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/sci/ - Science & Math


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15924439 No.15924439 [Reply] [Original]

Wisdom of the Occident edition
talk maths, formerly >>15898763

>> No.15924447
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15924447

4chan should separate sci and math into two boards

>> No.15924450 [DELETED] 
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15924450

>>15924439
Planning on starting to learn math...Where do I start? I don't know the basics. I'm black btw.

>> No.15924451

>>15924450
Cute bunny

>> No.15924471

>>15924447
empirifags need someone to keep them humble

>> No.15924613

>>15924471
that never actually plays out

>> No.15924621
File: 186 KB, 307x409, Screen Shot 2023-12-15 at 7.58.17.png [View same] [iqdb] [saucenao] [google]
15924621

relearning maths starting from arithmetics -> algebra -> geometry/trigonometry -> linear algebra -> calculus so i can start game development

just finishing studying arithmetics using quick arithmetic a self teaching guide and about to buy pic related.

anyway as i understand it linear algebra and trigonometry are basically pre-calculus, is "algebra 2" also part of the pre-calculus meme?

>> No.15924676

>>15924621
Pre calculus is trig, algebra 1 and algebra 2. Linear algebra is a separate subject from calc 1 and calc 2 but becomes relevant in calc 3

>> No.15924719

>>15924439
>Wisdom
wisdom only means realizing the true depths of human depravity and corruption.

>> No.15924909

i graduated summer of 2023 with a published paper under my belt, but decided to look for a job for a year before maybe considering going back for a phd. it's now almost the new year and i have yet to find a job because it is basically the worst market for math -> software dev in the past like 5 years. what are my chances of getting into a decent phd program for next fall at this point? will i be significantly less competitive because i didn't immediately apply for a phd?

because of a clerical error, i technically didn't graduate until fall 2023, but i haven't taken any classes since summer because i finished my degree reqs then.

>> No.15924920
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15924920

i own eevee, because of this my DNA doesn't split - i live on.

>> No.15924921

FACT: The axiom of choice is false in the true universe of sets and only holds in models which are missing most of the real numbers.

>> No.15925242
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15925242

>>15924921
FACT: I don't give a crap. I VILL use ZFC and I VILL be happy.

>> No.15925265

wow this general has become absolute trash after like 2022. what happened to all the regs?

>> No.15925273

>>15925242
Look at large cardinals for long enough and you'll see the error of your ways.

>> No.15925300

>>15924909
I graduated in 2017 and started my PhD in 2021. Got into a decent enough school for me. I don't think most institutions care too much unless you're shooting for a top 20 school (assuming you're in the US).

>> No.15925304

>>15925300
thanks for the data point! how's your phd going?

>> No.15925306

>>15924921
AC is true in L which is a subclass that can be built from ZF minus AC. It is just the place where all actual mathematics take place.

>> No.15925326

>>15925306
L is very obviously not the true universe of sets, it's a minimal model with as many sets missing as possible while still satisfying the axioms of ZF.
Try thinking about why power set holds in a model consisting only of the Von Neumann naturals, even though union fails.

>> No.15925379

>>15925326
>Try thinking about why power set holds in a model consisting only of the Von Neumann naturals, even though union fails.
If you're referring about the Ackermann model, it satisfies all ZF with the infinity axiom removed, including the union axiom.

>> No.15925402

>>15925379
>If you're referring about the Ackermann model
I am not.
The model I'm referring to looks like:
{{}, {{}}, {{},{{}}}, {{},{{}},{{},{{}}}}...}
and doesn't contain, say, {{},{{},{{}}}}

>> No.15925508

>>15924676
>Linear algebra is a separate subject
lmao, I think I can handle y=mx+c
do g*rmans really have to take a whole course on it?

>> No.15925524

>>15925508
Linear algebra is about vector spaces and linear transformations, not the analytic equation of a line

>> No.15925531

>>15925273
what do you mean? I simply don't give a shit about set theory, I just recognize it is a practical framework where I can dump all my autism energy into.

>> No.15925536

>>15925524
>t. smoothbrained europoor who can't understand mathematics

>> No.15925761

Fellas i think i bombed both my topology and analysis finals. i had been doing great on assignments but i showed up to the exams and just couldnt hack the technique. hopefully the prof wont think im stupid i wanted to work with him.

>> No.15925766
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15925766

Need a textbook recommendation for college algebra. I recently failed my college algebra class because the entire class was the final and the final was all proofs. Because the prof didn't assign a textbook and didn't have any insight on a textbook besides "google it", I'm looking for a textbook recommendation. I already know a lot of trig and algebra up to logarithims and graphing, I can divide a polynominal and work with radians but anything involving a series/choose/choice and inductive proofs is hard for me. I can deconstruct an expression using partial fraction decomposition but I can't do the proof. I'm leaning towards the openstax algebra book since reddit approves of it, but I was wondering if anything better existed. I already have a couple precalculus books and I know most of the material in them, but they don't cover proofs or any of the actual algebraic operations except as it pertains to calculus.

I need to know this because my junior college won't let me take calculus until I can do this.

>> No.15925771

>>15925508

americans have to take an entire course on it otherwise this >>15925766 happens

being able to graph an absolute value equation is not the same as writing a proof to show that two fractions multiplied by the same number equal a singular value

>> No.15925779

>>15925766
>the final was all proofs
wtf? I never took college algebra, but they made you do proofs in it? In my college I was convinced it was some slightly more difficult version of highschool algebra 2.

>> No.15925813

>>15925779

My specific professor is also the math dept head and he was annoyed that none of his MV calc, linear algebra and latex students knew how to prove cramer's rule, the power rule, the binomial theorem et cetera so when his turn came to do all the entry level math students this was what he decided he was going to teach. The entire final is just 10 questions of proofs which he spent the previous four months lecturing about. I was retarded because it's an easy final, it is literally the same lecture questions but on the exam, and he told everyone the best way to study was to copy it down verbatim because we'd be using it in calculus. There was some graphing but only absolute value, inequalities and piecewise graphs because his calc students keep getting that wrong too.

He also hates the internet and doesn't post anything to canvas either which annoys the other professors, but he's their boss.

>> No.15925819

>>15925813
I can't fathom how that was allowed. Was this was some non-western country? Some extra for students or classes doing well, and if they're students pursuing degrees with more advanced work later, in a similar situation I'd consider introducing people to proofs and giving extra credit for work on them. Particularly for students already getting A's who need more advanced material.

But the entire class? With just rote memorization for the final? You should've done well if you followed instruction to just rote memorize, so yes you were stupid for just not bothering, but holy shit that professor is retarded on so many levels. Is this a situation where you can inform the dean or superiors of how extremely off subject the algebra class material was? That can't possibly be allowable. But you may just be screwed and should've made some complaint about how irrelevant the material was for the class level to his superior once it began to go off the rails.

>> No.15925878

>>15925819
>young saars please doing the needful to reproduce these canned proofs in exam
>be sure to study every day to remember for great success
>thanking u my basterds
He's got to be Bharatiya. Nobody else would have such a bizarre view of teaching proofs in math education.

>> No.15925887

>>15925878
Well I have no personal experience with India. From proxy stories I was thinking India or a professor from India, as I've heard many horror stories of the elitist sort of rote memorization walls they put up to filter people who can afford tutors from those who can't among other things. It's such a fucked up system from what I generally have heard but I haven't looked into it beyond those stories so I've no idea how true or common any of it is.

If it was India and it's as bad as I hear then yeah his only option may be rote memorization.

>> No.15925894

>>15925887
Even worse it produces graduates who don't know shit about fuck, and can't actually prove anything they don't already know.

>> No.15925897
File: 1.10 MB, 476x267, 93200664usa-flag-amercan-waving-gif.gif [View same] [iqdb] [saucenao] [google]
15925897

>>15925819
>>15925878

I'm white and the prof is white too, but the professor does all the higher-level math classes and didn't want to do the entry level math. However, the school requires all the math teachers to do some amount of entry/lower division level instruction per semester to reduce burnout. I don't know how it is in Europe but the college didn't want to hire 3+ high school math teachers just to teach 1st semester college math because it'd mean paying a teacher a professor's wage or it'd mean making a professor teach retards. He also takes all the complaints when the professors whine about how they're given the dumb math class/students who are just taking it for a non-transfer degree. The prof in question has written several books on algebra and used to work at a real university.

This is probably a USA thing, again this is only a junior college and not even a state school. My class started with 35 people, half early college teenagers who tested out of highschool at 16 the other half being working adults. Only 15 made it to the final, and since he's curving the grades I'll probably still pass with a D+ despite getting every question wrong on the final because even a 15/100 is higher than a 0.

It's a weird situation and doesn't make sense in countries that aren't America, because in the US all of us would have been pushed into a technical school instead.

>> No.15925899

>>15925894
China does something similar from what I have been told. I talk to a fairly diverse group of people from many countries as relates to mathematics and almost every non-western country seems to do that. Mathematics especially seems to be used as an elitist rote memory filter to get rid of people instead of teaching mathematics. Makes their national scores and academics look better than they really are for ignorant westerners by just trying to get rid of lower or middle grade students, and helps reinforce social structures by making it exceedingly difficult for anybody poor to get prior training for tests and courses. I believe some western countries also do this too but it is more dependent on the institution than entire country like it is in China.

India is a very large place with many provinces and I genuinely do not know how widespread that issue is there. China is also a bit like that with many separate provinces but I believe they have more strict and enforced standards. Would be interested for input on that as I mainly hear stories from people. In my country things are a bit crazy so you just end up with some professors who are crazy or very few private institutions who do that to keep the poor people out.

>> No.15925900

>>15925897
Oh my apologies. Yes in the USA it would probably be on a professor or institution basis too. Your story just sounded so much like every story I have heard from India that I agreed with the other anon it very much sounded like India. I meant nothing by it other than that despite how people around here tend to post.

I think in the USA you should have been able to appeal to the dean, but now may be too late. You should talk to whoever the student representatives are and even if it is too late then make the school administration aware of it and the problems he is causing for teaching such off-topic material in such a stupid way.

>> No.15925904

>>15925900

Appeal? lmfao I already paid the school $400 to sit through this. It's actually not plausible for me to even sit in the classroom without paying in full first. No chargebacks. They legit do not care after they got my money. I don't know how it is outside the US but there's nobody to appeal to but the dept head and the person I am talking about is the dept head. At least where I am nobody else has the ability to change grades and even then the only way to change a grade is either "pass" or "no pass" not a letter grade, which means no transfer credit is earned. Doesn't matter for people who are only getting 2-year degrees but anyone taking a 4-year degree would have to retake the class regardless as state schools don't recognize a P grade.

When confronted about this by the other professors his response is "find another place to work at" which is amusing because it's the exact same response my highschool math teacher got when he ran out of scantrons and complained to his boss. Which is probably why I have to take algebra in college in the first place lol.

>> No.15925908

>>15925904
What kind of institution is this? Even community colleges have a dean and administrators. Head of department is not where the buck stops normally unless this is something strange like a prep school or private school? I have not heard of a department head just being able to fuck everyone up.

>> No.15925916
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15925916

>>15925908

It's a junior college. The dean exists to accept complaints from students and be student HR. They protect the school from lawsuits when students are kicked out for poor grades, not paying, smoking weed on campus or bipping cars. The dean isn't there to babysit students' problems with their professors unless it involves their race, gender, or sexuality. The dean legit doesn't care if students fail because the exam is stupid, especially with a large amount of the professors grade on a bell curve anyway to cap off the amount of students going into the higher courses. This is extremely common at the state schools where I live, they get so many applicants the school wouldn't be able to handle 10,000+ advanced math students in a single semester so they automatically cut off the lowest 50% based on the average final grade. Over three semesters this reduces the class size by 75%. The state has been sued over this and the courts say it's okay. No refunds even if the student otherwise had a passing score and retaking the class means another $5,000 on general pre-registration tuition. I personally know people who have had to blow an extra $10-15,000 retaking classes just to up their C+ to a B so they can be in the upper half and advance to the next level.

Which, obviously, is why so many people choose junior colleges now because tuition is only $100ish and $45 per unit/credit and not $5,000 base plus $450 per credit. But there's no higher division courses.

>> No.15925930

>>15925908
>>15925916

Also, he's not fucking everyone up, since there's no standardized testing at the college level therefore there is no way to determine if one student is better or worse than another. The whole state is doing away with the SAT/ACTs for traditional essays again anyway, and my college doesn't even require a GED to attend although a GED is required for transfers. The only thing that matters is the final transfer rate, and the ultimate success rate from the state school where the starting class size is about 100. But for students that are NOT transferring, and just want to be plumbers or some other labor job, this is irrelevant and he knows that.

Having professors that give a shit about you personally only begins at the 3rd or 4th year where the class size is under 20 and a single student dropping out because of a single class will make the prof look much worse. Or at least that's what my friends say, since it's impossible for a professor to track his ~500 or so lower division students and why would he when they are useless for his own research work.

>> No.15925935

>>15925916
Oh so it is a kind of weird ugly-duckling thing. Sounds like a weird almost-prep-but-not hybrid institution. Well your grade may be fucked then either way but it can't hurt to lodge a complaint, and to make the complaint hold more force explicitly state you are not complaining to change the grade but complaining about the course material irrelevance and professor as you've noted. There's also professor rating websites which can, at times, help people avoid horrible and abusive professors.

My only advice to you would be to try and avoid professors like this in future by checking who will be teaching your classes, and even withdrawing to pick another professor if need be if you get stuck with a bad one. If you can't change anything the only thing you can do is use everything you can to avoid this happening again for future classes, and that may mean changing to a state community college or other alternative too if it is unavoidable.

Also being you are learning or supposed to learn low level material, Khan Academy would be a great place to start. That way even though you are getting screwed by bad professors you can still learn what you are supposed to learn so you don't fail once you are in regular college classes. If you wish to learn about proofs I have found this channel listed here and I think very easy even for starting undergrads to begin with
https://www.youtube.com/@brightsideofmaths/playlists
This is the starting playlist that begins at sets and logic https://www.youtube.com/watch?v=N-X1EU7tHVo&list=PLBh2i93oe2qtbygdXz4u6Mkh7c_hMLBA8

So if you are stuck for the time being at least you can learn what the proofs and symbols all mean and may be able to muddle forward that way. Definitely use Khan Academy, too, to learn what you were supposed to be learning.

>> No.15925937

>>15925930
Trust me your professor really is just an asshole. It is not about size of class he's just a total prick. You do not have to care about each student individually to realize you're fucking people over by teaching them wrong material for their level and even making good students develop maths anxiety with stupid shit like that.

>> No.15925943

>>15925935

It's a junior college. A local community college. A discount county college that feeds the local state college and that provides various types of skilled degrees. For example, my JC also trains all the local firemen, nurses and pipefitters because that's what the county could get money for. I'm not sure if there is an equivalent to this in Europe, since Europeans have technical schools instead. Here, that's all private like Wyotech (which is arguably better, but only does car repair related degrees and nothing else). And no I can't avoid him because he's the only math prof willing to do the night class and I work during the day.

I'm just looking for a textbook recombination.

>> No.15925972

>>15925306
Nah, L(R) fits mathematical practice better.

>> No.15926123

>modal logic
is this actually something used in mathematics or is it just some philosophy shit?
we already have implication for necessity, wtf is a possibility supposed to mean?

>> No.15926138

>>15926123
It's usually interpreted through possible world semantics, i.e. p is possible if there is a possible world where x holds and p is necessary if it holds in all possible worlds. It is not mathematically relevant, more a thing for philosophers indeed. They also usually don't care about material conditionals and instead think about implications modally like this: a implies b if in every possible world where a holds, b holds as well.

>> No.15926151

Do you guys find discrete or continuous stuff more fun

>> No.15926225
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15926225

>>15926151
Whichever one makes it easier to calculate the optimal point of assmad you have to make a (surprisingly retarded) janny for them to archive a thread just to try and get the last word in. pic rel

But in general usually continuous stuff. Discrete just always felt like a chore.

>> No.15926238

>>15926123
It has simple semantics for instance in the case if S4 logic you can interpret [math]\box F[/math] as the interior of (the interpretation of) [math]F[/math].

>> No.15926241

>>15926238
>It has simple semantics
in terms of the subsets of a given topological space.

>> No.15926254
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15926254

Are all linear functionals of germs of smooth functions automatically derivations, or for tangent vectors do we only consider those functionals that satisfy Leibniz rule? I ask because ive seen tangent vectors defined both ways. I think the first case is correct if we consider the purpose of tangent vectors and how the derivative should behave but idk

>> No.15926271

>>15926254
Nevermind im an idiot, just consider v(f)=f(p). Not a derivation.

>> No.15926302

How do I convert How do I convert [math] \left( - \frac 1 2, \frac 1 2 \sqrt 3 \right)[/math] to polar?
The usual formula gives [math] (1, -\pi/3) [/math] which is the negative of the cartesian coordinate. What is going wrong?
[math] \pi/3 = \arctan ( (\sqrt3/2) / (-1/2) )[/math] but [math] \cos (-\pi /3) = 1/2 [/math]

>> No.15926317

Tried showing [math] S^2 [/math] cannot be the total space of a circle bundle [math] S^1\rightarrow S^2\rightarrow B [/math] is this the right way to go about it?
Basically look at the long exact sequence of homotopy groups, simplify it down and get [eqn] \cdots \rightarrow 0\rightarrow\mathbb{Z}\rightarrow\pi_2(B)\rightarrow\mathbb{Z}\rightarrow0\rightarrow\pi_1(B)\rightarrow\cdots [/eqn] So [math] \pi_2(B) [/math] must be isomorphic to [math] \mathbb{Z} [/math] right? But the total space [math] S^2 [/math] is 2-dimensional and the fibers [math] S^1 [/math] are 1-dimensional, so the dimension of the base space [math] B [/math] must also be 1-dimensional, right? So it must be essentially either a line or a circle, both of which have trivial 2nd homotopy group, right and so we get a contradiction?
I dont know shit about algebraic topology just wondering if the logic is sound.

>> No.15926321

>>15926302
atan gives the same value for (x,y) as it will for (-x,-y), so it can't do all of the work on its own. You have to adjust your answer based upon which quadrant it's in.
similarly, cos is an even function, so it can't distinguish between a positive and a negative input

>> No.15926384

>>15926302
This is why math programming lib have atan2(y,x)

>> No.15926409

Are there any concepts that can't be defined?

>> No.15926469

>>15926409
Truth can't be defined.

>> No.15926488

>>15926469
what.

>> No.15926535

>>15926488
He's referring to Tarski's undefinability theorem

>> No.15926559

>>15926535
ok but y tho
incompleteness != can't be defined at all
As a meme?

>> No.15926938
File: 50 KB, 1117x602, Screen Shot 2023-12-16 at 5.00.19 PM.png [View same] [iqdb] [saucenao] [google]
15926938

>>15924439
Consider the following formula.
[eqn]
\sum_{n=1}^\infty \left(
\left( \frac{r}{1+r^2} \right)^{2n-1} \sum_{j=1}^n \frac{(2n-1)!}{(2n-1-j)! j! b^{(2n+1-2j)/2}}
-\left( \frac{r}{1+r^2} \right)^{2n} \sum_{k=1}^n \frac{(2n)!}{(2n-k)! k! b^{(2n+2-2k)/2}}
\right)
[/eqn]
with [math] r = \frac{-1}{\sqrt{3}} [/math] and [math] b = 3 [/math]. As you can see from the included image, my plot on Desmos strongly suggests that this infinite sum adds up to [math] \frac{-11}{9} [/math]. However, a plot is not a proof. Can anyone please prove (or disprove) my hunch?

>> No.15926984

>>15926938
How in God’s name did you even arrive at that series? That’d be pretty useful to explain, if you’re gonna ask a question as convoluted as that.

>> No.15926985

>>15926409
>Are there any concepts that can't be defined?
No. x=x works for all concepts. A dog is a dog might be a shitty definition, but it's a definition, nonetheless. And if you try to remove self-reference from the definition of a definition, a lot of shit breaks, so you can't.

>> No.15927079

>>15926984
>>15926938

Well anyways, I was bored. The inner series quickly becomes sum(1/infinty), which is 0. So the outer series must converge to something, since every term after like the sixth is multiplied by 0. I tried sterling's on it, and even thought about bringing out the product log and the maclaurin crap, but my best guess is turning the sum into an integral, and using Wolfram or integralcalc to crunch it. (be sure to add 1/2 to each of the series' bounds for its integral form, so the integral approximates the series)

/calculator/ 9zltczn0ne

>> No.15927114

>>15926984
It's kinda convoluted to explain, and I don't think it would add much if I did. In fact, the actual series I want to work with is a little different from this one. But as far as I can tell it doesn't converge to a nice rational number like -11/9. I'm pretty rusty on calculating what sums converge to so I thought I would try and get an explanation here and make sure I understand it, then apply similar methods to other sums I am interested in as well.

>>15927079
Sterling's approximation is a good idea, I will try that. I am worried that it will add in inaccuracies for low n terms though. What do you mean by the following?
>bringing out the product log
>and the maclaurin rap
>turning the sum into an integral

>> No.15927138

>>15926938
Note this is equivalent to the double sum [eqn] -\sum_{n=1}^\infty 16^{-n} \sum_{j=1}^n 3^{j-1}\bigg( 4\binom{2n-1}{j}+\binom{2n}{j} \bigg) [/eqn] try to evaluate each double sum separately, or maybe there are some combinatorial sums, you know like finite sums of binomial coefficient stuff.

>> No.15927152

>>15927138
Btw to get this form i just plugged in your r and b values and combined powers of n, very simple. But calculating binomial sums is well studied and this form is a lot easier on the eyes, no?

>> No.15927189

>>15927114
calculator/1kt4q2kvkp

Ok, here's a crappy proof that it converges to a constant, specifically a function of the form -x/e^x +C, which has the x part eliminated as 1/e^x goes to infinty.

Series which involve combinations, usually behave linearly for whatever reason. I approximated the linear function's slope (still crudely but close enough), which left behind a simple series. I separated the series and turned them into integrals, ignored all those silly constants, and showed it ultimately turns into -x/e^x + something constant.

>> No.15927204

>>15927114
Turning a sum into an integral is actually very simple. Sums and integrals are the same idea, just an area below a curve. A sum starts out a bit further ahead or behind, compared to it's integral form, cuz sums are all blocky shaped.

A maclaurin series is turning a function into a sum of derivatives, in order to obtain a more manageable, approximate function ( that totally breaks down at some chosen arbitrary point far away. )

The product logarithm is the inverse of x * e^x, similar to how a logarithm is the inverse of e^x. It can be expressed as a series in a lot of ways. I had wondered if I could recognize that series, as a product log series. Then I would've used wolfram alpha to help me with that.

>> No.15927348

>>15927114
/calculator/4usp0jkitl

I got a better approximation this time. You can do the honors of punching that final integral into wolfram alpha and seeing the true form of your vanishing function, and how those constants turn into -11/9.

It doesn't much matter to me.

>> No.15927366

>>15927079
>>15927189
>>15927348

>/calculator/ 9zltczn0ne
>calculator/1kt4q2kvkp
>/calculator/4usp0jkitl

I'm not sure what these mean.

>> No.15927370

That final trick was that, every curve is approximately a parabola, when you're really close to the curve's maximum/minimum.

Cuz any function's expansion series can be written like A+cx(m)+cx^2(mm/2)+cx^3(mmm/6).... where m is arbitrary, very very small number, which makes each term always smaller than the previous term.

But since A doesnt matter, and cx is zero around minimum and maximum, the largest term that describes the function is cx^2(mm/2), a parabola.

>> No.15927376

>>15927366
They're desmos links. I was worried I would be flagged for posting an actual link.

https://www.desmos.com/calculator/4usp0jkitl

>> No.15927377

>>15927376
>9zltczn0ne
Oh ok, gotcha. Thanks.

>> No.15927492
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15927492

https://www.desmos.com/calculator/z6nz3srykq

in case you weren't able to figure out how to use the integral calculators

seems that an input of 1/16 cancels a lot of stuff out.

>> No.15927496

>>15927138
>>15927152
A lot of the stuff I'm finding on binomial sums seems to be based on the binomial theorem, which in this case would start with j=0 rather than j=1. I suppose I could account for that by just adding in the 0th term and then subtracting it out. But the bigger issue is that the top part of the binomial here is 2n-1 or 2n, rather than simply n. I'm not sure how to handle that, since the index of the inner sum still only goes up to n.

>> No.15927521

Basically 47/-64 times 5/3. -64 is a constant, and 5/3 and 47 are functions of the input. 5/3 has a known function, but 47 does not. 47 comes from the slope of the parabola used to approximate the inner series. This slope should still scale linearly with the 3, and it does seems to be 3*(T)^2

>> No.15927668

Can someone tell me how I'm supposed to write A U B in terms of A, B, and *. With A * B = (E - A) ∩ (E - B)
The answer is supposed to be (A * B) * (A * B) but I have no idea how I'm supposed to arrive at that. This is in chapter 0 of the book and all I have are informal set operator definitions

>> No.15927672

>>15927668
Also, working backwards so far I've figured out that E - (E - A) = E ∩ A, but I don't know how to get union from anything

>> No.15927706

>>15927672
draw a diagram

>> No.15927711

>>15927706
Only brainlets need diagrams. Anon needs to reason purely symbolically if he wants to make it.

>> No.15927717

>>15927706
Thanks but I figured it out with facts and logic. Drawing is cheating. What do you do when your sets are in n dimensions? I don't want to rely on a crutch

>> No.15927721

>>15927711
Hell yeah. However I would have been cometely lost without the solution so I guess I ended up cheating anyway. starting from the solution is a reasonable problem but going the other way is pulling expressions out of nowhere. This is motherfucking chapter ZERO

>> No.15927736

>>15927717
the diagram is for inspiration

>> No.15927773

>>15927736
I was joking, I did use some diagrams. I can't imagine thinking about sets without visualizing at all

>> No.15927993

Is there a sub branch for, not sure what to call it, "discrete topology?" Question in mind is about how a cube has 8 corners and 6 faces, but a pyramid on a 7-sided base has 8 corners and 8 faces.

>> No.15928002

>>15927993
https://en.wikipedia.org/wiki/Polyhedral_combinatorics

v − e + f = 2

>> No.15928026
File: 162 KB, 1440x494, Screenshot_20231217-030301_Firefox.jpg [View same] [iqdb] [saucenao] [google]
15928026

>>15928002
Thanks. And lol at this formula, I had run into a similar conundrum before, I didn't know it was acceptable to just include extra instructions with an asterisk

>> No.15928159

>>15925524
>assuming y and x are single dimension variables
lmaoing at your single dimension brain rn

>> No.15928227

Any good abstract algebra materials?

>> No.15928355

>>15928227
Aluffi.

>> No.15929612

>>15928227
Rotman

>> No.15929667
File: 169 KB, 1420x1188, b4ebbd2136b57371741c71d7c82f606e[1].jpg [View same] [iqdb] [saucenao] [google]
15929667

It's all just sets and mapping (which are also sets).

>> No.15929812

>>15929667
It's the other way around. It's all mappings and relations between sets (which are mappings).

>> No.15929826

>>15929812
What is the empty set a mapping of?

>> No.15930166

let R be a relation on a set X:
1)for all x,y,z \in X if xRy and xRz , then yRz
2)for all x \in X there exist b,c \in X such that bRc and bRx
How do I prove this is an ER?

>> No.15930195

>>15930166
First you prove it reflexive. Then you prove its symmetric and as a last step you prove it's transitive.

>> No.15930203

>>15930195
Yeah I don't know how to

>> No.15930212

>>15930203
>Reflexivity
Take any x in X. Then by 2) there is a b in X with bRx.
Since you know bRx and bRx with 1) you get xRx.
>Symmetry
Suppose that xRy. By reflexivity you know that xRx so with 1) you get yRx.
>Transitivity
Suppose that xRy and yRz. By symmetry you know yRx.
Since yRx and yRz with 1) you get xRz.

>> No.15930281

>>15930166
Why are you convinced this is an ER? Do we assume b,c are distinct elements? X could contain one element with a reflexive relation, satisfy this, and not be an ER.

>> No.15930472

>>15929826
It's a map with no domain

>> No.15930633

If a topological space is compact and non-orientable, does it necessarily mean there is an embedding of the Mobius band into it? I know this is true for 2-dimensional manifolds.

>> No.15930734 [DELETED] 
File: 33 KB, 849x747, riddle.png [View same] [iqdb] [saucenao] [google]
15930734

Here's a brain melting geometric riddle that I came up with.

In picrel, the line or the circle rolls (without skidding) so that the points A and B will come to contact with each other. Now imagine that the line in picrelated was horizontal, and the circle rolled on it and the point A traced a curve (a cycloid to be specific). Then imagine an alternate situation where the circle stayed stationary where it is, and it was the line that rolled along the circle, so that the point B (a fixed point on the line) would now trace a curve. The riddle is this: are the two curves the same curve, or are they two different types of curves?

I came up with this by playing around with Desmos and it took me a really long time to wrap my head around this.

>> No.15930745
File: 33 KB, 849x747, riddle.png [View same] [iqdb] [saucenao] [google]
15930745

Here's a brain melting puzzle that I came up with.

In picrel, The point B is a fixed point on the line and point A a fixed point on the circumference of the circle. They are positioned so that if the circle rolled on the line (without skidding), the points would come to contact with each other.

Now imagine that the line in picrelated was horizontal tangent on top of the circle, and the circle rolled on it so that the point A traced a curve (a cycloid to be specific). Then imagine an alternate situation where the circle stayed stationary where it is, and it was the line that rolled along the circle, so that the point B would now trace a curve. The riddle is this: are the two curves the same, or are they two different curves?

I came up with this by playing around with Desmos and it took me a really long time to wrap my head around this.

>> No.15930953

>>15930633
We usually don't talk about orientation for anything other than manifolds. When the dimension is three or more, the Möbius band embeds in any such manifold, since you can embed it in a ball.

>> No.15931253

>>15930953
That makes sense, thanks anon

>> No.15931704 [DELETED] 

>>15924439

Let us say a given planar set A has property P if for every equilateral triangle T contained in A , if C is the circle circumscribing T (i.e., C is the circle through the vertices of T) , then C is also contained in A .

Is there an open connected nonempty planar set A which has property P , but is not an open disk?

>> No.15931705 [DELETED] 

>>15931704
I'm in the jaws of the beast.

>> No.15931720 [DELETED] 

>>15931705
??

>> No.15931726 [DELETED] 

>>15930633
I'm think if a smooth manifold M is non-orientable, then there is a smooth map (probably can be an embedding) [math] \gamma : S^1 \rightarrow M [/math] such that [math] \gamma^*TM [/math] conatins a Mobius strip as a line subbundle. I don't think we need M to be compact either.

>> No.15931727

>>15924621
start from propositional logic

>> No.15931734

>>15930633
I think if a smooth manifold M is non-orientable, then there is a smooth map (probably can also be an embedding) [math] \gamma : S^1 \rightarrow M [/math] such that [math] \gamma^* TM [/math] contains a Mobius strip as a line subbundle. I don't think we need M to be compact either.

>> No.15931741

>>15924439

Let us say a given planar set A has property P if for any equilateral triangle T contained in A , if C is the circle circumscribing T (i.e., C is the circle through the vertices of T) , then C is also contained in A .

Question: Is there an open connected nonempty planar set A which has property P , but is not an open disk?


Not sure if this is a stupid question, but I've thought about this for a bit and I can't find a counterexample. I'm also not sure how one would prove A must be an open disk, though.

>> No.15931768

>>15931741
Actually wait, I don't think an open disk has property P.

Then my question is: if an open planar set A has property P, then is A necessarily the entire plane?

>> No.15931771

>>15924447
HE
WILL NOT
DIVIDE US

>> No.15931979

>>15931768
Open disks don't have that property? For a closed disk, any equilateral triangle with two vertices on the disk will have the circumscribed circle contained in the closed disk so long as the side length is less than [math]r\sqrt{3}[/math], which it has to be or the triangle won't fit in the disk. Then any equilateral triangle with only one point on the circle is the same as taking some triangle with two and rolling it around the boundary of the disk, which keeps the circumscribed circle inside. And if the triangle has no points on the boundary then the condition is obvious. So that part holds by some pretty basic planar geometry. Then do some argument about limits with closed disks of increasing radii filling the open disk to argue that this implies the result for the open disk and you're done. Claiming that all such regions must be open disks seems like the hard direction to me.

>> No.15931981

>>15931979
Oh there was no need to be vague actually the limiting argument is super easy. Take any equilateral triangle contained in the open disk and choose some radius [math]r'[/math] smaller than the radius of the open disk such that the closed disk centered at the same point as the open disk also contains the triangle ([math]r'[/math] can just be the maximum of the distances between the center and each vertex of the triangle).

>> No.15932129

>>15931741
>>15931979
Thinking more about this. I have a hunch you'll want any such set to have a boundary with strictly positive curvature. If there's any point with zero curvature then I think you can sneak a triangle close enough to the boundary that the circle around it will cross through. Requires more work but I think that might be a good direction.

>> No.15932198

>>15931979
>For a closed disk, any equilateral triangle with two vertices on the disk will have the circumscribed circle contained in the closed disk so long as the side length is less than r3‾√
, which it has to be or the triangle won't fit in the disk.
I think that fails in general; see this Desmos graph:
https://www.desmos.com/calculator/w1fznmvkmr

>> No.15932204

>>15932198
Also I actually asked this question on reddit: https://www.reddit.com/r/askmath/comments/18mcjh9/is_there_a_proper_nonempty_open_subset_of_the/

See the top comment on that post; it turns out any open (& nonempty) set with the property P described in >>15931741 must be the entire plane.

>> No.15932869

Is there a model in which CH fails strictly more than continuum many times?

>> No.15932948

>>15932869
>>15932869
Do you mean gch fails at more than continuum many cardinals or there are more than continuum many cardinal below the continuum? First is possible by easton's theorem, you can even have it fail at every regular cardinal. Second one is an easy no using the wellordering of cardinals, but you can have exactly continuum many cardinals below the continuum, for instance by making the continuum the aleph-oneth aleph fixed point.

>> No.15932976

>>15932948
>Do you mean gch fails at more than continuum many cardinals or there are more than continuum many cardinal below the continuum?
The latter.
>using the wellordering of cardinals
What about in ZF without C? Obviously Alephs kind of fall apart in that context, so I'll fall back on the original formulation of CH and say that what I mean is "Can there be more than continuum many intermediate sets between aleph-null and c under some prewellordering?".

>> No.15933161

can anyone recommend a book on computer algebra? not on a particular system, but a book covering theory of computer algebra and common algorithms

>> No.15933172

>>15924447
No, it should ban those that relate science and math to school

>> No.15933335

Why are there so many pajeets making math vids? Is this punishment for being an illiterate retard?

>> No.15933351

>>15933335
It's to piss you off.

>> No.15934029

What is an example of a field of characteristic 2 that is not isomorphic Z/2Z?

>> No.15934033

>>15934029
[eqn]\mathbb{F}_2((x))[/eqn]

>> No.15934035
File: 36 KB, 841x695, 2023-12-20-195040_841x695_scrot.png [View same] [iqdb] [saucenao] [google]
15934035

[math]\mathbf{M\text{-}Set}[/math] is the category of actions of a monoid [math]\mathbf{M}[/math] on a set. This category is a topos. It has a subobject classifier [math]\Omega=(L_M, \omega)[/math] where [math]L_M[/math] is the set of left ideals of [math]\mathbf{M}[/math], and [math]\omega[/math] is the action-map [math]\omega : M \times L_M \rightarrow L_M[/math] where [math]\omega(m, B) = \{n : n * m \in B\}[/math].
Maps between objects in [math]\mathbf{M\text{-}Set}[/math] are equivariant morphisms, that is, they preserve monoid actions such that they are invariant to application order (before or after traversing the map).
The truth-values of this topos are then the maps [math]1 \rightarrow \Omega[/math] which are just the left ideals.
The characteristic morphism describes the extent to which some object is equivalent to another object. In the context of [math]\mathbf{M\text{-}Set}[/math] say we have an inclusion [math](X, \lambda)\rightarrow (Y, \mu)[/math] where the first element is a set and the second its associated monoid action. Then the characteristic morphism describing [math]X[/math]'s membership in [math]Y[/math] is [math]\chi_f : Y \rightarrow L_M[/math] where [math]\chi_f(y)=\{m:\mu(m, y)\in X\}[/math].

This is in fact the only morphism that will commute the diagram shown to the right. Why?
Right answer gets infinite tendies in heaven

>> No.15934038

>>15934033
Is that polynomials in Z/2Z?

>> No.15934040

>>15934033
that is isomorphic to Z/2Z is it not?

φ(0) = [0]
φ(1) = [1]

for any a, b in F2
φ(a + b) = φ(a) + φ(b) (mod 2)
φ(a * b) = φ(a) * φ(b) (mod 2)

>> No.15934042
File: 138 KB, 2074x230, Screenshot 2023-12-21 at 9.26.16 AM.png [View same] [iqdb] [saucenao] [google]
15934042

What does linearly independent mean?

>> No.15934046

>>15934042
basically "cannot be combined to form each other"

>> No.15934049

>>15934046
What does combine mean?

>> No.15934050

>>15934038
No, it's the fraction field of the ring of power series with coefficients in Z/2Z.
>>15934040
A field is ismorphic to Z/2Z if and only if it has exactly 2 elements.

>> No.15934051

>>15927993
There is computational algebraic geometry

>> No.15934057

>>15934049
show that there is no way to add those numbers along with coefficients from [math]\mathbb{Q}[/math] to form 0 (sans the trivial combination where all coefficients are 0)

>> No.15934060

>>15934042
Any linear combination of them, with coefficients in Q, that equals zero, has coefficients all zero.

>> No.15934061

>>15934035
somebody answer me I know one of you faggots is smart enough. or has /sci/ fallen this far that it can only jerk off to undergraduate field theory all day

>> No.15934118

Is the universe Turing complete? You can make logic gates in it, and you can read and write data. And since it can simulate a turing complete system, does that mean that it is itself Turing complete?

>> No.15934133

>>15934061
I don't know enough about monoid actions to be sure of what [math]k: X \to Y[/math] is supposed to be, but I've seen enough undergraduate category theory to know that you can't go wrong by saying
>This is a consequence of unwrapping the definition of the universal property of the subobject classifier in a topos.

>> No.15934205

>>15934118
Yes in the same sense that cellular automata are often Turing complete.

>> No.15934362

Consider that category of contable metric spaces and cauchy-continuous functions. Equip it with the canonical topology to get a site, and form the topos of sheaves on this site. Does the internal logic of this topos satisfy markov's principle?

>> No.15934599

π=e
Math is so le beautiful...

>> No.15934707

Not sure if this is the place to ask, but maybe the math heads can help me understand this.

How do single player games against machine work on levels of difficulty?

I mean, I can understand a computer analyzing a chess board and figuring out what's the most beneficial move the best it can. But how do they make a chess machine that is 6/10 or 3/10 or 9/10 in terms of difficulty? Do they see less moves ahead than 10/10? Is there a random attribute that makes it commit mistakes every once in a while?

>> No.15934783

>>15934707
>Is there a random attribute that makes it commit mistakes every once in a while?
yeah that's pretty much it
it still knows what the best move is, but it'll deliberately play another from time to time. the harder the difficulty, the less frequently this happens and the better the suboptimal move will be
you can see this with especially "weak" chess bots, who'll play reasonably for a few moves and then just make a braindead blunder, and then go back to playing reasonably

>> No.15934898

>>15934042
why do you not know this? go read a linear algebra book before you continue with that

>> No.15934937

I'll be starting my masters next year at a top 50 university.

I am currently doing some independent research and will likely produce some interesting results by the time my classes start. What's the best way to use my new professors to have a shot at publishing at top tier journals?

Does anyone have any experience with this? Just to be clear, I would likely write the paper entirely on my own and would require no contribution from the professor. I would just want to use their connections in academia to reach higher tier journals. Is that a thing? Do these professors tend to have such connections? Enough to fast-track a publication from someone who otherwise is just a retarded first year master student (me)?

>> No.15934966

>>15934783
Hm, that makes sense, specially what you said about the weakest chess levels. Thanks.

>> No.15935031

>>15934898
>read linear algebra before algebra

>> No.15935067

is there such thing as seven sided dice where each face has a equal chance of being landed on?

>> No.15935077

>>15935067
Just take an 8-sided die and reroll whenever you roll an 8.

>> No.15935082
File: 126 KB, 315x306, gato-dice-120-sided-single-dice-lab.png [View same] [iqdb] [saucenao] [google]
15935082

>>15935067
There's a stackexchange question on it along with video of its fairness experiments:

https://rpg.stackexchange.com/questions/123545/is-gamesciences-non-cylindrical-seven-sided-die-a-balanced-fair-die

https://www.youtube.com/watch?v=mmbapsJudG4

Otherwise, the D120 (pic) is the die with the most sides possible that remains fair when rolled.

>> No.15935136

I'm reading some lecture notes, and the author says that if R is the ring of integers of a number field, and p is a nonzero prime ideal of R, then R/p is a finite field.

Can anyone explain why this is (or point to a book/source explaining it)?

>> No.15935160

>>15935136
To elaborate further, I understand why R/p is a field (because R is a Dedekind domain), but I don't see why R/p has to be a *finite* field?

>> No.15935164

>>15935136
>>15935160
https://math.stackexchange.com/questions/4180573/quotient-of-the-ring-of-integers-by-a-prime-ideal
better explained than I could and I can't be assed to copy it over

>> No.15935171

>>15935164
Perfect, thanks anon

>> No.15935494

>>15927366
>>15927376
Alright I've looked at these Desmos links for a while now, sorry about the late response as I've been busy with other stuff. But I've gotta say I still don't really understand why it converges to -11/9 specifically. I get that it must converge, but why that value in particular?

>> No.15935525

>>15935031
yes.

>> No.15935554

>slip up on a fairly obvious exercise
>mutter "good for nothing fucking retard can't do shit right you moron"
>mood ruined for the rest of the day
how do I stop doing this

>> No.15935702

Let F be a field. Let R,S be subrings of F such that FracR = FracS = F .

Then does Frac(R ∩ S) = F also hold?

>> No.15935739

>>15935702
Apparently the answer is no; I'll leave it as an exercise for y'all to find a counterexample

>> No.15935841

>>15935554
learn to love yourself anon

>> No.15935856

>>15935554
Stop making mistakes

>> No.15936428 [DELETED] 

If we have a simple parabola y^2=2px, is the equation in polar coordinates just
[math]\begin{cases}r=\frac{2p\cos \phi}{(\sin\phi)^2} \\ r=0 \text{, if } \phi= \pi /2 \end{cases}
Can someone verify this please?[/math]

>> No.15936430

>>15936428
[math]\begin{cases}r=\frac{2p\cos \phi}{(\sin\phi)^2} \\ r=0 \text{, if } \phi= \pi /2 \end{cases}[/math]
can someone verify this?

>> No.15936435 [DELETED] 

>>15936430
where the parabola is y^2=2px and the polar axis is the same as the x-axis.

>> No.15936457

>>15936430
where the polar axis is the axis and the parabola is in the form y^2=2px

>> No.15936458

>>15936457
x-axis*.

>> No.15936600

>>15934599
it is just as good as any other maths with them since when you round them to one decimal place, it's true. And you can't know all the decimal places of them, and you can't do pi+e and such things so rounding and approximations are the only thing you can do. When you are rounding, you can do it to any number of decimal places, depending on what you need. But if you want only 1 then it's true for them rounded to 1 decimal place.

>> No.15936613

how would you express this using formal notation:
https://en.wikipedia.org/wiki/Pigeonhole_principle

I want to make clear the mapping of more than one element of the domain to the codomain

>> No.15936898

>>15933172
based

>> No.15936911 [DELETED] 

If a permutation [math] \pi^2 = 1 [/math], is it necessarily a product of disjoint transpositions?

>> No.15937199

Let A and B be finite sets, #B > #A.
If f:A -> B is a function, then there exists b belonging o B such that, for some a1 and a2, a1 =/= a2, f(a1) = f(a2) = b

>> No.15937234

>>15937199
In other words, f can't be injective.

>> No.15937571

>>15935525
Do you read functional analysis before analysis as well? Retard.

>> No.15937575

>>15936613
my half assed attempt.
let A and B be sets. if |A|>|B| and f: A -> B then there exists x and y in A and z in B where x =/= y and f(x) = f(y) = z.
it's probably wrong and/or superfluous

>> No.15937583

>>15937571
Well obviously the fucking algebra book you're reading is expecting you to know basic concepts of linear algebra. Because people generally do learn linear algebra before they learn field theory and various results in field theory are convenient to put in linear-algebraic terms.

>> No.15937590
File: 93 KB, 1014x226, IMG_20231223_141607.jpg [View same] [iqdb] [saucenao] [google]
15937590

Discuss topological algebra, how do I prove this problem?

>> No.15937605

>>15937590
The screenshot is taken from an article called
>Some Open Problems in Topological Algebra

The reason those problems are called open is that nobody knows how to prove or disprove them.

>> No.15937616
File: 178 KB, 573x495, overt.png [View same] [iqdb] [saucenao] [google]
15937616

>There will never be an explicit well-ordering of the reals

>> No.15937618

>>15937616
Obviously, as there is no well-ordering of the reals, period.

>> No.15937619

>>15937618
Uhh yes there is. Just use Zorn's Lemma

>> No.15937621

>>15937616
There is if V=HOD

>> No.15937623

>>15937621
E x p l a i n

>> No.15937629

>>15937623
There is an explicit well-ordering of the whole class HOD (the Hereditarily Ordinal Definable sets). It's possible that all sets are in HOD, and if so then that well-ordering orders all reals. Opinions are divided on whether this is likely to be true in the true model of set theory (and even on whether such a thing exists), but I believe Woodin's "V=Ultimate L" proposal would imply it, as would the less popular "V=L". It's possible to construct models where it holds and also models where it doesn't, so it can't be provable from ZFC.

>> No.15937638
File: 217 KB, 733x595, bed.jpg [View same] [iqdb] [saucenao] [google]
15937638

>>15937629
Huh, so what is the well-ordering then? is it just the first uncountable ordinal?

>> No.15937640

>>15937583
>various results in field theory are convenient to put in linear-algebraic terms.
Literally the opposite.

>> No.15937644

>>15937638
HOD doesn't add any restrictions on the size of the reals, it can still be any cardinal of uncountable cardinality. "V=L" implies its the first uncountable cardinal, and I believe "V=Ultimate L" does too.

>> No.15937648

>>15937640
How's the first year of your undergrad been so far?

>> No.15937654

>>15937648
Bet you were postgrad before grad lmao

>> No.15938810

bump

>> No.15938937
File: 18 KB, 751x247, img1.jpg [View same] [iqdb] [saucenao] [google]
15938937

(1/2)

>> No.15938938
File: 76 KB, 793x693, img2.jpg [View same] [iqdb] [saucenao] [google]
15938938

(2.2)
Why do we approximate the function, and what is important about C = -1 being the only solution. Solution for what exactly?

>> No.15938939

"This limit" being 1. What about this? f(x) = (...) is the solution. Is this not the end? What is interesting about "this limit"?

>> No.15938947

Does this have to do anything with b being a free variable?

>> No.15939296

∈ does NOT mean 'is a', it does NOT mean 'is in', it does NOT mean 'belongs to', it ONLY means 'is an element of'
Why? because all of the other terms are standard english, and non-technical, are trying to define them to mean what ∈ means in set theory is just wrong, because that is not what those words mean in english.

>> No.15939299

>>15937616
then ZFC is wrong so scrap it.

>> No.15939321

>>15939296
Those are just informal descriptions. ∈ is not defined at all in set theory.

>> No.15939445

>>15937629
how do you well order HOD? it is possible to order OD (class of objects defined by a formula and an ordinal, by indexing formulas with ON) however HOD(x) is just "x and all objects in its the transitive closure are in OD". Under V = L we know that every x is in OD hence there is a well ordering of the whole universe. But under bare HOD?

>> No.15939450

>>15937629
>>15939445

Also (see Krivine set theory book), the statement "V = HOD", relativized to HOD, is not necessarily true.

>> No.15939460

>>15939445
>>15937629
>how do you well order HOD?
I mean: you might well order HOD as a sublclass of OD, in the larger universe however from "within HOD" it is not possible to "detect" this well order, although HOD satisfies AC (hope my point is understandable).

>> No.15939464

>>15939445
HOD is a subclass of OD. So just restrict the ordering from OD.
>>15939450
Yes, and in those cases there isn't an explicit well ordering of HOD in HOD. But I was talking about the case where V does satisfy V=HOD.

>> No.15939660

>>15939321
Primitive notions still have definitions inside the mind.

>> No.15939819

>>15938938
>Since P and Q are continous on the interval (0, infinity)
I just plotted P and Q and they are not continous, wtf?

>> No.15939824

>>15939819
At which point were they discontinuous?

>> No.15939834

>>15939824
P is discontinous at x = 1
Now I think Q is continous tho, I messed that up

>> No.15939875

>>15939834
Let [math]\varepsilon > 0[/math] pick [math]\delta = \min(\frac{1}{2}, \frac{\varepsilon}{2}) [/math] then
[eqn]|x - 1| < \delta \implies |P(x) - P(1)| = \left| \frac{1}{x} - 1 \right| = \frac{|1 - x|}{|x|} < \varepsilon [/eqn]

>> No.15940198

How do you prepare for Calc 1 in a few weeks?

>> No.15940201

>>15940198
Have an IQ above room temp

>> No.15940333

>>15940201
I was thinking watching a few playlists to kinda recall all the important stuff and refresh my memory.

>> No.15940347

Is Khan Academy a legit good resource to learn math?
Btw I'm still fucking bafled that we still don't have some kind of government funded, public, free learning service that would teach the most basic shit. I mean fuck you gotta pay on those faggot websites just to see explanations for calculations.

>> No.15940436
File: 518 KB, 2312x1594, file.png [View same] [iqdb] [saucenao] [google]
15940436

>got through high school with straight As
>did a math major with a 4.0 GPA
>never learned about the triangle inequality until I started working in algorithms
what the fuck

>> No.15940438

>>15940347
are you in uni? your school might have a wolfram alpha license

>> No.15940462

>>15940436
how do you not just know this intuitively, are you a neanderthal?
of course going in a straight line to somewhere is the shortest route

>> No.15940469

>>15940462
obviously. but I've never used it in my life until I got to research in algorithms and saw it referenced.

>> No.15940479
File: 68 KB, 745x460, munkres.jpg [View same] [iqdb] [saucenao] [google]
15940479

>>15940469
The triangle inequality is essentially *the* property that characterizes metric spaces, did your syllabus just restrict analysis to vector spaces on R/C or something?

>> No.15940481

>>15940479
I didn't learn that until I started taking grad school classes (and associated research)

>> No.15940502

>>15939660
Why is "belongs to" an invalid "definition inside the mind" but "is an element of" is valid?

>> No.15940505
File: 66 KB, 720x720, crowfly.jpg [View same] [iqdb] [saucenao] [google]
15940505

>>15940481
>being a literal birdbrain
fucking hell anon

>> No.15940566

>>15937644
>I believe "V=Ultimate L" does too.
Are you sure? I'm pretty sure we don't know much about the structure of Ultimate L if it even exists but in L(R) (which I would naively intuit is closer to Ultimate L than L is) the reals are inaccessible and incomparable with the first uncountable cardinal. If you mean that the continuum hypothesis holds, that's a slightly different claim.

>> No.15940637

Are there statements, that can be proved by strong induction, but not by weak?

>> No.15940649
File: 454 KB, 1580x2183, a55b0c31fdf53110aa6313d9cbccf119.jpg [View same] [iqdb] [saucenao] [google]
15940649

>>15940436
desu, I've used this for years and never realized that it's literally a triangle

>> No.15940791

>>15940649
How is this even possible, math tranny?
The triangle inequality is one of the main inequalities in analysis, a key axiom to *do* analysis, and in every textbook I've seen it introduced it has been accompanied by the intuition of the actual triangle.

If you got the one book that didn't bring in such intuition, why did you think that was an axiom? Why would it otherwise be an axiom? Did you think mathematicians just said "fuck it" and added that extra inequality for fun?

>> No.15940841

>>15940566
I checked and now I am sure, see here: https://mathoverflow.net/questions/288723/does-v-textitultimate-l-imply-gch

You might be thinking of the older proposal V=L(R)^PMax, which implies the continuum is aleph-2. (It definitely can't be incomparable because I'm talking about extensions of ZFC, which means the cardinalities are well-ordered.)

>> No.15940884 [DELETED] 

>>15940841
>It definitely can't be incomparable because I'm talking about extensions of ZFC
V=L is really an extension of ZF in the sense that L consists only of the sets that can be proven to exist from ZF alone, it just so happens that Choice holds when one limits V to only contain them. Ultimate L is an extension of L.
The post you're replying to seems to think that either L(R) is also an extension of L (it is not) or that Ultimate L is relatively constructible to some set/class not in L (it is not).

>> No.15940902

>>15940436
first of all this is such an obvious and intuitive fact
second of all your program much have been awful to not offer proof based analysis classes. Your value as a mathematician is worthless.

>> No.15940904

>>15940347
Just figure it out yourself.
Also there are python libraries that will do your math elementary homework for you.

>> No.15940914

>>15940649
>>15940436
Ok retard kuns, I'm going to explain this once and only once
Imagine you had a triangle, regardless of orientation, relative to one leg, the projection of the other legs onto that leg will always be equal to the projected leg, this is trivial and obvious. Therefore the sum of the lengths of the other legs are greater than or equal because when you rotate the leg to be parallel to its projection it will be greater than or equal to it.
L1 >= P(L1), (let P define the projection)
L2 >= P(L2)
=>
L1 + L2 >= P(L1) + P(L2) = L0
without loss of generality.
I'm 20 and I finished my bachelor's two years ago, but I've known this obvious fact for much longer.
And if you really need me to explain why the projection is smaller then you need to go back to america where the retards lives
Also
>algorithms
ishiggydiggy

>> No.15940917

>>15940914
*greater than or equal
sorry I'm part american

>> No.15940959
File: 30 KB, 500x228, mongol-bow-construction[1].jpg [View same] [iqdb] [saucenao] [google]
15940959

PBB (Proof by bowstring)

>> No.15940985
File: 125 KB, 2560x1024, Sine_cosine_one_period.svg.png [View same] [iqdb] [saucenao] [google]
15940985

I am not a math major, so I may lack the correct terminology to describe what I mean:
I am trying to develop a 2-dimensional chart, where one axis (let's say x) does not detail something "spectacular", you can just assume it to model the reals. The other axis, however, uses a kind of different dynamic. I want the two points y = 1 and y = -1 to detail a point that semantically entails "total difference from a y-value at y = 0". This same relation would apply for any further movement along this axis, meaning "y = 2" would describe "total difference from y = 1 (and also y = 0)", and so on. As such, something like y = 1.5 would detail "this point is half-different from 1", or even "half different from 37".

So the dynamic is: [math][Degree of similarity] = (y' - y) (mod) 1[/math]

This is obviously somehow related to trigonometry and periodicity. You can envision the concrete y-values as the number of peaks that have elapsed on a sine graph.
But I need it to be a Cartesian field. The reason this is important because under the system I am trying to develop, consider e.g. the two tuples (22, 3) and (22,7). While both describe something where there is a "total difference" within the y dimension from e.g. (22,1) -- but the two tuples also would NOT point to an identical thing; as such, the separation is important. Indeed, that is precisely why I keep those numbers around -- to give some means to distinguish.
MY QUESTION IS: is there even a math or notation concept where it wouldn't seem arbitrary for such a Cartesian chart to have x and y axes with different functionalities? As one axis just tracks the reals, but the other axis somehow denotes a periodic function.

>> No.15941130

>>15940649
Groid?

>> No.15941141

>>15940914
I didn't say I didn't understand it, retard-chan. I said I never had to use it before. You are unironically retarded if your language comprehension is so low you didn't understand that.

>> No.15941146

>>15941141
Why should I waste my time trying to understand inferior beings to their desire?
My goal is to escape the disgusting stain of humanity.
It's fine though, you're an uneducated moron, but worse you were maleducated. You should probably sue your institution if it didn't even teach you primitive monkey shit.

>> No.15941147

>>15941146
You sound like a fag and your shit's all retarded.

>> No.15941151
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15941151

>>15941147
I see, you can't afford a lawyer then. Too bad, if you're american I'm sure you wasted a lot of money. Well at least try to read books and improve, if you wanted to just call people retards all day then you should start by not making embarrassing posts like in your original one.
I'm doing you a favor by even talking to you, even though I doubt an ape like you can really evolve as much as me. That is man's great gift, to evolve with his own will.
I realize by now you're in debt by my words, so I'll let you work it off. Instead of interest become a greater being, something worth the air around itself.

>> No.15941164

>>15941151
You have no idea what he just said

>> No.15941216
File: 261 KB, 800x928, 800px-Riemann_surface_log.svg.png [View same] [iqdb] [saucenao] [google]
15941216

>>15940985
>is there even a math or notation concept where it wouldn't seem arbitrary for such a Cartesian chart to have x and y axes with different functionalities? As one axis just tracks the reals, but the other axis somehow denotes a periodic function.
Sounds like you could be looking for Riemann surfaces, though without further details on your system it's hard to say for sure.

>> No.15941310
File: 83 KB, 720x892, dawg.jpg [View same] [iqdb] [saucenao] [google]
15941310

>>15924439
Why are numbers with 3 digits like 222 or 555 divisible by 3 and numbers with 9 digits such as 111111111 are divisible by 9, but the same doesn't apply to numbers with 2 digits being divisible by numbers like 33 or numbers with digits of 8 have you get the idea

>> No.15941331

>>15941310
It's easiest to start off by explaining why it works for 9.
I'll just arbitrarily pick 621 as an example. What "621" is really describing is the sum 600+20+1, or [math]6 \times 100 + 2 \times 10 + 1 \times 1[/math].
We can rewrite this as [math]6 \times (99+1) + 2 \times (9+1) + 1 \times (0+1)[/math]; the important thing is that we are able to split the powers of 10 into multiples of 9, plus 1. Regrouping so that we consider the multiples of 9 separately from the "plus 1", we find that [math]621 = ( 6 \times 99 + 2 \times 9 + 1 \times 0 ) + ( 6 + 2 + 1 ) [/math].
Notice that the grouping on the left has a bunch of multiples of 9 added together, so it will obviously be divisible by 9. The grouping on the right consists of the digits of the number we started with, multiplied by the 1 we separated from the power of 10. So, if the sum of the digits is divisible by 9, the entire thing is divisible by 9. And, conversely, if the entire thing is divisible by 9, then the sum of the digits has to be divisible by 9 as well.

This works for 3 just as well because 9, 99, 999, etc., are all multiples of 3, too.
It doesn't work for any other numbers (in base 10) because they'd skip some of these positions. You can't split 10 into 1 + a multiple of 33, for example.

>> No.15941379

>>15941331
>You can't split 10 into 1 + a multiple of 33, for example.
Not a disagreement so much as an extension, but you can split 100 into 1 + a multiple of 33, so you do have something analogous for divisibility by 33 but with pairs of digits rather than single digits itself.

For example, 33 * 621 = 20493, and (after inserting a leading zero) we have that 02 + 04 + 93 = 99, and conversely any number for which the pairs of digits sum to a multiple of 33 is itself a multiple of 33.
You should be able to do something similar (though with more digits) for any number that isn't divisible by 2 or 5, I believe.

>> No.15941432

>>15924439

What's an example of a field F such that every element of F has an nth root for every positive integer n, but F is not algebraically closed?

>> No.15941439

>>15941432
field of rational functions k(x), where k is your favorite algebraically closed field

>> No.15941449

>>15941439
sorry I'm a dumbass. I thought you asked for a field that had all roots of the integers.
To answer your question, take the finite field with 2 elements.

>> No.15941491

>>15941331
Thank you anon. I've been stuck wrestling with this question all day. Also does this trick works with numbers that're only constituted by 3 such as 27 or 81 too? Also where did you learn all of this information?

>> No.15941501 [DELETED] 

>>15941491
>Also does this trick works with numbers that're only constituted by 3 such as 27 or 81 too?
Not in the same way as for 3 and 9, since they don't divide 9. But like >>15941379 said, you can find a similar pattern for most integers.
Specifically, for a given x, you need to find the lowest y such that 10^y-1 is divisible by x; this number will always exist given, again, that x is not divisible by 2 or 5. That doesn't mean it'll be pretty: if x is some prime 3 more than a multiple of 4, for example (7, 11, 19, 23, etc.), y will be x-1. So in the case of 7, you'd need to group your digits into sets of six... and thus you'd need to remember all multiples of 7 up to 6 digits. And it'd only get worse from there.

There are far easier divisibility rules out there that don't follow that exact pattern, though (e.g. for 11 you add the digits up, but make every other digit negative).
Anyway, all of this stuff is fairly elementary number theory, if you want something to look into.

>> No.15941504

>>15941491
>Also does this trick works with numbers that're only constituted by 3 such as 27 or 81 too?
Not in the same way as for 3 and 9, since they don't divide 9. But like >>15941379 said, you can find a similar pattern for any positive integer not divisible by 2 or 5. (And it can't be 1, obviously.) That doesn't mean it'll be pretty: for example, in the case of 7, you'd need to group your digits into sets of six... and thus you'd need to remember all multiples of 7 up to 6 digits. And it'd only get worse from there.

There are far easier divisibility rules out there that don't follow that exact pattern, though (e.g. for 11 you add the digits up, but make every other digit negative).
Anyway, all of this stuff is fairly elementary number theory, if you want something to look into.

>> No.15941510

>>15941491
I am ofc referring to the numbers of 27 and 81 in terms of grander numbers that're in the thousands and 10s of thousands instead of the 100s place.

>> No.15941512

>>15941504
Oh I c
Danke schoen

>> No.15941517

>>15941510
In that case, yes, and not quite.
27 divides 999, so it would be the same deal, just with groups of 3 digits.
But in the case of 81, you wouldn't be looking at 4-digit groups. No, no, the first repdigit composed of 9s that 81 divides is 999999999. so that's... well, at that point, it'd probably be quicker to divide your candidate by 9 and then see if the quotient is also divisible by 9.

>> No.15941539

>>15941449
Ah I see, of course.

Then as a follow up: what's an example of a *zero-characteristic field F, such that every element of F has nth roots for every n?

>> No.15941619

>>15940904
Are they also gonna do your in class exams? C'mon dude.

>> No.15941640

>>15941432
The field of complex numbers that can be constructed by radicals. It's not algebraically closed because it doesn't have a solution to x^5-x-1=0 (by the Abel–Ruffini theorem)

>> No.15941652

>>15941640
is it a field THOUGH

>> No.15941655

>>15941640
But when talking about radical extensions you only allow to adjoint finitely many roots. By adding finitely many elements to Q you will never get a field that includes all n-th roots of 2 for example.

>> No.15941794

>>15940637
Weak induction is equivalent to strong induction, so no.

>> No.15942078

>>15941655
It's common to only add finitely many roots, and easier to understand, but it's not actually impossible to just add in all of the roots.

>> No.15942088

>>15941655
he didnt call it a radical extension
it is a subfield of all complex algebraic numbers

>> No.15942376

Is Euclid’s elements a good place to start if I want to teach myself geometry? Any other recommendations?

>> No.15942692

Okay /sci/, I have been tinkering with this casually for a minute and I'm finally at my breaking point.
Given an NxN matrix M, and an N-dimensional vector g, both with the constraints that all of their elements are >= 0, how do I find a vector i such that o = Mi, and every element of o is >= the corresponding one of g. I also need all elements of i to similarly be >= 0. Bonus points if |g-o|^2 is minimized, and the same with the sum of elements in i.
I've done it by hand for the N=2 case, and I'm trying to wrangle the higher dimensional cases, but it's getting unwieldy fast.

>> No.15942694

>>15942692
Before anyone says it, M^-1 isn't a general solution, as it can yield negative values for i

>> No.15942695

>>15942692
>>15942694
Oh, one more thing, det(M) > 0

>> No.15942821

>>15942692
Find the largest eigenvalue of M and take i to be the corresponding eigenvector, scaled to be large enough to satisfy the inequality you want. By the Perron-Frobenius theorem, this i will have positive entries so long as your matrix M is irreducible. Otherwise, its rows and columns can be permuted to turn it into a block triangular form consisting of irreducible matrices, so you can just obtain an i for each block and then stitch the components together.

>> No.15943131

Is my reasoning for the integration by parts formula correct?

By our presumptions uv is differentiable so that means it's continuous and (uv)'=u'v+v'u also we know that there exists the the integral S (integral sign) u'(x)v(x)dx since this integral exists it means that the function u'(x)v(x) is also continuous. v'u=(uv)'-u'v and since (uv)' and u'v are continuous functions then their difference is also continious which means v'u is continuous and that the integral S v'(x)u(x) exists

>> No.15943579

>>15940462
neanderthals knew how to cook so they could probably understand the triangle inequality

>> No.15943721

Is there some mathematical object A(i) where for some i_1, i_2, A(i_1) is real analysis and A(i_2) is complex analysis?
I've done some differential geometry but it doesn't seem to take a big picture look at what analysis can be.

>> No.15943725

>>15943721
What a silly question. Just define A(i_1) := real analysis, and A(i_2) := complex analysis .

>> No.15943752

I posted this in the stupid questions thread but didn't get much help:

I'm trying to determine what the surface [math] S=\{ (x,y) \in \mathbb{C}\times\mathbb{C} \,:\, x^2+y^2=1\}[/math] looks like topologically?
For example, is it homeomorphic to some genus g surface with some n≥0 points removed, or some other known surface?

One idea I had was to look at the complex projectivization of S, which is [math] S'=\{[X:Y:Z] \in \mathbb{C}\mathbb{P}^2 \,:\, X^2+Y^2=Z^2 \} [/math];
unless I'm mistaken this is a smooth surface,
and by the standard degree-genus formula this should have genus 0, so it is a sphere;
and the in the set [math] \{[X:Y:Z] \in \mathbb{C}\mathbb{P}^2 \,:\, Z=0 \} [/math], S' has 2 points, namely [1:i:0] and [1:-i:0];
and [math] \{[X:Y:Z] \in \mathbb{C}\mathbb{P}^2 \,:\, X^2+Y^2=Z^2 \,\&\, Z\neq0 \} [/math] should give us our original surface S.
So then S is homeomorphic to a sphere with 2 points removed (so, homeomorphic to an open annulus/cylinder).

However, is there a way to do this without using the degree-genus formula?

>> No.15943787

>>15943725
What I mean is that universal algebra studies algebraic structures instead of their contents - why isn't there a universal analysis?

>> No.15943815

>>15943787
Algebra is predisposed to abstraction by virtue of the formal nature of its subject matter, while analysis is not.

>> No.15943879

>>15943787
you can do analysis on abstract Banach spaces if you want to

>> No.15943922

>>15943787
Lectures and Exercises on Functional Analysis - Helemskii

This is one place to look. Another would be de Rahm cohomology.

>> No.15943940

>>15943787
You can do analysis on more general normed fields, or even more generally, topological fields.

>>15943922
>Another would be de Rahm cohomology.
But what does this have to do with "universal analysis" which is what >>15943787 was asking about

>> No.15944087

Maybe some people here would know but don't check SQT,
>>15943989

I'm trying to find generalized treatment of provability and conditions for unprovable systems or results of systems. I am not talking about undecidability but it is related (just not synonymous). I suspect it has to do with the information entropy of the system and kolmogorov randomness but I can't find anyone dealing with proof theory and mathematical proofs (deduction) in relation to it without equivocating to unrelated things or merely SPECIFIC "impossibility proofs". Rather I am trying to find generalized treatment, the generalized set of "unprovable systems" or unprovable solution sets from systems, limits of proof theory as a generalized thing.

All I can find goes right back to Godel's incompleteness and that is not what I am trying to find. That and a whole lot of other equivocations. Even where I find people asking this exact question all the answers are equivocating to what the question isn't asking, or falsely claiming there's no such thing. So I know I'm not the only person who has had this notion but I can't find any actual material dealing with it.

>> No.15944173

>>15944087
>unprovable systems
No idea what this is supposed to be, but if you're talking about logical systems in general, the main result that characterizes what they can/cannot prove will be called a "completeness theorem" for that logic (often paired with a "soundness theorem"). The details of how to prove the completeness theorem would, in full generality, depend on how the syntax and semantics of the logical system is defined, although the only ones I'm familiar with are minor variations of the completeness/soundness theorem for classical predicate logic.

>> No.15944194

>>15944173
>exclude completeness
>get completeness
If you aren't going to read the post why did you bother replying? This is not about completeness this is about axiom schema and whether a given problem can be proved i.e. from a finitely axiomatized system. I'm asking a generalizable case that's most closely considered in model theory but not quite, or if it is I can't find people or works considering it.

Let me waste more time repeating myself 20 different ways since this is a recurring problem I've found in all posts asking the same thing: I know what completeness theorems are. I know what completeness theorems are. I am not talking about completeness. I am not talking about whether any given current set of axioms are complete. I am talking about the inversion of completeness as in ***can be / cannot be represented via finitely many axioms***

Such that: Given some generating formula there exists **no** set of axioms that could prove it short of reproducing the whole thing, i.e. a system whose results are not generalizable even if properties of the system generated inductively appear to have certain consistent results. Analogous to algorithmic complexity and incompressibility but not with respect to specific values but patterns or tendencies of systems producing those values.

So for the billionth time ***I am not talking about completeness***.
Completeness involves "given some set of axioms"
I am talking about "given the set of all possible axioms". That is, things not deducible via systems of finite axiomatization.

The nearest thing abstractly about model theory I know is about finite axiomatization. I am, specifically, talking about ***that which cannot be axiomatized finitely*** and ***that is not about completeness***.

>> No.15944253

>>15924439
why do LGBTQ+ marxist autists love grothendieck so much

>> No.15944257

>>15926123
idk but as a mathematical theory in its own right it is particularly beautiful, particularly the semantics side of it - bisimulations, game semantics, the connections to topology via stone duality, and so on and so forth. plus you then get to reap the philosophical benefits if you are that way inclined

>> No.15944261

>>15926138
>It is not mathematically relevant
if you're interested in things like (finite) model theory, categorical logic, and even computational complexity, it has more and more mathematical relevance these days. e.g. take a look at this bad boy https://arxiv.org/abs/2310.12068

>> No.15944270

>>15930745
one is a cycloid, as you said, and the other is a circle involute, they appear in the design of mechanical gears as they have some nice properties

>> No.15944296
File: 61 KB, 800x800, image0.jpg [View same] [iqdb] [saucenao] [google]
15944296

>>15944087
>>15944194
You are terrible at explaining. Just say what it is you want already in clearly defined terms.

>> No.15944503

why does this equality hold when a1 and a2 are collinear and pointed the same direction?
[math]|\vec{a}_{2}|\vec{a}_{1}=|\vec{a}_{1}|\vec{a}_{2}[/math]

>> No.15944519

>>15944503
Well where do you think those are pointed? And what is their norm?

>> No.15944538

>>15944253
They like to grothen on his dieck obviously.

>> No.15944552

>>15944503
Rewrite ai as ciei, where ci is a positive scalar and ei is a unit vector. Since they are pointed in the same direction, e1=e2. Then |a2|a1=|a1|a2 becomes c2c1e1=c1c2e2. You should be able to finish from here.

>> No.15944634

>>15942821
It appears the Perron-Frobenius theorem applies to matrices with columns that sum to 1, that isn't something I can guarantee

>> No.15944686

>>15944296
>You are terrible at explaining
You sucking at reading is not my sucking at explaining.
>Just say what it is you want already in clearly defined terms.
Case in point. Explicitly referenced "model theory" and the relevant term I am negating i.e. "finite axiomatization" in "study of things and properties of such that are not finitely aziomatizable". If there is a more relevant term that is what I am asking about.
>Use terms you don't know to ask about those terms
You've got brain problems anon.

>> No.15944721
File: 48 KB, 1080x826, Screenshot_2023-12-17-12-46-39-68_572064f74bd5f9fa804b05334aa4f912.jpg [View same] [iqdb] [saucenao] [google]
15944721

>>15944686
The incompleteness results rely on the axioms being R-decidable. So if you negate any incompleteness result, it immediately implies your axioms are not R-decidable, which is worse than not being finitely axiomatizable.

>> No.15944745

>>15944721
I know I'm far afield in the weeds compared to usual discussions. Hell most people don't engage with metalogics and "model theory" is kind of the abstracted abstraction of abstractions the crazy one in the basement occasionally publishes on. Probably I will just have to keep digging into model theory and what I'm already working on with complexity/information theory and either realize I'm being an idiot or onto something.

My schizophrenia aside, I like the cat. Thank you for including cat. Is very good cat. +1 internets to you sir and if you've any further remarks I welcome more cat.

>> No.15944753
File: 42 KB, 661x1000, quant_book.jpg [View same] [iqdb] [saucenao] [google]
15944753

Anyone here a quant or tried to become a quant?

>> No.15944803

>>15944753
I was a quant (i.e. actual trading) for about a year. I didn't do bad (I had a positive return) but the stress was something I couldn't handle, even literally sending me to the doctor a few times.

Now I am just quant-adjacent. Using quantitative tools but not in trading. I do trade my own money, partly with the old algorithms I used to employ, but that's not the same as actually trading the big $'s.

>> No.15944841

>>15944721
After intentionally hammering away at google-fu and searching portals on research in relevant fields I might've found something closer to what I am looking for https://en.wikipedia.org/wiki/Descriptive_complexity_theory

and there's probably intersection conceptually with expressive power and complexity. Issue is this arena seems pretty fucking dead but it's something in case a random person finds the thread trying to ask the same things I am asking about.

>> No.15944852

I have to either take:
>differential geometry
>algebra
>or TWO of the classes: fourier analysis, differential equations, complex analysis
Thoughts? My favorite classes so far were topology, probability theory and stats. I work as a software dev.

>> No.15944858

>>15944852
Is "algebra" in this case linear algebra, or something along the lines of group/ring theory, or something else?

>> No.15944860

>>15944858
Group and ring theory.
I've taken linear algebra.

>> No.15944869

>>15944253
Everyone loves grothendieck, maybe because he was a total genius who contributed a great deal to mathematics

>> No.15944873

finished the last class for my PhD. now its just research. ama

>> No.15944874

Can anyone give me any hints to find a non recursive general term of the equation
x_(n+1) = (n+1)*x_n + n
The answer I'm supposed to get is
x_n = n! - 1
but I've no idea how to get there, this section of the textbook wasn't very good.

>> No.15944877

>>15944874
are you sure that's what youre supposed to get? do you have x0?

>> No.15944881

>>15944874
You should post this in the stupid questions thread, /sqt/

Use induction:

x_(n+1) = (n+1)! - 1 = (n+1)*n! - 1 = (n+1)*n! - (n+1) + n = (n+1)*(n! - 1) + n = (n+1)*x_n + n

>> No.15944887
File: 11 KB, 303x288, eq1.png [View same] [iqdb] [saucenao] [google]
15944887

>>15944877
Well, I suppose I should be able to get the solution for any possible x0.
But here in the book I see the solution they gave only works for x0 = 0.

>> No.15944889

>>15944873
What's your research interests?
Mine are in topological algebra or anything about primes

>> No.15944892

>>15944889
mostly convex optimization, i'm more on the applied side of the department

>> No.15944893

>>15944892
Ahh you applied city slickers always get all the attention

>> No.15944909

>>15944881
I'm not asking how to prove the factorial is the answer.
In the exam I would have to come up with the solution from scratch in general for any equation of the form
x_(n+1) = a(n) * x_n + b(t)
The way they tell me to do it is to first find a solution for the complete equation multiplied by an arbitrary constant and a solution for the homogeneous equation and then add them to get the general solution.

>> No.15944919

>>15944881
But in any case, I don't understand this step. Why are they equal?
>(n+1)*(n! - 1) + n = (n+1)*x_n + n

>> No.15944923

>>15944874
To find the general solution for arbitrary x_0 :

Let u_n = x_n + 1 for each n.

Then x_{n+1} = (n+1)x_n + n
becomes
u_{n+1} - 1 = (n+1)(u_n - 1) + n
which simplifies to just
u_{n+1} = (n+1)u_n
which we can solve directly; it has the general solution
u_n = (n!)*b , where b=u_0 .

So x_n = -1 + (n!)*(a+1) , where a=x_0 .

>> No.15944935

>>15944923
To elaborate further: the way I figured out that u_n = x_n + 1 would be a useful substitution is as follows:

I basically just guessed the ansatz u_n = x_n + c for some unknown constant c (independent of n).
Then the recursion relation x_{n+1} = (n+1)x_n + n becomes
u_{n+1} - c = (n+1)(u_n - c) + n
which simplifies to
u_{n+1} = (n+1)u_n + n(-c+1) ,
which suggests c=1 should work.

>> No.15944941

>>15944869
he's no hermann weyl

>> No.15944953

>>15944803
That's kind of what I was suspecting. I've heard the work life balance isn't actually too bad (at least for quant researchers) but I feel like I'd die if something I did ended up losing millions of dollars.

>> No.15944992
File: 105 KB, 184x217, Capture2.png [View same] [iqdb] [saucenao] [google]
15944992

>>15926317
>So [math]\pi_2(B)[/math]
must be isomorphic to Z
right?
no, as we have a short exact sequence [math]\0\mathbb{Z}\to \pi_2(B)\to\mathbb{Z}\to 0[/math] all we know about [math]\pi_2(B) [/math] is that [math]\\pi_2(B)/\mathbb{Z}=\mathbb{Z}$$ please note that this does not imply that [math]\\pi_2(B)=\mathbb{Z}\oplus\mathbb{Z}[/math].
of course, this cannot be non-trivial so your reasoning is correct other than this

>> No.15945009

sorry for the terrible formatting, and i meant to say that it means [math]\pi_2(B)[/math] cannot be trivial.

i'm too drunk to correct it so gl

>> No.15945061

>>15944992
If A --> B --> C is a Serre fibration of finite CW complexes, then is it always the case that dimA + dimC = dimB ?
If yes, where could I find a proof of this?

>> No.15945342
File: 23 KB, 600x600, 61Sxfhpv7ML._AC_UL600_SR600,600_.jpg [View same] [iqdb] [saucenao] [google]
15945342

>>15944753
I'm trying to :). In a stats phd program right now.

>> No.15945364

>>15924439
Trying to learn basic integration by substitution, I know once you sub in u, you have to adjust the rest of the function to equal du. Is there a way to do this algebraically?

>> No.15945365
File: 77 KB, 937x737, eq2.png [View same] [iqdb] [saucenao] [google]
15945365

>>15944923
>>15944935
Thanks, playing around with Bing chat the only time I got it to actually figure out a working solution, it was the same as yours.
So I guess it really is the simplest way to solve it.
Fuck, neither the textbook nor the lectures said anything about finding variable changes.

>> No.15945375

>>15945364
What do you mean "adjust the rest of the function to equal du"?
u substitution is completely mechanical if you have the original integral in the right form ( ∫ f(g(x))*g'(x) dx ).
https://www.mathsisfun.com/calculus/integration-by-substitution.html

>> No.15945679

Hey I had a weird dream about the relationship between the harmonic series and imaginary numbers.
Have any of you heard of such a thing?
Ever since I read Rudin I've been having these bizarre dreams and visions of mathematics.
How do I learn more about this?
I think I heard about a relationship between the harmonic series and the zeta function, but I don't know much about it.
Where should I go after baby rudin? Someone recommended me basic math and baby rudin and so I read them.

>> No.15945692
File: 336 KB, 850x638, 1702226308682886.png [View same] [iqdb] [saucenao] [google]
15945692

>>15945061
If you have a serre fibration [math]B\to C[/math] with constant fiber [math]A[/math], then this is a fiber bundle and the relation is obviously satisfied

however in general there is no constant fiber A.

im not entirely sure where proofs related to these dimensionality relations are provided, but for general discussion on fibrations and fiber bundles, "A Concise Course in Algebraic Topology" by J.P May is a favourite of mine

>> No.15945693

>>15945679
Take a look at the definition of the Riemann zeta function ζ(s) for Re(s)>1 .
For s=1 , it would give the harmonic series, which diverges (in the classical sense).

In fact, s=1 gives a complex pole of the Riemann zeta function, after it is analytically continued beyond Re(s)>1.

Interestingly, if you look at say ζ(s=-1), you get -1/12, suggesting the idea that we can assign a "regularized" value of -1/12 to the (classically divergent) series 1+2+3+4+5+...

>> No.15945718

>>15945693
How that possible, shouldn't intuition make it obviously wrong somehow?

>> No.15945745

>>15945718
You can get a lot of unintuitive results when you play around with infinite sums, especially if you ignore your radius of convergence like that result requires (which, incidentally, is why it's wrong).
If you want another fun example of an infinite series gone wrong, consider the Riemann series theorem - that is, if you take a series that converges conditionally but not absolutely (i.e. it converges, but if you make every term the same sign it diverges), you can rearrange the terms to get whatever sum you like.

As an example, consider the alternating harmonic series, where every term with an even denominator is subtracted: [math]1 - \frac{1}{2} + \frac {1}{3} - \frac{1}{4}...[/math]
This converges to [math]ln(2)[/math], which shouldn't be hard to convince yourself of if you check the sum after the first few terms.
But if we rearrange it so that every positive term is followed by two negative terms, so we get [math]1 - \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{6} - \frac{1}{8}...[/math], we can simplify that to [math]\frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \frac{1}{8}...[/math]

Which, you will notice, is half of the sum of the original, despite us not actually adding or subtracting anything to the original series. Just a rearrangement of terms.

>> No.15945765

>>15945745
incredible, and shouldn't those terms evaporate because of how small they get?
I think I've seen something similar with 1s and -1s to get different numbers, but that was not the same as this.
This has to just be an error though right? Like from pushing terms around we've probably made the behavior of the tail change like pushing a ketchup packet until it explodes.

>> No.15945776

>>15945765
>Like from pushing terms around we've probably made the behavior of the tail change like pushing a ketchup packet until it explodes.
Something like that? I think? I'm not actually entirely sure what you mean by that but I'll pretend I am anyway.
Riemann's explanation was something as follows.

>A conditionally-convergent series can be split into two subseries, one of strictly positive terms (A) and one of strictly negative terms (B).
>Consider the case where both of these subseries converge to some finite sum. In that case, if we make every term of B positive, then its sum is still finite, and so A and B together have a finite sum. But in that case our series would be absolutely convergent, not conditionally convergent. So if our series is conditionally convergent, at least one of these subseries must diverge.
>But if only one diverged, and the other converged to some finite value, then the entire series would diverge, and would thus not be conditionally convergent. As such, it must be the case that both subseries diverge independently.
>So, we pick an arbitrary value we want to be our sum, and add enough terms from the subseries of the correct sign until we're just past it. Then we add terms from the other subseries until we've gone back. Then we add terms from the first subseries until we pass it again, and then...

The end result, as mentioned, is that you can find a rearrangement of any conditionally convergent series to any value you like, including either diverging to infinity or diverging to no value at all

>> No.15946218

>>15945342
I went through the Introduction to statistical learning book this year but its a lot lower level than ESL.

>> No.15946238

New >>15946236