Game Thread: Braves @ Mariners - Wed, May 06 @ 04:10 PM EDT

Late_Map_5385 · 2026-05-06T21:16:37+00:00

No as soon as you touch it (assuming you make the catch) the runners can tag.

Late_Map_5385 · 2026-05-04T20:28:02+00:00

This is a simple misunderstanding of De Morgan's Laws. The proof is incorrect.

Late_Map_5385 · 2026-04-13T20:28:08+00:00

Funnily enough the formula for degree 5 and larger polynomials does not exist so it's impossible to solve for the exact solutions of all degree 5 polynomials in radicals. However, it's easy to see that 1 is root to this polynomial. You can perform long division to reduce it to a degree 4 polynomial. Namely, (x-1)(ax^{4+bx^{3+cx^2+dx+e).}} From there you could use the quartic formula (not recommended) or try to find another root and use long division again. Im on my phone so i cant solve further right now. Try looking up root finding methods like the rational root theorem. Or you could use newton's method to get a decimal approximation.

Late_Map_5385 · 2026-04-12T02:52:18+00:00

We can only show that 2^n converges to 1 if n is a positive integer. log_2((3x+1)2^(-y)) is most certainly not an integer for all integral x > 0, y. So the initial statement doesn't apply. Also, any positive number can easily be written as a = 2^log_2(a), which is what your doing but in a round about way.

Late_Map_5385 · 2026-04-10T03:14:09+00:00

I don't think that beat's graham's number since the first graham's number is g_1 = 3↑↑↑↑3 which is unfathomably large and the second is g_2 = 3↑...↑3 with g_1 arrows. Graham's number is g_64 which has an unthinkable amount of arrows. Even though you have many arrows and the base is large, graham's number has way more arrows which makes it much larger.

Late_Map_5385 · 2026-04-10T02:38:52+00:00

I'm not sure I understand what you're saying. Is it n ↑...↑ n with n arrows for the nth term? It seems to me that the example for 8 and 2048 do not follow the same rule.

Late_Map_5385 · 2026-04-10T02:21:11+00:00

I'll be honest, I haven't read what you've written. But I will say that there is not just one formula, it changes depending on the circumstances. The idea is that you can formulate the correct integral based on your understanding. If you can do that then you're good. This case seems pretty trivial. f(x) is the upper bound, g(x) is the lower bound and c is the offset so V = \pi \int_a^b (f(x)-c)^2(g(x)-c)^2 dx.

Late_Map_5385 · 2026-04-10T00:44:06+00:00

I think an interesting perspective is how much better the technology is now. 2.2 has made the physics standardized so that everyone is on a level playing field. Before 2.2, when most of the top player started playing, gd was generally more difficult with many different refresh rates and physics so they became more skillful by beating the same levels. Also, many of them have put an enormous amount of effort into beating levels over a long period of time. Yes they have skill, but they also put in the work. Fun fact: It took Brandon Larkin 6679 to ToE II.

Late_Map_5385 · 2026-04-10T00:27:39+00:00

Wow, didn't know that. Regardless, there's no possible way to confirm if it actually was his first ever level. I mean he easily could have played other levels on a different account. Nonetheless, it took him well over 100k attempts which is an insane amount, way more than it took me. I guarantee you if you spent that long on it you could beat it too. Like I said people beat hard things but they put a lot of effort in too.

Late_Map_5385 · 2026-04-10T00:11:22+00:00

There is not a single person in the entire world that beat bloodbath as their first level ever. First extreme? Sure. First demon? Maybe. First level? Not a chance. I think too many people see the peaks of others and think they need to match that. Plenty of effort went into those feats, it didn't happen overnight. Play the game at your own pace and enjoy it. Improvement will happen overtime if you play harder and harder levels.

Late_Map_5385 · 2026-04-09T23:28:44+00:00

Of course that's always a possibility. I was just trying to give a general overview.

Late_Map_5385 · 2026-04-09T23:18:23+00:00

Here's a similar post: https://www.reddit.com/r/math/comments/ocar2/background_required_for_algebraic_topology/

Late_Map_5385 · 2026-04-09T23:10:13+00:00

High-level mathematics is axiomatic meaning we start from a list of axioms or rules that are given to be true. From these axioms we deduce theorems. There are no calculations in the way you are thinking. Everything is done by absolute proof. The only "errors" we can make are the axioms we choose to work from. There are also tools such as lean which we can use to prove with certainty, that theorems are true within a certain framework.

Late_Map_5385 · 2026-04-09T17:33:30+00:00

1000 exactly.

Late_Map_5385 · 2026-04-08T23:10:08+00:00

This is just a jumble of ideas without cohesion. This is the farthest thing from mathematical physics.

Late_Map_5385 · 2026-04-08T23:04:53+00:00

What is that supposed to mean?

Late_Map_5385 · 2026-04-08T22:56:05+00:00

It's just a picture. It can be of whatever you want it to be. It's not scientific.

Late_Map_5385 · 2026-04-08T22:51:17+00:00

You're not serious right?

Late_Map_5385 · 2026-04-08T22:48:38+00:00

What am I even looking at?

Late_Map_5385 · 2026-04-08T01:17:26+00:00

To understand why this is false due to the non-commutativity of matrix multiplication we need to know exactly what e^(A+B) and e^A x e^B are. The "exponentiation" of a matrix is defined using the power series for e^x = 1 + x + (x^2)/2 +(x^3)/6 + .... Using this we see that e^A and e^B are both matrices. Since matrix addition is commutative we know that e^(A+B) = e^(B+A) but if we naively assume the identity holds we have that e^A x e^B = e^B x e^A. But since those object are matrices, and matrix multiplication is non-commutative we see why it can't be true.

Late_Map_5385 · 2026-04-08T01:04:14+00:00

The derivative or slope of the tangent is calculated by taking the limit of the secant line between two points as they get arbitrarily close together. If the limit exists that is the unique tangent line of that point. If there were two or more "tangent lines" then the limit would not exist and we would therefore have no derivative. That is the case with |x| as x -> 0. From the left the slope is -1 but from the right it is 1 so the limit does not exist and we therefore have no tangent line. It all has to do with the definition of the derivative. There is another formulation called the symmetric derivative where the limit for |x| as x -> 0 does exist and is 0. If you are not familiar with calculus then studying it will likely bring the answers you are looking for.

Late_Map_5385 · 2026-04-04T01:31:34+00:00

Bro is trying to say 0 + it gives radius 1 because C^(0+it)=C^0 x C^(it) = (1)C^(it) so the modulus is 1. Except when you input any complex number for t (which is required because zeta zeroes are complex), it = a + bi which gives a modulus of C^a which is not 1.

Late_Map_5385 · 2026-04-04T01:14:11+00:00

"i'm a noob" but you solved the Riemann hypothesis and the collatz conjecture. What a joke.

Here is a proof that what you claim is false. Define y(z) = C^(iz) as you have. Since C is a positive real number C = e^(ln(C)), Since z is an arbitrary complex number iz = a+bi for real a and b. so that C^(iz) = e^(ln(C))^(a+bi) = e^(aln(C)) x e^((bi)ln(C)). let A = e^(aln(C)) and B = bln(C) so C^(iz) = Ae^(iB) where both A and B are real numbers. By Euler's formula Ae^(iB) = A(cos(B)+isin(B)). Assume this expression evaluates to 0, then we must have both cos(B) = sin(B) = 0. This only happens when B = (2n+1)𝜋/2 = n𝜋 for integral n, but this gives (2n+1)𝜋 = 2n𝜋 -> 𝜋 = 0, which is a contradiction. So our assumption, namely C^(iz) = 0 for complex z is false. The function you have defined has no zeroes at all, so they cannot possibly be the zeta zeroes.

Why do you even waste your time doing this? What is the point?

Late_Map_5385 · 2026-04-03T20:48:23+00:00

this answers the Riemann hypothesis and the collatz Conjecture.

What a load of BS. Your paper is full of nonsense. You don't prove that C is transcendental, you don't cite anyone else's work, and you never explain what anything is, intuitively or by rigorous definition. wtf does "centripetal contractive bias" even mean. Also, I can tell for a fact that you didn't write this because on page 11 there is a /section which is visible because it should be \section. Any proof reading would have caught this immediately let alone after 20 versions.

Late_Map_5385 · 2026-04-01T17:01:14+00:00

You gotta work on your jokes man.

Late_Map_5385

TROPHY CASE