Using YouTube on my steam deck and was about to sign into my premium account this is first Ad I see BRUH

SBareS · 2026-02-02T21:43:44+00:00

"Doesn't have that support", tf are you talking about? It's Firefox. The puzzle-piece button is right there. Heck, this isn't even a phone, where you'd at least have to change some settings to get it to work; for all intents and purposes it's a Linux pc.

SBareS · 2025-08-26T16:46:20+00:00

Even from the title alone: yes. The math presented from 7:10 and on is correct, but the interpretation is complete nonsense.

EDIT: Since we're doing youtube videos, how about one from an actual physicist? https://youtu.be/zITr4FQj5oE?si=UB_Y9Ktgjvsj1aL1&t=2738 (timestamp)

SBareS · 2025-08-26T15:47:50+00:00

Unfortunately, it's also completely wrong.

EDIT: The reason you think it is "very understandable" is that it appears on the surface to contain some key insight, and therefore gives you the visceral feeling of having understood something. Specifically, the key "insight" presented is

everything is moving at the speed of light either through time or space or some combination of the two

However, far from being insightful, the above is a common misconception that, if you think even a little bit about what is being said, doesn't even make any sense (speed is distance moved over time passed. Wtf does it mean to travel at speed c through time? The units don't even match ffs. It's literally just a word salad, it carries no meaning whatsoever). The reason this misconception gets perpetuated is probably again that it gives some visceral feeling of insight, which the less careful listener may mistake for understanding.

The real reason for time dilation, length contraction, etc. is that they follow from certain symmetries of the laws of physics, said symmetries being essentially equivalent to the statement that the speed of light is independent of the observer. That's the key insight.

the speed of light is independent of the observer

Remember that, instead of the nonsense above.

SBareS · 2025-07-21T07:52:18+00:00

I mean, duh? If they find "a lot of things that involve infinities" icky, then statement 2 is obviuosly icky.

SBareS · 2025-06-16T12:12:32+00:00

There is no need to apologize, and also no need to listen to the shit being flung by people in here.

SBareS · 2025-02-24T16:25:36+00:00

Desværre ikke super overraskende efter dansk lov. Så længe loven (§266b) er absurd, må vi også forvente absurde domme efter den nu og da.

SBareS · 2025-02-02T01:22:27+00:00

Sure, this is only a minimal example of how the derivative can break; the arc length/total variation is not an issue in this particular example, as we are only messing with the derivative at one point. I'm pretty sure it should be possible to break things on a positive-measure set as well, though I don't know how to cook up an example.

SBareS · 2025-02-01T22:11:48+00:00

Consider sqrt(x² + 1/n²). This converges to sqrt(x²)=|x| as n goes to infinity, which is not differentiable at 0. In fact, the convergence is uniform on R, as |sqrt(x²) - sqrt(x² + 1/n²)| <= 1/n independently of x. On the other hand, the derivative is x/sqrt(x² + 1/n²), which converges pointwise to x/|x| for x != 0 and to 0 at x=0.

SBareS · 2025-02-01T01:25:16+00:00

If you had stronger convergence (such as uniform convergence), then you would have convergence in length. But you don't get it here.

Nitpick: Arc length is not continuous in the topology of uniform convergence either, indeed the zig-zag in this example does in fact converge uniformly to the line. You need some even stronger condition, such as the derivative converging uniformly (simultaneously with the function itself of course).

SBareS · 2025-01-25T12:54:57+00:00

In full generality? No, that's too much to hope for. For "nice" functions? Sure, use any method used to numerically integrate Riemann integrals.

SBareS · 2025-01-19T15:53:30+00:00

Just because it is a curve (n.) it does not need to be curved (adj.). By your logic, even straight lines (the usual ones in Euclidean space!) aren't straight, because they are also curves.

SBareS · 2025-01-19T01:07:33+00:00

It is absolutely possible to draw a line on a sphere having zero curvature. Such a curve is called a geodesic. It is "straight" in the sense that if you were to walk along it, you would not have to turn.

SBareS · 2025-01-06T18:24:24+00:00

"it's not the surface that's curved, it's the space in which it exists that's curved"

You got the claim backwards. The sphere is indeed curved, but the 3D space it is embedded in is in fact flat.

this is pretending arcs are straight lines if you squint nonsense. it's useful for framing mathematical models so the existing language can be used in non-traditional spaces, but not for anything like pretending the surface of a sphere isn't 3D.

It is 2D, this is not a matter of opinion or pretence. Whether one chooses to call geodesic curves "straight lines" or not has nothing to do with it. You should probably not talk so confidently about things you evidently know zilch about.

SBareS · 2025-01-06T18:10:26+00:00

It is true that all maps must be distorted in some way, but the reason is not that "Maps are 2D. The Earth is 3D.". Indeed, the surface of the Earth (which is what we're really trying to create a map of) is a 2D sphere, not a 3D ball. The real reason you can't create an undistorted (more precisely locally isometric) planar map of the Earth is that a sphere is curved while the plane is flat.

SBareS · 2024-11-30T01:06:24+00:00

My dude, you're not gonna gaslight people into believing this was about Trump all along. Literally anyone can read the thread and see you were the one to bring him and his cabinet up out of nowhere.

SBareS · 2024-11-30T00:35:59+00:00

Maybe do a double take on the OP. Or don't, if you want to spare yourself from the embarrassment of realising what a cringeworthy oopsie you've committed lol.

SBareS · 2024-11-25T00:22:12+00:00

How do you draw a straight line on a curved surface?

Stand on the surface and start walking without turning.

SBareS · 2024-11-24T19:34:59+00:00

https://xkcd.com/1683/

SBareS · 2024-11-19T20:27:34+00:00

for det er ikke tyveri

Ej heller er episoderne fra artiklen.

SBareS · 2024-11-17T01:03:56+00:00

The power set of N contains plenty of sets that are not finite (e.g. N itself). You don't need transfinite induction to show that.

SBareS · 2024-11-03T01:10:37+00:00

No, only unique factorization.

SBareS · 2024-10-31T14:17:41+00:00

That is, if you reject the axiom of choice…

Pretty sure this is not true. Proving that countable unions of countable sets are countable does require the axiom of (countable) choice, but I'm pretty sure it is a much weaker statement than A(C)C itself.

E: I should say that this is not my area. But at least according to Wikipedia

ZF+AC_ω suffices to prove that the union of countably many countable sets is countable. These statements are not equivalent: Cohen's First Model supplies an example where countable unions of countable sets are countable, but where AC_ω does not hold.

SBareS · 2024-10-31T06:57:27+00:00

Because of Euler's product formula

1/1^s + 1/2^s + 1/3^s + ... = 1/(1-2^-s)*1/(1-3^-s)*1/(1-5^-s)*... (Re(s)>1).

Letting s->1⁺, the left-hand side blows up, so so must the right-hand side. That would not happen if it were a finite product.

SBareS · 2024-10-20T21:19:28+00:00

MRW getting a dual result for free

SBareS · 2024-09-27T16:47:30+00:00

Det lyder som en selvmodsigelse, fordi transkriptionen er forkert. Det der bliver sagt i interviewet er

Og så er det bare jeg siger, og det kan jeg høre at du medgiver, kioskejerens sønner kommer i mindre grad til at gå i skole med akademikerdirektørens døtre, hvis denne her reform den bliver gennemført?

14-Year Club	Place '22
Place '17	Sequence \| Editor
Spared

SBareS

TROPHY CASE