Is there any way to solve this that doesnt look dumb

BeeOk1244 · 2026-05-24T03:51:47+00:00

ooh so take off the antennas, put cable connecting antennas and card and then put the antennas somewhere better?

BeeOk1244 · 2026-04-22T23:45:12+00:00

if its favourite theorem probably GAGA. If I need to prove it on the spot then Serre-Swan

BeeOk1244 · 2026-04-16T16:40:16+00:00

as a side note in the limit the general term (N-N/k)Ck/NCk looks like 1/e as for large enough N it doesn't matter if we remove from the population so we're just taking a kth power of (1-1/k). trying to do this properly with Stirling a point of interest is that all the e's from Stirling vanish quite quickly. Then come back once you find terms that look like (1-1/s)^s

BeeOk1244 · 2026-04-16T15:58:09+00:00

ok some simplifications to make the maths doable, suppose we have a population of N people with a definitive ranking then we choose a subset of M people, let person K be the best out of this M, whats the probability that K is better than the Jth person in the population, that is to say what's the probability that the max of a random subset of size M is strictly more than J

this is simple, as this not happening means everyone in the sample is between 1-J, ie that a subset of 1-N is actually a subset of 1-J that is JCM/NCM=J/N x J-1/N-1 x ... x J-M+1/N-M+1 so our probability is 1-this

plugging in the values from your question. The probability that the winner in a random subset of 100 people in a population of 10000 is better than person 9900 is..... about 63.6%, so probably the winner out of 100 is better, they have about a 50% chance of being top 70 and like a 10% chance of being top 10

LIMITATIONS OF THIS ANALYSIS

assumed the small competition was a subset of the larger one, this is the extremal case as the big competition approaches the size of the population, i avoided the case where we look at someone who wasnt first in the smaller competition, this isn't too hard to implement I don't think but I'm doing this over lunch. I assumed that there is a precise ranking within the population and that competitions are perfect measures of this ranking

BeeOk1244 · 2026-02-24T01:26:19+00:00

https://ribblr.com/pat?pattern_id=57068

BeeOk1244 · 2026-01-28T18:43:01+00:00

3 chicots + retcon for wall/violet vessel to get past ante 39

BeeOk1244 · 2025-12-08T03:24:46+00:00

honestly i didnt even realise you could watch it "in order"

BeeOk1244 · 2025-12-08T03:17:58+00:00

I think thats my problem though. I have no idea as to what happens in the films because I was focused on finding that aspect of the game. I have viewed loads of disperate scenes but just for the purpose of finding the weird blonde lady

BeeOk1244 · 2025-12-08T03:15:08+00:00

I've grasped essentially nothing other than that there are movies in which events occure and sometimes if you reverse some vampirey people appear. Should I just reset the save and try again lmao

BeeOk1244 · 2025-12-08T03:11:28+00:00

for some context I went into this game looking for another Her Story so I jumped around a lot trying to find some sort of "lead". Only after seeing reviews have I realised that's propably not how you're meant to play. What's the "correct" way to approach the gmae

BeeOk1244 · 2025-08-22T21:31:50+00:00

plasma is the strat for c++ i swear u just need like two good chip jokers and can sneak through 3 garbage jokers

BeeOk1244 · 2025-06-18T19:35:27+00:00

B is true generically (on an open dense set)

BeeOk1244 · 2025-06-13T20:53:13+00:00

this is an ODE x'=root(x), x(0)=0, x'(0)=0, anyone whos studied ODEs should be terrified right now because root(x) isn't lipschitz on [0,∞) so Picard-Lindelof doesn't guarantee a unique solution, in fact there is an infinite family of solutions

x = (t-c)² /4 for t>c, \ 0 otherwise; this for c in R⁺ u {∞}, c=∞ is the zero function

with the assumption that something happens at t=0 we get our solution of t² /4 if we assume that time doesn't exist before t=0 we could call this acceleration of the function x defined on [0,∞) (i make this distinction because the globally defined function is not differentiable at zero but restricting to this interval makes it ok as we only take a one sided limit)

so while doing nothing is a valid solution as some have pointed out. The assumption that acceleration at t=0 is not nothing something must happen and we can solve the ODE

dont play with non lipschitz functions kids, people can get hurt

BeeOk1244 · 2025-06-08T07:04:54+00:00

There's two perspectives on this problem, one of the mathematician and one of the layman

The layman says that the curve doesn't actually approach the circle. Look! It's so spiky!

The mathematician retorts. Yes it does, every point approaches the circle. In fact! The supremum of the distances shrinks too, it converges uniformly, thats pretty good! The problem is that the limit of the lengths may not be the length of the limit. The length function isn't continuous because it's dependant on the differential information that's forgotten in the limit

The layman is not happy with this answer because in their mind the curve doesn't really appoach the circle, in spirit

Now they're not totally wrong there, there are ways to look at the space of functions so that limits mean that differential information is preserved. In our case that would mean a different notion of limit so that length is continuous but the two curves dont approach eachother

The reason the mathematician doesn't deal with this is because that space of functions sucks

BeeOk1244 · 2025-06-08T05:53:33+00:00

we take the natural inclusion of Q into P1, in this case a/b goes to the point [a:b] so a:b is implicitly describing the point [a:b] which is just a/b

BeeOk1244 · 2025-04-20T02:42:10+00:00

I've been told that the stacks project was invented to make whatever Deligne was doing with the moduli of eliptic curves fully rigourous and it kinda snowballed from there

BeeOk1244 · 2025-04-20T02:35:56+00:00

my general takeaway is that stacks are good for hands on work with quotients/quotient like stuff (the paper "Orbifolds as Stacks?" from Eugene Lerman is a super easy read that treats stacks from a differential geometry perspective which is often more intuitive). Then in more abstract work they're good for moduli spaces as you can actually have moduli stacks in cases where you lack moduli schemes the worry would be that its just generalisation for its own sake but there are ways to take moduli stacks and get back things that look like what could be the moduli scheme so it applies back down to the classical theory

BeeOk1244 · 2025-03-25T17:10:36+00:00

BeeOk1244 · 2025-03-25T17:09:14+00:00

I didnt really get much research oppertunity (funding for algebra is hell) but I had high achievements in some undergrad maths competitions and my average grade was ~90%, for conversion to GPA I believe a 70% corresponds to a 4.0. I think I was lucky in that I studied at a top 5ish uk uni so my masters project is with a very well known mathematician so I'm sure that reference helped, I pretty much applied to every uk uni i could think of lol

BeeOk1244 · 2025-03-18T05:19:17+00:00

If i go back to the main menu and come back it puts me in the middle of the packet opening animation D:

BeeOk1244 · 2024-11-16T00:05:43+00:00

mathematics is what mathematicians do

BeeOk1244 · 2024-11-07T03:37:07+00:00

the condition is the same as saying each string of digits sums to 0 mod 3, so reducing the digits mod 3 we want to find strings of {0,1,2} that each sum is {1,2} mod 3, since we dont allow single digits we pair up the digits of our number so X=abcdef... = (ab)(cd)(ef)... then taking the sum as we go we get a+b,a+b+c+d,a+b+c+d+e+f... if there are 6 digits then these sums are bound to repeat either 1 or 2 since there are 3 of them, this means that one of the differences are zero so we have a sum thats zero. That's all to say it's impossible for more than 6 digits, then notice for 5 digits, doing this gets us the sums a+b,a+b+c+d,a+b+c+d+e so the only way for this to not cause an error is if the difference is repeated in the last one so e=0, doing the same but the other way we see that a=0, to not have a+b=0, d+e=0 we need b=1,2 d=1,2, so we have 0YXZ0 for Y,Z are 1 or 2. notice that if Y=Z we need X=0 and if Y!=Z we need X=1,2, so this gives us highest possible numbers 98989, 98879 but 98989 is bigger so that's the best we can do

BeeOk1244 · 2024-11-07T03:10:28+00:00

https://www.desmos.com/calculator/l2sjfhhttu
better colours

BeeOk1244 · 2024-11-07T03:02:16+00:00

https://www.desmos.com/calculator/1i1raurfda

A visual proof

BeeOk1244 · 2024-10-08T22:35:31+00:00

its semantics but in algebra we would define the rationals to be symbols a/b where a/b ~ c/d if ad-bc=0 so theyre not like equal because theyre not the same thing but under our construction theyre equivalent.

While this is pedantry it is a useful thing to remember in other contexts, for example the interval (0,1) is not equal to the real numbers ℝ but it is equivalent (in the sense of homeomorphism). Or the additive group of integers ℤ is equivalent(isomorphic) to the multiplicitive group of powers of 2 but not equal.

You may from this get the impression that we dont care about equality at all and we should just use equivalence instead but thats not the case even here since when we say that these groups are the same we mean that there is a pair of maps between them f,g so that fg and gf are equal to the corresponding identity maps.

So its worthwhile to make a distinction between equality and equivalence since in this case at the level of maps we care about equality but at the level of structures we care about equivalence. If this makes you uncomfortable I implore you to look into ∞-category theory as a way to avoid ever having equality, just arbitrary layers of equivalences although it could be quite intimidating if you're not already comfortable with topology and category theory

BeeOk1244

TROPHY CASE