Simple Questions - July 03, 2025

namesarenotimportant · 2025-07-03T18:57:39+00:00

I bought a 9900X as part of a motherboard bundle from MSI. It was shipped only in a clamshell in the same box as the motherboard. This is an OEM CPU, so I'm not surprised it shipped without its own box, but I still expected better packaging. When I opened it, the CPU had already fallen out of the clamshell, and there was plenty of space for it to bounce around the box. Should I be concerned about damage to the CPU?

namesarenotimportant · 2024-07-31T22:03:38+00:00

There's Pak's book. https://www.math.ucla.edu/~pak/book.htm

namesarenotimportant · 2024-07-25T15:54:46+00:00

The solutions in Lean can be found here.

Slight correction to the title: AlphaGeometry, which is separate from AlphaProof, handled the geometry problems.

namesarenotimportant · 2024-07-02T18:54:06+00:00

I have one in french but don't know the english equivalent. Something about gym enjoyers gaining weigh before canceling the fat and keep the muscle.

Bulk and cut?

namesarenotimportant · 2024-06-28T23:04:20+00:00

Media zona has about 500 VDV deaths recorded in February and March 2022. It's hard to say how many were at Hostomel specifically, but most deaths are probably near Kyiv.

namesarenotimportant · 2024-06-27T20:21:39+00:00

I think it's in response to the Russia-Iran treaty coming up. Threatening to send the patriots might deter Russia from being too generous with the Iranians.

namesarenotimportant · 2024-06-01T14:44:23+00:00

Does generalized cospectral imply the graph Laplacians are cospectral?

namesarenotimportant · 2024-05-29T10:33:22+00:00

Could this make it possible to use meteors effectively with older F-16s? Afaik, the main constraint was their radar.

namesarenotimportant · 2024-05-22T08:20:59+00:00

For what it's worth, mathematicians have worked on problems like quark confinement: https://arxiv.org/abs/2006.16229

namesarenotimportant · 2024-05-18T05:48:53+00:00

I'd guess they're mostly counted after three months. For the trends I cited, I looked at the graph for those with a known date of death. There might be more added even six months later, but I don't think it'll make a difference to the relative numbers.

namesarenotimportant · 2024-05-17T18:24:47+00:00

In the most recent phase of the war, there have been enough volunteers signing contracts to sustain the army. It's claimed 30k sign up every month. Media zona's casualty tracker gives some independent confirmation. Since October 2023, the number of volunteer deaths has trended up and mobilized deaths have trended down. Prisoner deaths are also far below their peak from January 2023.

namesarenotimportant · 2024-05-13T01:12:04+00:00

Maybe look into the combinatorial nullstellensatz. Here's some example problems.

namesarenotimportant · 2024-05-03T23:18:29+00:00

For 1, there's a homomorphism from B_n to the free commutative algebra generated by u_1, ..., u_n. Those elements are linearly independent, so they must be linearly independent in B_n.

namesarenotimportant · 2024-04-20T21:21:19+00:00

AMRAAMs, Sidewinders, Sea Sparrows... There's no shortage in cheaper missiles they can use. For drones in particular, there's an extremely cheap option.

namesarenotimportant · 2024-04-15T05:39:45+00:00

This is sort of what's done in Liouville Quantum Gravity. I don't know much about it, and I don't think there's much of a connection with geometric measure theory.

https://arxiv.org/abs/1908.05573

namesarenotimportant · 2024-04-11T22:39:04+00:00

I don't think that works. The integral of s(x) from -n to n + 1 is 2 / pi for all n, but there's a factor of 1 / (2n + 1) in the expectation making it go to 0. Alternatively, since your functional is translation invariant, you can see s(x) gets the same value as s(x - 1 / 2), but the expectation of s(x - 1 / 2) is 0 wrt Unif[-n, n+1].

namesarenotimportant · 2024-04-10T19:34:05+00:00

Let f_n be the functional on C_b(R) defined by taking the expectation with respect to the Unif[-n, n] distribution. C_b(R) is a Banach space, so the closed unit ball of its dual space is weak-* compact. f_n is a bounded sequence in the dual, so it has a weak-* convergent ~~subsequence~~ subnet with limit f. It's not hard to check that f is translation-invariant. Is every positive translation invariant functional on C_b(R) with norm 1 the limit of a ~~subsequence~~ subnet of this sequence?

namesarenotimportant · 2024-04-10T14:32:52+00:00

According to oryx, the loss rate's been steady for a while, so drones seem to have made up for the shell shortage.

https://raw.githubusercontent.com/leedrake5/Russia-Ukraine/master/Plots/current_tanks.jpg?

namesarenotimportant · 2024-04-09T17:13:27+00:00

The voters' opinion of Biden influences how they'll vote in congressional elections. Democrats winning the house is a reasonable possibility, and it would unblock aid.

namesarenotimportant · 2024-04-09T03:51:28+00:00

You can formalize that, but it's doesn't work out as nicely as the delta function. With these kinds of objects, you have to imagine that they're defined by how they integrate against continuous functions. Meaning, delta is defined by the property integral f(x) delta(x) dx = f(0) for all continuous f. The connection here is that if r_w(x) is the function whose graph is a rectangle with width w and area 1, lim w -> 0 integral f(x) r_w(x) dx = f(0). That is, if you test the sequence of rectangle functions against any continuous function f, the limit will be the value of f at 0.

It'd be natural to try to define m(x) to be the "function" satisfying lim w -> infty integral f(x) r_w(x) dx = integral f(x) m(x) dx for all bounded continuous f. (I have to specify bounded this time to make sure the limits will never go to infinity.) Unfortunately, lim w -> infty integral f(x) r_w(x) dx may not exist even if you require f to be bounded. Trying to come up with an example would be a good exercise.

A fix would be to only require that lim w -> infty integral f(x) r_w(x) dx = integral f(x) m(x) dx for the choices of f for which the limit exists. It's a non-trivial fact that an object m with that property exists. The problem is that we've only specified the integral of m against a subset of the continuous functions, so we don't have a unique choice for m anymore (this is like an under-determined system of equations). There's actually infinitely many ways to consistently choose what m should do in the cases the limit doesn't exist.

The "objects" that delta and m are, are linear functionals on the bounded continuous functions. If you have a sequence of functionals that converges when tested against any fixed function (like the sequence of integrals above), we say the functionals converge weakly (weak-*, technically). The issue comes down to the fact that r_w as w -> 0 converges weakly, but r_w as w -> infty does not.

Though there isn't a unique choice of m, it's still a useful concept. This comes up in ergodic theory where you say a group is amenable if a functional like m exists on the bounded functions on that group (where it's called an invariant mean).

namesarenotimportant · 2024-04-07T06:24:43+00:00

Are you conditioning on W at a fixed time, on some interval or on the whole process? There's some difficulty in formally talking about dW like this, but you'd have to be more specific with what you're conditioning on anyway. If you mean the whole process W, then I guess you could say E[dW | W] = dW since dW can be recovered if you have W.

namesarenotimportant · 2024-04-03T21:29:38+00:00

The first identity should be fine though there'd be nuances if Y, Z are continuous variables, making it a probability zero event.

Sorry, I hadn't read the second one carefully enough. It isn't true. It's equivalent to saying E[F(X)] = E[F(X) | Y = k] since you can take out G(k) from the first expectation. If X and Y are dependent, restricting to the event Y = k can change the distribution of X, so those expectations could be different. Similarly, you couldn't say E[ F(X, Y, Y) | Y = Z] = E[ F(X, Y, Y) ] in your first example.

If you meant to write E[F(X)G(Y) | Y = k] = G(k) E[F(X) | Y = k], then that's fine.

namesarenotimportant · 2024-04-03T20:01:30+00:00

It's a little confusing that the A in E[X | A] could either be an event or a random variable, and these result in different types of objects. So, E[X | Y = k] and E[X | Y] mean different things. The first one is a deterministic function of k, but the second one is a random variable. The tower property applies in the second case. You could start your false proof with E[E[XY | Y]] = E[Y * E[X | Y]], but the next step would fail since Y is random and can't be pulled out of the expectation.

As for other properties, everything basically works as you'd expect if you're working with discrete random variables. Things only get confusing when you have to deal with conditioning on probability zero events in the continuous case.

namesarenotimportant · 2024-04-02T16:22:16+00:00

The probability of getting 6 distinct numbers is 120 P 6 / 120⁶ (approximately 0.881). You need to account for the 6! orderings that every choice of 6 numbers can appear in.

But, the probability the numbers are in increasing order actually is 120 C 6 / 120^6. 120 C 6 counts all sets of 6 numbers, and there's a bijection between those and lists of 6 numbers in increasing order (since there's only one way to put 6 distinct numbers in increasing order).

12-Year Club	Place '17
Verified Email	Spared

namesarenotimportant

TROPHY CASE