The person I heard the last puzzle (P122636) from attributed it to https://x.com/cshearer41 although I haven't found the tweet to confirm. Here's one of her most popular ones.

Question: What's the total shaded area?

12 = R*cos α for some given radius R of the big circle and angle α.
Radius of the small circle is r = (R*sin α)/2.

Area of 1/4 of the big circle is just π*R^2/4.
Area of the small circle is π*r^2 = π*(R*sin α)^2/4.
Area of shaded area is π*R^2/4-π*(R*sin α)^2/4 = π*R^2/4*(1-sin^2 α) = π*R^2/4*cos^2 α = π*12^2/4 = 36*π.

At the very least it seems to check out for the edge case where R = 12.
In that case there's no small circle and area is just the whole 1/4 of the radius 12 circle, which is 36*π.

P124173
Yep, that's correct.

https://teorth.github.io/equational_theories/blueprint.pdf

What do you think about this?

Looks like a cool project. Might be good to link
https://terrytao.wordpress.com/2024/09/25/a-pilot-project-in-universal-algebra-to-explore-new-ways-to-collaborate-and-use-machine-assistance/
https://terrytao.wordpress.com/2024/10/12/the-equational-theories-project-a-brief-tour/
https://github.com/teorth/equational_theories
for a more friendly explanation of what this is all about. When I first looked at the PDF linked I spent a bit of time slogging through the formal description of the deductive system and the sketched proofs that it's sound and complete, which is interesting but not really needed to describe the project.

One of the Putnam problems mentioned.

P123410
Putnam problem?

Let's try a new problem!
Three squares are arranged as shown in the diagram. The green square has an area of 12. What's the area of the yellow square?

im going regret this im unsure with the histdrawing style where the overlay bars should go and stop so ill round at the last digit after trying a modification to remove them
yellows 123370
greens 31050
so 31050 greens is 12 that means 2587.5 pixels is 1
theres 123370 yellow pixels and using the new 1 you get a size rounded up to 48
o look at that its the same as multiplying by 4 if rounded up otherwise its 47.67922***
P122662
P122667
lol

it's well known that HK = AJ, FK = HJ when AHFI is a square. then BJ = HJ = FK, so FK = CK. let AD = a, EG = b, then HJ = b / 2, AJ = a + b / 2, BE = b - a, AB = a. Since ABE is similar to AJH, AJ / AB = JH / BE, therefore b ^ 2 = 2 * a ^2, and the area of the yellow square is 24.

P122721
Congratulations!

P122662
The shitty drawing style was intentional so that you can't just get the answer by P122691. The problem is such that you can redraw it accurately if and only if you can solve the problem.

P123145 (samefag yous)
see i regret it also dont draw like historcialfag theres already enough histy content to go around

P123158
Just leave him as he'd like to be a handsome artist like me.

I like playing around with formal logic, but when proving things about synthetic geometry in the usual kinds of formal logic, you sometimes have to spend time proving things that would be obvious from a diagram. It's often stated that diagrams can't be part of a formal proof because they can deceive you into thinking you've proved false things, such as the infamous proof that all triangles are isosceles. But in all the false proofs using diagrams I've seen, the problem was that the diagram didn't consider all possible cases. There's no reason a system of logic using diagrams couldn't be made just as sound as regular predicate logic.

So I've started reading PDF related, which describes such a system of diagrams for proving things about geometry. It's a short book made from an edited version of his PhD thesis. Apparently people have also been studying systems of diagrammatic reasoning not specific to geometry such as Venn diagrams and similar systems, including proving various soundness results about them. One criticism I've seen about the system in the book is that the number of cases you have to draw can be very large, and you can see one instance of that in the book in Figure 9. But at least the operation that leads to Figure 9 is not something you'd normally do in a proof, so I don't know yet how much of a problem case explosion is in practice. I'm hopeful that this system can and will be improved upon, and that in general diagrams will become a more common tool in formal proofs.

Using formal logic only works when you already correctly know the relationships between things.

1+1=2 works for apples, but that logic doesn't work for things like radians that are cyclic instead of linear, it doesn't work for shadows, it doesn't work for a lot of stuff.

So logics are specific to objects, and to learn the logic of an object, it's relationship with other things, you have to observe it first to know it applies.

Science is testing hypothesis about relationships, but the actual formulation of relationships still relies on obeservations to deduce a pattern of relationships between objects. The reason formulas in math and physics seem to always work is just because those are the ones that were tested and worked repeatedly in different situations, not because there are some unviolatable rules that can be extrapolated indefinitely to deduce everything.

Formal logic requires you to get your premises correct, to get the correct conclusion. I also imagine it's pretty limited, but it's such a boring subject I didn't look deep enough into it to find out. I don't remember being able to describe time dependent relationships in formal logic, for instance.

All the formulas and rules we make up to describe the relationships we see have the possibility of hidden variables or factors that aren't accounted for that could change outcomes in a given situation. This always exists as a possibility because in real systems we can't test every possible situation.

Maybe if you have the right quest item in your inventory, an object will behave differently, because there was a hidden variable.

The way that our brains work, taking repeatedly observed things as the norm, you probably wouldn't notice when a relationship between objects was unusual, then it would become an accepted scientific fact. Magnets used to be magic rocks and extremely abnormal in the natural world. Magnetism was a hidden variable that has become an accepted scientific fact.

Rubbing iron with a magnet can make the iron magnetic, then tapping the iron a few times causes it to loose its magnetism. You don't think that's an usual relationship? What other things can you change the state of and transfer properties to like that? You can't rub a piece of wood with a rock to make it heavier, then tap it to make it lighter.

Magnetism only makes sense because people repeatedly observed it and assembled some rules to reliably describe its behavior, but for most of mankind's existence it was a hidden variable.

And magnetism can change the outcomes of other systems.

Magnetize a few parts in a mechanical clock and your formulas for the mechanics of a mechanical clock won't work anymore.

Mechanical clocks don't work the same if you stick a magnet in them and exert forces on the parts, magnetism was a hidden variable.

When he starts covering the SAS theorem you start seeing case explosion become a real problem. Pic related is just some of the diagrams that occur in one step of a proof of one case of the theorem. You can't even state the theorem in his system without breaking it down into a lot of needless cases, because the theorem involves two triangles and you have to draw every possible way the two triangles can intersect.

But it's also pretty clear that this could be massively improved upon. An obvious improvement would be to add a way to say that two figures, each depicted in its own diagram, exist in a plane without caring about how or whether the figures intersect. One improvement he suggests himself but does not implement is combining his diagrams with the usual sentential systems of logic. Making his diagrams into propositions that could be combined with logical connectives could be a good way to implement the capability I want. The way his system works, multiple diagrams always represent different cases, like propositions connected with "or". If we could connect them with "and" instead, we could represent the two triangles on different diagrams and say that both exist somewhere in the plane.

As for the sort of case explosion shown in the attached pic, it seems suboptimal even by his existing rules. You would get fewer cases if you constructed the new triangle using multiple steps. You'd still need to be able to depict two figures without saying how they intersect to get the cases down to a reasonable level.

Please advise the literature for studying the theory of formal languages, especially the theory of automata (preferably in Russian)

P119923
>(preferably in Russian)
Ask North Korea

The bunkbed conjecture is false https://arxiv.org/abs/2410.02545
How can a path being connected up to the last post not be independent of the 2 possible paths being connected following it?

It's definitely counterintuitive, I'll have to try to read the full paper to say more.

>How can a path being connected up to the last post not be independent of the 2 possible paths being connected following it?
That makes sense for the probability of being connected via a particular path (there are complications in the case where the path from u to v corresponding to a path from u to v' intersects itself, but that should work in favor of the path from u to v), but I'm not convinced we can get from there to an inequality about the probability of being connected via any path.

Pic related is already pretty counterintuitive, and it's simple enough even for people not into math to comprehend, so I'll try to explain it without using any more math language than I have to.

Think of this thing as a two-level maze. There's no way to get from one level to another without using a staircase; trying to jump will get you killed from fall damage. And there's no ground beneath the maze; if you fall off it, you keep falling into the void. Each circle is a pair of platforms, one at the top level and one at the bottom level. Only for the gray circles are the two platforms connected by stairs. Each of the triangles is a movable platform that was randomly moved to either the top or bottom level before you entered the maze. Each triangular platform connects three circular platforms at the level it was moved to; that is, provided they're all at the same level, you can walk from a circular platform onto an adjacent triangular platform and from there walk to either of the other two adjacent circular platforms.

The entrance to the maze is at u1 on the top level, and there are two exits at u10, one at the top level and one at the bottom level. The question is, given a random setting of the triangular platforms, is it more likely that it will be possible to reach the top exit or the bottom exit? Without calculating anything, I would have expected either the top exit to be easier to reach or for the exits to be equally likely to be reachable. But it's easy to check all the possible settings of the triangular platforms and see that it's in fact slightly more likely that the bottom platform will be reachable.

P118300
There are five reasonable paths you could take through this thing. I'm omitting paths that reach the {u3,u6,u9} triangle and don't go straight to the {u8,u9,u10} triangle. I made a chart showing which configurations allow reaching the exit along each of the five paths. Blue triangles are raised and brown triangles are lowered. I colored the square blue in the cases where the top exit is reachable and orange in the cases where the bottom exit is reachable. I didn't show the 32 cases where the {u1,u2,u3} triangle is lowered, since you clearly can't reach either exit in any of those cases.

A notable fact is that the chance to reach the top and bottom exits along any particular path is equal, but the chance that any path will allow you to reach the bottom exit is greater than the chance you can reach the top exit.

https://www.erdosproblems.com/start

I think we can solve some of them if we tackle enough.

Tao's work on them: https://terrytao.wordpress.com/2024/08/19/erdos-problem-385-the-parity-problem-and-siegel-zeroes/


GL!

Pic related is a typical high school geometry exercise to discover the sum of the interior angles of a polygon. To their credit, the authors only claim that what they have shown works for convex polygons. It is left to the reader to wonder if the trick can be adapted to polygons in general.

Without looking up the answer, can you show with proof whether any n-sided polygon can be decomposed into n-2 triangles?

P105586


i smell old yank algebra 1 & 2 / geometry textbooks

Yes I can, most people can what is the point of this thread anyway, we have a good thread and then retarded waste of oxygen *****s like you, shit bait and kill yourself.

P106052
You can't, can you?

It has been proven that the longest-running 5-state 2-symbol Turing machine which halts is the one that halts in 47,176,870 steps. All 5-state 2-symbol Turing machines which run longer than this have been proven to run forever.

The halting theorem famously proves that there's no general algorithm that can determine whether a given computer program halts. Some programs we know halt because we can run them to completion. Others enter obvious infinite loops. But for some programs, it's hard to tell if they eventually halt or just run for a very long time. It's tantalizing to wonder where the boundary point is between simple programs whose halting status we can figure out and those programs we can't.

Research into the busy beaver problem probes and pushes that boundary. When I posted about it a year ago in P52827, there were several remaining 5-symbol 2-state Turing machines whose halting status was unknown. They were conjectured to run forever, but it was not proven. Now it's been proven, and the frontier has been pushed out to BB(6).

Homepage of the collaboration that proved it:
https://bbchallenge.org/
Announcement of the discovery:
https://discuss.bbchallenge.org/t/july-2nd-2024-we-have-proved-bb-5-47-176-870/237
Machine-verifiable formal proof of BB(5) = 47,176,870 in Coq:
https://github.com/ccz181078/Coq-BB5
A well-written news story on the feat (sadly behind Cloudflare):
https://www.quantamagazine.org/amateur-mathematicians-find-fifth-busy-beaver-turing-machine-20240702/

Some interesting machines at the new frontier (BB(6) and others):
https://wiki.bbchallenge.org/wiki/Cryptids

P102051
P102053
This is an achievements without a doubt. However, how viable is actually solving for higher numbers?
https://www.sligocki.com/2022/06/21/bb-6-2-t15.html
I don't think we will ever have a solution to the halting problem, and thus the team will have to continue their work manually. Which from what I'm seeing is
>reduce the number of possible machines
>get rid of a few large ones
>brute force the way through the rest

I know the community is made up of mathematicans, but I do wonder if the search for BB(6) or BB(7) will end up at least partially using something like Minecraft@Home. There is only a one line wiki page about the idea so far
https://wiki.bbchallenge.org/wiki/Accelerated_Simulator

Is there any practical use to proving the busy beavers?

P102104
Knowing BB(5) will not make anyone's number of dollars go up. As for if either of those have meaningful value, I don't know.

Let A be finite.
Consider a uniformly distributed random function f: A->A.
How does [tex: N(n) := E[|f^n(A)|]] vary with n and |A|?

Is f^n f composed with itself n times?

>One nice fact is that in the limit for large |A|, [tex: \frac{|A\f(A)|}{|A|} = \frac{1}{e}].
Sounds similar to derangements (not the schizo kind).

I think I have a simple way to calculate it for the case [tex: n \geq |A|-1]. This is the case where you are guaranteed to have reached a loop. The idea is that you can just multiply |A| by the probability any given point [tex: x \in A] is part of a loop.

You have a probability of [tex: \frac{1}{|A|}] that f(x) = x and a probability [tex: \frac{|A|-1}{|A|}] that it maps to some other point.

In the case that [tex: f(x) \neq x], you have a probability [tex: \frac{1}{|A|}] that f(f(x)) = x, giving you a loop of which x is a part, a probability [tex: \frac{1}{|A|}] that f(f(x)) = f(x), giving you a loop of which x is not a part, and a probability [tex: \frac{|A|-2}{|A|}] that f(f(x)) goes to a third point.

We can continue this: In the case that x, ... [tex: f^k(x)] are all distinct, there is a probability [tex: \frac{1}{|A|}] that [tex: f^{k+1}(x) = x], making x part of the loop, a probability [tex: \frac{k}{|A|}] that [tex: f^{k+1}(x)] is some prior point, making a loop of which x is not a part, and a probability [tex: \frac{|A|-(k+1)}{|A|}] of going to yet another point not visited yet.

For |A| = 4 the probability x is part of a loop is
[tex: \frac{1}{4} + \frac{3}{4} \frac{1}{4} + \frac{3}{4} \frac{2}{4} \frac{1}{4} + \frac{3}{4} \frac{2}{4} \frac{1}{4} \frac{1}{4}]
and therefore the expected number of points that are part of a loop is
[tex: 1 + \frac{3}{4} + \frac{3}{4} \frac{2}{4} + \frac{3}{4} \frac{2}{4} \frac{1}{4}]
which can be rewritten in the form
[tex: 1 + \frac{3}{4} (1 + \frac{2}{4} (1 + \frac{1}{4}))].

You can calculate it with

def f(a):
e = 1
for k in range(1,a):
e = e * Fraction(k,a) + 1
return e

or the equivalent float version if you just want a fast approximation. For large |A| it seems to be proportional to [tex: \sqrt(|A|)].

https://oeis.org/A001865 is related; it is this function multiplied by [tex: |A|^{|A|-1}].

P101242
>or the equivalent float version if you just want a fast approximation. For large |A| it seems to be proportional to [tex: \sqrt(|A|)].
Makes sense, it is like a birthday problem.
Given this, I expect the distribution becomes mostly loops quickly, taking near [tex: ln(|A|)] steps.

I wrote a little script to calculate some of the values (times [tex: |A|^{|A|}]):

-----------------------------------------
from itertools import product

for m in range(1,8):
count = [0]*m
for f in product(range(m), repeat=m):
xs = set(range(m))
for n in range(m):
count[n] += len(xs)
xs = set(f[x] for x in xs)
print(count)
-----------------------------------------
[1]
[8, 6]
[81, 57, 51]
[1024, 700, 592, 568]
[15625, 10505, 8625, 7965, 7845]
[279936, 186186, 150096, 134856, 130176, 129456]
[5764801, 3805249, 3029257, 2670367, 2528407, 2490607, 2485567]
-----------------------------------------

With enough transformations I was able to find something related in OEIS:
https://oeis.org/A225213

-----------------------------------------
from itertools import product
from fractions import Fraction

def div(x,y):
return x//y if x%y==0 else Fraction(x,y)

for m in range(1,8):
count = [0]*m
for f in product(range(m), repeat=m):
xs = set(range(m))
for n in range(m):
count[n] += len(xs)
xs = set(f[x] for x in xs)
diffs = [a-b for a,b in zip(count[:-1],count[1:])]
print([div(d,m*(m-1)) for d in diffs])
-----------------------------------------
[]
[1]
[4, 1]
[27, 9, 2]
[256, 94, 33, 6]
[3125, 1203, 508, 156, 24]
[46656, 18476, 8545, 3380, 900, 120]
-----------------------------------------

Is lambda smart enough for Lean's Natural Numbers Game? Or anyone here already completed it (it's not too hard tbh)?

https://www.ma.imperial.ac.uk/~buzzard/xena/natural_number_game/index2.html

Basically, you program in Lean, a programming "language"/proof assistant used by autists and mathematicians. Your goal is to prove some profound shit like "a + b = b + a", starting from nothing except peano axioms. It's more fun than it sounds like. And is a good introduction to formal logic.

Yes, it works in TB but it's *****ing slow to load (some 20MB of library stuff to download). And you need to allow canvas access for the main menu to work.

PS: there's also a new version updated to Lean 4 but it's cringe and dumbed down and gay and doesn't even work with tor. Trust me, ignore it and just try this one.

It's a neat game and probably a good way to get introduced to Lean, but ultimately it's leading the player down a set path. I wanted to try something different. In particular, it bugs me that the usual ways to prove basic arithmetic facts to a computer don't seem to bear much resemblance to the way you would typically explain them to a human. So I'm working on ways to prove arithmetic facts using arguments about the sizes of sets. So far I've shown that if type A has size a and type B has size b, then their tagged union has size a+b. I'm using Coq instead of Lean because I'm more familiar with it, and right now I want to waste time autistically proving theorems about basic math rather than autistically learning details of another programming language. But I don't expect that to make much difference.

P99402
>It's a neat game and probably a good way to get introduced to Lean
tbh all you need
codeine cough syrup and a and a three 6 mafia cd ngl

P99402
I spent a lot of time trying to prove the following statement, and now I've finally done it:
>If I have a bijection between type A plus one extra element and type B plus one extra element, then I can construct a bijection between types A and B.
It was harder than I thought it would be, partly due to errors I still don't understand getting in the way. The culmination of that struggle is the function drop_right, which given an element of a tagged union and a proof that it isn't the second type, gives me the original element of the first type. The proof that this function works correctly went through where the corresponding proofs for similar functions, which I now realize were needlessly complicated, did not go through. With that ingredient in place, I was able to construct the bijection I wanted. From here on, it ought to be relatively straightforward to prove the critical statement
>If type A both has size m and size n, then m = n.
which will allow me to finally get to business proving things about arithmetic. If you include all the work leading up to that statement, it's definitely more complicated than just proving the theorems inductively, but my hope is that most of the complexity is in proving that statement, and in a system where we took it as an axiom, the proofs will be relatively simple. But I'll have to see. There may also be much an easier proof of the statement that I'm just not seeing.

P99402
P99695
I have some theorems about addition now!
>n + 0 = n
>a b c, a + (b + c) = (a + b) + c
>a + b = b + a

P99402 .. P99717
Added multiplication.

>A vector space just generalizes a field.
<No? Vectors are a whole separate set of objects you introduce. It's not just a generalization.
>But if you make the vector space the same field as the scalars it's the same.
<But vectors are different from scalars.
>But it's the same object.
<But they're different types. We write vectors in bold.
>Math doesn't have types.

>The next simplest example is the field F itself. Vector addition is just field addition, and scalar multiplication is just field multiplication. This property can be used to prove that a field is a vector space. Any non-zero element of F serves as a basis so F is a 1-dimensional vector space over itself.
...
Surely it causes problems if you can't assume vectors and scalars are distinct?

This is everything I hate about mathematics in one example. Are mathematicians not capable even of using tagged unions? Imagine if the vector space contained the objects used to construct the scalar field. It's insane and will make mathematics implode under special cases eventually. It's like math isn't made to find mathematical truths directly but so some "wiggers" (jewish) can go "UM ACKCHUALLY" and use intentionally hidden structure to make some important seeming abstractions in a way that somehow avoids helping anyone else understand. It's utterly disgusting.
Post other examples of this stuff itt, if you want.

Vector space? You mean module over a field?

Isn't a vector that can be a scalar called tensor?

P90602
no

P90596
lolz

Where is this log from? I want to bully mathlets (and maybe learn something) too.

>use the [quadratic] formula as a last resort
/watch?v=ij01aenZIfY

The video is a typical example of this bad advice. She tells her students to try to use integer factor searching to solve [tex: 3x^2 - 4x - 4 = 0], [tex: 5x^2 - 2x + 3 = 0], and [tex: 2x^2 - 3x - 7 = 0] before proceeding to the quadratic formula. Even with her contrived examples (the factor search works for one of them) you can see she's wasting time. But the harm isn't the wasted time; it's that some of her students will eventually see an equation like [tex: x^2 - 5.6x - 3.784 = 0], think that they're supposed to solve it by factoring, and declare "I don't know how to do this" rather than moving on to the quadratic formula as she intended.

P98926
That's just your lack of charisma/authority. In middle school, I was in "the worst class ever" [spoiler:across 4 years and 2 schools. Somehow I think they were exaggerating], yet some of the teachers were still respected. They didn't need to be violent or menacing. Mostly they respected us (e.g. calling us by our surname, joking with us when entering class...), while still being somewhat firm (e.g. giving out a test when the class became rowdy). Everyone was openly dissing them at the time (among the students, not to their face), but all the people I met with afterward agreed that they were good teachers and remembered them fondly, including the "troublemakers".
On the other hands we had teachers that didn't seem to give a damn when the class was noisy, teachers who yelled at the first occasion, teachers who regularly arrived late, teachers who didn't prepare their lessons well. Those weren't respected and the class was insufferable with them.
I know that things changed for the worse in the last decade, but I'm pretty sure not all your colleagues have the same problems you have.

P98944
>I was in "the worst class ever"
kek, me too, so much that the older students from other classes who had to repeat a year would ask to be transferred to our class

P98944
I see the teachers that you describe at my workplace. Both the ones who are fair but strict, and the apathetic ones who have given up.

I consider what you call as authority as just fear. Fear of consequences, and that is something I consider useless. I want students to be motivated to study something, but that motivation often comes only when their final exams come and by that time it's too late.

Until then, they just behave like horny shits with an attitude that they can study whatever it may be by watching videos like in OP without the teacher's help.
They have told me that they watch these videos by skipping or watching it in 2x when they cram. It's that bad. There is also an option in youtube where you can skip when the speaker is not saying anything. These channels mainly monetize this demand for cramming before exams and their aim is not to transfer true knowledge but to help students cram.

"Fear of consequences", a.k.a. responsibility, is probably the most important thing you can teach your students.
>that motivation often comes only when their final exams come and by that time it's too late.
Give them shorter tests more often then, ideally without telling them beforehand when there will be one. Are you allowed to give homework still? Ask a random student to give the solutions rather than doing it yourself.
>watching videos like in OP without the teacher's help.
Recommend them better one. Stuff like 3Blue1Brown for example.
>they watch these videos by skipping or watching it in 2x when they cram
Makes sense. Do you expect them to read every word from your handouts as well? Skipping to the information you need is the way to go.

P98991
What I want to teach is motivation, not responsibility driven through fear. Fear can be useful but fear is external.
I would be happy if the students were actually interested to go to school.

>Are you allowed to give homework still? Ask a random student to give the solutions rather than doing it yourself.
Yes, I can do that. In fact they schedule a test every two weeks. Problem is they don't care about these tests so they just hand over blank sheets.

>Recommend them better one.
They suck at english. Even my english sucks but it's even worse.

i never understood all the buzz around the barber paradox
>the barber shaves all those and only those who do not shave themselves, and everybody must be shaved
obviously the barber shaves because thats his job as the barber
however, when he gets home, he is no longer performing his job
and he cannot go to his shop bc by the time his shift ends, the barbershop is closed
so he shaves himself like the other men who do not go to the barbershop
wheres the paradox? are mathematicians stupid or just extremely *****ing autistic?

P98780
Ok this chick is hot (quit with the subliminals)
Anyway your still gae and cringe tho tbh ngl

>We have transcended societies of control
We did what?
>Work does not stop once you get out of your workplace
Denpa, you only ever worked on one single job in your entire life. Why do you automatically assume that all careers are like this. This is very narcissistic and egocentric behavior
>I am sure that the barber would have a phone and he would be expected to be on his toes any time his phone rings
Most barbers i know are either the sole employees of their own shop, or are part timers that don't even work full week.
Only the sole employees get called off-work
and i doubt that they are on their toes or anything
>once you go on higher levels of work, your phones or your emails will be bombarded with tasks, orders, things to do tomorrow, sudden meetings, voice messages shouting at you to do your work immediately
White collar issues, consider contributing to society
>A barber stays a barber even if he's at home
>He doesn't need to specifically be in a barber shop to do his work
If he'll do his work outside of the shop he'll violate 17 OSHA articles and at least 4 federal laws after breeding not only siberian plague but also HIV, hepatitis and rotavirus.
Although on-call barbers that come to your house also exist but i never seen those so might as well be non-existent

P98833
I'm just trying to say that a barber does not stop being a barber just because he got out of the shop. He's not going to stop being a barber even if he got fired from his job.

Our societies used to be disciplinary, and worked like a prison. Prisoners lived and were disciplined in prison. Same way, workers worked in factories like cogs in the wheel, and students went to educational institutions to study. Now, students are expected to study all the time, and workers worked at home. Societies where activities were only confined within buildings under the supervision of authorities (managers, teachers, prison wardens) have now shifted to something that offers freedom, but the control extends on a societal level.

ig theres no more effective way to blackpill someone than just letting them touch grass
years of trying to get into denpa's empty skull didnt yield as good a result as just him getting a job

P98882
*****y feet and calves btw

It has been a long time since we've had one of these, so here is an easy puzzle:
How can you arrange 12 squares to form a net for a single cube? You may not cut them, superpose them, or leave any unused.

□□
□□
□□
□□
□□
□□

One square for each edge of the cube. Fold along the dotted lines.

P95410
Can you show how you fold it? I don't think this works

P95478
That is correct. As a subsidiary question, how many such nets are there? Nets that are rotations or reflections of each-other only count once.

P95486
>Can you show how you fold it? I don't think this works
+-+-+
| 0 |
+-+1+-+
| 2 |
+-+3+-+
| 4 |
+-+5+-+
| 6 |
+-+7+-+
| 8 |
+-+9+-+
| A |
+-+-+
>How can you arrange 12 squares to form a net
Two squares will be superposed onto each face when folded into a cube, but not as a flat net.

P95277
Does this thread relate to black qube moon matrix or QubesOS?

how do you factor something like 6x^2 + 12 x + 6 (from (2x + 2) * (3x + 3))
every time i learn this i forget it :( too lazy to open my textbook rn

6x^2 + 12x + 6
6(x^2 + 2x + 1)
6(x + 1)(x + 1)
6(x + 1)^2

is how you factor the first expression you mentioned

>the autist scared off the girl
buuuuuuuuuuuuuuuuwAHAHAHAHAHAHAHAHA
i bet it was jd too

If the title made you think if of having a stroke, this thread is to discuss two of the problems presented in this year's high-school exit exams. In specific, for higher level mathematics (the proper one, not social science mathematics). So you should be able to solve these. Also it's worth mentioning that the problems have been slightly refitted. Rest assured, the contents are exactly the same.

Problem 1:

Take [tex: f(x) = (1-ax)^{\frac{-1}{2}}], where ax<0, [tex: a \neq 0]

a, Prove using induction that
[tex: f^{(n)}(x) = \frac{a^n(2n-1)!(1-ax)^{\frac{2n+1}{2}}}{2^{2n-1}(n-1)!}], [tex: n \in \mathbb{Z}^+]

b, Show that the Maclaurin series (not the Taylor series) for f(x) up to an including the [tex: x^2] term is
[tex: 1 + \frac{1}{2}ax + \frac{3}{8}a^2x^2]

c, Show that
[tex: (1-2x)^{-\frac{1}{2}}(1-4x)^{-\frac{1}{2}} \approx \frac{2+6x+19x^2}{2}]

d, Knowing that the expansion of the Maclaurin series for f(x) is convergent for |ax|<1, indicate what restriction must be put on x for the previous approximation to be valid.

e, use [tex: x=\frac{1}{10}] to make an approximation for the value of [tex: \sqrt x] expressed in terms of [tex: \frac{c}{d}] where [tex: c,d \in \mathbb{Z}^+]

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

Problem 2:

Consider the following ODE:

[tex: \frac{dy}{dx} - y cosec(2x) = \sqrt(tan x)]

a, Use Euler's method with a step size of [tex: \frac{\pi}{12}] to make an approximation for the value of y when [tex: x = \frac{5\pi}{12}]. For added spice, you may not use your calculator here.

b, Show that
[tex: \frac{d}{dx}(\frac{1}{2}ln(cotg x)) = -cosec 2x]

c, Show that [tex: \sqrt(cotg x)] is an integrating factor for this ODE

d, Solve the ODE for y

e, Consider the results from (d) and (a), and that (d) is valid for [tex: 0 < x < \frac{\pi}{2}].
(i) Calculate the exact value of y for [tex: x = \frac{5\pi}{12}]
(ii) Considering the slope of the curve, indicate why (a) is not a good approximation
(iii) Write the reason why the approximation for y is lower than the actual value

f, Using the ODE, deduce why the solution has a positive slope for [tex: 0 < x < \frac{\pi}{2}]

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

I think both are fun to do. Problem 1 focuses on a practical application for a primitive tool for approximation of square roots. Problem 2 is exploring an ODE and testing the student's knowledge on when can approximations be applied.

Problem 1.

Why even have a gay "a" parameter?

Take f(t) = (1-t)^(-1/2)

a, we have to prove instead that the n-th derivative is (2n-1)!/(2^(2n-1)*(n-1)!)*(1-t)^(-(2n+1)/2)

First derivative is just (1-t)^(-3/2)/2 and base case for n = 1 is 1!/(2^1*0!)*(1-t)^(-3/2) which is obviously the same.

Taking the derivative of (2n-1)!/(2^(2n-1)*(n-1)!)*(1-t)^(-(2n+1)/2), and conveniently multiplying and dividing by 2n, one gets (2n-1)!/(2^(2n-1)*(n-1)!)*(2n+1)/2*(2n)/(2n)*(1-t)^(-(2n+3)/2) = (2n+1)!/(2^(2n+1)*n!)*(1-t)^(-(2n+3)/2) which is what one gets if plugging n+1 into the original formula.

b, for t up to t^2 it'd be 1/0!+1/2*1/1!*t+3/4*1/2!*t^2 = 1+1/2*t+3/8*t^2. Substituting t=ax gets the desired result.

c, substituting above for t=2x and t=4x it's just a matter of multiplying (1+x+3/2*x^2)*(1+2x+6x^2). Expanding and ignoring the terms with exponents > 2 leads to 1+2x+6x^2+x+2x^2+3/2*x^2 = 1+3x+19/2x^2 = (2+6x+19x^2)/2

d, both |2x| < 1 and |4x| < 1 must be true so |x| < 1/4

e, if t = 1/10, then 1/sqrt(1-1/10) = 1/sqrt(9/10) = sqrt(10) / 3
Using b, it can be approximated as 1+1/2*1/10+3/8*1/100 = 843/100, so sqrt(10) is approximately 3*843/100 = 2529/100 = 3.16125
[spoiler: sqrt(10) = 3.1622776...]

Too lazy for problem 2.

>Why even have a gay "a" parameter?
No idea to be honest. I suppose it's for part (e). In which I made a little typo. It was supposed to be calculating the value of [tex: \sqrt 3]. However, the overall solution is the same:
[tex: 3^{\frac{1}{2}} = (1-\frac{a}{10})^{-\frac{1}{2}}]
So [tex: a = \frac{20}{3}]
Substitute into (b) to get [tex: 1 + \frac{1}{2}\frac{20}{3}\frac{1}{10} + \frac{3}{8}\frac{400}{9}\frac{1}{100} = 1.5 \approx 1.7320508...]
And in case anyone's curious, With terms up to x^3, we get [tex: \frac{43}{27} \approx 1.593], for x^4, [tex: \frac{1067}{648} \approx 1.647], so it is indeed slowly converging.

Though this was my mistake. The rest seems fine, except for (a), where I don't exactly understand where the a^n went and where did the minus come into t's exponent.

P94920
>I don't exactly understand where the a^n
Right, I kind of ignored that when I decided to just use t and not ax but a similar reasoning gives you the missing a^n.
>and where did the minus come into t's exponent.
u wot
(1-t)^x -> -(1-t)^(x-1).
You start with an exponent of -1/2, you're only going to get lower and lower values as n increases.
I didn't say anything because I thought it was kind of obvious but you actually made a typo with (1-ax)^((2n+1)/2) by missing the minus sign.

P94925
>Right, I kind of ignored that when I decided to just use t and not ax but a similar reasoning gives you the missing a^n.
Alright.
>u wot
To my excuse, where I have this on, the gap between the minus sign and the fraction is close to being a print error. Checked problem 2 and I don't see any other typos, other than some ignored constraints in the definition. This is embarrassing, should I redo the OP?

P94927
>This is embarrassing, should I redo the OP?
No, embrace the cringe.

I've been playing a bit with ternary logic and arithmetic. The initial reason I started thinking about this is that it annoyed me how in binary a n-bit signed integer can represent the range [- 2^(n-1) ; 2^(n-1) - 1], which isn't symmetrical. Of course that's only a convention (although one that is very convenient for several reasons) and it could as well represent [- 2^(n-1) + 1/2 ; 2^(n-1) - 1/2], but then 0 wouldn't be represented.

In ternary, a n-"tit" signed integer can represent all the integers in [-(3^(n-1) - 1)/2 ; (3^(n-1) - 1)/2]. So a 3-tit signed integer can represent the 27 numbers between -13 and +13, for example.

Let's see how to construct such a numbering system. We could use 3's complement for negative numbers. For example, with 2-tit numbers:
12 = -4
20 = -3
21 = -2
22 = -1
00 = 0
01 = 1
02 = 2
10 = 3
11 = 4
Now that's very ugly, and there is no obvious way to determine whether a number is negative or positive from it's ternary representation. I don't like it.

So instead of assigning to each tit a value in {0, 1, 2}, we can instead assign them a value in {-1, 0, 1}. For convenience and legibility, I will note the different symbols as {1̃, 0, 1}, but remember that 1̃ has a value of -1. With those values, and a classic positional system, we can easily represent all natural numbers.
The positive numbers are:
0, 1, 11̃, 10, 11, 11̃1̃, 11̃0, ...
(0, 1, 2, 3, 4, 5, 6, ...)
The negative numbers are:
1̃, 1̃1, 1̃0, 1̃1̃, 1̃11, 1̃10, 1̃11̃, ...
(-1, -2, -3, -4, -5, -6, -7, ...)

This system has some nice properties:
1. All natural numbers have a unique representation
2. The range that can be represented by an n-tit number is symmetrical with regards to 0
3. The leftmost non-0 tit contains the sign information
4. The opposite of a number can be easily computed by replacing all 1s by 1̃s and all 1̃s by 1s
5. It's a positional system, so usual algorithms for addition and multiplication apply.
6. A n-tit variable can represent ~60% more values than a n-bit variable

We can also have a gray-like code. Here's how to generate a sequence where each consecutive term only differs by one tit:
1. Start with the sequence S ← {1̃, 0 1}
2. T and U both take the elements of S
3. S is reversed
4. Each element t of T has the tit 1̃ appended to its left
5. Each element s of S has the tit 0 appended to its left
6. Each element u of U has the tit 1 appended to its left
7. S ← T | S | U where "|" represents the concatenation operator.
8. Redo steps 2 to 7 until the sequence is long enough.

For example, with 2-tit numbers, we get:
1̃1̃ 1̃0 1̃1 01 00 01̃ 11̃ 10 11


Now let's do some logic!
We could easily emulate binary logic, for example by interpreting 0 as a boolean 0 and 1̃ and 1 as a boolean 1. That's not fun though, and wouldn't allow us to compute on ternary numbers, so let's invent toolean logic.
I tried coming up with a few logic gates, as extensions to the ones we're used to with booleans.

"and", it's a ternary multiplication. Notice how the lower-right quadrant is the same as with the boolean and. I'll note the operator as ⋅
 a⋅b | 1̃ 0 1
----+-------
1̃ | 1 0 1̃
0 | 0 0 0
1 | 1̃ 0 1
That one has some nice properties:
(a ⋅ b) ⋅ c = a ⋅ (b ⋅ c) = a ⋅ b ⋅ c
a ⋅ a = abs(a) (that is, 1̃→1, 0→0 and 1→1)
1̃ ⋅ a = neg(a) (that is, 1̃→1, 0→0 and 1→1̃)
1 ⋅ a = a
0 ⋅ a = 0
a ⋅ b = neg(a) ⋅ neg(b)

"or", it's a ternary saturating addition. Again the lower-right quadrant is the same as with the boolean or. I'll note the operator as +
a+b | 1̃ 0 1
----+-------
1̃ | 1̃ 1̃ 0
0 | 1̃ 0 1
1 | 0 1 1
That one isn't as nice, so I would be happy if you have better suggestions. In particular, I hate that (a + b) + c ≠ a + (b + c). Still, a few interesting properties:
neg(a) + neg(b) = neg(a + b)
a + a = a
0 + a = a
a + neg(a) = 0

"xor", isn't an overflowing addition (half adder), because I wanted the lower-right cadrant to stay the same as the boolean xor. I'll note the operator as ⊕
 a⊕b | 1̃ 0 1
----+-------
1̃ | 0 1̃ 0
0 | 1̃ 0 1
1 | 0 1 0
Again, I hate that (a ⊕ b) ⊕ c ≠ a ⊕ (b ⊕ c)
Some properties:
a ⊕ b = (a + b) + ((a + b) ⋅ a ⋅ b) (can this expression be simplified?)
a ⊕ a = 0
0 ⊕ a = a
a ⊕ neg(a) = 0

There is no "not", because why would not(0) be 1 over 1̃? In any case both variant can be implemented as (a ⊕ 1) and (a ⊕ 1̃) respectively.


Can the logic operators do arithmetic? Yes they can!
Here's the truth table for a half adder
| 1̃ 0 1
----+-------
1̃ | 1 1̃ 0
0 | 1̃ 0 1
1 | 0 1 1̃
That is, (a ⊕ b) + ((a + b) ⋅ neg(a ⋅ b)). Can this expression be simplified?

And here's the truth table for a full adder
| 1̃1̃ 1̃0 1̃1 01 00 01̃ 11̃ 10 11
----+------------------------------
1̃ | 1̃0 1̃1 01̃ 00 01̃ 1̃1 01̃ 00 01
0 | 1̃1 01̃ 00 01 00 01̃ 00 01 11̃
1 | 01 00 01 11̃ 01 00 01 11̃ 10
Now I haven't got a clue as to where to start to get a toolean expression out of this.


What do you think about all this? Did I make any mistake?
Some more things I want to look into:
1. Other properties of toolean logic, especially to help find and simplify expressions
2. Find an expression for the full adder
3. Design ternary CMOS circuits
4. Check out what people smarter than me have written, but I want to play around a bit more before.

P91162
WTF
In boolean they are trivial, like
y(x1, x2, x3, x4) = !x1 & x2 & x3 & !x4 for y(x...)={1 if x==0110; 0 otherwise}

P91174
>Can't read french.
Lol, git gud. But yeah

P91176
Probably not, but I haven't ran an exhaustive search for those.
Thanks for the code, I see I made a dum-dum now. I assumed "Not(a+b) === Not(a)*Not(b)" and "Not(a*b) === Not(a)+Not(b)" were equivalent, which they are not, since not(not(a))≠a. What language is that btw? Looks convenient.


If I didn't make any other mistake, there is no logic that verify:
1. "or" and "and" are associative and commutative
2. None of the operators is a constant
3. The logic satisfies de Morgan's laws
4. "and" distributes over "or"
5. The set of operators is complete, in that it can be used to generate all binary operators by composition.

I can find some if I loosen any of those conditions though. Attached are the logics I found by removing the distributy constraints. Obviously, there are duplicates then.

P91183
> What language is that btw?
Why, it's Haskell.

> I can find some if I loosen any of those conditions though. Attached are the logics I found by removing the distributy constraints.
I would like to remove 3 (De Morgan) instead. Those look like being ad hoc for boolean logic.

P91229
>Why, it's Haskell.
I really should use more functional programming, it makes the code so much clearer.

>I would like to remove 3 (De Morgan) instead.
Fair enough. I'm running the search for those.

P91176
I've found 333 of them. I should have optimised my code better, because it took 12h.
Results and code attached. "ternary_ops.py" is a bunch of helper functions, "find_nands.py" is the script that runs the search.

>>P91261
> 12h
Dude, just use CVC.

henlo frens
wat r sum god places 2 git sick epic classic math books 4 cheap?
asking 4 fren of fren

https://annas-archive.org/
It has everything from libgen, z-library and others.

P90160
doesnt have asstr tbh

P90115
no u XD
P90131
heyo that,s kekkou neato desu
i didn,nt know aboutthat
P90132
P90141
P90160
P90166
sory frens i shoould have clarified i was looking 2 build a colection of physical (paper books

P90615
>i was looking 2 build a colection of physical (paper books
1. buy printer
2. download files from P90132 or P90160
3. print
4. punch holes in the paper
5. put in binder

First, try writing like a non-retard.

Can someone explain where the following mathy things are used?
I'm interested in where it is used mainly, and also how it's used.

>Fourier Series
I know it's used to solve heat conduction equation and that's why the autist Fourier came up with all this but I'm sure there's more uses to this
>Partial Differential Equations
>Analytic Functions
>Complex Interpretations
>Vector Calculus

>Fourier Series
Spectral analysis
>Analytic Functions
>Complex Interpretations
>Vector Calculus
Almost everywhere. It's a basis of applied maths.
>Partial Differential Equations
Same. They are more obscure though, like string/membrane oscillations or gas flow.

>Fourier Series
Aren't the used all the time in signal processing? Anything to do with a superposition of waves really. Visible light, radio, sound, etc. Heard there are some more niche uses as well.
>PDE
I think these are used to death in finance. Don't ask me why. Something something modelling unknown functions.
>Analytic Functions
Aren't they related to doing approximations? If so, comp sci is a large field of use.
>Complex Interpretations
AFAIK they are used for getting really weird integrals, quantum mechanics, and solving polynomial equations. Also fractals.
>Vector Calculus
Fields. Gravity, electromagnetism, different flows. Makes sense since in these fields you are working with 3d spaces, movement, intensities, etc.

P89351
P89353
I'm actually looking for where they're used in Physics, also Physics that does not involve Electricity, Magnetism, Optics and all those electron or light related topics but things like force, heat, etc.

>Gravity
>different flows
>string/membrane oscillations or gas flow
Good, this is fine. I'll read more into this.

P89453
Those topics are ubiquitous. F(vector)=m d2/dt2 x(vector), well, the whole theoretical mechanics is a bunch of vectors and their integrals. Heat dissipation may be modelled by PDE. What else? I don't know.