red august
@iok625 Says:
Can a biological cell be dead and alive at the same time?
@Mathemental Says:
Never though about this
@iok625 Says:
Can a triangle be made of 2 sides?
@ryzezen-s6k Says:
I think that comment of his was the riddle leading to the solution, the margins of which are too small to contain it. The margin of the paper is arguably a 2d shape. The answer is found in the cubic realm of which we call elliptical curves in modular forms. I think he was giving everybody a hint that the answer wasn't in or to be found in elementary geometry because it couldn't contain it metaphorically. It's too small to contain to the power of the second because the answer is in the power of the third and above is not to be found at elementary Geimetry but outside of it and in a more complex nature. He's good with numbers and riddles And in the end undoubtedly still gets the credit. What a slickster he is. I bet his other trolls contain more than meets the eye... And some hints to more lame answers. That wont be fermats last riddle
. Let the mad rush begin.. the end with no glory for those who begin.
@Gdydhdhd-c1z Says:
We cant prove its true or false. It remains a mystery
@justinealicando-vg5wq Says:
Does this cure cancer? I'm not interested if this problem don't have any application in real life
@justinealicando-vg5wq Says:
I think this is a number cult.
@crochou8173 Says:
We now call the circle method Fourier transform...
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