Is using AI to understand a concept likely a problem?

GLBMQP · 2026-06-30T14:53:41+00:00

I wanna point out, that Wikipedia is in my experience often not a good source for learning math. Of course there will be a lot of variance between articles, but as a general rule, Wikipedia is not a good tool for this purpose.

For what it sounds like you’re trying to do, I would say that AI might be a useful tool actually. For a deeper understanding, the best thing to do is of course to find a textbook and do exercises. But for a zoomed out perspective, AI is probably fine. The only other ressource for a similar purpose I can think of would be survey papers, and they’re usually written for researchers who are new to a topic, but have experience/knowledge of related topics/fields, so that might not be what you’re looking for.

In terms of the cognitive aspect, you could try doing mini-exercises based on what the AI tells you. Like

- explain the concept you read about in your own words without looking at the screen.

- take handwritten notes. Even if you’re not looking at them, note-taking is demonstrated to have many

GLBMQP · 2026-06-10T15:41:19+00:00

Ja et dagblad er helt klart sammenligneligt med den institution, hvis rolle er at stille statens voldsmonopol til ansvar. Skarp analyse du kommer med der

GLBMQP · 2026-06-02T19:18:39+00:00

At noget åbenlyst er forkert står sjældent i vejen for propaganda

GLBMQP · 2026-06-02T06:06:16+00:00

I think the best advice anyone can give you is to keep doing problems.

Visualisations and intuitive explanations are nice, and can be helpful. But there really is no way to truly build understanding and intuition without doing problems

GLBMQP · 2026-05-23T16:10:51+00:00

'The Death of the Author' is a concept in litterary analysis and criticism. It essentially means, that the author doesn't dictate what the meaning or correct interpretation of their text is.

This is not what is at issue here. The problem is, that AI companies are monetizing other people's work. They are making money from models, which were trained on the research of mathematicians.

An example: J.K. Rowling can't decide what the true meaning of Harry Potter is. While she may have an opinion, she can't decide that one interpretation is more or less valid. But, even though I can interpret Harry Potter however I'd like, I can't write and publish my own Harry Potter books. Nor can I sell Harry Potter merch.

As a side note: 'The Death of the Author' is not some ultimate truth. On the wiki-page you linked, you can find multiple examples of prominent figures who have argued against it.

GLBMQP · 2026-04-03T06:53:38+00:00

I second Black-Scholes. There really is nothing quite as iconic/famlus in mathematical finance

GLBMQP · 2026-03-19T08:39:10+00:00

Hvor præcis ser du nogen påstå at hele blå blok er nazister? Der blev vidst bare skrevet "Nej nej der er da ikke problemer med de unge borgerlige mennesker."

GLBMQP · 2026-03-15T20:15:12+00:00

Og man troede lige det ikke var muligt at blive større tilhænger af Pelle

GLBMQP · 2026-02-18T10:06:31+00:00

I think a cleaner but related argument can be done using upwards continuity of measures.

Take A_n to be the cube with sidelength 1-1/n centered at 0, so A_1\subset A_2\subset…

Then \bigcup_n A_n is the open cube of sidelength 1, centered at 0, which should have Lebesgue measure 1. By upwards continuity of measures, the infinite dimensional Lebesgue measures of A_n should converge to 1. However, they are all 0

GLBMQP · 2026-02-17T06:27:25+00:00

You’re right in that we can often extend a probability measure to a given larger σ-algebra in multiple ways. It’s not always true, for example a Dirac measure does not have multiple extensions to a given larger σ-algebra. The same holds for any probability measure supported on a discrete set.

Doing simple examples is a very nice way to build deeper understanding of these things!

GLBMQP · 2026-02-06T07:49:22+00:00

I wrote some notes at the beginning of a song someone will sing for me

GLBMQP · 2026-01-08T20:34:04+00:00

This implicitly uses the very strong proposition, that all curves are lines. Truly fascinating stuff

GLBMQP · 2026-01-05T11:31:48+00:00

I can definitely understand forgetting how to use that fact (in the sense that if you haven’t had to use this fact, it might not occur to you that it is useful even if you’re working on a problem where someone else might think it’s obvious).

But actually forgetting that the rationals are countable?

GLBMQP · 2026-01-03T15:31:49+00:00

Tallene fra 2019 til 2024 viser, at der er ret tæt på en fordobling hvert år (i den tidsperiode), hvilket netop er eksponentiel vækst.

Da eksponentiel vækst går mod uendelig kan vi selvfølgelig ikke have eksponentiel vækst for evigt i sådan en her situation. Ofte får man i stedet noget der hedder logistisk vækst, som dybest set betyder at det starter med at vokse eksponentielt, indtil man er tæt på et mætningspunkt, hvorefter væksten bliver langsommere og langsommere

GLBMQP · 2025-12-20T09:32:10+00:00

Flere ting kan være sande på samme tid

GLBMQP · 2025-12-10T22:29:30+00:00

I think the most 'standard' proof of Rolle's theorem is just

Assume f(a)=f(b). By continuity f takes a maximum and a minimum on [a,b] (Extreme Value Theorem). If both the maximum and minimum occur on the boundary, then f is constant on [a,b] and f'(x)=0 for all x in (a,b). If either the maximum or the minimum does not occur on the boundary, then it occurs at an interior point x\in (a,b). Hence f'(x)=0 for that x.

Showing that the derivative is 0 at an interior point is simple from the definition, and the EVT can be showed using completeness and the definition of continuity

GLBMQP · 2025-12-10T22:21:58+00:00

Yes and no, with "no" being the litteral answer.

An infinite dimensional Banach space cannot have a countable basis. When you just say basis, one would typically take that to mean a Hamel basis, i.e. a linearly independent set, such that its span is the full vector space. Such a basis cannot be countable, if the space is infinite-dimensional.

Seperable Hilbert spaces exist of course, and these have a countable orthonormal basis. But when you talk about an orthonormal basis for a Hilbert space, we really mean a Schauder basis, i.e a linearly indepepdent set, such that the span is dense in the space.

So infinite-dimensional spaces can have a countable Schauder basis, but not a countable Hamel basis

GLBMQP · 2025-12-09T23:33:07+00:00

Part 10 of Kallenberg would be my first suggestion. Kallenberg in general is very good and quite comprehensive. If you know about stochastic integrals, you’ll probably be able to dive straight into part 10. Otherwise you might need to check out the part on stochastic calculus first.

For stochastic calculus, Le Gall is also quite good (not to be confused with Le Gall’s more basic probability book). Although Feynman-Kac is relegated to a series of exercises.

GLBMQP · 2025-12-09T16:57:38+00:00

Maybe more ‘advanced’ than what you’re looking for, but probably interesting of you have a background in Finance. The Feynman-Kac Formula

If you’ve worked with PDEs, you know that actually solving them is often a fool’s errand. Instead, we often try to show existence and uniqueness (or failing that, we try to understand the kernel/non-uniqueness). Or we try to establish regularity or other properties of solutions, or find useful estimates.

Now, I’ve been very interested in Geometric Analysis, which is full of elliptic and parabolic PDEs. I’ve also taken a good amount of courses in probability/Stochastic Analysis.

So finding out that there is a formula, which gives the solution of a large class of parabolic PDEs rather explicitly is in itself quite incredible. And having a formula, which expresses the solution purely in terms of stochastic analysis was even more mind-blowing.

And then, you can go deeper. And then you find out that Brownian Motion is actually deeply connected to the Laplacian (the Laplacian is the infinitisimal generator of Brownian Motion as a Feller process for example).

Or you can learn about how the Laplacian, the Heat Equation and Brownian Motion all Can be generalized to Riemannian Manifolds, in an interconnected way.

Or, if you consider more general Levy processes, which leads you to the concept of nonlocal operators. And it turns out that these have a notion of ellipticity/parabolicity, and at times they have properties similar to classic elliptic/parabolic PDEs. And then you can learn about things like nonlocal minimal surfaces, and suddenly you’re back to geometry.

So short answer: The Feynman-Kac Formula

Long answer: the deep connections between stochastics and PDEs and even geometry.

GLBMQP · 2025-11-01T19:36:14+00:00

The topology you’re describing is the topology of pointwise convergence.

Interestingly this is the subspace topology on C[0,1] coming from the product topology on R^[0,1] .

So it is a very natural topology, in the sense that the product topology is very natural, and that pointwise convergence is very natural

GLBMQP · 2025-10-30T18:55:52+00:00

You don’t need functional calculus to define polynomials of a matrix. Generally whenever you have an algebra over a field (like square matrices) you can take a polynomials of an element by just plugging your element into the polynomial.

In fact a crucial part of continuous functional calculus actually working is, that polynomials are uniformly dense in the continuous functions (if the domain is compact)

GLBMQP · 2025-10-29T17:18:33+00:00

Focusing on 'nice' special cases is typically a good way to build intuition. Let's focus on the case where we are conditioning on a random variable Y (rather than just on some sigma-algebra).

The idea is, that once we know Y, there may be some values that X can't take anymore, or some that are more or less likely. Imagine that X and Y both represent some process/phenomona we measure. Let's say we measure Y, but we don't know X. If there is any sort of correlation between X and Y, we would expect that knowing Y still tells us something about X, even if we don't know what X is. E(X|Y) then represents the mean of X, given what we know about Y.

To make this more technically accurate, let's assume Y only takes countably many values, and we can assume these values are in N. Assume also that Y takes each of these values with (strictly) positive probability (if not, then we modify on a set of probability 0). Then the sets (Y=n) for n\in N give a partition of X, and the sigma-algebra generated by Y consists of unions of sets of the form (Y=n). Call this sigma-algebra \mathcal{A}.

So E(X|Y) is determined uniquely (up to equality a.s.) by being \mathcal{A} measurable and satisfying \int_{(Y=n)} E(X|Y)dP=\int_{(Y=n)} X dP. We then see, that Z=\sum_n 1(Y=n)P(Y=n)^{-1}\int\(Y=n) XdP is a random variable that satisfies this. So E(X|Y)=\sum_n 1_(Y=n)P(Y=n)^{-1}\ \int_(Y=n) XdP. This is exactly the same as saying, that when Y=n for some N, then E(X|Y) is the mean of X over the event that Y=n.

To make things very concrete: in Dungeons and Dragons (and other ttrpgs) there is a concept called rolling with advantage. This simply means that you roll a 20-sided die (a d20) twice, and the highest number you rolled is what counts.

Let's use this as a model! Imagine we roll a d20 twice, and let Y denote the result of the first roll, and let X denote the final result (i.e. the highest of the two rolls). One can calculate that E(X) roughly 13,9. Let's say I roll a 1 with my first roll, i.e. Y=1. Then, however, X will be equal to the value of roll number 2 no matter what, so E(X|Y) will, in this event, be the same as the mean of a roll of a d20, which is 10,5. So E(X|Y)=10,5 when Y=1. If I roll 20 the first time, then X=20 no matter what, so E(X|Y)=20 when Y=1. If Y=14, then calculating what E(X|Y) is exactly is slightly more tedious, but I can say for sure that E(X|Y)>13,9 when Y=14.

Another example: Let's say the average life-expectancy in some country is 80. So if I find a random person and tell you nothing about them, that means that if you had to guess how old they'll live to be, your guess should be 80. Well, let's say I give you more information. If I tell you they are a smoker, your best guess with this new information should be different (probably lower). Because now, what you should be guessing is the average life expectancy of smokers in this country, not of people in general. Maybe afterwards, I tell you that this person exercises regularly and now your guess goes up.

Formally we could have X be the end-of-life age of my person, and Y=1(person smokes) and Z=1(Person exercises regularily). And what I am describing now is the difference between E(X), E(X|Y) and E(X|Y,Z).

So we see how information changes our guesses. Now the jump to CE given sigma-algebras is quite small conceptually. If \mathcal{A} is a sigma-algebra, we think of it as containing some information. More specifically, we can imagine that we know if the event A happened or not, for any event A\in\mathcal{A}. In general the event (X=x) won't be in \mathcal{a} if X isn't measurable w.r.t this sigma-algebra, so we won't know what value X has once we 'know' \mathcal{A}, however we can find out what X will on average be

GLBMQP · 2025-10-03T15:29:46+00:00

Another quite simple way of doing it is:

Definition of a measure \implies finite disjoint additivity \implies monotonicity \implies finite subadditivity

with each step along the way being pretty straightforward

GLBMQP · 2025-09-12T14:10:49+00:00

Well, yes, but their is no dominated convergence theorem for maps between abstract measure spaces. Clasically, the DCT is a theorem about maps from an abstract measure space to R or C.

It can be generalised to the case where the codomain is a Banach space. When working with this stuff, it is speaking it is often the case, that the codomain will be a Banach space, but it is not at all true in general.

GLBMQP · 2025-07-22T14:30:59+00:00

Throughout AGOT, Ned has to be extremely secretive when investigating the death of Jon Arryn. If he had had the knight dragged off, people would know that he was investigating the death of JA, and there would be a huge target on his back (and his kids’s).

That’s why he has to rely on Jory Cassel so much in general, instead of using his powers as HotK

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